FINITE ELEMENT METHOD IN MECHANICAL DESIGN W. J. Anderson July 19, 1982 July 19, 1982

te'nr\-^0CT

TABLE OF CONTENTS LECTURE PAGE Introduction, Historical Review, Notation..... 1 Line Element, Assembly Process........... 17 Constant Strain Triangle............. 24 Virtual Work Theorem, Potential Energy Theorem... 31 Derivation of a Finite Element by the Virtual Work Theorem.................. 37 Derivation of Line Element by Energy Expression. 42 Beam Bendinq Element................46 Subroutines DCOMP and SQOE, ^,,,,,,,,,, 55 Modification of Equilibrium Equations Prior to Solution........... 62 Removing Rigid-Body Modes.... 67 Bandwidth Concepts........... 7j4 Convergence A Special Line Element Derived Using Matrix Notation......... 75 Comparison of Various Line Elements... 78 Argyris' Natural Mode Method............ 81 Interpretation of Rigid Body Modes Using Displacement Functions........ 87 Coordinate Transformations............90 Thermal.Stress 9ha Electrical & Fluid Networks. 95 Interpolation, Natural Coordinates......... 101 Gaussian Integration................ 110 Dynamics...................... 116 Nonlinear Problems................. 120 Steady State Heat Conduction........... 128 Example of a Heat Conduction Problem Using Line Elements................... 135 Variational Approach to Field Problems...... 138 Conclusion................. 145 Homework Problems........ 147 Sample Exams.................. 164 Appendices: Reprints. Matrices and Matrix Equations, Miscellaneous.

INTRODUCTION. HISTORICAL REVIEW. June 14, 1978 NOTATION. W.J.A. I. Introduction The finite element method has greatly helped many engineers and scientists in their career fields. It is a tool that allows methodical solution for systems with complicated geometries and made of many varying materials. The user of finite element methods develops intuition about the fundamentals of mechanics, because he is forced to think in terms of matrix mappings for all phenomena. After becoming comfortable with these linear transformations, one gains a tremendous organizing effect in the thought process. Many physical systems are discrete from the outset, e.g., truss systems, space frames, electrical and hydraulic circuits. These systems have been well handled by engineers in the past 50 years, are often called "networks" and can usually be studied by use of ordinary differential equations. Other physical systems are continuous in nature, e.g., plates, shells, flow problems and electrical fields. These systems have been handled with great difficulty in the past, are often called "field problems" and require the use of partial differential equations. The goal of most approximate engineering theories is to convert these field problems into something resembling a network problem, i.e. to discretize the field. The process of discretization can be done by global or local interpolations of the field variables. Many methods such as Galerkin's method or the classical Rayleigh-Ritz method represent the field variables by a series of global functions, each of which must satisfy certain boundary conditions. On the other hand, the finite element and finite difference methods attempt to model the field variable only over small discrete regions. The local approach has many advantages in terms of reducing the complexity of the final set of matrix equations (the "bandwidth" is often smaller, and the boundary conditions can be more easily applied). The finite element method is currently the most successful method for discretization of a broad range of field problems. The finite difference method is still the favored method in many field flow problems, such as problems where an unknown fluid-solid interface occurs due to phase change. Global methods are still 1

2 preferred, however, by many people. There are those who feel that the "boundary solution" process has a bright future. This method can be used in conjunction with local methods to solve problems with infinite domains. (1) The process of discretization can be done through the use of distorted or true modeling. There was a period during the 1950's and 1960's when civil engineers,in particular, attempted many distorted discrete models, e.g., a frame analogy for a slab. This was a dead end which probably delayed serious finite element modeling in civil engineering by several years. (Some reviewers believe distorted modeling to be a step on the way to finite element methods. This author, after having tried to do such modeling, feels it was a step backward.) Modern finite elements are usually based on true modeling, e.g., a plate element representing a plate. An exception is that some flat plate elements are yet being used to model curved shells. There is some question as to whether this is distorted modeling since, in the limit as elements become small, effects of curvature are included. Occasionally, one encounters in nature a system which is partially discrete and partially continuous. The finite element method (unlike the finite difference method) has no problem with the assembly of a discrete model of such a system. Beam, truss and spring elements are often combined with plate and solid elements in finite element solutions. II. Historical Review The finite element method has grown in step with the development of the digi(2) tal computer. Early theoretical work was done by Courant( on the torsion of (3) noncircular shafts in 1943 and by Argyris( on aircraft structures in the late'40's and early'50's. The field was not really "ripe" at that time, however. The paper which had the largest practical impact on the field was "Stiffness and Deflection Analysis of Complex Structures," by Turner, Clough, Martin and Topp (4) appearing in the September, 1956 issue of Journal of Aeronautical Sciences. This paper rejected the idea of further use of distorted models and instead developed two-dimensional elements to model beams and sheet. A result was the constant strain triangle and a rectangular element. The constant strain (Turner) triangle survives today and is still an excellent tool for teaching the theory.

This early work on modeling structural systems was primarily supported by the aircraft industry, which had a very difficult geometry and insufficient classical methods available. Other technical fields, however, were busy working with systems which were naturally discrete and therefore concentrated on supporting technology such as mathematical assembly of stiffness matrices, equation solvers and eigenvalue solutions. One should therefore distinguish between the modeling aspects inherent in finite elements, which are unique, in contrast to all the supporting methodology developed in other fields, some of which dates back to the 1800' s. As civil structural geometries became more complex, civil engineers turned to finite element models in the 1960's, with such ferocity that at the present time, more of the public domain work, computer programs and text books are due to civil engineers than any other technical field. Workers in field problems such as hydrodynamics, electricity and magnetism became serious about finite elements in the late 1960's and early 1970's. Each of these fields had already developed tremendous capability in numerical analysis and only needed to add the modeling ability. At present, the new frontier is fluid mechanics and the interaction between fluids and solid components. The finite element theory may provide a new way to look at fluids problems since it is strongly topological in nature, and the layout of the grid on the fluid field has an impact on the convergence of results. III. Examples of Finite Element Solutions Typical finite elements are shown in Fig. 1. In each case, one needs to assume the internal displacement field (or temperature, flow velocity, etc.) and develop a model relating forces at / the nodes to displacements at nodes. Fig. la. One-Dimensional Element. This is called the displacement method (Truss, Beam... and is the most common approach in finite elements. When the elements desired have been created, they are entered into an element "library" within a computer program. The user then assembles a model of a complicated structure by Fig. lb. Two-Dimensional Element. (Plane Stress, Plate...)

4 using combinations of elements. The load ramp in Figure 2 is an all- welded aluminum structure, with steel grating on the bed. The picture shown is a 3/4 view of the assembled plate and beam model. The use of such corn- puter graphics is extremely important to avoid errors in connecting the elements. The rest of te patic r Fig. lc. Three-Dimensional Element. ments. The result of the particular (Solid). study was to reduce the weight by 10% while maintaining the same load capacity. A stress analysis was desired for a cylindrical tube with two slots cut in it as shown in Figure 3. The axial load caused high stresses at the ends of the slots. A contour plot of maximum shear strain on a "flat" between the slots is given in Figure 4. This showed an unacceptably high stress at the radius of the slots. After carrying out a parameter study on the width of the slot and the radius, the design was abandoned. The total study cost about $4000, or approximately 1/2 as much as a photoelastic study would have cost. An interesting symmetry is present in the geometry, but the general purpose program SUPERB could not exploit it at the time, to allow solution of a smaller portion of the body. The program NASTRAN, MSC version, has "multiple point constraints, " which can take advantage of this symmetry. A problem for solidification of hot napthalene as it is drawn down a cooled tube is shown in Figure 5. solid-liquid interface hot water This work has been done by Antonio interath (98t w F) bath (98~F.) Valle for his doctoral dissertation in the Chemical Engineering Department at the University of Michigan. The finite elements used were very ice water complicated and included temperature bath (32~F.) and convection efforts. The grid used in Figure 6 for the first quadrant J resulted in the fluid convection shown solid napthalene drawn off in Figure 7. The solidification line involving a phase change is denoted. Fig. 5. Solidification of Napthalene.

,.~~ —--— ""l~~~'T~~~"TTT~~~"~I 1 (.. PP) C, i~, S^ ^ CL x t 1| ^$^ ^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~I, &^ ^Q~~~~~~~~~~~~~~~~~~~vr7 ^;~~~~~~~~~~~~~~~~~~c ^ /QN^3^

6 Fig. 3. Finite element mesh for a cylindrical tube v/ith clots.

n A CL _ rI - ~ cm 0 C 0 0, o e O.- 0 o o o o - a: c: i ~ L, o in o in o UN (a C O C( C C ) C3 C) a o C Si C) 4j C _ I- I- en z 2-Ct i'nLo co -......_.._ — ________... (' ('3 ClC C 0 C3 a= cc - 0 2 - Cie I)a: 0 a:.J — o CLi 4,-4

8 Z 0,2 1,0 Axisymmetric heat conduction and thermal convection mesh. 576 Elements. 3axiymt eat codcina n thra oveto eh 576 I Elements. 7

L;} *9 ~ ---- - ------ CONi 7CT IOn PATTERNS FOR olMESH 2 37 mm ID, STATIONARY, Thot = 98C Gr = 0.95E7 Fig. 7. Convection patterns for molten napthalene.

10 In comparing computer costs for this problem with an earlier finite difference solution, it was found that the finite element method was more expensive. The convenience of the finite element method may eventually make it more desirable, even for this type of problem, in which finite difference methods are known to be most competitive. Finally, on the light side, one can imagine the hero of finite element theory rushing to solve the next problem (Figure 8). After sketching this figure, the author recalled that he had participated in several studies modeling the human, including the human head, knee,. artery and heart. A three-dimensional model of the heart is shown in Figure 9. There will be much work in the future along these lines! Fig. 8. Finite Element Man to the Rescue!

11 ~ cf * / / ~Vent fricle Fig 9. Finite lement model of the heart. Heethaar, Pao and Ritman, Computer and Biomedical Research, 1977 (approx. ).

12 IV. Notation A. Vector and Matrix Notation The notation of Duncan, Frazer and Collar will be used for our matrix work: column vector {x} row vector LxJ square or rectangular matrix [A] transpose of a matrix [A] diagonal square matrix [B j -1 inverse of a square matrix [C] B. Elasticity (3-D). (u(x,y,z) displacement field {u(x,y,z)} - v(x,y,z) I w(x,y,z) J x(X,Y,Z) -(xy, z) ~:(X,YZ) strain field {e(xYz)} (x)y z) y (x,y,z) c (x,y,z) rZ (X, y3 Z) strain field {a(x,y,z)},y T (X,y,Z) Xy Ty (x,y,z) yz T (x,y,z)

13 C. Finite Element Notation nodal displacements for a single element: q q2 q. {q} - (.. qn nodal forces for a single element: Q Q2 {Q} - <. ). Q. n nodal displacements for an assembled system: r r {r} - | 2. r N nodal forces for an assembled system: R1 R2 {R1 1: {R} R N' generalized coordinates: If {a} -_.'t

14 V. References 1. Zienkiewicz, 0. C., "The Finite Element Method," 3rd ed., McGraw-Hill Book Co. (U.K.) Limited, London 1977. 2. Courant, R., "Variational Methods for the Solution of Problems of Equilibrium and Vibration," Bulletin of the American Mathematical Society, vol. 49, 1943 (pp. 1-23). 3. Argyris, J. H., "Energy Theorems and Structural Analysis," Aircraft Engineering, 1954-1955. 4. Turner, M. J., Clough, R. W., Martin, H. C. and Topp, L. J., "Stiffness and Deflection Analysis of Complex Structures," Journal of the Aeronautical Sciences, vol. 23, no. 9, September 1956 (pp. 805-823). VI. Textbooks 1. Desai & Abel, Introduction to Finite Element Analysis, Van Nostrand-Reinhold (1872). Call # TA 640.2 D48. 2. Zienkiewicz, The Finite Element Method in Engineering Sciences, McGraw-Hill, (1977), 3rd ed. Call # QA 640.2 Z66. 3. Huebner, The Finite Element Method for Engineers, Wiley-Interscience, 1975. 4. Tong & Rosettos, Finite Element Method, MIT Press (1977). Call # TA 347 F5 T66. 5. Bathe & Wilson, Numerical Methods in Finite Element Methods, Prentice-Hall, 1976. Call #. TA 347 F5 B37 6. Gallagher, Finite Element Analysis Fundamentals, Prentice-Hall, 1975. 7. Martin & Carey, Introduction to Finite Element Analysis, McGraw-Hill, 1973. Call # TA 335 M37. 8. Przimieniecki, Theory of Matrix Structural Analysis, McGraw-Hill, 1968. Call # TA 642 P77. 9. Strang & Fix, An Analysis of the Finite Element Method, Prentice-Hall, 1973. 10. Hamming, R. W., Numerical Methods for Scientists and Engineers, 2nd Ed., McGraw Hill, 1973.

15 o o u~ LA LA o LA 0 0 iL il o r- 0' LAu' L Ln 0s U) 0' "4 Tf4 " 00'-4 "4 0 0 "4~ 0 -4 N N N N f rN l N N N U'4 0c 4'4'4 44 t44' 0 044) a l 044 I ~~0 U).F. 44 2 ~; S.r," i i ^ 00. 0 o.0 44 0 U) ~~~~~~~~~~~~~~~-b U 0 4 14. 1S 4 U ) 0 Q24, 44 o o - *s -s.r (d - s s " 4~~ ~ ~~ C ~ 1. F 0 0.~~~~~~~~~~ ~0 0i) 4, o, 51 a | 1P ~- s I t r4 a O O~~~~~P 444 401..5'U0 U 0) &~~~U 4' 0 >u U, l * S 0 o 0~~~~~~~~~~~~~- 9 ~ ~~~ S 0 0 > U~^'0~ 0' H'4 0 "- o'0114, 4 >0 V ^' 14 >0 0. 0 14 V 0 14 1 4 " 1.4, 0 44, ^ 2 S iS-S~ ~ ii S'^'S S ^-S o to J5 W P W to 0 u EE 0 o. F k " Z ~~~~ 4400~~~ t ) O 44 0 0 "4 3 a 0 0 0 S4 0 a0 2 3 r BIBC4 ro ">'.. 0 - 0 U y ^I SE iS J S1 5'- ^ 5; J' r; S gs ti3 <u 0? * 14S r 14 0UtO k 04 C ~-*~ ~ ~'S <u **"~ Cd $30 0 0r 0 0 0 V 0 0V'4'4 014C0004~ U 0 Z 00E 44' H 0~~~~~~~~~~~~~~~W0 )z C 0 - * ( VA) C 0' s. o. s. 0 0 ^ 0 Q"40-< -< -< -< ^t ^< ^ -4 -< ~-< -r 4S4 u o4 "4 ). < Zo (4 0 0 ~ 0 i V "4 " - - i^-.a.oi (Q'" -s.44 rS' - *0 H o U) 14.t bo Z m 0g S Z 4 >4Us) >4 Ui 0 0 ^ 0 00 v.S 0 S' u.44>. 14 4 040t0d ~-( L ~ ) cU) U 0~~~~~~~ 0 -v "4 0 d IV 0 0 *r ~ ^ 2r~ ^~ S~ 2r~ - ~ 2~ 2~..2 <..00C 4' CIO r4 x.O~ ~k k40 4) U -: c f (uafl " f l C (d1 c4 to %ZS4 r4 0 0 S r " i 40V-4 t o4 V Ao F 0 00 rz 2 H i "I l I.U u r. O x "S 4 r4 J 44 4) 0 >0 ~' 0 I 0 I II F 4)(U4 o'v " uOD ~ i ~| ri9 si ^| k s^ ji n s i - ^ ~ a i <u c 7r a u clJlS' C' S ~ u ): L2 * c 2 4>4 0 >> >4 0 0". s j & g 4 <- e! u a o4u u 8)i 0 U V 44 U V 2 2 *- 0 V 00~4 co 0 14 0 0 14 0 44 (4 0 0 0 0 1 0l ( 0.. 14 4 14 14 a 1 > > O M ~ 5 > 40 J 44*' 4 J 44- 44 *' * * * 44. 44 44 44 44 *2ga U 4) u U t 44o (A CI ~ ~ ~~ IC 1~~~~ rC ~~~Ci ~ ~ ~ CI ~~~C Ic 0) *tfiM < cr k (d g~ O~ OI Q. Q~ Q~ 0~ OI OI Q~ 0` QI Cd 0 L4 O r4tn r0 0 4 cd..4 S500 F " U 0 ~.. 0 U ~ n H~' ^ )> 4 t 14 P 4 co 14A 0 U 00 4 4 0 P4~0 U 04 a IV L] 0 44'01 " S <U O S C 3 t) <a >*; N (T) ~;S)Sd1.0 F, 4 0 V 44 Ai 0 0 0 40 <y - 3N.W. 9 o 2 Q N SUoh oQ O Q S ^ D iS k oL 2 a) 4 (d rS: Id ~~~~~~~~ ~~ ed cb 404 o PO Sc C~~~~ c cs4d to r f 4)..4 0 0.. 4 0 0 0 o a, o la V Ai V *A 4, J o:~l U r ~e i; a a u t Sc ":; i 0 0 E E la q a 9 r 3 la 0 0 0 0 t5 or o 0 0 0 V tod W p) v ~ 14k r W >k k 4 ~ 0 A d W4 a 4i 4- b 4ia

16 o Lu o o o o o %n o o0 o sr os r > ( <4 C4< u. i. - I 0 0 0 0 u0 02 ~ 00 _ _ _ ax u0 0 0 0 ) E ~' 0 _ ~, 5,, ~,- E 1 B > * 4 6 E a, *5 = c | 0o C o 2o o 4- 4 v 4) 4 4'l*l "1 0'' ^' i- rr rtc 4a - i o 0 |5 0 u0 " E<U4 0 (0 I~ ~( (4. H U o axU U o. 0. 4'5 E S.c 8s 0 *c -S - -.2 5 g S 0 L< 0 0 < 0 02 02 S 0I u~ g 0. 6 O1 6 bla SZ.'0 c.0 r'.\ (402 0 O.0o 0i A 4i * 0 02 >b H (A 40 re 0)

LINE ELEMENT. May 30, 1978 ASSEMBLY PROCESS. W. J. A. I. LINE ELEMENT A two-node, constant area line L element is to be created. The ele- A ment has two nodal degrees of free- / dom ql and q with their associated__ _ forces Q1 ana Q2. (The general rule __ o W X. is that the product of nodal force and displacement must yield energy.) X This simple element can illustrate several important features of finite ele- Fig.. Two-node ine element. Fig. 1. Two-node line element. ment theory. Since we have not yet developed a general energy approach, we must at this point use equilibrium ideas instead. For instance, it is clear that 2 =-Q1 (1) from equilibrium in the x direction. We also have the one-dimensional stress strain law a =E ~ i /x a (2) X X and the one-dimensional strain-displacement law x d fd w Ere = dux) (gfe c icm) (3) where u(x) is the displacement field. If we assume that strain ~ is constant in the element, we can x approximate the strain-displacement law by AL x L q2-ql (4) This turns out to be exact for a constant area element. It also implies (with Eqn. 2) that stress is constant. r From equilibrium at the nodes, z 2 L 1 1 A Q - = -aA (5a) _ A.-.S Q = acA (5b) 2 = ~ A L i- ^A Fig. 2. Equilibrium. = A A? Ace 17

18 Rewriting this in terms of strain and finally displacement, one has Q1 = - AE(q2-ql)/L (6a) Q2 AE(q2-ql)/L (6b) This can be put in matrix form: rQ1 "1 q (7a) f AE lE Q2 ) L-1 1] 1 q22 (7b) This is the fundamental relation in finite element theory, the load-deflection law. The matrix relating load and deflection (including the scalar factor) is defined to be a stiffness [k]: - [k] = -- - 1] -Ves c\': 1 " -& (8) L -1 1 1 4 3 eI,." We were fortunate to obtain this answer by equilibrium methods. We were even more fortunate that the answer is exact for the constant area element. Note that the determinant of the stiffness matrix for an element is zero, i.e., the matrix is singular. This is always true. II. LINE ELEMENT. VARYING AREA. To push our luck a little, A we will attempt to find the stiff-/ ness matrix for a two-node line element which has varying area. *I -- In this case, we again have Q Q (9) 2 = - Q (9) from equilibrium. We again have the stress-strain law Fig. 3. Varying area line element. = E e (10) x x and the strain-displacement law du x dx For tlis element, we.vold assuming) constant strain a priori, because we know that it cannot be so. (We have a constant force transmitted through a varying area, which leads to varying stress and then to varying strain.) A general way to attack such a problem, and the reason the method is called the "displacement" method, is to assume a displacement field such as

19 2 3 u(x) = a + a2x + a3x + a4x +... (12) "displacement function" * or to interpolate the displacement between the nodes using u(x) = ql(l - ) + q2) (13) "shape functions" or "interpolation functions" When there are only two nodal coordinates q and q, it can be shown that the displacement function can have only two independent constants. From other reasoning (that a1 and a2 are important and cannot be discarded) one finally obtains u(x) = a1 + a2x (linear displacement model) (14) which leads to E(X) = d(a1 + aX) = a2 (constant strain model) (15) and we are back to a constant strain model whether we like it or not! Thhe solution will be only approximate. We are left with an apparent paradox. The assumed displacement field causes a uniform stress but a nonuniform force through the element (because of its varying area).. - If we attempt to define nodal loads < ^ ()A \, in a way to satisfy local equilibri- J ^. ^ A - um, we have a free body diagram as _ in Fig. 4. This leads to- \ Q1 x- a o )A1 (16) du(O Fig. 4. Attempt to satisfy dx 1 internal equilibrium. (Failure!) * It is possible to describe the internal displacement field either by displacement functions or by shape functions. Each has its advantages. There must always be a unique mapping from one relation to the other.

20 Using the shape function in Eqn. 13 to represent the internal displacement field, q2-ql Q E(-= )A SATISFIES INTERNAL 1 (7a) EQUILIBRIUM. VIO-,,~~~Likewise.LATES NODAL (GLOBAL) FORCE BALANCE. Q E..1)A (17b) 2 = E( AL A2 17b) UNACCEPTABLE! An alternate philosophy is to maintain nodal (global) equilibrium and to violate internal equilibrium. For instance, one can take Q = - A 1 - x average _- - " -q — __ _ = - L )A average (18a) _ _ q2-ql Q E( -)A (18b) 2 L average Fig. 5. Satisfaction of nodal equilibrium. This leads to EA I = average (19) Qtoj -1 1 q2 Q2 Surprisingly, this is a satisfactory model of the varying area line element, even though internal equilibrium of the element is violated. The displacement field leads to a stress field which is nowhere in equilibrium in the interior of the element. This is a common feature of elements to be developed by the displacement finite element method. QUESTIONS: 1. Write out the matrix form of Equations (17a) and (17b). What is the mathematical problem with [k]? 2. What does A mean? This question is cleared up for this simple element later y an exact solution using a more complicated (logarithmic) displacement field. 3. Evaluate a and a2 in Equation (14) in terms of ql and q2. You must recognize tiat u(O) = ql and u(L) = q2 to succeed. II1. ASSEMBLY OF CONSTANT AREA LINE ELEMENTS. The assembly process for finite element theory is straightforward and powerful. A good way to visualize the process is to look at two disjoint line elements and then to pin them together.

21 ASSMSLY -E L Fig. 6. Two disjoint line elements. Q 1 - 1 f -1 ql -1-E^ ~^2 <: 1. q (20) Q _ 1 - q2 L 2 L-1 1 q4 (21) 2Q 4 4 21) The way to assemble is to create a matrix problem large enough so that each element can be "imbedded" in it. Realizing that upon assembly one hasqq and that there wille only 3 independent coordinates, one writes -q2 1 8 ) l(22) 3 3x3 3 Equation 22 is a logical format for the assembled system. The equations for the first element can be written, by adding a trivial equation: 1r t~' ^ ~ ~- [-) ^Il- > a-D FQI E~A1/L -EA/L 0 ql Q2 = -EA/L E1A/L 0 q (23) 2 23 o 0. Lo 0 0 q4 Likewise, the second element is described by 0 _0 0 0 0 q Q3j-'-2 -- QL- -0-g tA kz 2Q ~ E2A2/L2 -E2A 2/ 1 2 (24) Q Q | -E2A2/L2 E2A2/L2 4 Q4 -I \

22 Equations (23) and (24) are of the mathematical form {X} = [P]{Z} (25a) {Y} = [Q]{Z} (25b) and can be added {X} + {Y} = ([P] + [Q]){Z} (26) Assembled Stiffness This is what people mean by "stiffnesses just add."'\- s. -'il-k+;)?z W In our example: 2=-^' l. +'Q1 E1 A1/L 1 -E1A1/L1 0 Q +Q3 - -EA1 /L1 (EA1 /L -E 2A/L2 q2 < (27) 2 31 1 1 111 272 +EA \ ) IL) / + E2A2/L2) Q4 0 ~-EA2L2 2A2/L2 q4 The stiffness matrix has assembled nicely and the only displacements which appear are the independent degrees of freedom. The force vector can be simplified by imagining a fictitious pin* at the joint. There is in general an ckternal force, on the pin, which satisfies = ^Q + Q (28) hence we define the new symbols [K] and {r}: — \ Ql - = [ CK] {r} (29) Fig. 7. Fictitious pin. 3/4 view. assembled independent d.o.f. assembled stiffness In most problems, the structure is continuous and this mental process of separating elements and then putting together again with a pin never occurs phys ically.

Wn W) cn rn => ~ot ='d ~., o~~ Y1v cl~~~~~~ l > H > > ^?c! ^ ^~> W ~z~~~ ~ 1-d I'd cn C~~~~~~~~~~~C ii a "as. IgI > tLI I | -.i^S^ &.!. s- o~ g. ~ ~ >sg= > W NV i? cri cd C en W r t. I o 0 ^ En t o o rt C) O O C) O CD O 3 C+ CD 4 nt (D C I S:' I jm ) O CD D -S 14 P o 0 U (D (D 4 C-t. CD c+ CD c H^ s C+ (D Cn CD CD~3 0 OQ cr9 P m ~ ~ ~ ~ ~ 1(D~~~~ 1-o - o o 0 W (D LnojO( 00 (D En QA - ^ ~ * -:3 --'4 u ~ I-A.:r rsj c-i o cCFJ U1 Ln -41 (D CDCa30 3t rt, t^ G9ti o.( O (D U^' ED CO ~ 0 O /,- 0^ -^f(l Lj 00 C^ V C ^ ~ ~ ~ ~~~~ eW (DD^ t! p0t 0 ^ 00 (O Y C 0 (D CD t rH "0:! I~'" r 0c1 Ln til W r C` y W~~~~~~~~~~~~~~~~~ 0 VJ S S-^ 4 S3 ^ S OC ~D O rn =f ~D, W lI? 0 CD O (P tl CY (D D O -J 4 - -I m~ m (3 ^ ^ Og ^ ^S~~~s ^ ^ S0?^m g iN*O P.,',=< aa En H -) sL nU dC I I (D > s~~~~~~~~~~~- I Ns LO U.) x &~ 0.T+ H O s) II 55 FL?s U1?^O tl O CD~~00 W tn CFN Cf O i,,: Y-P OOc C1~~ ~ C ~o O r, - U~~~~~~ ~~~~ ~ ~~~~~~~~~~~~~. W........ -, >,4 ~,_, < ( 0 0 0 (D o 0 > o (C 0 0 0 D 0 m En~,-t..~~~~^ —. 0 (D C. 0 r, jr} CL {P n P~f V+ ~P ~3 3 CD ~4] ~ C1 -J (DVrf ~~~~. to. (I E CE3 0ft rt7 VD. 0 %I 0 I 1 I I I I ~ ~ ~ ~~~~~~~~~~~~~~~i I, —a ED H (D-(D F*. O C} P0(D 0 F- C Pt W H. (D (Dc 0:3, {3~ 3c En.A P) D WCf O ~ u c~ (3 (D re I (Dt ~~~~~~~~~~~~~~~~~~~~~~~~~~~~D~~~~~~~~~~~~~~~~~~~~~~~~I.. I-"-' n3 r CD 3 ct r ct I ct (b O~~~~~~~~~~~~~~~~~~~~~~~~I — (P ci Y Y cn td Ut~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.m

23 Fortunately, the force vector consists of only externally imposed loads! This is a wonderful situation, because the finite element method then suppresses the internal forces and the need to know them. (This is comparable to Lagrange's equations in analytical dynamics which are a means of suppressing internal forces of constraint and which allow easier solution.) One finally defines the external load vector symbol {R} and writes: {R} [ K ]{r} (30) " tA\,,A( H. sf externally - a ssc! i applied r loads This simple matrix equation is the heart of most finite element solutions. The equation still needs to be modified before solution since the assembled stiffness matrix is singular at this stage. This will be discussed later. A similar assembly process can be done in electrical and fluid circuits and field problems in many branches of science and engineering. For each class of problem, the law used for assembly is appropriate to the variables, and includes Kirchhoff's current law in electrical circuits and conservation of energy in heat conduction problems.

May 21, 1979 CONSTANT STRAIN TRIANGLE* W. J. A. I. Review of Plane Stress/Plane Strain The theory of elasticity is concerned with tensor quantities and relations between tensors. For our purposes, we will be happy to deal with vectors and the linear transformations between them. In three-dimensional elasticity, for instance, the sketch below Ladicates some of the important concepts: = ^i fc n (ciy. (2) / I I — ~,/., Equilibrium Equat ions a = 0 (1) and z = Efcn(o, y) (2) so LlaaL the significant variables become: S x-S tX Reference: Turner, Clough, Martin and Topp, "Stiffness and Deflection Analysis of Complex Structures," Journal of the Aeronautical Sciences, Vol. 23, #9, September 1956, pp. 805-823. 24

25 Specifically, the transformations used to relate stresses, strains and displacements, are etook<S- A 1&JAKC L - 7fSV Or-Lv^f cj-vA2V~-5 Vrx E/1-V2 Ev/l-V2 0 x 0 TY = V/l- E /1-V2 / 2 0 L, ay y y [^J L~ 0 G [Y [^ -a xyj XY a l Comparable linear transformations exist for plane strain. Note that the stress/strain law involves constants, whereas the strain/displacement law involves derivatives. II. Development of the Stiffness Matrix for the Constant/Strain Triangle The case of plane stress discussed in Part A applies to many twodimensional situations for plate-like bodies. Let us consider the special plane stress problem where a triangular element has been cut from the plate. If we imagine nodal forces and displacements as shown for a general coordinate system, it is then helpful to consider these new quantities in relation to the internal (generic) quantities as follows: sM- ~ Coo,4 If we remember that, in general, stiffness matrices relate nodal force and QiODAL TCCS -V displacement vectors, then our job is to find whether linear transformations \U"W ox cEx r u(xy) Q2 E [ [?J q2 can be found in the two locations with question marks. One needs only then to proceed leapfrog-fashion with four successive transformations to relate nodal displacements to nodal forces.

26 The unknown transformation between nodal forces and internal stresses can be found by simple equilibrium arguments. First, strains are assumed constant. E~ = a - (,) (4a) E,~~ ~~= b ~(4b) Zy = b Y = c (4c) xy Then the triangle is considered to be imbedded in a uniform sheet of material under constant stress in one direction, say y. (The other stresses can be considered in turn and the results superimposed.) that would place the triangular element in equilibrium. A better approach half to each nearby node. is1 M' f t' The exploded diagram shows the triangular element with concentrated loads at the midpoint of each side. This is a possible type of concentrated loading that would place the triangular element in equilibrium. A better approach is to break each concentrated load into two equal parts and then to apply half to each nearby node. Tt ^- - -^t~~

27 Q' —'XJt It can be seen that Q4 ay(x-x3)h/2, for instance. One can carry this process out to find the other forces for this loading and then to include a and T stresses. The results is a linear relation: {Q} = [i]{a} (5) where the symbol & has been chosen to stand for "equilibrium." Now we must solve for the linear relation between generic displacements and nodal displacements. This is more difficult conceptually because an intermediate step develops. First of all, the strain/displacement relations are integrated: u(x,y) = E dx v(x,) = edy (6a) = a dx = Jb dy (6b) =ax + f(y) = by + g(x) (6c) When the unknown functions f(y) and g(x) are substituted into the shear equation / Du Dv 7 = c-_ (7) -- + -x = C L by 3x c 7 one obtains f'(y) + g'(x) = c (8) The derivatives f'(y) and g'(x) must therefore be constants. Arbitrarily choosing f'(y) - A yields g'(x) = c - A. One can summarize the results in matrix form: rul ~x 0 0 y 1 0 a LO Y x -x O 1_ b (9) A 6 as o_ Ate + A B [q]{a} (10)

28 This doesn't seem to help because we have merely defined a new vector of constant coefficients (three of which are the specified constant strains; the three constants of integration can be shown to represent rigid body modes). It is rather easy, however, to relate these constants to the nodal displacements, since q U(Xl Y) = a x1 + A Y + B q2 v(x,yl) = b y1 + (c-A)x+ C (11) 6 v(x3,y3) = b Y3 + (c-A)x3 + C In matrix form {q} = [A]{a} or {a} = [A] {q} (12a,b) Now, put all these results together {Q} = [ ]{o -stress-]{ strain - ]$[C][strain-]{u} = [ [Cdispl. = [ [C][[S-D][(]{a} {Q} = [ ][C][S-D][q][A] {q} (13) {Q} = [K]{q} (14) One can also define [B] = [S-D][f][A] 1 [B] [S-D][l], [N] E [[A]1 Concluding Comments: This equilibrium approach works for this case, but is difficult to extend to more complicated elements. Energy methods must be used instead. Turner, et al., did not use a general cartesian coordinate system (a global system), but rather used a system with the x axis lined up with the base of the triangle (a local system). III. Assembly of Triangular Elements A plate is cantilevered from a wall as shown, with load -. Consider a 3-element representation of the plate using the constant-strain triangle. There are 5 nodes and 10 degrees of freedom. For each D.O.F., we must specify a force or a displacement.

29 For each element, we have {Q}= [k]{q} where [k] is a 6X6 matrix. Assemble the stiffness matrix for the structure using symbols for the element stiffnesses as shown. I, r / I_ / X X' ^. i /T::s;x X a h _ _ _6 / [K] = [kl] + [k2] + [k3] = A X A A /, X ^r a ns^ ~~~~I/ xax x ci~ EXAMPLE OF TURNER TRIANGLE The stiffness matrix for the ) ) Turner triangle is found most easily by using a local coordinate system / shown. One defines X21 E X2 -X1l~o~C-lx x X X / I and obtains after much calculation:

30 Y3 x2 _ _ -vX x21 x21 x21 x21 2 Vk2 x 2 xX X x2 2 vx2 1 2 0 x2 x21 y3X21 X21 Y3X21 Y3 [ k ] Eh - 3 - 2 21 rx C U) r~ i=Eh -- ^ _ _J_ _ _2 _AVx Vx 2(1-v2) 21 21 x21 x21 4-J 2 u Vx X X2 Vx, x x I c x1 1 x2 x1 0 x21 y3X21 X21 Y3X21 3 0 0 0 0 0 0 X2 X1 x21 - v Y3 Y 3 Y3 x2 x2_ _ x_2 X2 X2 X1X2 X 2 X2 Y3x21 x21 Y3x21 21 Y3 2 Y3 - -1 3 x2 21 X21 x21 2 +Gh Xlx2 xl xl x! x 0 2 Y3X21 x21 y3x21 X21 Y ( _- y3 ___y_3. X2 Y3 X y O X21 x21 x21 x21 1 _ 21 _ 1 x1 1 21 0 x2 x1 x21 0 0 Y3 Y3

/g. 4 t' ) -4 IA x6 A- </ 6' v, (,..,,.i~ ri 2-'.X,,f' - P nE,'; -.e,. A t csrol Y clr - X!i -..;

VIRTUAL WORK THEOREM October 6, 1977 POTENTIAL ENERGY THEOREM W. J. A. I. Background Much of the finite element theory in solid mechanics is concerned with the stress-strain law (constitutive law), the strain-displacement law, and the load-deflection relation. These are symbolically sketched in Figures 1-3, where a scalar version of the vector quantities is given. These sketches are an arti- fice, but are very useful in sort- ing out concepts. Many finite element problems are linear, where each of the rela- Fig.l. Stress-strain law. (nonlinear elastic). tions in Figures 1-3 will be a e straight line. Also, most finite element problems have no prestress {oo}or prestrain {C } such that the lines in Figures 1-3 would all pass through the origin. (The un- loaded, undeflected structure is almost always available as a reference; therefore the load-deflection relation passing through the origin -Q in Figure 3 is the general case.) Let us not assume linearity yet; t us nt a e linrit y Fig.2. Strain-displacement law. (general case). but, rather, retain the general nonlinear case through the discussion on virtual work..J {it- -. rt i Fig.3. Load-deflection relation. (general case) 31

M o 4 32 i A. Stress-Strain Law (Constitutive Law) The general form of the nonlinear, elastic stress-strain law is of the form {o} = function ({E}) A linearized version of this law, valid in the neighborhood of a reference point I (Figure 1) is {a} = [c]({} - }) + {A 0 The standard practice is for this reference point to correspond to the origin in Figure 3; this means that {a } and {c } are prestress and prestrain present in the unloaded structure. Prestress and prestrain are zero in most structural problems. B. Strain/Displacement Law The relation between internal strain and nodal displacements is sketched in Figure 2. This is a geometric relation (kinematic) and is not a function of the material, but does depend on the total strain involved. Most engineering structures remain in the linear range. C. Equilibrium Position If a body is loaded with static loads {Q},. it deforms and reaches a static equilibrium at a point labelled II in Figure 3. The shaded area represents the work done by the external loads on the body, if the loads. are slowly applied. D. Virtual Displacement A virtual displacement is an, co, / infinitesimal displacement from position II, in a manner consistent with geometric constraints. i The body moves to position III and undergoes an increment of strain energy shown as the shaded area in Figure 4. The loads during this virtual displacement are assumed to remain constant. This virtual displacement is called {Aq}. It results in a virtual strain {Ac}. Because strain energy is approximated by the shaded area in Figure 4, the change in stress going from II to Fig. 4. Strains during a virtual Fig. 4. Strains during a virtual III is negligible. The strain energy due to virtual strain is displacement. energy due to virtual strain is

33 Q (3) AU = (stress)(A strain)dV v II II + III and the amount of work done by the / oac emi:n r c0/:external forces (shaded area in |'<yrq 4 Figure 5) is: / / N (4) AW = Z Q A qi i=l 1 II II + III fi___ o l_ The energies in Eqns. 3 and 4 will be equal only when the material is nondissipative, i.e., elastic. The concept of distributed external forces such as gravity has not yet been included. Fig. 5. Virtual displacement. E. Work Work is defined most fundamentally in an incremental manner < a- o 1-k~?od — (5) AW E (Force)(A Displacement) One must be on guard to never define work as a product of force times displacement and then to take an increment: W = Fd Beware!!! + d False!!! AW = FAd + AFd J Furthermore, the distance involved in this expression is the distance moved by the body on which the force acts, rather than the distance the force moves. This is very important with respect to sliding forces. F. Virtual Work r-s- (c ~o< Vec, O W ~e o\J 0o V.o) <Ao -A This is the work done, whether by external or internal forces, during cool!-l a virtual displacement, i.e., in moving from state II to state III. II. Virtual Work Theorem In mechanics, there are three fundamental statements of mechanical equilibrium which may be used interchangeably. These are Newton's laws of motion (principally the second law, F = ma), Hamilton's principle and the virtual work theorem, all of which can be used for statics and dynamics problems. The virtual work theorem applies to all of mechanics, but we will use a version valid for nonrelativistic, nonthermal problems.

34 Virtual Work Theorem: A body is in equilibrium if and only if the sum of all virtual work done during an arbitrary virtual displacement is zero. (increment of work (increment of work by internal for- + by external for- = 0 7 ces II - III) ces II + III) Fig. 6. Configurations of a i.e., body under load. (6) AW. + AWtn = 0 General Statement of Virtual Work in + IIl II + IIIa Theorem for Bodies in Figures 4, 5 and 6. An important subcase occurs when the material of the body is nondissipative, e.g., nonlinearly elastic. The work done by the internal forces (i.e., stresses and strains) can be shown to be the negative of the strain energy increment by an involved mathematical proof not done here. (7) AW. = - AU internal II + III II + III Inserting Equation 7 into Equation 6, one obtains: Modified Virtual Work Theorem: A nonlinearly elastic body is in equilbrium if and only if the increment of work done by external forces equals the change in strain energy during an arbitrary virtual displacement. i.... ~.........____ Virtual Work Theorem for Non(8) -AU + A Wexternal = 0 dissipative materials, typically IIII II II III nonlinear elastic with prestress and prestrain. Comments: i) The theorems to this point don't imply the existence of either a function W or a function U which mean anything. It only means one can calculate the increments of energy, i.e., the shaded areas in Sketches 4 and 5, for the case of energy-conserving materials..ii) Equation 8 ought to be intuitive to most readers because, for a nondissipative material, the work done by the external forces during any displacement ought to equal the energy stored in the body. This is really a conservation of energy idea, and some people prefer it to the more general statement of Equation 6 for that reason.

35 III. Potential Energy Theorem Finally, the potential energy theorem can be derived. This theorem is less general than virtual work and is restricted to linear elastic systems and to energy conserving external force fields. For the potential energy theorem, we need to use a variational operator 6. We will change our viewpoint of the A symbol used above through the identity (9) 6{u} - {Au} II II +* III In other words, we pass from an incremental usage of A to an operator form 6. The symbol 6 will retain the concept of a virtual displacement, i.e., a small change from the II position, with forces and stresses remaining unchanged. If the external force field is conservative (energy-conserving) and hence can be derived from a potential, i.e., (10) -i = -r L ~ "- work potential external forces] a generalized gradient operator which acts on all displacement-like quantities. then we can consider lto exist. Putting all of our "work" ideas together: AW x = (Forces) x (A displacements) external +1 II - III = (Forces) x 6(displacements) II II = 6(Forces) x (displacements) II II (11) = 5(-~ II where we now admit the existence of a force potential~?and use an operator 6 which acts only on displacement-like quantities. Also, since the system is linearly elastic, we can write symbolically AU = f (stress)(A strain)dv II -^ III v II II + III II +III v a = / (stiffness)(strain)(A strain)dv II II -+ III = f(stiffness) (strain) 6 (strain)dv II II I' 1 2 = 6 - f (stiffness) (strain) 2dv >1 k<- Fig.8. Virtual strain energy. (IEncrement in strain energy due to virtual strain).

36 (12) A U = 6 (U) II +- III II This, of course, implies a strain energy function U as a function of strains and an operator 6 that acts on only strains (which are displacement-like). We now combine results of Equations 8, 11 and 12 to get (13) - 6s - 6(U) = 0 II II (14) 6(U +W) = 0 II II or Potential Energy Theorem for linearly c(1 5 6y) 0=X elastic systems with conservative (I II,~~ 5) 6Cexternal forces. Potential Energy Theorem: The potential energy of a mechanical system made of linearly elastic elements and exposed to conservative forces is "stationary" (has a zero first variation) at the static equilibrium configuration. Comment: If one generalizes to systems with even mild nonlinearities, then the question of stability of this equilibrium position becomes important. Then the second variation of the potential energy is needed. This subject is studied in a more advanced finite element course.

DERIVATION OF A FINITE ELEMENT October 10, 1977 BY THE VIRTUAL WORK THEOREM W.J.A. I. DERIVATION FOR CONCENTRATED NODAL LOADS Consider a single finite element with only concentrated external forces acting on its nodes (Fig. 1). The three configurations are shown. Use the modified form of the virtual work theorem, but use only the lower case 6 to indicate both incremental and operational procedures. 6{u} - {6u} etc. Fig. 1. Typical Finite Element Recall that the flow chart for unknowns is: e c ~9} 3 D]f j'< ~3> (a) where the stress-strain matrix is named [C] and the equilibrium matrix is suppressed. The virtual work theorem becomes 6W 6U = 1) II +III I I + III or {6q}T{Q} - f { 6 }T{G}d = 2) I II dv = 0I II -+ III The virtual strain is found from the expression for strain: {E} = [B]{q} 3) by operating on both sides with 6: {(6} = [B]{6q} 4) Hence, setting {a} = [C][B]{q}, we have 37

38 {6q} {Q} - f {6q} [B] [C][B]{q}dv = 0 5) }T Since {6q} does not depend on the integration variables, it can be factored out, as well as {q}, to get {6q}T {Q} - f [B] [C][BIdv{q} - 0 6) Since {6q} is an arbitrary vector, the vector enclosed in the large parentheses must be zero. This leads to {Q} = f [B] [C][B]dv{q} 7) v or, upon defining the integral portion to be exactly the stiffness matrix, we have {Q} = [k]{q}. 8) This approach has defined the stiffness matrix without the need of an equilibrium matrix! Several forms of the stiffness matrix follow, for computational use: [k] i f. [B] [C][B]dv 9a) V = [N] [S-D] [C][S-D][N]dv 9b) T = [A]1 -l [] T[S-D]T[C][S-D][f]dv[A]r 9c) The last version is particularly good for computation when displacement functions are given; the next to last can be used when shape functions are given. II. DERIVATION INCLUDING EQUIVALENT NODAL LOADS Now, suppose the element has acting on it several kinds of distributed loads. These can be volumetric (gravity), surface (pressure) or line loads (as on a beam). In each case, a contribution to the virtual work will be made by a force moving the body through an increment of displacement. Although the force and the spatial dimensions vary, the relevant displacement in each case is {6u} = [N]{6q}. Hence, if {X(x,y,z)} = volume load I {T(x,y,z)} = surface load {A(x,y,z)f = line load

39 Then 6W - 6 U = 0 II + III II +- III becomes {6q} {Q} 6uT x z)d + { T{T(xy + f{u}{xz)} + {u dS{6u}T{x+(xy,z)}dv T X/ D V II II II II - / {6B}E{a dv = 0 II This is rewritten {6q}T ({O + fl[N T^da + fS[N] T dS + [N]T{}dv - [B]T[C][B]dv{q}) =: 0 The integrals all can be evaluated once the internal displacement fields (and hence [N], [f], [A], etc.) have been assigned. Each integral is given a name. {Q} l - f [IN]T {}dk Equivalent nodal load due to line load T {Q}e -n = Ifs[N] {T}dS Equivalent nodal load due to surface load e.n. 1. S T {Q} - f [N] {X}dv Equivalent nodal load due to volume load e., n.. v X [k] f [B] [C][B]dv Element stiffness and finally: {Q} + {Q} + {Q} + {Q = [k]{q} e.n.l. e.n.l. e.n.1. at T x is the general equation of equilibrium.

40 III. PRESTRAIN AND PRESTRESS Many physical problems have an initial strain (thermal problems) or initial stress. Also, in nonlinear problems, one often introduces a prestrain or prestress as an artifice to aid solution. In all of these cases, a linear stress-strain law is used: {a} = [C]({c} - { }) + {a } Then the virtual work would give ~ an additional increment of strain energy _ * I o E 6u - /{6}T{ea}dv f([B]{6q}) ([C]({} - { })+{ca })dvol 0 0 = {6q} f [B]T[C]{}dvol -{SqT} ol[B] [C]f{ }dvol + {6q} Tf [B]T { }dvol After "cancelling" {6q}T and combining with the enerby balance in the last section: {Q} + f [N] {}dv + f [N]T {T}dS + f [N] {T}d + f [B] [C]{E }dvol - f[B] {r }dvol V s JC V 0 0 = JIB]T[C] [B]dvol{q} V {Q} + { Q} +Q} + {Q + {Q + {Q = [k]{q} -{Q1 {Q}' e.n.1. n 1. e.n.l1. e.n.. e.n.l. =k]fql body surface line prestress prestrain forces forces loads IV. REVIEW OF EQUIVALENT NODAL LOAD CONCEPT The virtual work theorem has led to [k{q} = {Q}external + v[N]T{X}dv + f [N]T{T}d ++ f [N]T{ }d4 + f [B]T[C]{E }dv k external v s v o nodal loads - v[B]{ }dv.

41 We will define {Q} - [N]T{X}dv body forces {Q} -f [NT{T}dS surface forces {Q} - [N]T{}dd e.n.. /[N] line loads {Qe.n. f [BC] [C]{ }dv prestrain Q.n. 1. f[B] {a }dv prestress

DERIVATION OF LINE ELEMENT July 10, 1976 BY ENERGY EXPRESSION W.J.A. I. LINE ELEMENT. ENERGY FORMULATION (c c.. 9,, t. >""_ A concrete example is needed to tie down the preceding theory. Consider the constant area, two-node line element. We need to find the operators [C], [S-D], and [N] in the expressions for stiffness: T T [k] = f [N] [S-D] [C][S-D][N]dV [B] and for line load: {Q} fe I[N] {Z(x)}dx. e.n.l. 0 CmO^ general I ^ polynomials.. I.. 1 O, ^^W ^tcL L Fig. 2. Shape functions (general). Figure 1. Line Element The shape (interpolation) functions can be found by inspection under the knowledge that N1(O) = 1 N1(L) = 0 N2(0) = 0 N2(L) = 1 At this point, the shape functions might have the general shape given by {u(x)} = [N1(x) N2(x)] (qJ and sketched in Figure 2. The shape functions are, in fact, assumed to be linear, however, leading to 42

43 q {u(x)} = [1- x/L x/L] [N] We found earlier that the stiffness matrix [C]= [E] and the strain displacement matrix [S-D] = [d in one-dimensional elasticity. An intermediate matrix [B] is useful for calculations. [B] - [S-D][N] = [x-][1 - x/L x/L] [-/L 1/ /L] The stiffness matrix now becomes I,7 ikl =,. x L / rd-_. -, x- ~? L T [k] = A 0[B] [C][B]dx = 1/L = A - 1/L [E][- 1/L l/L]dx 1/L_ 1/L = A f/ [- E/L E/L]dx E/L2 E/L L2 = A /L dx E/L E/L j AE L - 1 1 which is the same result as obtained from equilibrium arguments, and is exact for the constant area case at hand. The equivalent nodal line load becomes

44 {Qe.n.l. f[N]T {(x)}dx k/^~ ~ load per unit length (1 - x/L) r(x) - L ( l dx x/L 0(x) f (1- x/L) a (x)dx fo x/L (x)dx II. EXAMPLE OF EQUIVALENT NODAL LOADS Choose a specific case, such as /(x) o(x/L). Then iiiJI Figure 3. Load Distribution Figure 4. Equivalent Nodal Loads. (Line-Load) s rL 2 0/ (1 - x/L)(x/L) dx {Q}e.n.l. Lf L 2 /. tZ^0 x/L(x/L) dx f [(x/L) - (x/L)3]dx /^(x/L) dJ

45 L L L2(30 L34 ) 1!i 0 L 0 4 oL < Does this represent faithfully the total running load on the element? Total L 2 1 1 3L L Load (x/L) dx = x = 03 Load 0 0 o 2 3 0 3 This checks. The equivalent nodal loads are statically equivalent to the original distributed load. ~n?.,~c ~~r'o~,!cL~~.^- \.,'~CI /_(';~v - jY (- 1.7'Cy -)'S. __ _..._-, /".! i7, ^ h- ~ $rr:;7 —-~-"~~~~~~~~~~~~~~~~~~~~~~~

....._ __. ~October 26, 1977 BEAM BENDING ELEMENT W.J.A. I. REVIEW OF CLASSICAL THEORY Consider a straight, slender elastic beam (Fig. 1). The beam is constrained to move in the x,z plane. Only lateral forces p(x), shearing forces Qz and bending moments My act on the beam. There are no axial forces. The Euler-Bernoulli- PNY) Navier approach for this "pure, — ) - bending" case is to study the motion of the neutral surface (elas-. tica) of the beam under the as- X X Q sumptions: 1) linear, elastic material 2) small deflections and small slopes ~~~~~slopes ~Fig. 1. Straight, slender beam in 3) plane sections initially per- x-z plane. pendicular to the beam axis remain perpendicular after deformation 4) there is no dependence of stress nor strain on the y coordinate. The strain-displacement relation for a beam comes immedi- \ ately from the "plane sections" assumption (Fig. 2). At a distance z from the neutral axis e = (z) - z(0* ) \ / R (x,z) = (R-z)AO - RAO RA \ / z R. The radius of curvature R is found Fig. 2. Beam, curved to circular from differential geometry to be arc. 46

47 a~w(xo) 22... I.. since the slope ax is small, 6' ~=~ | - = 3w(x, 0) 2J3 R 2 a x wax and;F -O 2 (x, O) Fig. 3. End view of cut beam. e (3) a = -z = = T on surface. x a x y z yz Because of the small lateral dimensions of the beam and because all surface stresses are zero except o on the top surface (Fig. 3), one can argue that a (x,z) m 0 (4) and 77 a (x,z) _ 0 (5) V T (x,) - 0 (6) e yZ Also, thue plane sections assumption immediately leads to =n t t ( F 4) (7 Fig. 4. Side view of plane sections zxz grid. = 0 rect stress in the problem is ss o and the stress-strain la4w of imporance sis X a (x,z) = E ~ (x,z) ~~~~~~~~~X XA~~~ZX x x but the plane sections assumption has _.._ made the corresponding shear strain _. Y vanish and therefore no energy is._ afsorbed in shearing action.) Fig. 5. Top view of beam with plane section grid. y = 0. xy

48 II. MODELING CONCEPTS FOR BEAM For a beam, which transfers X energy to neighboring beams through shearing forces acting through ver- tical boundary displacements and moments acting through boundary w rotations, one must model both deflections and rotations at external nodes. One needs, for a two-node J beam element, 4 nodal degrees of freedom: w1, el, w2,' 2' Other nodes could be added in the interior for a more refined element. Consider the two-node, four degree of freedom element. Let us set up the set of finite element vectors Fig. 6. Two node, four degree of and strike out the components known freedom beam element. to be zero: E] [S-D] x Ix Lx u(x,y,z) i M1, ~ 1 of \. \3 w2 (9) xy "Ixy |7y M 4 2 LM2 yp( Xyz w(xz) Fig. 7. Chart of vectors appearing in the finite element derivation. Remembering that the stiffness matrix depends on an integration of strain energy of the form f {}E T{c}dv (10) v one sees that only the product of a and ~ survives, i.e., 6u = (6x5) (x) (11)

49 and only the mapping of a into E (shown in Fig. 7) is important. Note that x is completely determined by the neutral surface deformation, { } = z {w(x,O)} (12) x _ dx 2 and this, therefore, provides the only necessary strain-displacement relation. Using the symbol w(x) for theneutral surface deflection, we retain only: P1 {o X(x,z)} {I(x,z)}} {w(x,0)} r1 M1 [C] [S-D] 2a P2 (13) ~~~~~2 3 [3 1 2 l21~~~~~~~~~~[^] 2 L2 III. CONSTRUCTION OF THE BEAM ELEMENT The only important internal displacement is the vertical displacement of the elastic axis w(x,O) - w(x). It must contain the first 3 generalized coordinates in the displacement function: w(x) = al + a2x + a3x + a4x in order to meet the convergence requirements, i.e., inclusion of rigid body modes and constant straining mode (constant curvature). The fourth generalized coordinate is needed to provide the fourth nodal D.O.F. mentioned above. The stress-strain law for uniaxial stress is: {a (x,zr} = [E]{S (x,z)} (14) The strain displacement law is: ~ (x,z) = [ - z ]{w(x)} (15) x dx2 Fig. 8. Beam.

50 The Q matrix is t21 2 3 {w(x)} = [1 x x x x ( (16) "4 The [A] matrix is found by using the definition e(x) dw dx and by choosing a local coordinate system x1=0, x2L o 0 1 0 0 Oa = 2 3 (17) w2 1 L L L o3 2 4 02 0 1 2L 3L2 O4 [A] Inverting [A] yields: L3 0 0 0 L 0 0 0. [A] 218) L -3L -2L 3L -L( 2 L -2 L The shape function matrix is [N] = [(] [A] 1 3 2 3 3 22 3 2 3 22 3 = [L-3Lx +2x3 L x-2L2x2+Lx3 3Lx -2x3 L2x2+Lx] (20) L The columns in the matrix [N] are "shape functions" and give the internal displacement field corresponding to a unit nodal displacement. Some authors call these "interpolation functions."

51 x Fig. 9. Shape functions for beam element. The [B] matrix is [B] - [S-D-][N] 2 = [- z 2] N] dx - - z 2N] - [-6L+12x, -4L 2+6Lx, 6L-12x, -2L2+6 Lx] (21) L The stiffness matrix is [k] rS /2 h/2 [B C] [B]dz dy dx (22) After integrating this, one defines - -1,3 I - b h 12 and gets 12 6L -12 6L 2 2 4L -6L 2L2 EI (23) [k] = 3 (symm) 12 -6L 4L2 The equation of equilibrium for the beam is:

52 {Q} + {Q}u = ]{q} (24) equiv. nodal loads {Q} 1 = [NJT{i}d space (25) e.n. 1. space 0,X,T The general laws for equivalent nodal loads worked earlier are valid. These can be evaluated for special cases when desired. IV. EXAMPLE Problem: Find the rotation at the ends of a beam with pinned ends when subjected to a line load of 100 lb/in. Use a single element. EI = 10 lb in L = 100 in. V = 0.3 Solution: The equations of equilibrium will be of the form ext. + {1equiv. [k]{q} ext. equiv. concen- nodal trated loads o o forces IOi } - Find [k]. Simply put in the appropriate constants. -^ 12 600 - 12 600 40000 -600 20000 lb/in. [k] = 10 12 -600 Fig. 10. Beam with constant load 40000- distribution. Find the equivalent nodal loads. 3 2 3 L - 3Lx + 2x 3 22 3 L x -2L x + Lx {Q} e {100}dx e.n.l. 0 3 2 3 lO}dx L 3Lx 2x _22 3 -L x + Lx

53 4 4 2 4 L L4 - + L 5 L5 5 L L L 2 + 2 3 4 100 3 24 L3 L4 L 5 5 L L 2 000 lb 12 10083300. in.lb. 3 4 2 5000 lb. + 83300. in.lb. 8, 8.00' n. b, 5000 lb. - 83300. in. lb. Fig. 11. Equivalent nodal loads. Sample problem. Note that substantial moments result. Also note that the vertical equivalent forces vary as L whereas the moments vary as L2. As L + 0, the vertical forces dominate and the rough concept of "lumped" loads (neglecting moments) becomes valid. Now, proceed with the solution: P1 5,000- 12 600 - 12 600 0 O 0 83,300 40000 - 600 20000 0 \ + 5 = 103 P 5,000 12 - 600 0 0 j - 83,300 (symm) 40000 02 ^*~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~,', ^ *

54 Rather than solving in place, write the first and third equations: P1 + 5000 = 6 x 105 0 + 6 x 105 02 5 5 P + 5000 = - 6 x 105 0 - 6 x 105 2 and the second and fourth equations: 83,300 = 4 x 107 0 + 2 x 107 2 - 83,300 = 2 x 107 1 + 4 x 107 e2 Solving for 01 and 02 in the latter equations: 112 re )ofo. 00417 / 2 = / 00471radians i 4 92 00417,.,-t 17rO(r l417 Then find the external loads P1 and P2 (0o.4~) [P 1 = 5000 / \ = ( ) lb. Fig. 12. Beam deflection. Sample P 2 - 25000J problem. The complete equation of equilibrium is: - 5000 5,000 0 0 83,300 ] 0. 00417 - 5000 5,000 0 0 - 83,300 -0.00417 0 o i 4Q.E.D J3 CO 4C) *1 Hg4 am a co ex o th al o e f e d o The above answer can be compared to classical Euler-Bernoulli theory and is found to be exact. This means that the cubic shape functions used for the ba arsuiintoeclmoeteuiomldnca.They are also exact fo r c oncentrated loads at the nodes.

EQUUATIOA SOL\VERS T. TATIC A. Li.vear c 13. No,,,.tqr C. E'gcviNVlue. Pro6Iletes Ac. xt~j( i

SThrATIC L- UtE^. EIu SOLEERS G ENeTAL FokH o0F e:quNtrO 3: n ^^unq lavv Io^NMW MNE R.,L gCOJCkPT Fe SOLCI1OAJL u'TT'TnI 1IS JOdT CT UALLy tOJ)E O)D LAniE tMA-Te\CE.E TOO EXKPEXSIUE D —RECT PhETOP S ELI M tIOTiA (CH Ih SCHOOL CR:HRtS RULE (COLLE.E FRESHNM^) GAUSS ELtMfAJTTroAj (COLLEGE SEU(oR) k iGAusS-toltrTLE FtTot-eA'toJ AKil L,ILLT SYvMM. ITErATToXJ METhOtSQ $-, GAUSS -SEI DEL SOUTHW E'LL

|_AVAu B4EQO F TrA^UdlLAR RORM M i C C 1(S) 0 7? 2 3 o 5 Y =8 Ix =7 rxn =.. -~ -^.~...8-...7. ETC +sM on 18_Lt8 a.. iO N s o l ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ 0*

C. 8 Y z^.) 55 0 Ito I t o to r j. \ ( o:eo l;3(:,l;2.^,, q^~ 2 0 VOW, z(^3 =

Vat 42,,d, 4 0 = Rztu~l~Q<-<l^4, ^..^ IeAwce I'L ( | o o ~( o o'^ i ~ 2. oa0\2a 1 0( o I 0 E38oJ Co\V<.;n ^e. ^eytoad c!~'ec'^o^.. i o nr,0 oZ &-Mo bi 0 o s r0 22 ^o^ kaefc.?<^ ~.hCtI Z r,~+~C=Sr

IRM c (e [,i) (tR r a" K0.t~~~~~~~~~~~ t AHnce (s+i)A (Sl Ad sutk iS oSt eJ iAU(qt. /Uva^tse:. at t ^ reti^ +r4W3u(<i^

Ca? -o rA 1 z, I5 \kA 5 Gi'

31 January 1973 W. J. Anderson, SUbROUTINE "DCOMP" a' la W. Weaver Application of the finite element method typically leads to -sets of linear, algebraic equations. One approach for solution is to decompose the matrix of coefficients and then to solve the equations in a successive fashion. This can be done with the subroutines "DCOMP" and "SOLVE". These follow the modiified Gauss decomposition recommended by Melosh in his paper "Manipulation Errors in Finite Element Analyses". To decompose an NxN real, symmetric, positive definite matrix[A], set LA LUD [uCo][u wheretU3is an upper triangular matrix with unit elements on the main diagonal, and [D1 is a diagonal matrix. This decomposition is unique. In component form, we have Aai, AL, d oom 0 ~z ~,o,1 U u.3 A^,, A,; 1 0 0. *.. ~ U, U1 o A I SUJ 5I~ *0. 4o S A[- 11 u.', I; Consider the Aij as known and solve for the elements Dii and Uij. Carry enough terms for a 4x4 matrix and try to generalize Lhe results. A,, Aix A3 A4 I 0 0 0~ L Dl, R UU D13 4 Ac, A" Axs U I U o |O ADM W4| - V A'. 44 = U(U3 ~ 0 0 0 D33 D4| 3A4 ULj+ + + 4I 00 0 o In the first row of Aij All: D11 hence Dl1 = All. (1) Alj s DllUlj hence Ulj = Alj/Dl lj/All (2) For the other rows i 1, we have first the diagonal case where i J j: A44. 11U14) 2+ 22(U24)2 + D33(U34)2 + D44 and this generalizes to i - 1 Di - Aii- L Dkk(Uki) (3) ii 55 55

56 For off-diagonal elements i 7 1 we have, for instance A34 11U14 + U23D22U24 + D33U34 Solve for the U34 eiement, assyming all other quantities previouslry found: U34,A34 - U k3DkkUk D33 This can be generalized to iAij -: UkiDkkUkj ij k=l (4) Dii The decomposition must proceed in an orderly way. The elements for D.and U are calculated and then stored in the same locations reserved for the A matrix. This saves the creation of another NxN matrix, which can be critical if N is of the order of lCOO! The applicable equations in each region: ~,. U8 1+,'A-. ^ "". A -- ^ ^,033 U31 3 ^ 1 b v As owf 1 1',,'....'I C,., The solution proceeds row by row from the top. It also proceeds from left to right, starting at the diagonal of each row. The newly coiputed quantities Dt t andU i. must replace the original elements A1., as they are calculated. Furthermore, the new elements must also be Aenamed Aij! Eqn. #1 Find D11 Dl = All Store as All A(1,1) A(1,1) Eqn. //2 Find( Uij Ulj = Alj/ll A(1,J) - A(1,J)/A(1,1) Eqn. #3 Find Dii i>l Dii Aii - Dkk(U ) AIkiI) - A (I (I) -. A(K,K)*A(K,I)-2 Eqn. #4 Find Ui for l<i(j;-I t I U _ Ai j UkiDkkUk A(I J) _ A(I,J) ^,A(K,I)A(K,K)JA ( Dii A(I,I)

57 W. J. Anderson SUBROUTINE NAME: DCOMP(N, A, *) This subroutine decomposes a matrix [A] into a diagonal matrix [D] and an T upper triangular matrix [U] such that [A] = [U] [D][U]. The matrix [U] has unit diagonal elements. The given matrix [A] must be real, NxN, & symmetric, pos. definite. The matrix [U] is stored in the upper right hand of the matrix [A] and the matrix [D] is stored on the diagonal of [A] at the completion of the calculation. I - - - ^; --- ~~ J=I N SUM=A(I, J) I K.= I-1 I. EQ. 1-K = 1, K1 C ONTINUES uM;M-^iSU M*=SUM-(K, I)A (K, K) \ETURN 1 -M. I F'~~~~~ A(I, J)= SUM/A(I, I) I A (I, J) =SUM I -- - -^. -_-_ - -CONTINUE

9/17/77 58 ~ ~,,.-,:, o C Assu-, tme i + — i t"v 4, i 2. I orrA ^. ^ & Lh. 2" M.?'. lxci 1 9U~, I I *\f t,3 1<'z*, ISu*-^ -t-^aZ T( a <' > (,, soz^ (or P^5 L, -"t - ^oy C CA Crul -CoA'hh 4i v C. co.. IJ or(: Bi. k2, v. 15 I. 31". A^^l t0 x - _,., U, B- t); I,, ( LP~~~~~~~~~~~~~~

E~kq. 5l f- _ k t sit;o x.-.,_s,,x, *N < b', U, X *^ W (- > X Nn hX- X., - S " ",.Kk X: -= X —--- - a ",

60 FLOW CHART FORt GUB:iOUTINE "S.OLVE" 24 May 1971 a'la W. Wea SUBROUTINE NAME; SOLVE(N,U,B1,X) This subroutine colipletes the solution of the matrix equation [AX1 = ID. The matrix [A is real, symmetric and positive definite. It has been decomposed and stored in the matrix U1l. Both [A and LU] are N x N. I r,o' " LT' ( /tN, M l M- - ( - - I - Kt I L -— S —E —------- r _ _ ___ L - - -w s K0^ LI - X = ^^ O xJ I~~~RT~i^

61 1 SUBROUTINE DCOMP( NA, ) 2 C THIS SUBROUTINE DECOMPOSES AN N BY N SYMMETRIC MATRIX INTO THE 3 C PRODUCT CF AN UPPER TRIANGULAR MATRIX PREMULTIPLIED BY ITS 4 C TRANSPOSE AND A DIAGCNAL MATRIX. A = U' D U 5 C A = ORIGINAL SYMMETRIC MATRIX 6 C U = UPPER TRIANGULAR MATRIX WITH UNITY ON MAIN DIAGONAL. 7 C O = DIAGONAL MATRIX 8 C THE MATRIX U IS RETURNED TO THE ORIGINAL PROGRAM IN THE UPPER 9 C RIGHT PORTION OF A(IJ). 10 C THE MATR IX D IS RETURNED TO THE MAIN PROGRAM ON THE DIAGONAL OF r12 b%^MmSiO A(o0,o) 13 DO 4 I=1,N 14 DO 4 J=I,N 15 SUM=A(I, J) 16 K1=I-L 17 IF (I.EQ.i) GO TO 2 18 DO 1 K=1,K1 19 1 SUM=SUM-A(K,I)*A(K,J)*A(K,K) 20 2 IF (J.NE.I) GO TO 3 21 IF (SUM.LE.0O) RETURN 1 22 A(I,J)=SUM 23 GO TO 4 24 3 A( I,J)=SUM/A(II) 25 4 CONTINUE 26 RETURN 2 7 ENC D0 OF FILE 1 SUBROUTINE SOLVE(N,U,B,X) 2 C THIS SUBROUTINE SOLVES THE LINEAR ALGEBRAIC SET OF EQUATIONS AX 3 C A IS AN NX N, f.EAL, SYMETERIC M;ATRIX. 4 C X AND A ARE VECTORS W ITH N ELE i'ENTS. 5 C THIS SUBROUTINE MAKES USE OF THE DECOMPGSITION OF A INTO A 6 C PRODUCT CF TWO TRIANGULAR AND ONE DIAGCNAL MATRIX. SEE SUBROUT 7 C DCOMP FOR DETAILS. 8 INTEGER K,K1, I,K2,N 9 REAL X(10), B(10),U( 110 ) 1 D00 2 =1,N 1 1 SUM=B( I) 12 K1=I-1 13 IF(I.EQ.1) GO TO 2 14 DO 1 K=I,KI 15 1 SUM=SUM-U(K,I)*X( K) 16 2 X(I)=SUM 17 DO 4 Il=l,N 18 I=N-I1+l,/ 19 SUM=X(I) 20 K2=1+1 21 IF(I.EQ.N) GO TO 4 22 00 3 K=K2,Nl 23 3 SUM=SUM-U(I tK)X(K)*U(I,I) 2/. t4 X(I)=SUtM/U( 1, ) 25 P E T.UP.. 26 END i;) r, F- II.E

October 26, 1977 Modification of Equilibrium Equations Prior to Solution W. J. Anderson I. PROBLEM STATEMENT Assuming that a finite element problem has been properly posed and the element equilibrium equations have been successfully assembled, one has a set of equations {R} = [K]{r} (1) assembled vector of degrees of freedom external loads, plus assembled equivalent nodal loads stiffness matrix Mathematically, 3 possibilities exist: (1) All components of {r} are known and {R} is to be found. (2) All components of {R} are known and {r} is to be found. (3) A mixture of components of {r} and {R} are known, and the remainder of each are to be found. The first case is easy to solve; the {R} vector is recovered by direct multiplication. This case almost never occurs in practice! The second case is the standard form desired for most equation solvers but is not obtained as a well-posed problem upon first assembly. Rigid body modes must be constrained by setting certain displacements zero, and this casts the second case into the third case., The only assembly of finite element equations of any practical interest and of any difficulty, is case 3. This is called a mixed boundaryvalue problem. If we concentrate on the mixed boundary value problem, case 3, we need to modify the equations in some way to recover the form of case 2 suitable for equation solvers. There are several methods available which reduce the set of equations in size or substitute a dummy equation for the equations with specified displacement. II. PARTITIONING Partitioning is a method of reducing the number of equations studied to only those where displacements are unknown. It proceeds more easily when there is a fortuitious or pre-planned numbering of degrees of freedom so that all specified displacements are left to the last (say): known -- R1 Kll K1 r - unknown ( 1 = 1.......1 (2) unknown - R J jK<K known The equations are then partitioned, where each submatrix follows the normal laws for matrix operations: 62

63 {R1} = [K l]{r} + [K12]{r (3) {R2} = [K21]{r} + [K22]{r (4) Equation 3 can be solved for the unknown displacements {r }: {r1} = [ -l]-lR1 ) -[K12]{r2}) (5) 1 [ 11] ( 1 12 2 known If the reactions at the supports (degrees of freedom where displacement is specified) are desired, then the final step is to use equation 4 to directly solve for {R2}, since the R.H.S. is known. III. SOLVING "IN-PLACE" (PAYNE & IRONS) One can solve the equations for a mixed boundary value problem, without * regard to the order in which they appear, by an artifice due to Payne & Irons. The set of equations is modified by converting the known displacements into unknowns, and then forcing them to the desired value by choosing a very large value on the diagonal as shown: ORIGINAL PROBLEM: K11 K12 K13 r 1 F4000 Ki1 22 K23 r2 = < 000. (6) jK31 K32 K33 27 R3 J MODIFIED PROBLEM: K11 K12 K13 r r 4000. Kll 12 13 K21 22 23 2 7000. (7) K31 K32 1012 x K3 r3 101 x 2.7 x K33 The modified equations are then solved in the usual way by versions of Gaussian elimination or by iteration. To see the effect of the large constant which is used, separate out the last equation from (7): K3rl + K32r + K3 x 1 101 x2.7 x K (8) 31x1 32.2 33 3( 8 * For engineers and mathematicians who have worked with the Simplex method of linear programming, this approach is similar to the "big M" method of driving a variable to its constrained value.

64 Since stiffness matrices are diagonally dominant, i.e., K3 is of the same order as K31 and K32, r3 2.7 (9) The value of r can be made as close to 2.7 as wanted by increasing the constant 1012 even further. In solving "in place", one inserts some dummy equations in the process and therefore must lose (at least temporarily) some information from the original set of equations. The information lost is the reaction force at each node, which can be recovered by a final step using direct summation, e.g. 3 R = ZK r (10) 3 j=l 3j j j=1 K- known from previous solution IV. SOLVING IN PLACE (WEAVER) An alternate method of solving in place is to extensively modify the stiffness matrix as shown, using the same original matrix in the last section: rKll K12 0O r 4000. - K13(2.7) K21 K22 0 r2 = 7 000. K23(2.7) (11) 0 0 1 Lr3 2.7 This method is exact. It involves more manipulation than the previous method, however. It is practical, and has been used in some smaller programs such as Weaver's "PS2." V. STATIC CONDENSATION Every finite element solution must use a modification procedure similar to those in sections II, III and IV to put the equations in the standard form for solution. There is a further way to eliminate a portion of the problem and reduce the number of equations further. This is called static condensation. It really is Gaussian elimination in a preferred order. A method described here is based on partitioning. Suppose we are dealing with a set of equations in which we are not interested in several degrees of freedom, i.e., we are willing to suppress them, or "condense them out." Let {r2} represent the undesired degrees of freedom, which have been numbered as the last degrees of freedom.

65 K11 K12 1 RI J1 a( I<,(We% K21,K22 2rlj = {. *necsired Partitioning the equations gives [Kll]{r } + [K12]{r} = {R1} (13) [K21]{r} + [K22]{r} = {R2} (14) One solves for {r } from (14) and inserts it in (13), thereby eliminating {r2}, as shown below: {r2} = [K22] ({R2} - [K21]{rl}) Hence (13) becomes [K ]{r } +K ][K [K[K ]{R} [K]{r } ) = {R} Collecting terms, one can define new stiffnesses and forces: ([K11] - [K12][K22] [K21]){rl } = {R1} - [K12][K22] -{R2 [K] {R} with the result that a new problem involving fewer degrees of freedom, but more complicated stiffness and force system, is defined: [K]{r1} = {R} Comments on static condensation: A. Partitioning is usually not possible because of the scrambling of desired and undesired degrees of freedom, but other methods are used. B. One can use static condensation to allow a better elastic representation of a body, particularly its interior, without putting external loads at those suppressed d.o.f., i.e. {R2} = 0. The practical reason for this is that the user does not want output data at these d.o.f. and likewise does not want to have to prescribe input data there either. C. Static condensation can be used for single elements to condense out unwanted nodes in the manner of comment B. VI. SUBSTRUCTURING This is a method of dividing large structures into substructures for easier solution. It is philosophically similar to static condensation except that the force vector {R2} is not usually zero.

66 VII. FINAL COMMENT The partitioning method for static condensation, substructuring, Gauss-Dolittle decomposition, Cholesky decomposition and high school elimination methods are all just variations on Gauss elimination using a preferred order for eliminating variables.

/8/80 W. J.A. REMOVAL OF RIGID BODY MODES I. DEFINITION OF RIGID-BODY MODES Many elastic bodies can translate freely or rotate freely because there are not enough constraints applied to fix the body to the earth. The body of a typical desk telephone often is free to translate in two directions and to rotate about a vertical axis. Each such distinct type of motion is called a rigid body mode. Mathematically, one says that a displacement field is possible which causes no strain energy, i.e., / 2{ r }[K]{ r } = 0 (1) This equation implies tt Fig. 1. Desk telephone. This equation implies that det[K] =' 0. Rigid body modes are legitimate in many analyses, particularly dynamic problems where the overall motion of a body is important. Their presence causes trouble in static stress analysis, however, because they prevent the solution of the equilibrium equations. Every student in finite element theory will sooner or later attempt to solve a stress problem [K]{r} = {R} (2) without constraining the rigid body modes and will get warning message s from the finite element program. The reason is that [K] is singular and cannot be triangularly factorized. The requirement for removal of rigid body modes is a nuisance in many stress problems which could reasonably be stated as if the body were in mid-air. For instance, stress in an inflated tire results primarily from 67

the internal pressure and vehicle weight whereas it should not depend on which direction the car is traveling. Nevertheless, for a unique solution, one must specify the orientation of the time in space to not allow rigid body modes to occur. In the theory of elasticity, there are: 6 rigid body modes in 3 dimensional space 3 rigid body modes in 2 dimensional space 1 rigid body mode in 1 dimensional space. For structural elements such as beams and plates, there are 6 rigid body modes in 3 dimensions. When imbedded in 2-D, however, one must decide each case separately. The opposite of a rigid body mode is an elastic mode, i. e., a displacement field which causes strain energy. II. HOW TO REMOVE RIGID BODY MODES In many finite element programs, the user must artificially constrain degrees of freedom to prevent rigid body translations and rotations. Other computer programs such as MSC/NASTRAN have automated procedures for this. Nevertheless, every user should understand the logic required to remove these modes. Generally, one must artificially constrain enough displacements to prevent rigid body translation or rotation but not enough to constrain elastic deformation between any nodes. Application of such an artificial constraint at a node should not introduce any external force at the node. Every three-dimensional body should have at least six displacement components specified (typically zero) to remove six rigid body modes. Likewise, two-dimensional bodies require three displacements specified, and one-dimensional bodies require one displacement specified. One rigid body mode is removed by constraint of each properly chosen displacement. 68

The original physical problem often has displacement constraints that fortuitously remove some rigid body modes. The remaining rigid body modes must be removed by the analyst. III. EXAMPLE Let us look at the stress around the circular hole in a sheet under uniform tension (Fig. 2). Note that the problem says nothing about orientation of the sheet because it has no effect on stresses. The engineer must ensure that the system does Fig. 2. Rectangular sheet with circular hole, under tension. not rotate or translate, Two-dimensional problem. however. If the full problem is to be solved (neglecting symmetry for now), one could use the extremely cruide grid in Fig. 3. To constrain the three rigid modes, one could set to zero: 10 11 (a) horizontal displacement at node 4 (b) vertical displacement 5 at node 4 (c) vertical displacement at node 8 Fig. 3. Node pattern. These constraints do not affect the elastic modes, whereas the following set does and is therefore unacceptable: (a) horizontal displacement at node 4 (b) vertical displacement at node 4 (c) horizontal displacement at node 8. 69

The latter set of constraints does not allow any cumulative elastic strain in the x direction between nodes 4 and 8. It also does not prevent small rotations of the body about node point 4 and is therefore a complete failure. Finally, if symmetry had been used, one would solve the problem in Fig. 3. In this case, however, symmetry sets to zero: (a) horizontal displacement at node 10 I). (b) horizontal displacement at node 12 12\ (c) vertical displacement at \ 78 node 7 (d) vertical displacement at node 8. Fig. 4. Fortuitous removal of rigid body modes by symmetry considerations. This is sufficient to remove rigid body modes, i. e., the physical problem contains none and therefore further artificial constraints are not needed. Symmetry also implies that the vertical forces on 10 and 12 are zero, and horizontal force on 7 is zero. Node 8 is unique because it is a loaded node on a plane of symmetry. The horizontal load due to the neighboring mirror image acting on 8 is zero by symmetry, so the only load on 8 is the equivalent nodal load resulting from the edge stress. 70

IV. MATHEMATICAL NOTE Consider the strain energy in a structure at equilibrium, state II. If the structure were at zero stress and strain at state I, then the increase in energy absorbed equals the work done: A U A W = I-'1II j'{ } {ii}dV - {r} {R} 2 II 2 T r = {r} T[K]{r} II II Fig. 5. Work and energy. Generally speaking, the strain energy must be greater than or equal to zero for any displacement {r}. Hence, {r} [K]{r} > 0. If, however, a nonzero set of displacements {r} exists for which the strain energy is identically zero: {r [K]{} = 0 then the displacement {r} is a rigid body displacement. Furthermore, the matrix [K] is then called positive semi-definite. 70a

October 22, 1976 BANDWIDTH CONCEPTS..___._......... W. J. A. Reference: Desai & Abel: pp. 19-21; pp. 162-163. The computational difficulty in solving large sets of linear, algebraic equations depends on the way the equations are mathematically coupled to each other. Three concepts are important: "bandwidth," "wavefront," and "pointer" methods. Equation solvers usually require that one or the other of these be minimized in order to reduce computer CPU time. Equation solvers of the banded (SAP4, SAP 6) or wavefront (SUPERB) types require careful ordering of nodes or elements, respectively. Equation solvers using pointers (NASTRAN) use sparse matrix methods that only store nonzero stiffness terms. o 0 o 0 0 0 0 I. BANDWIDTH O O Bandwidth refers to the width of the * ~ X) 0 a diagonal band made up of nonzero terms o o in a banded, sparse matrix. Typically, \ 5~MAb by numbering the nodes properly, the F.E. method leads to a stiffness matrix o 0 o a which has nonzero terms near the main diagonal and zero elsewhere. The bandwidth is given as a dimensionless number Bandwidth and semi-bandwidth. indicating how many terms wide this band is; no nonzero terms may fall outside the band. Bandwidth can be given in terms of compact or detailed notation. 4. The semi-band width, B, is more impor- ltant in symmetric matrices and is used almost exclusively.* FRS' SEMI-BANDWIDTH = (BANDWIDTH + 1) 2 5 &7 This means that the semi-bandwidth includes the main diagonal term. Consider a physical problem withX x o x X 0 0 nodes numbered in two different ways. In each case, it can be shown that ^ o B = (D+ l)f X 00 XK X 0 0 $ K where D is the greatest difference in X O o node number within an element, and f is the number of D.O.F. per node. In K 0 O K 0 the first case, cK Y O K X B = (5 + 1)(2) o s O O K = 12 Fig. 2. First nodal numbering scheme. (compact notation) * In fact, for most structural problems, authors will use the term bandwidth when semi-bandwidth is actually meant. This seems to cause little trouble in practice. 71

72 In the second case B = (3+1)(2) Hence, the bandwidth depends on the 5ECXO way the nodes are numbered. For Gaussian elimination types 2+ of solution, the CPU time required is proportional to the size of the matrix times the semi-bandwidth squared: F (TIM) ( (N)(B)2 ^ 0 0 TIME KX K K 0 0 0 In the example discussed, the running K.^ X K> 0 o time for the second case would be N B 2 0 K ^ (CPU) = 2 (CPU) O O. X. K = 2- (CPU) 12 1 - 0.44 (CPU)1 Fig. 3. Second nodal numbering scheme. 0U.44 (CU)1 (compact notation) and the CPU time is cut by more than half by renumbering the nodes! SAPIV and SAP6, as well as most other public domain finite element codes are dependent on bandwidth ideas. Note that the numbering scheme for elements is unimportant for bandwidth ideas. II. WAVEFRONT Another type of equation solver, also based on Gaussian elimination, is the "wavefront" approach. This was originally developed in England by Bruce Irons. In this approach, the element numbering is important, rather than the node numbering. The goal is to hold as few equations in high speed core as possible, putting the remaining terms on disk or other lower cost storage. The solution proceeds element by element. The equations of equilibrium for a node are activated when an element is first formed which contains that node. The equations of equilibrium for that node remain active until the last element containing the node is processed. That node is then deactivated. The solution is merely a Gaussian elimination in a preferred order.

73 An example is now given where the same problem is solved with two different element number schemes. FIRST CASE --— 1 _ 3 7 4 Element Active Node # of Active Nodes O " o Nodes Dropped 1 1,2,3,8 4 18 8 2 5 6 1,8 2 2,3,4,5,6,7 6 2. 23,4,5,6,7.6.. 4,6 Fig. 4. First element number3 2,3,5,7 4,5,7 ing scheme for wavefront solver. 2,3,5,7 SECOND CASE Element Active Node # of Ac- Nodes tive Nodes Dropped! 3 7 4 1 1,2,3,8 4 1,8 1 3 7 4 2 2,3,5,7 4 3 I 2,3 3 4,5,6,7 4 3 4,5,6,7 44,5,6,7 8 2 5 6 Fig. 5. Second element numbering scheme for wavefront solver. The second case has a smaller wavefront. Note that the numbering scheme or the nodes is unimportant for wavefront ideas. The proprietary program "SUPERB" has a wavefront solver. NASTRAN did not originally have a wavefront solver, but the recent proprietary version (MSC) does. III. "POINTER" METHODS FOR SPARSE MATRICES IV. DIFFICULTIES WITH CPU TIME FOR 3-D SOLUTIONS USING SOLID ELEMENTS Reference: Zienkiewicz, Edition 3, pp. 135-136. The run time for 3-D problems can be prohibitive if care is not taken. An example which is rather scary is presented in Zienkiewicz's book, where comparable sized l-D, 2-D, and 3-D problems are discussed. Suppose each problem has 20 elements in each direction as shown. It is known that CPU time T for the equation solver varies as: Number of S emi- )2 Equations Bandwidth

74 Note that the number of degrees of freedom. -.-....m... per node increases from 1-D to 3-D. The number of nodes and the bandwidth also increase strongly. Indeed:' 2 2.-i) TiD const(Z) (2) = const (84) 1-D; *** - T c = const(1, 866, 312) 2-D T3_D = const(27,783)(1392)2 10 = const(5.83 x 10 ) Even comparing the 2-D and the 3-D solution,....:.''. ~ 5.,,' 3-1> we see that the 3-D solution requires 24,600.' times the run time as the 2-D. This is a fantastic increase in complexity as dimen-'': sions increase... Fig. 6. Comparison of 1-, 2-, and 3IV. CONCLUSIONS & COMMENTS dimensional spaces Minimizing bandwidth (or wavefront) saves on CPU time and on core storage requirements. Core storage times typically represent 2/3 of the total cost of running F.E. programs. At present, then, one should attempt to reduce core storage as the highest priority. Three-dimensional solutions (solid elements) are extremely expensive if care is not taken. Effort in the direction of using fewer but more sophisticated elements, is the proper approach. Isoparametric elements appear to be the answer here.

Convergence Concepts in Finite Element Analysis by Robert E. Sandstrom, Ph.D. University of Michigan, 1981 The goal in any computational analysis is obtain solutions for a given problem as efficiently and accurately as possible. The purpose of this lecture is to expose the utility of convergence concepts as a tool to assist you in reaching this goal. Understanding these concepts will help you become a "smart" finite element user. Effort Meter Accuracy Meter Problem 0% 50% 100% \ % 50% 100% Statement === ===> \\ / \ Analysis / \ Results/ Convergence in finite element analysis generally includes those factors which prevent us from computing the exact answers to our "real world" problem. Some factors which effect the rate of convergence include: 1. Real world complications such as nonlinearities and ill defined or unknown parameters. 2. Modeling concepts, practices and accuracy. 3. Discretization of Continuum. 4. Solution algorithm. 5. Roundoff error. This lecture will deal with item 3. Despite the fact that items 1 and 2 are probably the most important factors affecting the accuracy and validity of the analysis; they will not be discussed in this lecture. They reflect broad consideration which is not resricted to finite element application. Items 4 and 5 will not be discussed here because most modern computer codes employ accurate procedures and which generally are not within the control of the user. On the other hand, the finite element user is directly involved with item 3, discretization of the continuum. Users must make decisions with regard to element type and mesh size. These decisions can drastically effect the quality and cost of the analysis. In this presentation the mathematical aspects of convergence will be reserved for the text books. Those concepts which will enhance our "smartness" as finite element users will be presented in an intuitive manner. 74a

Concepts Remember that the finite element method is basically a Rayleigh-Ritz procedure. To refresh your memory the Rayleigh-Ritz technique is used to solve structural problems in the following manner. First, we select (or invent) a set of admissible displacement fields. They may be continuous or piecewise continuous and they must satisfy the "geometric" boundary conditions on displacement and slope. We then tune the displacement fields in our set so that the potential energy of the structural system is a minimum. The "best" answer will produce the lowest value of potential energy. A basic characteristic of the Rayleigh-Ritz method is that it will: 1. Underestimate displacements. 2. Underestimate strains. 3. Underestimate stresses. 4. Overestimate stiffness. Furthermore, it is generally recognized that the Rayleigh-Ritz method may produce displacements of acceptable accuracy; however, the stresses and strains computed through differentiation of the admissible displacement fields may yield poor results. Users of the finite element method should be aware that stresses and strains computed from a finite element analysis may be very poor even though the displacements are accurate. The accuracy of a Rayleigh-Ritz analysis can be improved by doing any of the following: 1. Increase the number of admissible displacement fields. 2.. Increase the order of the displacement fields. In the finite element analysis the accuracy can be improved by refining the mesh and/or using higher order elements. Refining the mesh is equivalent to increasing the number of admissible piecewise displacement fields and using higher order elements is equivalent to increasing the order of the displacement fields. Both improvements may increase the cost of the analysis. One still may ask'What is the cost of a poor result?' Failure? Element Convergence Requirements Taken from Reference [1] Is it possible to converge to the "exact" result? The answer to this question is yes, provided that the following conditions are satisfied. 1. The displacement functions chosen should be such that it does not permit straining of an element to occur when the nodal displacements are caused by a rigid body displacement. 2. The displacement function has to be of such a form that if nodal displacements are compatible with a constant strain condition such a constant strain will in fact be obtained. 3. The displacement functions should be so chosen that the strains at the interface between elements are finite. 7 4b

When these conditions are satisfied, the finite element discretization will approach the exact answer in the limit as the mesh size is reduced. This idea is consistant with concepts used in the study of calculus. The point here is that we can achieve acceptable accuracy, but again it may be expensive. Convergence Rates How far must one go to obtain acceptable results? Zienkiewicz gives the following guidelines regarding solution errors and convergence rates. 1. The exact solution will be obtained if the displacement function for an element exactly fits the correct solution. 2. The solution error for displacements is given by O(hP+1) where h refers to the meshsize and p equals the order of the diplacement function used in the element. 3. The solution error for stresses and strains is given by O(hP+l-m) where m equals the mth derivative of the displacement field used to compute the stresses and strains. As an example the convergence rates for plane strain triangles are given below. Constant Strain Triangle: Shape function: Linear i/ < 0^ \Convergence Rates: 0/ o^ _^a~ ~Displacements: O(h2) Stress/strain: O(h) Linear Strain Triangle: Shape function: Quadratic / Js I Convergence Rates: I/ z / s \Displacements: O(h3) Stress/strain: O(h2) It can be seen from the convergence rates given above that if the mesh size is refined by a factor of two that the error in the stress results will be reduced by an order of O(h/2) for the CST element and 0(h2/4) for the LST element. It should be noted that the concept of convergence rate is relative. It will not measure the absolute error reduction; it will only indicate the relative error reduction one can obtain through mesh refinement and the use of higher order elements. 74c

Mesh Refinement verses Higher Order Elements Mesh refinement and the use of higher order elements should be governed by: 1. Desired output (displacement and/or stress strain data). 2. Stress-strain distribution throughout the region of interest. A simple point to remember is that constant strain elements ( linear shape functions) will represent the displacement field by a patchwork of discrete linear interpolation functions and the stress-strain field will appear as a histogram type distribution. This histogram analogy should be kept in mind when selecting element type and mesh size. Disregard for this concept may lead to serious modeling problems. Linear strain elements are more forgiving in this regard, relaxing ( but not deleting) the requirements on mesh size. Linear strain elements will model the stressstrain distribution by a patchwork of linear interpolation fields and the mesh size should be selected with this mind. These arguments could be extended for higher order elements. / Type of element Vertical Load of A Couple at AA' / B.A....... —/...... ---, —-i-....Max. defl. Max. stress Max. defl. Max.stress at AA' BB' at AA' BB'. -l __ | _.__ ~A' X 0-26 0-19 0-22 0-22;B —— ~ -—'-_ ___ ___' A' 0-65 0-56 0-67 0-67 A cantilever beam analyzed by several n- - plane stress elements. Note accuracy 0-53 0.51 0-52 0.55 improvement with higher order elements.. ( Fir:. / ) i 0.99 0-99 1-00 I-00 100 1-00 1-00 1-00 EXACT 100 1-00 1-00 1.00 A second point which should be kept in mind is that constant strain elements may produce acceptable displacement fields (provided that a proper mesh size is used); however, stress and strains may not be accurate for reasons given earlier. Higher order elements may not always significantly improve the displacement results, but they usually will improve the stressstrain results. Example The example shown of the next few pages demonstrates the improved stressstrain data obtainable from linear strain elements over constant strain elements. 74d

Suppose we want to compute the stresses in a bar due to its own weight. Let E = Young's modulus A = Cross sectional area wo = weight per unit length x Exact Solution Differential Equation: EA d2u = -wO Boundary Conditions: Displacement u(x=U) = 0 Strain ce = du(x=0) = 0 dx Displacements: u(x) = wo02 1 - j l2 2EA j Stresses: x = -wo A 74e

Finite element solution using two constant strain truss elements. Element Data -— > q Shape functions: Stiffness Matrix: -— > ql 1 —-> q2 N1 = 1 - 5 1 -1 o3 ~^....^~,........ oEA E=0 E=1 N2 = E -1 1 Global Solution Equivalent Nodal loads: Node 1 R1 = Wo 4 R2 = WoQ Element 1 2 2 x Boundary Conditions: r3 = 0 Node 2 9 Global Equations: R1i =2 1 -1 rI Element 2 i.= 2EA R2 -1 2 r2 -I I I Node 3 Nodal Displacements: rl = 1 Wo2 r2 = 3 W2 <====== These displacements are 2 EA 8 EA EXACT! Stress at node 3 of element 2: ax = EEx = E dN1 r2 + r3 <===== 25 ERROR! dE d- 4 A E 6=1 7 =1 74f

Finite element solution using one linear strain truss element. Element Data -— > 9, — > qj | —> q2 -— > q3 i=0 E=0.5 E=1 Shape functions: Stiffness Matrix: N1 = 2E2 - 3E + 1 7 -8 1 N2 = 4E - 4E2 EA -8 16 -8 3Q N3 = 2E2 - [ 1 -8 7 Global Solution Equivalent Nodal loads: Node 1 R1 = WOR 6 R2 = 2wo, 3 x Boundary Conditions: r3 = 0 Node 2 > Q Node 3 Nodal Displacements: rl = 1 w2 r2 = 3 wo22 <====== These displacements are 7 4g 2 EA 8 EA EXACTI Stress at x within the element: ax = EAx E A 1 r1 + d2 r2 + d3 r3 = -w^ <====== EXACT! 74g

References 1. Zienkiewicz, O. C., The Finite Element Method, Third Edition, McGraw-Hill, 1977, Chapter 2. 2. Desai, C. S. and Abel, J. F., Introduction to the Finite Element Method, Van Nostrand, 1972, Chapter 5. 74h

Oct. 17, 1977 A SPECIAL LINE ELEMENT W. J. A. DERIVED USING EQUILIBRIUM Equilibrium ideas will be used to derive the stiffness matrix for a special element: a line element with varying cross-sectional area A(x) = A (1 + Bx/L). The reference area A has dimensions of L and the dimensionless constant B is positive for an element with area increasing from left to right and negative otherwise. An exact solution can be found. This ele- ment will be useful for comparing with later approximate solutions to the varying area element. i I Q, Procedure We wish to fill in the missing operators in the general scheme: d 1i[ [E] [?] [?] < Q < {e } <- {u(x)} < {a <X X Q2 J The equilibrium matrix relating nodal loads and stresses is found by a trick: C (x) = Q2/A Q2 A (1+Bx/L) 0 or Q2 = A(l+Bx/IDc(x). Also Q - Q = - A (l+Bx/L)o(x). Putting this in matrix form, one has {Q A- A (l+Bx/L) 1 0 [ = Ox() Q2J AL (l+Bx/L) The relation between nodal displacements and u(x) is found from our knowledge of stress in the body. (This is similar to Turner's procedure of starting with known strains and then working out displacements.) 75

76 O (x) c (X) = x x E or u(x) = E(x)dx+ C >J integration constant -= g xE dx + C r Q2 dx E A (l+Bx/L) + f Q2 L A B ln(l+Bx/L) + o 1 "2 If one identifies the constants shown as generalized coordinates a., we can proceed. {u(x)} = [ln(l+Bx/L) 1] [4] rql t Ut_(O 0)' 0 1 ^q2 )~u(L) _ln(l+B) 1_ 2J [A] The [A] matrix is easily inverted to get: 1 1 ln(l+B) ln(l+B) [A]-1 We now have all the ingredients for the stiffness matrix, using the sequence of operators from nodal displacement to nodal forces:

77 [k] = [g][S-S][S-D][l][A1 r-A (i+Bx/1 1) 0- Ao (l+Bx/ LX I In (1+B) In (1+B) [E][ ][ln(l+Bx/L) 1] dx Ao (l+Bx/L)J i 1 o L 0 - A (+Bx/L) = [E][-d ] _ ln(l+Bx/L) ln(l+Bx/L) d ln(l+Bx/L) ln(l+Bx/L) A (1+Bx/L) o [N] The shape function matrix has been calculated as an intermediate step in the equation above. It is a set of interpolation functions, of course, with unit value at the appropriate node. Carrying out the derivative - A (l+Bx/L) L O (1+Bx/D In (1+B) (l+Bx/ l n (l+B) A (l+Bx/L)_ and, finally, E Ao B stiffness for varying[k] = Lln(I+B) area line element _- 1 Comments: A 1.) Note that the "effective area" of the line element is n(+B). If B = 1.0, In(l+B) for example, the effective area is 1.44 A, which is perhaps smaller than one would expect. The more flexible portion of the element dominates in this case. 2.) Other solutions for this element, which use approximate shape functions, will yield "stiffer" results, i.e., the stiffness matrices will have larger terms in general. This is because approximate solutions constrain displacements.

COMPARISON OF VARIOUS LINE November 9, 1977 ELEMENTS W.J.A. From several lectures and homework problems, a sequence of line elements has been generated which can be used to model the varying area link shown: In increasing order of accuracy, A 8 xA) we can use (A) A single, two node constant area line element. This is a constant strain element. One would choose the reference area, say at x/L 1/2 to get Fig. 1. Physical Case 1 -1 EA (l+B/2) [k] = EA1 1 +/U(x) u(x)1+a2x (B) A single, two-node area line element. This is again a constant strain emn in sptef the Fig. 2. Model A. Two nodes. element, in spite of the violation of internal equilibrium which results. One has, _) u(x) = a 1+a 2X EA 1 -1 [k] - (1+B/2TI L ] -JB t 1 1i Fig. 3. Model B. Two nodes. (C) A single, three-node, varying area line element. This is a linear strain element. Zi )u(x) = a1+a2x+ 3x EA~ (6+6B+B [kj - L +(+B)1 i 1 Fig. 4. Model C. Three nodes. 3(g+>t) 1 (D) The exact solution, yielding G u(x) = a ln(l+ BL a logarithmic displacement 1 function i Fig. 5. Model D. Two nodes, 78 logarithmic displacement.

79 EA _ 1 -1 [k] L ln(l+B) L-l 1 One can now compare the multiplication factors, using B=l, say, where [ k] = k EA f[k] L 1 1 (A) kf = 1. 5 (4. 0o high) (B) k = 1.5 (4.0% high) (C) k = 1.445 (0.1%o high) (D) k = 1.443 (exact) It will be left as a challenge to the student to do a model with two cons tarit-area, two- npte elements to see if the answer is better than model C, above The question is whether one parabola or two straight line segments more accurately model the true logarithmic curve. The question raised here is fundamental: Do you in general get more accuracy by using a larger number of simple elements, or by using a smaller number of more refined elements? Current belief is that fewer, but more refined elements are the "way to go." See the attached page for a review of a paper by I. Fried, in which this conclusion is reached in regard to round-off error in dynamics problems (This is different from the above case concerning modelling error in a static problem, but is the only definitive work on the general question at the moment.) X)! i C//Cs M C p Fig. 6. Comparison of Model C and proposed2 element solution.

80 BOUNDS ON THE EXTREMAL EIGENVALUES OF THE FINITE ELEMENT STIFFNESS AND MASS MATRIXES AND THEIR SPECTRAL CONDITION NUMBER Fried I. J. Sound and Vib. 22(4), 407-418 (June 22, 1972) 14 refs Refer to Abstract No. 72-1592 This paper deals with the conditioning of ma- A number of examples commonly known to be trixes in finite element approximations to eigen- ill-conditioned are studied and estimates for value problems. Much of the approach has to spectral condition number given. These cases do with the spectral condition number (the ratio include nearly incompressible solids, nearly of the maximum to minimum eigenvalue) of the inextensional rings and bending with shear. mass and stiffness matrixes. It has been previously established that roundoff error in computer solutions of eigenvalue problems d It is concluded that one obtains better total acon this spectral condition number. I curacy in a problem by using a smaller number of higher order elements (those with higher degree polynomials as shape functions) than using \ The eigenvalues of the assembled mass and la larger number of simpler elements. ) stiffness matrixes are first bounded using straightforward arguments such as Rayleigh's principle. These bounds depend on properties of the individual elements and the maximum The paper is well written and is recommended number of elements connected at a single node. for both researchers in finite element methods Several theorems result, yielding upper and as well as users of finite element programs. nontrivial lower bounds for the eigenvalues. One small typographical error is the omission The latter is important for the case of element of the symbol X 1 from R.H.S. of Eq. 36. stiffness because some zero eigenvalues are present (corresponding to rigid body modes). William J. Anderson Dept. Aerosp. Engr. The most important theorem (Theorem 4) shows Univ. Mich. dependence of the spectral condition number on Ann Arbor, Mich. 48105 the element length scale h and on the order of original field equation. Specifically, for a differential equation of order 2m, the spectral condition number varies as h-2m for the stiffness matrix and is independent of h for t he mass matrix. Hence, as element siza vanishes, the stiffness matrix becomes increasingly illconditioned. Reprint from THE SHOCK AND VIBRATION DIGEST, Vol.5, No.9

Nov. 13, 1978 ARGYRIS' NATURAL MODE METHOD..-. —-------—.....X W.J.A. Reference: (1) Argyris, J. H., "Recent Advances in Matrix Methods," vol. 4, Progress in Aeronautical Sciences, Pergamon Press, Oxford, 1964. (2) Weaver, William W., "Outline of Notes on Supplementary Topics," Unpublished Notes, 1969. I. BACKGROUND The natural mode method used in finite elements is attributed to Argyris (Ref. 1), but is closely related to the modal methods used in dynamics for many years. The general idea is to identify physical types (modes) of deformation according to whether the motion is of a rigid body type or a straining type. We know that rigid body modes are present in an element; the present discussion takes advantage of these modes by including them explicitly in a description of the displacement field. The "unknowns" in the problem then become the generalized coordinates rather than nodal coordinates. The end result of using natural modes is to have much simpler stiffness matrices when expressed in the natural form. One can also gain physical intuition from their use. Do not confuse natural modes with natural coordinates! Natural modes have to do with the field variables whereas the natural coordinates have to do with the underlying coordinate system. II. NATURAL MODES The generalized coordinates {Oc} in the expression {u} = [] {a} will represent certain specific modes of displacement if the c functions are carefully chosen. For the line elements, if we choose: {u) - [1 (-1 + 2 ] tri t rigid straining body mode mode 81

82 Then a1 is the magnitude of the rigid body translation, and a 2 is the magnitude of the straining mode. This can also be shown in the solution {q} = A ] a 2 where 1 i - 42 h 1 LY2 J 2 Note that post- multiplication by a and a 2 brings out the " column -wise' character of the matrix [A], as can be seen by rewriting: 9q 1, -1 = a1 +aa2 q2 1 1 t t rigid straining body mode mode The choice of natural modes is not unique because they all have arbitrary amplitude and because any amount of rigid body mode can be introduced into a straining mode with the results being still a straining mode. One must choose as many rigid body modes as required for the space in which the element is imbedded. One also must include all forms of constant strain. It is best to choose rigid body and straining modes which are as dissimilar (orthogonal) as possible. III. CASTING THE ENTIRE F. E. DERIVATION INTO NATURAL MODES Since the generalized coordinates {a } correspond to a displacement, we need a generalized force {Q } to correspond. [See Fig. 1.] We would like to cast the problem in terms of generalized forces and displacements.

83 r? ( {x -X)} [C] - [S-D] [] [A] - to ) { T} a U i ) v t work t t work t I work_ t L__ work L Fig. 1. Vector transformations. Let us define {Q } by requiring that the amount of external work done during a virtual displacement should be the same whether described by physical or generalized coordinates:' {da} { } = {dq}T{Q} ([A] {da})T {Q T T {da} [A] {Q} {(da} [A] T{Q} Since {da)} is arbitrary, we must have {Q } = [A] T{Q} 4 generalized forces This defines the generalized forces. Now look at the equilibrium equation to see how the generalized forces and displacements are related. In the original form {Q} = [k] {q} Replace the physical variables with generalized variables -1 [A] {Q } = [k] [A] {a } * This is not trivial because in general the increment of work done does depend on the coordinate system used, and differs for coordinate systems moving with respect to each other. Our two coordinate systems will be "at rest" with respect to each other.

84 T Premultiplying both sides by [A] {Q } = [A] [k] [A]{ a } This is exactly the relation needed for a generalized stiffness, [k ] [A] [k] [ A] < generalized a stiffness This generalized stiffness is actually much easier to calculate than the stiffness because T I ]T T[SD] -l [k ] - tK T 1 [qr] T[ [S-D] [C] [S-D] [ dV = [] T[SD] [C] [S-D] [C] dV V Generalized equivalent nodal loads are related to equivalent nodal loads the same way that ordinary nodal loads are. {Q } = [A] T{Q} a e.n. 1. e. n. 1. = [A] T [N] T{})dSpace space = [A] T ([4] [A] 1) T{f )dSpace space T [A] T [A] -1 [T] T{ }dSpace = f [q] T{ }dSpace < --- generalized e. n. 1. space It is possible to solve a problem in terms of these generalized coordinates; the procedure is of primary value for academic insight.

85 IV. EXAMPLE USING LINE ELEMENT For the line element, [] -= [ 1 -1+2] a x [C] = [E] T T [k] - J [] [S-D] [C] [S-D] [] dV V Fig. 2. Natural modes ) 1' 2 a ax / 1 0 0 0 0 2 Af [E] [0 - ]dx 0 LJ9J 2 o 0; 9 _ O OJ [k ] =A,J dx ~ 2 4AE 0 0 Note that the presence of a rigid body mode has made this stiffness matrix "sparse" (with many zeroes).

86 Because this stiffness matrix isolates the stiffnesses causing strain energy, it ought to be easy to calculate strain energy using [k ]. Strain energy in a multiple degree-of-freedom, discrete system is aidentified Students: Can you prove thi s I= 1 {} Tk ] { / relation from U = 1{a}k }{aI) T 2 aU = {q [k] {q}? )T 0 1 i 12 10 2 ~~2 a~4AE a 22 1 4AE 2 2 2 and, as expected, strain energy depends only on the generalized coordinate corresponding to the straining mode.

INTERPRETATION OF Oct. 27, 1978 RIGID BODY MODES USING DISPLACEMENT FUNCTIONS W.J.A. The presence of rigid body modes in a single element can best be seen from displacement functions, rather than shape functions. For instance, Turner' s original triangle element has a u(x, y) x 0 0 y I v(x, y) 0 y x -x 0 1 IC =^-^ v-^ —^ qS generalized coordinates straining rigid body modes modes where a, b and c are the constant strains and A, B and C cause rigid body rotation. In more detail: a ~ constant direct strain in x direction b constant direct strain in y direction c - constant shear strain A rigid body clockwise rotation about the x, y origin B - rigid body translation in x direction C - rigid body translation in y direction One concludes that the generalized coordinates and displacement functions used in this Turner triangle are equivalent to Argyris' natural modes. Another way to see the rigid body modes inherent in an element formulation is by the [A] matrix, if it is available. A convenient coordinate system such as one at the centroid (Fig. 1) is helpful but not necessary. 33 (i Fig. 1. Two-dimensional triangle. 87

88 7(X1,y17 0o Y1 X1 -X1 0 1 x2 0 0 y2 1 0 [A (Xl, ) I -- " X2 ~ ~ yZ 1 oi 2AS = =Y X2 -X2 1 0 y3 x3 -x3 0 1 \ / \x2 1 O' rigid body mode s The rotation is clockwise around the origin in this case. _ 4,-A ~A (a) Rotation about centroid (b) Horizontal translation (c) Vertical translation. Fig. 2. Rigid body modes for Turner triangle. A square two-dimensional element is considered next. An elastic body in 2-D must have 3 rigid ( -1, 1) (- - (1, 1) body modes as seen here: ) vu(x,y)i F[ 0 -y.... I i v(xy) iO0 1 x.x a 0 1 i rigid straining body modes (1 ) 3 Ymode s ^c~-~-c-i 3-. —m l ries- -1) Fig. 3. Plane stress square.

89 The [A] matrix may be formed to yield 4Q(xl' y19 1 0 -1.... 0 1 1 0 -1..' <|(x'y) Y s0 1 1..... 10 1 -1..... [A] -- *1 0 1..... i(x3J 3) } 0 3 3:,0o 1 -1..... s to Ji o 1..... | (x4, 4 0 1 1 rigid straining body modes mode s Note that the choice of modes is not unique. One could choose rigid body translations in two arbitrary, different directions. The rotation could occur about any other axis in addition to the origin. The magnitude of the modes is also arbitrary, since the generalized coordinates ca. adjust to whatever displacement functions are used.

Nov. 6, 1978 COORDINATE TRANSFORMATIONS.A _W. J. A. Reference: Desai and Abel, Section 6. 5, 6. 6 I. LOCAL VS. GLOBAL COORDINATES Thus far, we have not distinguished between "local" and "global" coordinate systems, and indeed, they are often identical. There are times, however, when certain elements are located in a structure in skewed positions such that a single coordinate system for the entire system is impractical. One then prefers to have a separate "local" coordinate system for each element and a "global" coordinate system for the assembled structure. In Fig. 1, the 3 line elements shown have been ~! imbedded in a 2-D space at various / angles. Each line element is well t _ understood in its own local system and might be easier to handle in a d local system before assembling the stiffnesses into a global system., Fig. 1. Several line elements in a 3-D structure. II. IMBEDDING A LINE ELEMENT INTO 2-D AND 3-D SPACE To generalize the earlier solution of a line element, let us imbed the line element in a 2-D space. It then becomes a 2-D truss member. Q1 EA/L 0 -EA/L 0 q T Q 0 0 0 0 q Q -EA/L 0 EA/L C q 3 3 Q 0 0 0 0 q4 - 4 4 Fig. 2. Line element embedded in 2-D. This step is particularly easy to do with an x axis lined up with the element axis - i. e., a local coordinate system. One can also imbed the 1-D line element into a 3-D local coordinate system: 90

91 Fig. 3. Line element imbedded in 3-D. aQ EA/L 0 -EA/L 00 q 1 1 sQll Q 0 00 0 0 q Q2 O O O O 2 2 Q 0 00 0 0 0 q -3 l3 (2) Q ( |-EA/L 0 0 EA/L 0 0 q 4 Q 0 00 0 0 0 q 5 5 Q 0 00 0 0~ q Q6J 1~~ ~ ~~ 6e One concludes that local coordinates systems can be very helpful in developing element stiffness matrices. III.'ABORTED ATTEMPT TO ASSEMBLE, USING LOCAL COORDINATES If one tries to assemble a stiff- ness matrix for the two truss elements shown using oc coordinates, the result is a disaster. The trouble is at the center node, where the stiffnesses lead to force components CX in 4 different directions. One can' t sum stiffnesses easily.s This can t Fig. 4. Attempt to assemble in local sum stiffnesses easily. This can be solved by developing a transfor- coordinates. mation of coordinates for each element before attempting assembly. Fig. 5. Difficulty with coordinates at center node.

92 IV. DISPLACEMENT TRANSFORMATION AT A NODE. 2-D. The approach to use in Fig. 4. is not to relate the displacements of ele- A ment A to those of element ), but rather to relate both local systems to a single global one. Furthermore, the problem is to concentrate on coordi- nate rotations, since translations of X coordinate systems don' t affect stiff- 4 nesses, displacement or forces. Fig. 6. Coordinate transformation. u Consider a centered displacement vector { } with its description { in the local system and { g) in the global system. Imagine that the vector { has been found. By projecting these components on the local coordinate system u = u cos c + v sina v = - u sin a + v cos a. ^g g one has: /uA Fcos a sin a u] -sin a cos a v [t] where [t] is a transformation matrix for the node in question. If this same process is repeated at each node in the element, one obtains: &- [t] [0]: [o] 1 2 -.o] I I i q - [0] [0] [tO] q l v2-i - I g [t T]

93. e., {q} = [T] {qg} Forces likewise transform: {Q } = [T] {Qg} The equilibrium equation for a single element has already been derived in a form that holds either for local or global coordinates. In local coordinate form: {Q} =- [k ] {q} Then using the transformations above, [T] {Qg} [k [T] {qg) {Q} = [T] [k] [T] {q } -1 T A curious feature of the rotation matrix [T] is that [ T] = [T]. (you can check this out on [T] for yourself.) This matrix [ T] i s said to be "orthogonal.1 Hence {Qg} =[T] [k] [T] {qg [kg] g Hence, at this point, we know how to transform displacements, forces and stiffnesses from local to global coordinates. Equivalent nodal loads will transform exactly as do the above concentrated nodal loads. A common user procedure is to lay out a problem in terms of a global coordinate system. The typical computer program will, however, create stiffnesses for elements such as beams and membranes in local coordinate systems and then transform those stiffnesses to global coordinates, as shown above. Assembly is done in the global system. V. SKEWED BOUNDARY CONDITIONS An exception to the above procedure is for a skewed boundary condition, such as an inclined roller sketched here. Most general purpose computer programs will separate out the assembled equations for that node and will write them in terms of a local coordinate system aligned with the skewed support. The

94 boundary conditions are then uncoupled in these nodal degrees of freedom, and the assembled equations can be solved. Such skewed boundary conditions arise often because of symmetry in structures, such as in the pie-shaped wedge removed from the cartridge chamber sketched in Fig. 7. Fig. 6. Inclined support. 00/ Fig. 7. Cartridge chamber. VI. FINAL COMMENTS Many students, upon working out their first sample problem on coordinate transformations, do it in the wrong order. In the common local-to-global transformation, you should develop a [T] matrix for elements one at a time and find the global stiffness for the elements one at a time. Assembly is then carried out. For the case of the inclined support or a plane of symmetry one can work with the assembled global stiffness matrix and use global transformation matrices of the form: [I] 0 0 0 transformation for O[I] 0 0 " the particular node o [a] = 0 [T - 0] 0 with skewed boun0 0 [t] 0 dary conditions. 0 0 0 [I] where the [t] matrix in this case rotates the global coordinates at a node back to a preferred local system. In general, then,use care as to whether transformations for a node, an element or assembled structure are required.

94a _ —-- 7 Nov. 1980 THERMAL STRESS W.J.A. References: 1) Zienkiewicz, 3rd edition, pp. 96-98 2) Desai & Abel, pp. 216-217 I. Equivalent Nodal Load We have shown in previous lectures that a prestrain { } causes an equivalent nodal load {Q} J [B]T[C]{ }dV. (1) e. n. 1. Vol 0 vol 0 o This came as a result of the stress-strain law L-} = [C]({E} - {E}) +{o}. (2) Let us concentrate on thermal strain. We will set {o } = 0 and use the law o }G = [C]({E} - {o}). (3) To develop feeling for the prestrain { e }, use a scalar example of a line element with left node fixed. Suppose the element is to be both loaded and heated. If the line element is initially unstressed at q2=0, then the loaddeflection and the stress-strain curve for zero temperature (reference level) would pass through the origin, as shown for T=0 in each figure. Once heated, however, a new load-deflection curve shown as T = +100 is used. One could interpret the new stress-strain curve as a prestress effect, Q, o/ ~,,[.-.-* 0/ o4 Fig. 1 L Em a' i d l Fig. 1 Line Element 2a, b, c. Heated line element under load.

94b namely as the stress needed to hold the heated line element in place with zero strain, but we have decided not to use prestress. (Note that under that interpretation, a negative stress is needed to hold the element in undeformed shape.) One does, however, identity e as the amount of strain caused by the free thermal expansion with no stress applied. I personally like this approach because one can keep signs straight - the thermal strain is positive when associated with a positive temperature. II. Thermal Strain in Various Situations In a 3-D solid, if one has local coordinates aligned with principal material directions Ex a 14 T 0 E a 2A T Yo E 3AT J~ = = (x~ 0 a (4) xy Yyz 0 ^0 ~zZX 0 where a 1' a Z and a 3are coefficients of thermal expansion. These a. may be flinctions of temperature and should account for all strain between the current temperature and the reference temperature (here taken as zero for convenience). In a 2-D, plane stress case x 1AT 0 {E E A ZaT y (5) o Yo Y 0 In a 2-D, plane strain case {E ) = E (l+v)* a ZAT (6) Y 2 ~xy L 0 ~J

94c If an element has orthotropic material properties, but the coordinate system used, say (x, y), does not lie in principal directions, one has for plane stress: _E - iAT x {Eo} = A1 A2T r (7) Y0xy 3 A T If a material has a temperature dependent coefficient of expansion, then, in each of the expressions above, one replaces the simple a AT with T j a i(T)dT, T o where T is a reference temperature. III. Example: Two-Node Line Element Consider an aluminum link 10" long. Model the link as a two-node line element. Properties are E =10 psi A y = 0.3 -= = 1. 23x 10-5 in (.j in. F. Fig. 3. Aluminum link. The link is to be heated at 100 F. and then subjected to various boundary con- ditions and loading situations. r / The stress-strain law becomes 49 / {r} = [E]({ }-{ O}) I / (8) 7 3 / = [10 psi]({e}- {1.23x10 }) -- - 7 / Fig. 4. Stress-strain curves for heated and unheated case.

94d The equations of equilibrium are Q1, Q 1 EA 1 - j e. n. 1. 0 where r Q T 1 \, = XJ [B] [C]{ edV (10) QVol e. n. 1. Ec 0 - 1/L L 0 =i L [E]{(rAT)Adx 1/L EAa A T EAaAT -12,300 lb - ^ ^ (11) 12, 300 lb We have, prior to specifying loads and boundary conditions, Ql + -12. 300(106 } 1 1 q} (1) Qk 12,300 in L q

94e The sketch of load vs. deflection is given. The initial loads (equivalent nodal loads due to initial strain) are absorbed into {Q} TOTAL Fig. 5. Combined loading. A. Constrained Element If the nodes are constrained, the element is held in place by the external concentrated forces. Solve Eqn. 12. Qz I-12, 300l 03 Q2 12, 31X 0o 12,50 l{b 6173o0 lb Hence rQll i 12, 3001 <= {.-,3 lb. 2QzJ:-1Z, 300 Fig. 6. Loads on constrained line elerme nt. The points on the stress strain diagram and the load deflection diagram are given. Point I on the load deflection diagram is unmotivated and is probably irrelevant. ^ 1 /I b Itress-St% _ Fig. 7. Stress-Strain and Load-Displacement for Constrained Line Element.

9h4 B. Freely Expanding Element If the same line element as above is heated 100 F. and allowed to expand freely, find the resulting displacements. The external nodal forces are zero. Set ql = 0 to remove the rigid body mode. Eqn. 12 gives 0 -12, 300 1 -1 0 { + O1 lb. = (106 psi) 0 +12,300 -1 1 q 2 From the second equation, q = 0.0123 in. e 1^~ {T0.61123 Fig. 8. Free expansion of aluminum link. -7>^5,, -^^, Fig. 9 Stress-Strain and Load-Displacement for Freely Expanding Line Element

3 Dec 1978 W.J.A. ELECTRICAL FLTUiID NETWORKS References: 1) Zienkiewicz, "The Finite Element ilethod", Ed 2,1971,Section 1,5. 2) Murphy, Glenn, "iiechanics of Fluids", Ed, Int. Textbook Co.,1952. I. Electrical Network Consider a network of resistors subjected to a.c. or d.c. voltage. (We will not include capacitors or inductors because that requires complex arithmetic.) There is an analogy I 7 between the electrical and the mechanical cases: [MECHANICAL ELECTRICAL. ( force at a note due to a certain member = current displacement at a node, voltg common to all members - b 1 FFig. 1. Electrical network stiffness of a member conductance ( / ) For the example shown, nodes are 1,2,...8 and elements are 1, 2,... 7 Consider a single element with no1i1*s i and j. Define voltage and current at a node,with current Re I' positive for flow into the elembnt. (Note a funda- mental difference between current in a resistort and force in a 1-D structure, in regard to sign 4 convention.) Pig. 2. Resistive element Jy continuity: T J =-E(1);.y Ohm's law: I -v _-, — V -Lt- R6-e~~~~ ~(2) LJ-=^ - (3) In matrix form: TTn i' nI (u)m In summation form: e e 1 -. V, (pll, t (5J) admnittance matrix term 95

96 Assembly is done by looking at a node and using Kirchhoff's current law, (current at a node) = 0. We do a slight departure by labeling the external current as positive into the node and the internal element current positive away from the node (into the-element): c, -_,fi.. (6) external current ( current away from node i into node " i" toward no-'e j through element e r6 m3 Fig. 3 Kirchhoff's current law This law is just like the case of a truss assembly. The assembled form is easy to see because of the scalar nature of current flow. Consider trie example shown in Fig. h. One can do this either from Eq. (6) or by inspection using the topology of the network. ~cas[3 o Vlr?^ a~/~ o V3 3 |(7) 44o o yo /g V4. The admittances "add" into the stiffness matri/: in the same way as stiffnesses do. The & are the external / currents fed in at each noie. I The internal currents do not appear in this matrix formu- lation, although they can be recovered, if desired. fig. h. enxample assembly

97 II. Fluid Network -Pine Flow Assume volume flow is proportional to ) ----. — pressure drop across the pipe, and assume volume flow is inversely proportional to a friction coefficient J'~P -',:~Fig. 5 Pipe element 9, e _ (8) friction coeff. Also p_.p Q " r" (9) In matrix form: S'y L-''2 UP2! (10) The sign convention on pressure and volume flow agree. Both tend to cause flow into the pipe. Again, as in the electrical network, this problem has only' limited vector character. It is inherently scalar (l-D) and is not a degenerate 2-D or 3-D problem. The pipe can be bent as in Fig. 6 or even tied in a knot! In general, the friction coefficient4 is a function of volume flow, but we will at first consider the case where 4 is constant (slow, viscous flow). Fig. 6 bent pipe Assembly Use an assembly process based Q4. Q on imbediing each element equation - < -- into an expanded matrix the full - P P - size of the assembly. In Fig. 7, f by continuity, at the junction:' Fig. 7 Pipe assembly (1I)

98 For the separate pipes: {@'2 = 4 -4 44 4 14 -4 -)'{R+/ (12X13) Imbedding these equations in a 3x3 matrix form, and setting P3 Pf: *QxtQ4 }' /,. ( i ) _ I PL (14) We -can now identify each of the volume flow terms on the left as the external flow into the node c{3 U< -----—, -—' —~ 3 -a I |(as above) - i (15) ^ 11 ~P3 | F;9 &. ASS^U^^- KYS+U.. (eobc4$ rfcu*\Vier~A) c a! t 3( external assembled nodal fluid sources "flow" matrix pressures only The script notation is intended to signify an assembled system, in contrast to capital letters used for single elements. III. Nonlinear Systems Actual pipe flow problems are nonlinear in the sense that the pressure drop is approximately proportional to the square of the discharge in turbulent flow, rather than proportional to the first power. The expression for head loss Hf due to friction is (44 - ) (16) where f is a friction coefficient, J/d is the pipe length to diameter ratio, V is the flow velocity and g is the acceleration of gravity. The friction coefficient has a mild dependence on flow velocity, as shown in standard Stanton diagrams (Ref. 2). The result is that pressure drop is prooortional to velocity slightly less than squared, perhaps to the 1.8 power.

In finite element theory, one casts nonlinear problems into quasi-linear ones, such as by absorbing the nonlinearity into the "stiffness" matrix. A typical relation between pressure drop and discharge might be a =^ 9 Q (17) constant for instance. One would factor out Q as shown aP (^ Q, )Q (18) The previous expression for flow in a pipe becomes ] _I 1 p f (19)'. Cl- " fi.Q IT t: One now views the flow matrix as a function of volume flow,ie. 1jklJ- Lk(g)I1 (20) Because the flow matrix for an element depends on the volume flow through the element, the problem is nonlinear. For an assembled system: (21) For such nonlinear problems, one typically uses an iteration solution. One type of iteration requires an initial guess for flow rates. The system flow matrix is then calculated, and the set of coupled, linear algebraic equations is solved for unknown Q and 0^. The process is repeated until convergence to proper flow rates and pressures is achieved. This is called a secant stiffness approach and is one of the oldest ways to solve nonlinear problems. The method is sometimes useful for smaller problems, but often does not converge. IV. Example of Nonlinear System Suppose pipe flow in the system of two pipes in Fig. 7, Section II follows a law AP: r^6. One cala show after some discussion that the assembled equations should be:

99'(,1 r > -' ___ I 1/ ~,.. j / Iov "t C -l_ o For larger problems with many nodes, it is not so easy. The problem is to identify flow rates within individual pipes, which is suppressed information for internal elements in a system. This corresponds to the structural situation in which internal forces are intentionally suppressed. Let us carry out a numerical solution where 4' -:L l.s*/4 (1P - too b/14 ~-~ = 1 " P -200 Ad, 5 SO and we wish to find the flow rates. (The matrix is singular, but we can solve the problem in this "direction" without reducing the matrix size.) A Fortran computer program has been written to iterate for the solution. A listing is given. For this specific problem, the results converge to 6( = -12* C. 6L = l.7 3 - 8. Convergence is moderately slow, requiring 29 iterations for 1% accuracy. V. Comments The physical problem discussed has been contrived in order to show a variable stiffness, secant iteration. The secant iteration does not converge for cases where the nonlinearity is strong. The dividing line for convergence, found by trial and error,is where the pressure head varies as velocity squared. For more pronounced nonlinearity, such as a velocity cubed dependence, the method diverges. The secant iteration can therefore fail.

100. C I::':'E 2. C T-HIIS F'PROGRAFM IHAS BIEEN CREIATED F31OR AERO 510 3 C 4 CIT IS S MEANT TO ILLIUSTRATE A VARIABLE..E STIFFNESS 5 C METHOD, USING FLOW THROUGIH FIPES AS AN EXAMI-I.LE. 6 C 7 DIMENSION Q(3100) 8 FREAL K1(3Y3) 9 READ (51 ) 1-Fl 1yF':Ly' P2 F;'3EY ER ERORF i'21 Y Q3y ITER 10 1 FORMATl (8F10.O/2F10 I + 5) 11 WRITE (6y4) Fl:l P 1::2 F'FE2 Y PI I Y ERI' OR yQ Q2y f ITER 12 4 FORMAT (10(F14+6/) I5) 13 I=0 14 2 I:= I+1 15K( I1Y 1)=-1 /F1/(ABS(QC)* E):16 K(3y3) -=./F2/ (ABS (3)**E) 17' K(1,2)= —K( 1 ) 18 K7 is, 3) =0. 19 K(2,1)=K(1 i2) 20 K(2,2)=K(1 1)+K(3y3) 21 K(273)= —K(3,3)'22 K (3Y1 ) =0. 23 K(3P2)=-K(3y3) 24 Ql=K (1,1 )*F:P1+,K(1 Y2)*FP2 25 Q2=K(2 1)*P1+K (2 2)*P2+K(2,3)3F' 26 Q3=K(3v2)*F'2+K(3y3)*P3 27 Q(1 I)=Q:l 28 Q(2I)c=Q2 29 Q(3 I) Q3 30 WRITE (6,3) I Qil YQ2YQ3 31 3 FORMAT (' THE' YISY' TH ITERATION GIVES A FLOW VOLUMEVECTOR'/ 32 1 F14+4/F:L4+4/F14+4) 33'IF (I EQo1) GO TO 999 34 ILAST= - 1J.35 ERR=ABSF:' Q ( 1C I) U- ( 1. ILAST ) +AIBSF (C ( 2 Y I ) -Q ( 2 2 ILAST) ) +ABSF ( 36 1 Q( 3I)-Q(3J: ILAST) ) 37 IF' (ERfR.*LE. E'RROR) GO TO 1000 38 999 IF (I. GE* ITER) GO TO 1000 39 GO TO 2 40 1000 CONTINUE 41 END

W.J.A. 13 Feb. 1976 J DYNAMICS I. Reversed Effective Forces Newton said: {Q} = [m]{ q } and got great fame. D' Alembert said: {Q} - [m]{f} = 0 and also got fame, fortune, etc. Moral: Look at things in a different way! You can treat dynamics problems as if they were static by including inertial forces as reversed effective forces. For lumped systems where finite dimensions are involved in equilibrium processes, reversed effective moments are also generated and must be applied about centers of gravity. II. Consistent Mass and Damping Matrices Generally speaking, one can imagine internal forces being generated in a structure due to increasing orders of derivatives of the internal dis - placement field {q}) {q} {qt, {}, {'q"}, / t t t t elastic damping inertial "jerk"? We have discussed elastic forces extensively up to this point. Damping and inertial forces are also familiar. No significant forces are generated by higher derivatives of the displacement field, although the third derivative causes physiological discomfort and is a factor in ride quality of vehicles. Let us develop equivalent nodal loads for damping and inertial forces in the identical way as for other body forces: (X} {X} externally + {} + {X}inertia damping caused damping Then Q} e. n. ol [N] T{ } dVol 5

A commonly used, but often artificial, type of damping is "viscous" damping in which a body force tends to oppose the instantaneous velocity of the particles: V {X}damping t -c } N0 } The inertial force is universally acclaimed as:'6 a' 2 {X}. - - p Q — {cv } inertial 2 C aat Equivalent nodal loads for these effects are: {Q} 1= K [N - c t-{u} )dV + J [NI (- c 2a {}dV V V_ t Discretization with shape functions {u} = [N]{q} - [N(x, y, z)]{q(t)} where at {-u} = [N]{q-} 2.{u} = [N](q-} at leads to Q)} 1 = - f c [N]T[N]dV{q} - I p[N]T[N]dV{q} V V EC] [m] Hence the equation of motion becomes: [K]{(q = {Q} - [C]{ - [m]{q} In standard form, one writes: [m]{q} + [C]{q} + [K]{q} = {Q} 6

The finite element theory is an excellent method for discretizing a field problem and reducing it to this standard form. Methods for solving this equation of motion are universal in the sense that other discretization methods such as lumped mass methods and finite differencing also produce this standard form. In this respect, the methods to follow, such as the normal mode method, apply equally well to problems generated in other ways. The method of developing mass and damping matrices above is based on virtual work. It gives matrices which depend upon the type of motion assumed through the shape functions, and hence the matrices are consistent with the displacement field. These are called "consistent" damping and mass matrices. II. Example of a Consistent Mass Matrix --— Beam Evaluation of the consistent mass matrix for a beam is not trivial, because the shape function is intricate. From earlier notes (29 Sept. 1975), the shape function for a 2-node, Euler-Bernoulli beam is: [N] = [L3 -3Lx2 + 2x3, - 2Lx2 + Lx3, 3Lx2 - 2x3, -L2x2 +Lx3] 3 If the mass density of the beam is constant over its volume, the consistent mass matrix is found after much manipulation to be: 156 22L 54 -13L 4L 13L -3L2 420 m 420 |156 -22L (Symm. 2 4L (See Przelmieniecki, Theory of Matrix Structural Analysis, page 297) 7

IV. Lumped Mass Matrix An old-fashioned but effective way to calculate the mass matrix in a discretized problem is to divide the structure into subdomains surrounding each degree -of-freedom (d. o. f. ) and arbitrarily assign a portion (lump) of mass to that degree of freedom. This makes sense with regard to translational degrees of freedom but is far less intuitive with rotations. A good example of a lumped mass matrix is for the Euler-Bernoulli beam, where half the mass of the beam is concentrated at each translational node and there is assigned no rotational (. 4 inertia at all. The latter assumption J has some basis in the fact that rotational H1 aV inertia is already neglected on a local basis in this beam theory, but is really not justified for finite length beams. The result is: 1 0 0 0 -0 0 0 0 [m] = AL 0 0 1 0 0 0 0 O V. Comparison of Consistent and Lumped Mass Matrices The consistent mass matrix is a much richer source of inertial effects than the lumped mass matrix. Each nodal displacement and rotation causes inertial effects (reactions) at all other degrees of freedom when the consistent mass matrix is used. One limiting case of comparison is useful: L -> 0. One might suspect that as the length of the beam element becomes infinitesimal, the distinction between the consistent and lumped masses might disappear. This is not so. Taking the consistent mass matrix, and letting L -* 0, and discarding terms of 8

order (L2) compared to those of order (L): 156 0 54 0 pAL 0 0 0 0 Lim [m] = L- consstent 420 54 0 156 0 0 0 0 0 This shows the coupling that remains between displacements at the two ends of the beam, even in the limit of small length. As a specific example of this behavior, consider a positive, unit acceleration in the first nodal coordinate. The reversed effective forces, as converted to equivalent nodal loads, are shown: ACCELERATION,.,....,q _ 1 156 pAL 420 54 2 10 AL 454o p AL 4 pAL 420 42t t Reversed effective forces Reversed effective forces from consistent mass ma- from lumped mass matrix. trix. VI. Types of Vibration Problems Vibration problems are divided into two major classes, free and for ced. 1. Free Vibration. No external forces act on the body. The motion of the elastic body is due only to a balance of internal forces, including elastic, damping and inertial effects. Mathematically, this is an eigenvalue problem, and reveals much of the inner character of an elastic system. If the system has no damping, or if it is neglected, one often speaks of 9

"natural vibration" in the so-called natural frequencies and modes. These quantities are often used as building blocks for understanding the next category (forced vibration) and so studies of this idealized system are very important. 2. Forced Vibration. In a linear system, the character of the force input largely determines the character of the response. Proceeding from the simplest to the most difficult: A) Harmonic t B) Periodic t C) Transient t D) Stationary Random \ t E) Random- t: We will spend most of our time on free vibration and the response of linear systems to harmonic or general forcing. Often the goal of the forced vibration solution is to reduce the problem to one of solving uncoupled ordinary differential equations with constant coefficients. 10

112 W. J. A. Categorization of Nonlinear W. J. 11 January 1977'-'J -~ A.P r o b le m s References: Section 7. 3 in Desai & Abel Nonlinearities enter solid mechanics through the constitutive law (stress-strain) and through the strain-displacement law. Nonlinearities destroy the uniqueness of solutions in elasticity and allow multiple solutions, bifurcations and jump instabilities. Graphically, GENERAL CASE Nonlinear Stress-Strain Nonlinear Strain-Displacement Time Dependent (High Speed Crash of Auto Frame) STATIC PROBLEMS DYNAMIC PROBLEMS Linear Stress- Linear Nonlinear Nonlinear / Strain; Stress- Stress- Stress- same entries Linear Strain- Strain; Strain; Strain; as at left Displacement. Nonlinear Linear Nonlinear |Classical Strain- Strain- StrainClassical Elasticity Displace. - Displ. Displ. Plastic Large / Material Deflection Nonlinear of PlastiElastic cally DeMaterials forming Bodies Linear Stress- Linear Stress-Strain Strain; Strongly Nonlinear Mildly Nonlinear Strain-Displacement Strain-Displ. Strain-Disp. Elastic Post-Buckling Euler column of Cylinders & StifBuckling; fened Plates; Bifurcation "Oil Canning" it Instabilitie s ( ~ —-----------— < ^ —-These are small strain & These are small strain & large displacement problems. small displacement but with a large, dominating load in a

November 14, 1977 INTERPOLATION. NATURAL COORDINATES. W.J.A. Reference: Desai & Abel, "Introduction to the Finite Element Method," Sections 5.4, 5.5 I. INTERPOLATION Reference: Hamming, "Numerical Methods for Scientists and Engineers," pp. 227-250. Interpolation is the basic feature ( of many engineering methods, including finite element methods. Let us concentrate on a scalar function of one independent variable, f(x), as sketched. I A linear interpolation of f(x) for l points lying between xl and x2 can be I l based on an average slope I I f(x ) - f(x ) x X X2 - X1 to get Fig. 1. General function. f(x2) - f(xl) f(x) = f(x ) + 2 (x-x ) x2 - x1 1 By rearranging terms, we get f(x) f(x1)[l- ( )] + f(x2)( x 2 1 2 1 and f(x) f(x l) + f(x) X - - - -1 x' 1 2 1 x>/ I I I X2X1 f ~x2)( I This particular formula samples the func- CW- - tion at two points (the endpoints) and interpolates by multiplying the weighting I functions which account for "how close" x I is to x1 and x2. We have already done I - this sort of linear interpolation in our x X constant strain line element and the Turner (constant strain) triangle, where for the line element Fig. 2. Linear interpolation using function values only. 101

102 u(x) = q1(l - x/L) + q2(x/L) (4) in local coordinates, for example. Hamming suggests that interpolation uses information about the function f(x) for interpolation in the form of: 1) function values only 2) values of derivatives 3) differences of function values 4) arbitrarily placed samples. Where functional values only are considered, the finite element expression previously used for the displacement field is adequate: u(x) = Z Ni(x)qi {u} = [ N(x) ]{q} (5) 1 1 x n n and can be used for higher order approximation, such as with quadratic shape functions (interpolation functions). II. NATURAL COORDINATE SYSTEMS: LINE ELEMENT. Reference: Huebner, "The Finite Element Method for Engineers," pp. 134-150. For a line element of length Q, lying on (xl,x2), define x -x 1= x2-x L = 2 1' 2 1 (6) x-x1 = x-x L - 1 2 1 2 2-X1 The normalized coordinates L and L2 satisfy: L +L = 1 (7) X 1 2 Note that L1 is unity when x is at the left node (x=x1) and L2 is unity when x is at the right node x=x2. This Fig. 3. One dimensional domain. means that the "natural coordinates" L and L are themselves interpolation functions (like Nl(x) and N2(x) used previously) and can be used to express the internal displacement field: u(x) = q1L1 + q2L2 (8)

103 On the other hand, L and L define the Cartesian coordinate x in a similar interpolation: x = x L1 + x L (9) This can be proven:? x -x x-x 1 x21 x = x,( — + x 1 x2- 1 2 -X1? Xlx2 - X X + x2x - x2x1 x2 - x1? x(x2-x1) (x -x ) = x This checks! The fact that the natural coordinates L. can be used to interpolate both the dependent and the independent physical variables is important. In higher dimensional spaces, this allows curved boundary elements (and not just straight) to go with higher order displacement polynomials (not just linear) in isoparametric element theory. Now consider the mapping between x and the L.. Put the relations x = fcn (L1,L2) and L1,L2 = fcns(x) into matrix form. Equations 6 yields, directly from the definition of the natural cqordinates: CL1 ~ x2/P - l/R: A r (/Ql 1 ) (10) L2 l/ 1/ x and Equations 8 and 9 give: 1i\ 1 1 2\ } = L 3 ( } (11) Equations 10 and 11 form an unlikely transformation "pair." There is only one Cartesian coordinate x and really only one independent L.. The relations could.1 be simplified to single equations if desired, but that is not helpful for computations. The chain rule for differentiation is:

104 d dL1x + dL (12) dx aL1 dx L2 dx and using Equation 7 for the derivatives, w.r.t. x: d 1 + 1 a +dx Q UL- + Q L (13) 1 2 Integration in the natural coordinate system for arbitrary polynomial terms becomes X -P p! q! ~ rX1 L1 2 qQ)! (14) x L L dx =- (p + q + 1)! Example 1: Two node, constant area line element. Assume constant strain in the element, due to a linear displacement field: {u(x)} - [L1 L2] q2 Then x +1 [K] = TN] [d ] T[E] [d ] [N]dx _ x1 dx dx r^i fI' 3 dLdL1 dL = r 1 dx [E][ dL2 Fig. 4. Two node, constant area X [ d2 d x dx line element. dx _ L x 1 E E x1 [ ]dx EA 1 1 T1 1 i This checks our earlier solution with Cartesian coordinates.

105 Example 2: Three node, variable area line element. Given: x = lL + xL2 A we will choose: u(x) = q12Ll(L1- ) x —x + q24LL2 A = A( + B 1) + q2L2(L2- = A (1 + BL2) 3 22 o 2 In matrix form this becomes: {u(x)} = [2L(L1- 2) 4L1L2 2L2(L2 2 q2 q3 [K] x1 [NT[ d ]T d [K] = [fx [E][EN]A (l+BL2)dx 1) (,L^) 1 2x d, d(2L -LL) Fig. 5. Three-node, variable area line element. x +k fX+1 d 2 2 d x(4LlL2) [E][d (2L-L) x(4L1L2) dx (2-L2) Ao (l+BL2)dx X | 1 42 dx 1 1[- 1 - (4L2)(21 2-l)]A2(l+ ( Q- L2 -1) 1 (4-l) 4-(4-1)(L -L Ll-i) -^ L1- 1) 1 i 1 4 1 = EA f -(4L (L ) (L 1 1 (i4L2-l)]A(l +BL2)dx o X 2 + 4 [- (4L1-1), -(2- 1 2- - (4L2 - 1) 1 2 4Ll 4 -1 -(4L -1) 2 (4L -1) (4- -1) j2 (4L2 1-2L2-L1)l 2(L2-L1)(4L2-1) (l+BL2)dx - 1 (4L 2-)(4L1-l) r',j(4L2-l)(L2-L1) R2

106 Admittedly, for this relatively simple problem, the integration above is equally as difficult as a Cartesian integration would be. The advantage of the method lies in higher dimensioned domains. Typical calculations are: 1 2 k = EA f/ (16L 8L + 1)(1+ BL )dx 11 o 0 2 1 1 2 EA x 2 2 = - [16L + 16BL L -8L - 8BL1L2 + 1 + BL]dx o2 x1 1 12 1 2 Now, 2 2! 0! 2 _= 1 I 1 ~ 3! i ~ 3.2 3 fL1L2dZ 2! i 2 - 1 L2d 4! = 4.3.2 12 f Ll d = 1! 0! = 1 2! 2 1f L L d 1! l! 1 f1 2 3! 6 f 1 dQ = Lod~ 1 I R1 f L dQ =- 2 k 16( ) - 16B - 8(2- ) - 8B + 1 + B 1. ~2 [(3 12 2 6 +2 EA o- 16 16 8 1 - 4 + 1) + B( + ) 91~ 3 ~12 6 2 EA o 7 1 - [ + B]. The total stiffness becomes: 14 + 3B - 16 - 4B 2 + B EA [K] = - 16 - 4B 32 + 16B - 16 - 12B 2 + B - 16 - 12B 14 + 11B In this element, the independent variable x was given a linear representation in Li, L2 whereas the displacement field is given a quadratic interpolation. Because the independent variable is interpolated with a lower power polynomial than the dependent (field) variable, the element is said to be subparametric. There are also superparametric elements where the converse is true, and isoparametric when both are interpolated to the same order. Isoparametric elements are most common of the three.

107 III. NATURAL COORDINATE SYSTEM: 2-D, AREA COORDINATES. For a triangular element, one can interpolate x = xL + xL + x 3L3 y = YL1 + Y2L2 + Y 3L3 (15) 1 = L1 +L2 + L3 The quantities L. can be found from inverting this relation to obtain L1(x,Y) [a1 + bx + cly] C L2(x,y) = ^[a2 + b2x + c2y] (16) L (x,y) = [a3 + b3x + c3y] where A - area of triangle and a1 = x2Y3 - x3Y2.-. —. Fig. 6. Area coordinates 1 Y2 (17) for a triangle. c1 = 2 1 The pair of matrix equations 16 and 17 define the L. once given the linear interpolation model, or one can start with the linear surface such as Ll(x,y) sketched and then prove the interpolation. The integration for the area coordinates can be shown to be f LP dA - q! r! 2A 1 2 3 (p+q+r+2)! IV. NATURAL COORDINATE SYSTEM: 3-D, VOLUME COORDINATES. The volume coordinates for a tetrahedron are shown in TABLE I. One cannot plot the domain of the (L1,L2,L3,L4) quadruple because it is a surface in a four-dimensional space. Again, one is able to invert the relation between x,y,z and the L. to obtain L. = function (x,y,z) i=1,2,3 so that the chain rule for differentiation can be applied.

108 V. "MAPPING" COORDINATES. THE SERENDIPITY FAMILY. TABLE X shows a system of coordinates in which the N. skhpe Of^+,', are related to a,n,C system. The mapping is done through the use of the "serendipity" interpolation formulae; e.g., for a hexahedron in 3-D: 1 8 Ni 8 E Z + (1 + i ti)(1 + n ni)(l + r Si) i=l and where (~,n,I) are coordinates with nodes (., n.i, i.) lying at the eight corners of a cube. The cube has sides two units long each, extending from (-1,1) on each axis. This means that the original body is mapped into a cube.

109 TABLE I | TABLE OF NATURAL COORDINATES | Length Coordinate: L., Ll1 L22 2 ILdKX 1 L1 +L2,( ) i 2 Xl XL (.0 Area Coordinates: (suitable for triangle) x L1X1 + L22 + L3X3 y -Llyl + L2Y2 + L3y3 1 L1 + L2 + L3 -- Volume Coordinates: (suitable for tetrahedra) x = LX1 + L2X2 + L3x3 + L4x4 y = Llyl + L2y2 + L3y3 + L4y4 L3 L+ r3) z LZ1 + L2z + L3z + L 94Y4 ^ (5:c.fce 1 = L+ L2 + L3 + L4 4TAttt. fI IF sgE>b-n PrY, i,wE't"....MAPP.u C00tb.TgS.. 2-t: Quadrilateral Coordinates: x = ([(1 )(l -n))xl + (1+) (1-T)x2 y + (1+)(l+rn)x3 + (1-0)(l+n)x4] I/ I = z4 (i+Uti)(i+nni)xi - - - K1' ^f 1 I ^ /'>-^'/. 4 ~ ~6i = Xr Mx where N = (i+rr)(+nn ) ( i) -'i) I i1 4 (+':1i y = Z Niyi 3-t' Hexahedral Coordinates: x = Z Mi Xi = Z Mi Yi - = X)i i where Ni = d(1 + rr(i )(l + n i) +,i) i'a )_/y_ -" — cZ NI. J - ^ ^ -^ -N, y~~~~~~~~~~

November 12, 1976 GAUSSIAN INTEGRATION W. J. A. I. POLYNOMIAL APPROXIMATION - CLASSICAL THEORY Reference: Hamming - "Numerical Methods for Scientists and Engineers, " Chapters 14-19. A. Motivation O We often need to integrate a scalar function (such as strain energy) over a volume. * Can develop the theory in terms of scalar function of one variable, e.g., strain energy in a line element. * The function being integrated is often not available in closed form. In the case of isoparametric elements, the Jacobian is very complicated - possibly the determinant of a 32 x 32 matrix. B. Interpolation Hamming gives four basic ways of finding a functional value f(x) at a Ad) general point x when the function is not known everywhere analytically: - -I 1) Methods which use tabulated \ data from points near f(x), I say f(A) and f(B) and to esti- mate a value between, as one I does in log tables. - lA;_.... *-. 2) Methods using differences of A functional values, e.g., the straight line approximation: f(x) = f(A) + (- A)(f(B)-f(A)) 3) Methods involving derivatives e. g., Newton' s method, f(x) f(A) + dIA(x -A) 4) Gaussian quadrature using arbitrary points on the x axis. 110

111 II. GAUSSIAN QUADRATURE Reference: Zienkiewicz, The Finite Element Method, vol. 2, pages 144-147. A. We will discuss the symmetric form of Gaussian quadrature. The goal is to integrate the area under a curve for -1 < x < 1, where the function can be sampled at any point. One proposes a forni of solutions: 1 N f - f(x)dx w (k) -1 k=l In other words, one replaces the ( integral with a finite sum of terms, - each consisting of the product of a "weight factor" w and the function evaluated at an'abscissa" xk. This is a remarkably bold step in- X asmuch as not only the weight factors, but the abscissa must be found during the process itself. B. Defining Equations Reference: Hamming, Chapter 19 Suppose we wish to use a two-term approximation to the symmetric integration: f f(x)dx = w1 f(xl) + w f(x2). -1 This means there are four unknowns, w, w2, xl and x2. We assume f(x) is available in some form for numerical sampling, that is, we can pop a value of x into f(x) and get out a number, even though we can' t write out f(x) analytically (f(x) is a "black box"). We can impose four conditions on the problems, in equation form, and hopefully have four equations for the unknowns. One way to proceed is to imagine that f(x) is a polynomial (it often is). If f(x) were cubic, 2 3 f(x) = a0 + alx + a x + ax, we could insure that our integration process would be exact, by demanding that it separately be exact for a constant, a linear term, a parabolic and a cubic.

112 If f(x) = 1, then the integration formula becomes f (l)dx = wl(l) + wZ(1) -1 or 2 = W + w | 1st defining equation If f(x) = x, 1 f x dx = wX1 + w x2 or 0 = Wl1 + WZX2 - 2nd defining equation 2 If f(x) = x 1 2 2 2 f x dx = wx +w x 21 2 11 2 2 I3 = w1x + w2xj f 3rd defining equation and if f(x) = x 13 3 3 f x dx = w x 3 +W x I1 12 1- 2 2 3 3 o = w1x + w 2x 4th defining equation We have created 4 defining equations and can solve for all w and x The expression f(x). wlf(xl) + w2f(x2) — l can be used to integrate an arbitrary function f(x); furthermore, it will exactly integrate polynomials up to cubics. The latter idea is confirmed by writing out the special case

113 1 1 Z 3 f f(x)dx f (a + alx + a2x + x )dx -1 -1 3 1 1 12 13 = a f (l)dx + a f x dx + a f x dx + a J x dx -1 x1 -1 x-13 and our defining equations assured us that each of these integrals would be exact. This is linearity of one type. The equations for the wk and xk are certainly not linear, however! III. EXAMPLE OF GAUSSIAN INTEGRATION IN ONE DIMENSION Reference: Desai and Abel, page 147, Ex. 5-11. This example is taken from a textbook but recast into symmetric form. Let f(x) = 0. 5 - 0. 5x and use the 3 point Gaussian integration 3 f f(x)dx = w. f(x.). T ) -1 i=l'1 Assume someone else has already solved the defining equations to give t x = -0.775 w = 0.556 x = 0 2 = 0.889 xj = 0.775 w3 0.556 K Xt X c one has: Jf f(x)dx = (.556)f(-.775) + (.889)f(0) + (.556)f(. 775) -1 = (.556)(. 8875) + (. 889)(. 500) + (. 556)(.1125) = 1.0005 This is an approximate check for what one expects for the area, but it should be exact! (Because f(x) is a quintic or less). The error is due to round-off of the w. and x. given by Desai and Abel. 1 1 Redoing this problem with more exact values from the "Handbook of Math. Functions, " by Abramowitz and Stegun, one has f f Odx= (0.55555.. )f(-0. 7745966692) + (. 88888.. )f() + (. 55.. )f(0. 7745966692] 1 1. 000000000 (accurate to 10 significant figures).

W. J. A. January 26, 1976 NONLINEAR PROBLEMS I. General We have some numerical methods that require solution of [ k({q}, {Q} )] {q} = {Q} (1) and in which the secant stiffness is used (by definition). Other problems are of the form [k({q},{Q} )] {q} = {Q} (2) in which the tangent stiffness is used for the most accurate solution, but in which the elastic stiffness or the secant stiffness can be used as approximations. Finally, because the stiffnesses in (1) and (2) are generally very complicated to recalculate, we might want to get the nonlinear effect by, replacing (1) by [ke] {q = {Q} + n +. +{Q} e.n.l. (3) E r0 0 0 or by replacing (2) by [ke]{ q} = {AQ} +{AQ}en + A.n.. (4) O O 0 0 (Usually only one of initial stress or initial strain is used at a time.) II. Virtual Work It greatly helps understanding of the different approaches if one returns to the virtual work formulation and includes the nonlinearity. Suppose that, indeed, both the stress-strain law {} = [C({E},{u} ] {} (5) and the strain-displacement law are nonlinear: {E} = [B({q, Q )] {q} (6) Note that the nonlinear quantity has been factored in a way to remove the linear part, i. e., the vectors {E} and {q} appear to the first power on the right hand side. 113

ct I Fig. 1. Nonlinear stress-strain Fig. 2. Nonlinear strainlaw. displacement law. The modified virtual work expression, for total work done during a virtual displacement is; - (6U) + (6w t ) 0 (7) external {6~}Tac>~ dv+{6q} T f-dv + T q> 0 (8) J f{ag} }d +^{(6q} {Q}1 + S {6u} T} dv+ f {6u} T{ ds -= (8) V V S strain cone. body surface energy loads forces forces This expression is valid for nonlinear elastic material cases and many nonlinear geometry cases. We are interested in cases with no physical prestress or prestrain. A. Material Nonlinearities Let us consider the easiest nonlinearity, material, alone: {o} = [C ({E},{(o} )] {} (9) {E} = [B]{q} (10) t constant * For strong nonlinear geometry, one needs to adjust the coordinate systems somewhat. 114

115 V. RADAU AND LOBATTO INTEGRA TION Reference: Section 19. 7 in Hamming A minor variation in Gaussian quadrature is to take one or two sample points to be fixed, typically at the endpoints of the region (nodes of an element). If you choose one sample point a priori to be at the end of the region, you have Radau integration whereas choosing two such points gives Labotto integration. Specification of a sample point provides more accurate information at that point; however, it also eliminates a defining equation and reduces the overall accuracy of the method. There is clearly a trade-off involved in the choice of methods.

114 The example 3 1 1 f f(x)dx - 5 (0.5 - 0.5x) = w. f(x.) -1 -1 i= 1 1 has been integrated exactly since 6 defining equations were used. The 3-point Gaussian quadrature can integrate a 5th-degree polynomial exactly. IV. COMMENTS ABOUT GAUSS' QUADRATURE (a) If you know the degree of a polynomial to be integrated, it does not pay to use more integration points than necessary for an exact answer. Typical examples are in plate and shell elements. Even Zienkiewicz had some trouble with this several years back. (b) Abramowitz and Stegun give tables for "abscissas" and weight factors for Gaussian integration for up to 96 sample points! They truncate the values to 15 decimal places. (c) The solution of the defining equations in general involves finding zeroes of a polynomial called the Legendre polynomial. (d) There are multiple solutions for the abscissa and weight factors, which is reasonable because of the nonlinearity of the defining equations. There are 3 sets for cases with less than 13 sample points (n < 13) and 4 sets for 16 < n < 96. There may be additional solutions, but we discard them because we desire real numbers and desire the abscissae lie in the integration region. (e) Desai and Abel show a nonsymmetric form of Gaussian quadrature: b n J f(y)dy (b-) w. f(Y) a i=l where b-a b+a Yi- ( )X + (z) This allows one to convert from the symmetric, normalized data pairs (w., x.) given in math tables to apply to a given physical 1 problem. problem.

The virtual work expression becomes: -{6q}T [B]T[Cs({E},{c} )][B] dv{q} +{6q} TQ} +{6q}T J [N] {x} dv v v strain stress T T+{6q} J [N] {T} ds = 0 (11) Note that the 6 operator acts only on displacement-like quantities, and not on the [C] matrix. There is a common, nonzero leading vector {6 q: {6q}) } = 0 (12) therefore the second vector must be identically zero, leading to [B] [c({q},{Q} )] [B] dv{q} = {Q} + / [N]} dv + [N] ds (13) V v S or [k ]{q} {Q} + { Qe.n.. + e.. (13') body surface forces forces We have arrived at the happy conclusion that [k] = f[B] [({q},{Q} )][B]dv (5.45a) s v Desai & Abel which would be an intuitive guess, but is not a trival result. This allows us to solve equation (1) by the variable stiffness approach. We now need to know how to solve the incremental equation (2) for nonlinear material. We can do this by a variational process in which we have the equation [ks]{q} -{Q} = (14) but wish to study small displacements away from the current equilibrium solution, without violating equilibrium. Using the A symbol as an operator (not the same as for a virtual displacement), it is claimed that the tangent stiffness is defined by 115

/ Fg —-- --- -/ Fig. 3. Increment taken along true load-deflection curve. {aQ} - [kT]{ Lq} (15) Data used to find [kT] comes from material tests in the incremental form: {LI} K[CT(E, )l]{AE} = [CT][B]{} (16) (In other words, material tangent moduli will be used to calculate structural tangent stiffness. Rewriting our old expression for load-deflection: {Q} = [ks] {q} [B] T[C(q, Q)]B] dv{q} v [B] T{} dv (17) v PI, nonlinear stress constant One can take the variation 116

{AQ} = J [B]hAo } dv v = J [B] [CTr A} dv V v -J f[B]T[C ][ B] dv{Aq} (18) V or {AQ} = [kT]{ Aq} (19) where [kT] is the tangent stiffness, and is defined: [kT] j[B]T[ CT(E )][B]dv (20) This could be true for any nonlinear elastic material, or for elastic-plastic behavior as given in equations 7.19 and 7. 20 in Desai and Abel. We have now found secant and tangent stiffnesses for nonlinear material problems. B. Geometrical Nonlinearity (Zienkiewicz, pp. 413-417) Now suppose that {0} = [C] {} {u} = [N]{q} (21, 22)' constant { 6u} = [N] {6q} (23) { E = [B({q, Q)] {q}) (24) Furthermore, if[B] is defined so that {65 } [B({q,Q} )] {6q} (25) we are set to study the nonlinear geometry problem. First of all, for large strain, the virtual work expression, equation (8) must have proper interpretation. The stresses involved must be Kirchhoff-Piola stress and the volume and surface integrals must be in the deformed configuration (State II). We are interested in both [k ] and in the tangent stiffness form [kT]: {AQ} _ [kT]{Aq} (repeated) (26) both of which we can get from equation (8) after much manipulation. 117 it

-{6 q) f [B] T{} dv + (q Q} {) q}T [N]rf dv v +{6q}T S [N]T{T} ds = 0 (27) S Again, factoring out{6q} Tand realizing that it is arbitrary, f [B] T[C][B]dv q} = {Q} +{Q}e.n.l. + {Q}e.n.. (28) V - X T From here, we see that the secant stiffness is [k ] _ j [B] [C][B] dv (9) V Note that both [B] and [B] enter this expression. Again, we wish to vary equation (28) so as to move along the true load deflection path. Combine all loads into a single {Q} vector for this argument and again use the A operator. A J [B] {fr} dv = A{Q} (30) f [tAB]({} dv + J [B]{^a} dv = {aQ} v J [AB] T{} dv + f [B][C] { A} dv = {Q} v f [AB] {fo}dv + [BT[C] [B] dv{Aq} = {AQ} (31) v Now, [B] can always be written [B] = [B] + [BL(q)] (32) constant where [BL(q)] is a linear function of {q}. Equation (31) then becomes 118

1Q l/ / + j[B.T[C] [B. dv,-.(~> dv 3 [k~](hq) (4I / I / v i Fig. 4. Increment taken along true load-deflection curve. (As in Fig. 3.) J[B] { } dv + [[B]T[C][B]dV + J [BL] C][ Bo dv + j [BL [C] [BL] dv] {q} = {AQ} (33) v The first term is rewritten, as it always can be done (need to work on specific cases to see): J [ABj {cr} dv [k] {Aq} (34) v V_ geometric (mild) and not a function of {q}. and the other terms are defined f [B]T[Cc][BL] dv = [k] (35) J[BLIT[C][B]dv + J[B]T[C][BL] dv+ J[ BL] [ C][ LBdv [k (36) rv v so that equation (33) becomes 119

([k] +[k] +[ [kL] ){q} = {Q} (37) and by definition, the tangent stiffness is [kT] =[k] + [ke] + [kL] [T] ke] +- [k] large displacement matrix elastic ( (strong nonlinearities) geometric matrix (mildly nonlinear) This is the tangent stiffness used in the Newton-Raphson iteration process. Q.E.D. 120

Example of a Nonlinear Problem Turbulent Pipe Flow We previously developed (Lecture 25, Aero 510) equations for pipe flow. Assuming a uniform, cylindrical pipe as shown, the pressure drop PI Q. is related to the volume flow as P I-P Q - 1 2 1 fe(Qe) It was pointed out that the constant f depends on the flow rate. For turbulent flow, one might take fe(Qe) f eQ08, where the superscript "e" on the volume flow Q has been dropped for simplicity. This relation leads to: 1_ 1 ~1 ~ ~e 0.8 fe Q0.8 1 1 1P Qz fe0.8 e 0. 8 2 The difficulty here is the dependence of the 2 X 2 flow matrix upon the flow rate itself. Numerical Example of Turbulent Pipe Flow Suppose 2 fe 10 psi-sec f = 10 P 2 gallon P = 5 psi P = 3 psi Then 121

0.1. 0.1 Q8 Q 0.1 0.1 3 I - Q~0.8 0.8 Q Q One can solve for Q and Q by making an initial guess for Q calculating the values of Q andQ, -and iterating. This is a ivariable stiffness" approach 1nd is 2''and eratng and is classified as an iterative method using a secant stiffness. More often, the stiffness matrix is a function of the vector on the R. H. S. 122

Nov. 24, 26, 1976 STEADY STATE I W.J.A.' HEAT CONDUCTION i References: 1) Desai & Abel, Chapter 12. 2) Zienkiewicz, Edition 2, Chapter 15. Many field problems involve the Poisson equation, an elliptic, secondorder P. D. E. with nonhomogeneous terms: In the heat conduction problem, i is the temperature, k, k, and k x y z are thermal conductivities in those directions and Q is a heat flux per unit volume. The k, k, k and Q are all fun'ctions of spatial x y z coordinates. Boundary conditions for Equation 1 are ~ ~Q$ is a heat source.n ^28 it -0 - i o GfP, +^, ad* J **0- "+ ~ (~ = ~ 53S (tv R k O; i Ccrh\ ^ port5 t) / ^^ \ 128~

129 where I, I, are direction cosines of the normal to the surface, x y z directed outward, l(x, y, z) is a prescribed temperature on the surface S1, q(x, y, z) is a prescribed flux into the body on surface S and c(q(x, y, z) - o (x, y, z)) is a flux due to convection on surface S3, (positive outward). The heat transfer coefficient a is a function of space. The concentrated heat flux Rk is needed for later discretization work and needs to be included as a concept at this stage. Other types of boundary conditions due to radiation and conduction do not fit easily into the linear form needed. here and will not be discussed. Let us restate the field problem as a global conservation of thermal energy for the body. This can be done by minimizing the following functional with respect to the temperature field l(x, y, z). A ~{-k - ( k{ gcV A(Sl) - k' { +,(w ] Q t ~~~~~~K ~(6) - S "^ s + (- Js- O t6 where there are a total of K concentrated heat fluxes. We could recover the differential equation and boundary conditions by carrying out a general variation F.A(4i) = 0, but this is not our goal. Rather, we wish to discretize the system. To discretize, lay out a grid of N nodes and E elements. Include all previously assigned concentrated flux locations in the larger, nodal pattern. Consider shape functions within each element i Ne to be sufficiently smooth, and the elements to be conforming so that no heat energy is lost in the cracks between elements.* One can then'subdivide the functional: Ap) = A^ e-A-sI (7) (Assign the concentrated flux Rk to only one element so that it does not appear repeatedly. ) This requires that the temperature field be continuous across element boundaries but allows discontinuity of flux (slope) across boundaries. We have seen this type of discontinuity in the Turner triangle.

130 For each element, use a shape function N _. Drop the superscript under the condition that different shape functions can still be used if necessary. YW = L-N 4___ -1 r- | L3 >- r *l 1' II r js Nr Tr. The functional depends on all N nodal temperatures, so the variation can be rewritten: 6AW) | v,. +-.; Wai, d t (9) This statement must hold true for all values of 4J, hence the derivatives must be separately zero: (10) But by partitioning A, w -e~ (I-'1 ) (11) Interchaanging differentiation and summation: E e. aAe We now need to form A and determine a. This is the longest and most difficult step of the process. Once the individual derivatives are

131 found, we must assemble them in the form of Equation 12 to get the matrix equation of equilibrium. The functional is discretized: Ae -_ xo i N U^ Cll "kz^)r"A X TY, Ni Ssr r t Vot ^2; e t ^ ( ^<w-e dS - | R r ( o) (3) where the concentrated fluxes Rk are acting on element e and are assigned to only one element (i.e., not counted twice). Differentiation yields: p = 1, 2,..., N free index e = 1, 2,..., E free index ~s ~ (^,^-w,)Nrs -^ C^o (14) r =i, j,.., s dummy index p = 1, 2,.., N free index e = 1, 2,..., E free index

132 Note that there are actually NxE such quantities defined; the functional within each element is differentiated by each nodal temperature. Now interchange. and S and factor ir out of the integral: P N p" Yol r 5 Is Si -Rp (+ b) (15) r is a dummy index summed up to s p =, 2,..., N e = 1, 2,..., E In matrix form, ban<3P il uer h4W t -~e- tr (16) internal convection nodal flux, energy flux, outflow outflow

133 where typical terms are written in summation form ^r 5.,(1~ e C WV. t NIP C). Nv- L p e W vr -1 2 O( k+~aa~r "Y Ct A +t k ta (17) he S NpNN.(S (18) 7\/ 1Npr P5 C< P 10 ~d -(19) e/l - kI The vector fluxes are rather easily written in vector notation. (S^N is - - S NilASOLQJ$V - 4L&?S - 5N LNj -M (20) VoW S..S, The heat conduction and convection matrices are given in D and A as Equation 12. 4c. For assembly, one should view Equation 16 as a relatively sparse set of equations imbedded in an NxN matrix format. If each element is so developed, and all equations summed properly, the assembled system'become s -M Mtl }+ 1-'3 Y X- O ~ first variation L41) L4L) j - 1of energy (21) functional In this case of thermal equilibrium, there is some question as to which side of the equation the various nodal fluxes and equivalent nodal fluxes should appear. Separate out the prescribed nodal fluxes {R} and label the rest:.trer^ ^ e^^~i (obsvctOi dvil MW1) iolo^ i4^ w4.Io, dutoNr ^ $3 ^olo'csvc 4^c a. Dowse

134 The {R} vector is defined at every node and plays the same role as the concentrated external force in the elasticity problem. The three {f) e. n. f. terms are then assembled versions of the first three terms on the R. H. S. of Equation 20, but with minus signs dropped. Equation 22 requires discussion. Basically, it is a sum of heat fluxes. The first term represents the thermal state of the material in the body, and all the remaining terms are heat flow into on out of the body. Some students would prefer to see all the flux terms on the R.H.S., in which case, the thermal state of the body is in response to all heat fluxes into the body. The homogeneous convection terms [ H] {} will always be lumped into the "internal energy" terms for solution, however. II. LINE ELEMENT FOR HEAT CONDUCTION Given a rod of constant cross section A, -- length L, and constant thermal conductivity k, develop the heat conduction matrix [hi for a, L 2-node element. Assume 1 V - L! /L -- ) \ A Vt. ^S^C/^Ax. A~lL ML " I —*

1 Dec. 1976 W. J. A. EXAMPLE OF A HEAT CONDUCTION PROBLEM USING LINE ELEMENTS A pair of pliers is often used as a heat sink when soldering semiconductor diodes into a circuit. By grasping Xto the diodewire between the diode and the heated joint, a secondary heat path is given and the diode hopefully does not receive as much heat. This is in fact a transient heat conduction problem, but we will look at the heat flow before r ltOE the diode and pliers change temperature, i.e., a short time, quasi static Fig. 1. Physical problem. approach. Study the idealized model shown and find the difference in heat flow into the diode (at node 4) with and without the heat sink. Neglect the heat capacity of the wires. Physical parameters: The pliers and diode are at 72 F. The iron is at 500 F. Jaws are as- J sumed 1/4" D and 11/2" long. Wire Z 1. segments are 0.015" D. and 1" long. Fig. 2. Idealized model. -4 B TU k 6.00 x 10 sec.OF. in (heat conductivity for steel) pliers sec. F. in -3 BTU kire 5. 05 x 10 o (heat conductivity for copper) wire sec. F. in The element heat conduction matrices are: V"1-1 -41 1i T p A- k [ - (rr/4)(.25) 6x 10 [- 1 -1 [hp].5 i 4 1.96x10- ( -r / 4 )11 21, 1 - 1 [hw] (/4)(.0152 (5.05x 10 ) - K The steel jaws clearly conduct heat better, primarily because of their larger The steel jaws clearly conduct heat better, primarily because of their larger size. The matrix equations will first be found, including the pliers. 135

136 The assembled heat conduction matrix, by inspection, is 2hW + hP _hW -h -h -hW hW 0 0 [ - -h 0 hp 0 -hw 0 0 hw 21.4 -.892 - 19.6 -.892 -= 6 -.892 +.892 0 0 10 - 19.6 0 19.6 0 -.892 0 0 +.892 Solve for the required heat flux, R4, defined positive into the circuit at node 4. 21.4 -.892 -19.6 -.892 1 0 500 R Unknown temperature (s) are found first. In this case, only one temperature is unknown. 21.4 1 -.892(500) - 19.6(72) -.892(72) = 0 -1 = 89. 8 F. - TEMP at junction of pliers and lead. Now get R2. R4 = 106[(-.892)(89.8) + (.892)(72)] - 1. 59 x 105 BTU < HEAT flow out of diode at 72 F. sec

137 For the case without a heat sink, merely set k = 0, and resolve. 2kW -k (500)k (72) = 0 500 + 72 111 = -2 d1 = 2860. TEMP at midpoint of wirewithout pliers Solve for R4. -6 -6 R = -.892 x 10 (286) + (.892 x 10 )(72) -4 =TU o - 1.02 x 10- BTU HEAT flow out of diode at 72 F. sec Knowing the heat capacity of the diode, one could find the rate of temperature rise in the diode. Q. E. D. 720 WITH SI~~~~HEAK~T | 0. 0000159 BT SINK sec 0 0 500~ 6 _72 89. 80 WITHOUT I HEAT SINK BTU SINK * 0. 000102 sec 500~ 286~ 720 Fig. 3. Schematic of temperatures with and without heat sink.

12 April 1978 VARIATIONAL APPROACH TO W. J. A. FIELD PROBLEMS References: (1) Zienkiewicz, O. C., "The Finite Element Method," vol. 2, McGraw-Hill, 1971 (2) Desai, C. S. & Abel, J. F., "Introduction to the Finite Element Method," Van Nostrand-Reinhold, Chapter 12, 1972. I. INTRODUCTION Variational methods are important to finite element methods; indeed, we have already seen virtual work and potential energy. The current lecture attempts to back off a bit and look at the variational process itself in more detail. The examples chosen are energy conserving structures and the function is the potential energy of such a structure, but the applicability is broader - to steady state field problems in torsion, heat conduction, electric fields and magnetic fields. These all represent energy conserving systems which possess an underlying variational statement. II. ALTERNATIVE USES OF VARIATIONAL METHODS There are two basic variational approaches, general and direct. Both require a functional such as potential energy. A. General Approach - This method leaves the dependent field variables in functional form (such as ~(x,y,z)). A variation of the dependent variables is then performed. This leads to a differential equation (Euler equation) and associated boundary conditions. B. Direct Variation - A solution is assumed in series form before the variation is taken. A variation with respect to discrete variables (generalized coordinates) is then done. In the application to finite elements, the assumed series do not satisfy boundary conditions; rather, the boundary conditions are applied as a last step to an assembled, discrete system. III. GENERAL VARIATIONAL APPROACH Many field problems have an energy conservation law which states that a certain functional r(f) has a minimum at the solution O(x,y,z) = 0*. This means that 6r(0) = 0 at 0 = 0*. A typical functional, which leads to Poisson's equation, is: = f k 2 + k a) + k () ]- Q }dVol 2 x ox y ay z(-z vol (q- 1 2 - / (q < 2 a -)dSurf (1) surf 138

139 The dependent field variable l(x,y,z) might represent a displacement or potential function. (For structural problems, 0 is a displacement and 7T is potential energy). The variation with respect to (w.r.t.) $ yields 6~ = f {2 [2k y3x 6)x2 ( ) + 2k (60)] - Q&6} dVol 2 x DX DX y Dy 3y z () +2 vol - f (q6O - ac 6))dSurf = 0 (2) surf The terms involving derivatives of 6d must be transformed. Use a version of the unsymmetric Green's formula* to obtain 6 T= -f {[- k + - k + a k k ] - Q}6bdVol x x ax Dy y Dy Dz z Dz vol + ~ [k -x + ky q + k - - q + a4]6 dSurf = 0 (3) surf Since 60 is an arbitrary variation of 4, we will argue that the integrands in the volume and the surface integrals must separately vanish. If the curly-bracketed quantity in the volume integral did not vanish identically at some point in space, then the arbitrary variation 60 could be chosen as a delta function centered at that point and the law would be violated. This leads to the differential equation (Euler equation) 9 9D 9 9 (4 3 O - -k -+-k -+-k -- Q(xyz) = 0 (4) k + k — + k ax x Dx y y 9y 9 z Z - Q(xy ( Likewise, the surface integral must vanish, but this can be accomplished in two ways, by the square-bracketed quantity vanishing on a part of the surface, say S1: k x + k a y + kz t - q(xy,z) + c = 0 (on S) (5) X x yy y zz z'zz and by the variation vanishing on the remaining surface, say S2: 60(x,y,z) = 0 or O(x,y,z) = $B (on S2) (6) This completes the theory, in a sketchy way, for a typical "general" variation. Fig. 1. Surface regions * The standard form of Green's formula is given in Wiley, C. R., "Advanced Engineering Mathematics," 4th Ed., McGraw-Hill, 1975, p. 676, Eq. 8.

140 IV. EXAMPLE. LINE ELEMENT WITH ELASTIC SPRINGS. GENERAL VARIATION. Consider a line element with springs at each end connected to fixed supports. The coordinate system is adjusted so that the springs are unstretched at O(0) o and' (L). Concentrated loads act at _ -' x=O and x=L. A distributed line load ~(x) acts along the length of the element. The element has varying area ( and modulus of elasticity. The Q1 springs have constants a and a2. The situation is intended to represent a field problem for I(x) in the region 0 < x < L. The general variational approach will be used to develop the differential equation and the boundary conditions. We have the familiar line element _ cast in the role of a field problem, L with the inclusion of springs. The springs are treated in the common. Fig. 2. Simple field problem. potential energy manner as elastic ele- Line element. ments within the system rather than as external forces. Potential energy for the system (wall to wall) is 7rr() - (STORED STRAIN ENERGY) + (WORK POTENTIAL OF EXTERNAL FORCES) (7) Remember that the work potential W is given a sign convention from the classical physics notation where {Q} =- {V)IW and where {V} is a generalized gradient operator. (In this case Force = - d ( for instance). Hence; 1L 0 7(T) = 7' A(x){c }T[C]{c}dx *+ {(} 1 0x a0 f..,......',........................ STRAIN ENERGY + [- 4' (x)t(x)dx - {Q}T{I}] (8) WORK POTENTIAL

141 where L (0) W{a~} - )(9) L (L) J {Q} = (10) Q 2 {ax} (dx } (11) { } = dx) [C] = [E] (12) The variation with respect to 0 is carried out: L d d 6S(r) = f A(x)E(x)(d-x (x))(- 64(x))dx + a1 (0) 6H(0) + a2((L) 6D(L) 0 dx 2 2 L — ~x) 1 - 40 (x) 6~(x) dx - ( (13) J2 Q 6~(L) Integrating the first term on the right hand side (R.H.S.) by parts and collecting term's, one has L d d - 7'T() = - [ AE dx +;(x)]6$(x.dx (14) d d P(O) + [AE -x O(L) + a2o(L)-Q2S16(L) + [-AE d( + c1(0)-Q]6((0) ) We want this to vanish. The integral term is easy, because we want 6&(x) to be arbitrary in the interior of the link. Hence, the Euler equation is d- (AE d (x) + (x) Equation of Equilibrium (15) dx dx The boundary conditions are tougher. At the left end, we can either specify a natural (force) or a geometric boundary condition: - AE d(0) + a1(0) - Q1 = 0 (16a) or d6(0) = 0 (16b) respectively.

142 Likewise at the right end of the link: AE d (L) + a2 (L) - Q = 0 (16c) or 6~(L) = 0 (16d) The force condition corresponds to an equilibration of forces on the pin. Fig. 3. Interpretation of boundary conditions Fig. 3. Interpretation of boundary conditions. Hence, at each end, we may either set the pin in equilibrium or we may hold the pin at a prescribed displacement. The latter causes 6d to be zero during the variation. The general variation yields a differential equation and consistent boundary conditions. Perhaps this problem is sufficiently intricate to show that the boundary conditions are not easy to see from equilibrium. (Note that we did not use equilibrium to get the sign sense of the elastic forces - they come from the variation.) V. DIRECT VARIATIONAL APPROACH The functional in Eqn. (1) is to be minimized. We have previously shown a general variation w.r.t. the scalar function I(x). We can now take a different route, to discretize the function O(x) and to take a variation w.r.t. the discrete values. In other words, <(x) is sampled at a certain number of x points (nodes) and those values of D(xi) used as new variables. This is the Rayleigh-Ritz approach. Ifet us conceive "global shape functions," which are nonzero within only one element. For a scalar field variable D(x), and for a total of M nodes, one has

143 M W(x) = Z N.(x)>i (17) field nodal values variable 2 of (I(x) = LN1(X), N2(x),... NM(X) (18) When x lies in an element e, the shape functions N.(x) switch to a set of shape functions zero everiwhere except in that element.* Instead of a scalar variation a~r 67T = D7^6, we now need a vector variation: 6 = +63 661 + (19) a(2 jM M In summation notation, we have M aTr 67 = E 6 (20) In vector form, using Zienkiewicz's shorthand notation for the gradient vector fr lT 6: a{ } {6D} (21) This, of course, is the process on which the entire finite element theory is based. One can view Eqn. (20) as an approximation to the full Taylor series expansion: * This is a difficult concept for engineers but seems easy for mathematicians to visualize. In any event, the generation of such global shape functions means the assembly is accomplished automatically, and, in fact, is not needed. The shape functions must still allow for inter-element compatibility of derivatives on P up to order one less than the highest derivative appearing in the function (Eq. (1)). See Zienkiewicz, Ed. 2, pp. 24-26 for details.

144 MM M 2 A, = Z 65i 2 3 &p 6 + -.. i=l i I i=l j=l qi j - J "Differentialil j"l ~ij { } "Differential"

145 CONCLUSION Aug. 2 5, 1980 W. J. A. We have seen the fundamentals of a new approach to problem solving, the finite element method. In teaching finite elements for ten years at the University of Michigan, the author has developed an increasing enthusiasm for the method. Master's level students have entered a year-long program with no knowledge of the method. After studying for two terms, they have been able to start work immediately on doctoral dissertations. There is possibly no other training tool which has increased intuition and knowledge of mechanics as much as the finite element method. Upon reviewing the various branches of mechanics and physics which are studied in undergraduate engineering courses, one sees that many courses could be redone using finite element methods. At the first blush of enthusiasm, one might wish to recast all undergraduate teaching into a finite element mold. One such attempt was made in solid mechanics at the Aerospace Department of the University of Texas. It did not work and the department eventually backed away from this. One can conclude that although finite element methods are powerful and can solve many problems, they cannot replace classical methods because classical methods are required to create individual elements. Indeed, the interpolation scheme used in the displacement method is mathematically equivalent to the Rayleigh-Ritz method. Galerkin's method is often used to create elements, particularly fluid flow and other nonconservative elements. It appears that the finite element method must be a strong partner with the classical methods of analysis rather than supplanting them.

146 OEWORK

HOMEWORK PROBLEMSI June 23, 1978 W. J. A. Problem 1 Find the exact stiffness matrix [k ] for a two-node line element with varying area A = A (1+Bx/L). Use C,f an equilibrium method, relying on the fact that the line element is statically determinate. Problem 2 Find an approximate stiffness matrix [k] for the two-node line element in Problem 1 using the energy approach. Assume constant strain. Problem 3 Find the stiffness matrix for a plane stress, constant strain triangle with local coordinates <C shown. Use the equilibrium approach with assumed displacement function: u(x,y) = a 1 + ax + a 3 v(x,y) = ca +a x+ ay (-. i X 4 5 6 (oo) ( oC The fact that the vertices have unit coordinate values will simplify the problem greatly. Use the expression for[k] involving [A3'1 outside the integral. Problem 4 Assemble the stiffness matrix for the following structure using symbols A and G- to represent the stiffness terms. Number nodes as you wish. Use the compact notation. Problem 5 For the beam element based on Euler-Bernoulli theory and with displacement function w(x) = aa + 3a 2 x + 3x 4x3, develop the equivalent nodal loads for the loading shown. L 1|5i-d | P(x) = Po(X/L) 147

148 Problem 6 A thin plate sketched in Figure 1 is to be subjected to various inplane loads. A f \ finite element study is to be made in which — ^-r " *"*two rectangular and two triangular elements will be used. The rectangles each have eight nodal degrees of freedom and the triangles each have six d. o. f. J/y~~~~ \ ~ (a) Number the nodal degrees of freedom in such a way as to minimize the bandwidth of the assembled stiffness matrix. (b) Which of the elements in the assembled \_______ stiffness matrix are zero? (c) Explain in words how a distributed loading along one edge of an element can be handled. What are the principles involved? Problem 7 (a) A line element has changing crosssectional area as shown. Develop the stiffness matrix for a 3-node element, choosing the interior midpoint as a node. Use a quadratic displacement function. (b) Reduce the element's stiffness to a Z x 2 stiffness matrix by "condensing out"''"i 9-, <^ j^ the middle node. Present this matrix in its simplest form. (c) Compare the 2 x 2 stiffness matrix with A = A (I+Bx/L) o that obtained earlier for a uniform prismatic link and with that obtained in Problems 1 & 2.

149 Problem 8 A) Solve the following set of equations with the use of a Gauss-Dolittle decomposition: 1 2 1 x1 2 2 5 0 xZ = 5 1 0 10 x3 12 B) Solve the equations by a Choleski decomposition method. Problem 9 Solve the set of equations given in Problem 8 by using a Gauss-Seidel iteration. The method converges slowly; a programmable calculator may be needed to get satisfactory convergence. Start with an initial guess: ( (0) 1) x Problem 10 Write out the equations necessary to program in FORTRAN the solution by Gauss-Dolittle factorization of the standard equation: [K] {r} = {R} Assume the matrix [ K] is an n x n, real, symmetric, positive-definite matrix. The vector {R} is known and {r} is unknown. Assume [K] is a full matrix and not banded. Use the lower triangular factorization. Problem 11 Develop a FORTRAN subroutine for the Gauss-Dolittle decomposition required in Problem 10. Name the subroutine DCOMP(N, K, *), and write the decomposed portions [L1] and [D] back on the lower triangular part of [K], in order to save space. Use the * location for an error return. Problem 12 Develop a FORTRAN subroutine for the factorization solution of the equation studied in Problem 10. [L1 ] [D] [ L1] T {r = {R) (CONTINUED)

150 Problem 12 Continued Assuming the [L1 ] and [D] are available in the lower triangular portion and diagonal portion respectively of [D]. Name the subroutine SOLVE(N, K, R, D) where R(I) is the reaction vector and D(I) is the displacement vector. Carry out both the forward solution and the back substitution. Problem 13 This question is intended to test your knowledge of the stiffness matrix, its definition and physical interpretation. A linearly elastic coil spring is to be treated as a single finite element. Nodal degrees of freedom are the axial displacement ul, uZ and the rotations v, vZ. The spring is 12" long and 2'1 outside diameter. It is made of steel wire of 0. 050" diameter. U9 I4, V/ Nodal Displacements Two experiments have been carried out to provide stiffness data. The left end of the spring is clamped firmly. A displacement u2 of 1" causes forces U1 = -100 lb and V1 = -50 in. lb. (v2 = 0). A rotation v2 of 1 radian causes forces U1 = -50 lb. and V1 = -100 in. lb. (u2=0). (Assume the spring remains within the linear range during this rotation). (a) Using equilibrium concepts, construct the stiffness matrix for this spring. (b) If two such identical spring elements are joined in series, find the stiffness matrix for the assembled system. Use letters to represent any stiffness components not found in (a). Problem 14 Hia.d ^ ^A truss is made of two elements as shown. iAs~ i-c //\\ A horizontal force' is applied at the'L-//4, \\60 ~center node (the vertex). Find the displace~/-~Y/45t \t ments at this node in terms of the length L, r \\ ~L^ ~ areas Ai, moduli Ei and force'. Use a i~~r4\\ 4global system aligned with the horizontal to express your results. L- \\777r

151 Problem 15 An electrical network is shown. (a) Set up the equation for a single element I4^ / \ with nodes i and j. ( b i\ R1C (b) In analogy with the structural stiffness problem, does the voltage at node V. correspond to force or displacement? To what ___AA A _^ ~ does the current I. correspond? $ ^V'3 (c) Give the assembled equations in matrix form for the network. Do by inspection. Problem 16 (Weaver) Consider the rectangular element for plane stress and plane strain with generic and nodal displacements as shown. Assume the displacement functions for u(, r1) and v(,rn) to be u(Y,r7) = 1 +aZ +a 34+ 4 a4 v($,rl) =5 + 66 7+ 7+a8 it in which g = x/a and r7 = y/b. Also let 3 = b/a. l L i... The stress-strain matrix is ei v 0 [D] E v e 0 [D] (lv)e Fe 0 0 e where plane stress: e =1, e2 = l-v, e3 = e/2 plane strain: e, 1-v, e2 = l-v, c3 (e/ Derive the following items for plane stress and plane strain: a. Stiffness matrix [k] of order 8x8. l>. cEquivalent nodal loads Fe n. 1 for a linearly varying force normal to edge 1-2 and oi intensity p1 at node 1 and intensity pZ at node 2. c. Equivalent nodal loads for uniform gravity loading pg (per unit volume), acting in the -y direction.

152 Problem 17 What are the equivalent nodal loads for a 10" long cylindrical bar due to a surface traction 2 -- {q(x,y z )} = {100 lb/in } if the bar is one inch in diameter? A two-node line element is-used. Problem 18 A linear elastic rod is loaded as shown. _______ -____ F It is of uniform cross section and mechanical E. properties. _.... -..... (a) Carry out a one-element and a twoelement solution for displacement of the rod using the 2 D. O. F. line element. igLo) |(b) Make comparisons of strain energies, potential energies and displacements for the two cases. Present this in graphical form as displacement field u(x) versus x o3;. L and energy at state II (equilibrium) vs. number of degrees of freedom.,!* — 0 z:3 Problem 19 Use Lagrange multiplier s to impose the constraint of assembly D3 = =2 on the two cz z. ) -—.. u. -Z prismatic links shown. The result is the,'d —>0 4t~ <f-c ~4-4' ek+> matrix equation of equilibrium with 5 VP T1 Lunknowns.

153 Problem 20 Find the equations of equilibrium for a finite element for a prismatic link, using Galerkin's method. p(x) Diff. Eqn.: CW. -X. — -^~-I A d u(x) _ —- -__ ) —-- W, EA d —u + p(x) = O Marl By L 9~Sb~ ~ dx2 The relation between nodal loads and strains at nodes is: du Q = -EA 1 dx 0 du Q = EA- I 2 dx L Problem 21 Cast the entire prismatic link development into the natural mode approach. Assume a uniform cross section and constant mechanical properties. < —- _ H e — L(a) Define the two modes needed, one a straining and one a rigid body mode. (b) Develop the relation between nodal displacement and generalized coordinates {q} = [A] a} (c) Develop the relation between the generalized forces and the given nodal loads {Q } = [A] {Q} (d) Develop the stiffness matrix [k ] which relates the generalized coordinates and the nodal action vector. {Q } = [k ] {a}e a a

154 Problem 22 Zienkiewicz claims in Section 3. 5 of "The Finite Element Method, " lthat minimizing the functional x [( ) + ) C(Id(surf) surf + (qc+ 2 a 2)d (bound) boundary is equivalent to solving the differential equation 2 2 2 2 + 2+ C _ 0 ax ay Prove this by carrying out a general variation of X and show the resulting Euler equation and consistent boundary conditions.

155 Problem 23. (Estimated time: 1 1/2 hour) Develop the matrix which relates internal stresses to nodal forces for the constant-strain, plane - stress triangle. This is the [ ] matrix mentioned in the Lecture 4. Use a global coordinate system. q6 / 1. A Problem 24. Develop the stiffness matrix for the constant-strain, plane-stress triangle in the local coordinates shown. You may drop the subscript 2 on the coordinates for the calculation, since it is obvious that a local system is implied. Use the direct method, with the equilibrium matrix developed in Problem 23. Comment: The toughest part of this problem is the inversion of the [ A] matrix. Work with another student on this to avoid error. Also',t; e;itr +v use, H+e 4ov CXrSSie for Jk3 oi pyue 8;uI v|A Co fAT 5( )CVrAl^ (X, ) a^ ( >^ o)

156 Problem 25. Fluid Flow in Pipes A pipe assembly is shown. If e f Q1.8 1 2 Q in gallons/sec Q,,p, AP in psi and if Q;= 2 gallons/sec J Q3 = - 5 gallons/sec Q4 3 gallons/sec - -, 1.8 2 1.8 f 0. 5 lb. sec /(in.gal ), find the fluid pressure at each node. Problem 26. 2 3 If f(y) = 1 y - y, evaluate 5 J f(y)dy 1 using Gaussian quadrature. Use the minimum number of sample points.

157 Problem 27...... 2 3 4 A sheet of metal has been modelled with an assembly of plane strain quadrilaterals and triangles as shown. Using "compact" notation, - the half bandwidth is _. - the maximum wavefront is _ - the number of zero terms in the assembled stiffness matrix is If an optimum number of nodes had been done, the half bandwidth could have been reduced to If an optimum numbering of elements had been done, the maximum wavefront would have been Problem 28 Two identical line elements are joined at the center and fixed at the ends as shown, with zero initial strain. A concentrated load of 500 lbs. is applied to the right at the connection, and the bodies are heated 100~ F. Using finite element theory and notation, find the displacement at the center connection and the reactions at the fixed ends. 6 E =30 x 10 psi AT + 100F. (l = —-+500 Ib. L= 10 in.._ L=10 in.......'7 b.... -6 o 0 a = 7 x 10 in/in. F. -- L -- L - Problem 29 Consider a coordinate system with origin located at the center of a line element. Use a displacement function -- L - u(x) = al +az L -L Find the equivalent nodal reactions for a distributed load p(x) = pO(X/L)2 where p0 has dimensions of force per unit length.

158 Problem 30 A concrete dam is to be constructed on rock as shown. Tle heavy lines indicate boundaries between materials (rock-concrete, concrete-water, etc). A plane-strain solution using triangular elements is to be found. The rock will be modelled only as far from the dam as felt necessary to give accurate stresses in the dam. /< T-~r 0 3COt%- bMA' W AT^Rt A-VA (1) How many nodes will have nonzero equivalent nodal loads due to water pressure? (a) 4 (b) 5 (c) 6 (d) II (2) What boundary conditions should be applied at the bottom most nodes in the rock? (a) zero forces (b) zero displacements (c) zero strains (d) finite forces (3) Which of the following would be impossible to model with today' s state of the art? (a) thermal prestrain due to heat in concrete (b) earthquake effects (c) water height which causes crack initiation at base of dam. (d) natural modes of vibration of dam (e) none of the above are beyond the state of the art

159 (4) What do you see as the major weakness of the given approach? (a) (a) There are not enough small elements at the base of of the dam where cracking in the rock might initiate. (b) A rectangular element would be better for rock. (c) Dam problems are really plane stress. (d) It is difficult to include equivalent nodal loads for air pressure. Problem 31 A plane-stress problem has been posed. A thin sheet of metal is to be acted upon by four loads as shown. In addition, gravity acts on the plate. As presented, no constraints on displacement are given. Assume that the external forces place body in force equilibrium. (1) How many elements does the subdivided structure have? 8 16 24 (2) How many degrees of freedom are there in the assembled structure SugTDlfuVbt 5Sn.ucTuRiE if the Turner Triangle is used? 13 26 36 (3) How many degrees of freedom must be constrained in order to eliminate rigid body modes? 3 4 6 (4) Would the following be an acceptable way to eliminate rigid body modes..... without introducing additional stresses? yes no 7 1/ IF/7

160 Problem 32 ^ | —-\ —---- A cantilevered pole is modelled by Euler-Bernoulli beam elements as shown. Each element is of constant cross-sectional properties and has four nodal degrees of freedom. Including the node at the support, 1) How many nodes are there in the assembled structure? 2) How many D. 0, F. (detailed notation) in the assembled structure? 3) How many equations of equilibrium in assembled form before applying boundary conditions (detailed notation)? 4) What is the minimum half bandwidth of the assembled stiffness matrix? Yes No 5) Isa it possible to solve for the unknown displacements in this problem without finding the unknown forces? Why? I | ^~~~ 6) Suppose a more refined element were used which involved 4J w lx)a== Of, + a -x + X" +axy+x + ax4+ X but still assumed Q W(X.(1 2 a3x + a x constant area. The refined element would lead to a solution with smaller or greater deflections under the load?? | I

161 Problem 33 The integrand for a certain isoparametric element integration involves quartic polynomials. The numerical integration is to be done with Gaussian quadrature. Write out the necessary defining equations used to generate the we ighting factors and sample points, using the minimum number of sample points for an exact integration. Use the symmetric form of integration. You don't need to solve the equations. f(x) -1 1 Problem 34 For the quartic above, if one used Radau integration where one fixes one sample point (at x = -1, say), how many sample points would be needed for an exact integration, including the fixed point? Problem 35 A constant temperature gradient triangle for use in steady-state heat conduction is to be developed. It will be the equivalent of a Turner triangle for the heat problem. A right triangle with a local coordinate system will be considered (see sketch). y -The nodes are numbered as shown. The first shape function is given: 3 _ (0, b) N = 1 - x/a-y/b. 1'\ (0, o) (a, 0) 1. Find N2 and N3. 2. Calculate the heat conduction term h22' 3. Set up the expression for equivalent nodal flux due to a surface flux source q(x, y) = qx2y2 You need not carry out the integral.

162 Problem 36 Two identical steel bars have been purchased and holes have been drilled at each end for pins. The stiffness EA/L is 1000 lb/in. for each bar. The bars are cooled ten degrees Fahrenheit from room temperature and then assembled in such a way as to stretch them to their original room temperaoOO, Ature length. This results in a tension of 100 lb in each bar. A load of 1000 lb. to the right is then applied at the center pin. Use an initial strain concept to find: 1 2R {0}E,,O{r.},{R} 0~ 0 Problem 37 2 A truss structure is assembled as shown. The nodes are numbered 1-3. The ele- / ments are numbered with Roman numerals. Suppose the structure has been assembled: U k1 k12 k1 k k k6 1 11 12 13 14 15 16 Ul V1 k22 k23 k24 k25 k26 V1 U23| | (ym y| |3 V2 V "2 U3 (symmetry) u3 V3 k66V3 1) What sign does k have? Why? 2) What sign does (k33k4 - k3) have? Why?

137 SAMPLE EXAMS FROM "FINITE ELEMENTS IN MECHANICS, I" (Aero 510, M.E. 557, AM/ES 510)

2 Hour AERO 510/ME 557/AM 510 22 Dec. 1981 Honor Code FINAL EXAM W.J.A. Open Book 1) An assembly has a plane of symmetry at y = 0. A beam passes through this plane as shown, with axis in the y direction. Using global coordinates, and standard MSC/NASTRAN notation, which of the following degrees of freedom should be constrained at the plane of symmetry: (a) 2,4, 6 (b) 1, 3, 5 (c) 2,5, 6 (d) 1,2, 3 (e) 3,5, 6 2) Two straight beams are welded together at an angle of 300 as shown. A scribed line is drawn along the neutral axis. The body is then loaded. An analysis predicts that a rotation of 10~ should occur at node 2. After loading, then, what is the local angle 3 shown in the deformed body, as indicated by the scribed lines? (a) 20~ (b) 3003 (c) 40~ (d) None of the above i --- (e) There is not enough information to tell. 3, 4, 5) A heat conduction rectangle has shape function: [N] = [( -x/a)(l -y/b) x/a(l-y/b) (l-x/a)y/b] ab 4 3 (0o)) C(o) What is the equivalent nodal flux into node 3 due to an input of -V N ~ mm/mm/sec along the top edge? The element has thickness h and heat conductivity (a) (1/3)/a (b) (1/4)Aa2 (c) (1/2)kfa (d) ab (e) (1/2)^a

2 6) A slightly curved line element lying in the x, y plane is to be mapped onto a double unit line segment in an attempt to create an isoparametric element. How many Gauss points are needed to exactly integrate the physical length of the element given by the formula L = + 2 dx (a) 1 (b) 2 (c) 3 (d) 4... ^ (e) Transcendental functions can' t be exactly integrated. 7) A solid, tetrahedron element is to be generated with displacement functions. How many rigid-body modes does it have? (a) 1 (b) 2 (c) 3 / (d) 4 (e) 6 Lj I 8) How many straining modes does it have? (a) 4 (b) 6 (c) 10 (d) 12 (e) 18 9, 10) A uniform line element has length L and cross-sectional area A. What should the consistent mass terms m11 and m22 be? (a) A p L/2 (b) A p/2 (c) p AL/3' (d) p AL/4 j (e) p AL/6 11) What are the lumped mass terms mll and m for the same uniform line element of the previous problem? (a) p AL/6 (b) p AL/2 (c) p AL/4 (d) p AL/8 (e) None of the above. 12) Suppose you were asked to develop a Gauss integration procedure for integration in a triangular region in the x, y plane. The Gauss formula would look like 3 A f(x, y)dA = wi f(xii). i= 1 (continued on page 3)

3 12) (continued) What functional values would you choose to develop the required defining equations? 2 2 (a) 1, x, y,x, xy, y (b) l, x,y,x+y (c), x, y (d) x, y, z (e) l,x,y,z,x,y 13, 14) A plate problem has been solved using 6 QUAD4 elements as shown. It is desired to redo the problem with the same number of QUAD8 elements. If a Gauss elimination solution is used, by what factor will the computer CPU time increase for the equation solving portion? (a) 2 (b) 3 (c) 5 (d) 10 (e) 20 ________ 15, 16) A hypothetical structural element with stiffness [k] has a set of loads applied which give an equilibrium displacement {q*}. A subsequent virtual displacement {Aq} is given. How much virtual work is done by the external forces? Units are Newtons and mm. [k]= 10[ 1 4 1 {q} = {Aq}= 0 2 1 52.01 (a) - 5 N mm (b) 0 N mm (c) 10 N mm (d) 50 N mm (e) 100 N mm 17) Design sensitivity parameters can be as easily found in a static stress problem as an additional i (a) set of boundary conditions (b) material property (c) generation of element stiffnesses (d) natural frequency (e) none of the above. 18) Electrical and fluid pipe networks have a sign convention for electrical and fluid flow that is: (a) the same as for trusses in 2-D (b) based on flow into a node (c) depends on the underlying coordinate system (d) not valid for curved pipes (e) none of the above.

4 19) DMAP procedures in MSC/NASTRAN can be used to (a) exit early in a solution sequence (b) insert additional calculations (c) resequence nodes (d) print out additional information (e) all of the above. 20) A beam has a concentrated load P at a point 2L/3 from its left end. What fraction of this load should be applied to the right end of the beam as an equivalent nodal load, of only one beam element is used? (a) 0.75 p (b) 0.71 (c) 0.66 4 (d) 0.60 (e) 0. 50 Note: If the previous question is too difficult mathematically for you, it has the same answer as this question (which is viewed as an alternate question, only - not as extra credit): The Direct Matrix Insertion feature of MSC/NASTRAN allows a person to (a) Insert real and complex matrices and vectors (b) Allows only insertion of real matrices (c) Allows insertion of only real and complex vectors (d) Insert data throughthe executive deck portion of the deck (e) None of the above. 21) The Turner (constant strain) triangle) has linear interpolation functions. This means (a) no stresses can be recovered (b) the element is more accurate than 6 -noded triangles (c) the cr stress is the same throughout the element (d) the cr and cr stresses equal each other throughout the element x Y (e) none of the above. 22) A line element with 3 nodes (a) requires quadratic shape functions (b) has 3 generalized coordinates (c) has one rigid body mode (d) all of the above. 23) In any linear, elastic structure, K.. = K.. because (a) forces in the x direction cancel those in the y direction (b) positive strain energy must be generated by any displacement (c) the system can create energy (d) angles are small (e) all of the above.

5 24) Equivalent nodal loads (a) are always smaller than lumped loads (b) depend on the shape functions (c) are found by differentiation (d) are easier to calculate than lumped loads (e) none of the above. 25) Gauss integration is to be used to integrate a function on a domain [-10, 10]. The function is believed to be a quintic (fifth degree polynomial). How few sample points are required for an exact integration? (a) 1 (b) 2 f( (c) 3 (d) 4 (e) 5. K.. —-----—.- AX lre10\

t2 1, /IO/^HMES So/0/ E SS7?1 2+4c lqE I FAL tX^Mh - ISuTmRUCTORS SOLUIO4. to PQOB. ~4. (Q.). 2,e> ~' c o TA s',, rll e. fos. Z. (b) 30o TkfCe;. no rfc L\, reot-fc CLz,ecfic+'). PW. 3,. (e) [f; \ rt1- - ~+Xj;[^t c cZlt t6ie. 7. (e) Pr^s. 8. (b) fP(6.,lO. (c) I _,, S IA,),eAcK = CA S - /X = eA S E -2-'x+X/ )K = 3^AL Pos. ((. (b) Pos. 11, (a)'Aof. I13,1 (d) cp. -..i... ~ - I5 - - 60 S 9O CPUO- tl 4 (ss s; =3) tLS 4.,, IZ * ^ Lt c~ur, pr F-"- k 5 _ - - 5 " 26 q t I-s' =It. - 55

?\o /t? PP..o,,. (C8. (6) it r~8, t. I. ).= h13.!po&. 217 (e) tito({e.n.A. (b) [N& K = 1' flBsa. 173 (be r~e2. 4o (c) Pto S. I S.. bdS)M i eefm~~~~~os,~~~~ ^ 24 1, d IthC bf V%,TP 8, =p [3124'\ 21 Pfroe. 22. ( ) r&. Is (6b1 PeF", 2.+ (b) Pao e. zs Cc.)

Midterm Exam Midterm Exm AerQ 510 ME 557 21 Oct. 1981 50 Minutes 50 Mnutes AM 510 WJA Honor Code Problem 1. (10 points) For the constant strain (Turner) 3 *. 3) triangle shown, find the shape function N3(x, y). Note: The form of the 2-D shape function is: ful N1 0 Nz 0 N3 0~ q1 v C N1 0o 0 ~ N q3 l 2 36 Problem 2. (30 points) Two identical beam elements are joined as shown to form a cantilever beam. If the free end of the beam is deflected 1 mm upward and rotated 0. 2 rad counterclockwise, and the midpoint is 2 deflected 0.4 mm upward and 0. 1 rad clockwise, what is the vertical (shear)'J - force on the left end of the beam? 8 EI = 10 N mm Comments: Use finite element concepts. Do not derive any stiffness L = 100 mm matrices or equivalent nodal load vectors from first principles. 3 Problem 3. (20 points) t- L L- -- A rectangular sheet of metal has a square hole cut from its center as shown. The sheet is clamped at its left and upper edges and is free on its right and lower edges. A 100 N force is applied downward at its lower right corner as shown. Write down a physical description of all boundary conditions that need to be explictly entered into a general purpose code such as MSC/NASTRAN.,- 3 X A hypothetical example is given in l 8 tabular form: Node Displ. Force 27 1,4,6 F = 250 N VV z 38.- F = -100 N y

Problem 4. (10 points) 8 f6 An elastic plane-stress quadrilat- eral is shown. The active load is a 3 5?0 N SOO 503 N 500 N force at node 3. The stiffness kl 5 is 1000 N/mm ^ (detailed notation) and the sheet is 0.5 mm thick. How much virtual t work is done by the external forces if node 3 undergoes a 1 mm virtual displacement.? Problem 5. (20 points) A 3-noded line element has a linearlyvarying line load acting: (x) = 0 x/L The total value of the line load is ____ *L/2. If the nodes are equally L spaced, how much of this line load should be applied to node 2 as an equivalent nodal load? Nl(x) (1 - 2x/L)(1 - x/L) 4x N (x) = L ( - x/L) Zx N (x) L(1- 2x/L) 3 L Problem 6. (10 points) Carry out a Gauss-Dolittle factorization of the following matrix: 4 a 2 7 1

2 Hours Aero 510/M.E. 557/A.M. 510 W. J. Anderson Open Book FINAL EXAM 19 December 1980 Honor Code There are 9 problems totaling 200 points. Please put the solutions in order in your bluebook or paper. Problem 1. (10 points) Find the semi-bandwidth, in compact notation, for the given axisymmetric body. 3 Problem 2. (15 points) Find the shape function N3(x,y) for the given constant strain triangle../ \ 9( (3)+)e(6e+) Problem 3. (40 points) A clamped-clamped beam is loaded at its center with a 100 lb. load as shown. Calculate the stored strain energy and the potential energy at equilibrium for the following constants:'oo EI = 10 lb.in2 I L = 50 in. -- -.

2 Problem 4. (20 points) An electrical circuit is shown. It consists of 7 equal resistors. Using the notation C - 1/R (conductance), write out the assembled conductance matrix for the system. Problem 5. (20 points) ( )) Assume that Gaussian integration is to be developed in two dimensions for quadrilateral elements. Write out the first three (lower degree polynomial) defining equations. A total of four sample points are to be used, as shown. ( \ 2___(X4) Problem 6. (30 points) (a) How many rigid body modes does an axisymmetric element have? (b) A certain line element problem has 10 elements connected end-to-end in a straight line. By what factor will the CPU time for the equation solver (Gaussian elimination) be increased if the problem is re-solved with 30 elements? (c) A single plate element is clamped along one edge as shown. How many degrees of freedom will be unconstrained in the solution of the problem, either with SAP6 or with NASTRAN's QUAD4 element? (d) Is Euler buckling a geometric or a material nonlinearity? (e) Near equilibrium, a virtual displacement causes no virtual work. Does potential energy change much with the same virtual displacement? Problem 7. (20 points) Write down the appropriate nodal force and displacement vectors to be used in a plane stress solution of acantilever beam, as shown. {R} =? {r} =? \ _,~r^ - ilo tb

3 Problem 8. (35 points) A copper wire is modelled with a single quadratic element. The ends of the element are held at 0 degrees F. and a soldering iron at 450 degrees F. is _ firmly held at the center of the wire as shown. Using & quadratic element, how 45 D much heat flux into the wire occurs at the soldering iron contact point? A = 0.001 in2 L k = 5.05 x 10 BTU/(in.sec. F.) L = 4 in. Problem 9. (30 points) An 8-noded square element is given as shown. A constant pressure p acts on the right face. Calculate the equivalent horizontal nodal load at node 4 due to this pressure. The shape functions evaluated on the line x = a must be parabolic. The element has thickness h. u).(x,y) 0 N2(x,Y) 0 q Iv 0 N1(x,y) 0 N2(x,y).. ql6 S +,, 3 (.~A)R aQ) (O)m ( s) > (v)_-^

ME 557 AERO 510 AM 510 W. J. Anderson HOUR E Nov. 24, 1980 Place all answers on the computer-graded form provided. Your final score will be the total number of right answers. Wrong answers do not count against you, so you should guess answers to any unsolved problems. Choose the answer supplied which is closest to the correct answer. It is not intended that the exact answer is always provided. Each answer will count 4 points. Some of the questions are meant to carry 8 or 12 points credit and require repeating the answer on several lines on the answer sheet. Please print your name and sign the honor code on the answer sheet. Make no marks on the R.H. margin of the answer sheet. You should not hand in the exam sheets, only the answer sheet. 1. Which approach would yield the most accurate model of a 6" long, constant-area, line element with only end loads? (a) one two-noded element with constant area. (b) one three-noded element with constant area. (c) two two-noded elements with varying area. (d) two three-noded elements with constant area. (e) all of the above. 2. Generalized coordinates (a) ftiust have the same dimensions as physical coordinates. (b) must be the same in number as physical coordinates. (c) must not be used for beams. (d) are used in shape functions. (e) cannot cause rigid body motion. 3. The plane stress rectangle shown has a [f] matrix with a rotational mode in it. Which column corresponds to a rotation? (2A) (++) j 0 x 0 x -y x2 x [ =] - 1 0 y y x y2 (a) column 3 (b) column 4 (c) column 5 (d) column 6 (e) column 7

2 4. Which column of the [M] matrix in problem 3 corresponds to a uniform expansion (dilitation) of the element? (a) column 3 (b) column 4 (c) column 5 (d) column 6 (e) column 8 5. The set of displacement functions in [)] of problem 3 has a problem because (a) three of the displacement fields are not linearly independent. (b) there arenot enough rigid body modes. (c) the displacements are not geometrically isotropic. (d) all of the above. (e) the displacements are better for a triangle than a rectangle. 6. A line element is aligned 45 from the horizontal with a roller support at 30~ from the horizontal as shown. The > proper local system for the inclined support at node 2 yields (a) u, =.707 u +.707 v (b) v =.500 u +.866 v 5 (c) ug =.866 u +.500 vg 7 T g (d) u2 =0.'Ow.. X (e) v - 0. g 7. The pinned support at node 1 of problem 6 requires boundary conditions (a) u = v = 1 g g (b) u v = 0 g g (c) uv = vR =1 (d) u +v = 0 g g (e) none of the above. 8, One assembles the global stiffness matrix (a) in local coordinates. (b) in global coordinates. (c) in global coordinates after eliminating nodes with forces. (d) only after converting to local coordinates.

3 9, A heat-conduction rectangle 10, is to be developed for the 11. rectangle shown. The shape function includes a term N xy 3 2 The element has thickness h. (o,) ({2, The heat conduction coefficient ___ _ k is constant. What is 3 the value of h33? (a) - kh (b) - kh 1 2 12 xy (c) 9 kh CO) C2 o) X 12 (d) - kh (e) 2 kh. 12, If a heat flux of 10 BTU/inch /sec 13. is applied into the lower boundary (y=0), how much equivalent nodal flux goes into node 3? (a) zero (b) 5h. BTU/sec ttttt t t (c) 5 kh BTU/sec (d) 20h BTU/sec (e) -5h BTU/sec 14, If a uniform heat flux of 100 15. BTU/inch3/sec is applied to the entire body (2h in'), how / much equivalent nodal flux goes into node 3? (a) -lOh BTU/sec'! (b) zero (c) +10h BTU/sec2 (d) +50h BTU/sec (e) +100h BTU/sec. 16. The consistent mass matrix (a) is always better than a lumped mass matrix. (b) is always worse than a lumped mass matrix. (c) has more nonzero terms than a lumped mass matrix. (d) is not physically meaningful. (e) has dimensions of FT2/L4.

4 17. Tangent stiffness can be used (a) to calculate stress. (b) to investigate stability of a body. (c) to determine the integration volume. (d) to develop the [B] matrix. (e) all of the above. 18. The nonlinear geometric law given in the [B] matrix (a) is something like a tangent law. (b) is at most linear in the displacements. (c) contributes to the geometric stiffness. (d) all of the above. (e) is a material property. 19. Suppose one wishes to numerically integrate the strain energy in a 4-noded line element, using Gaussian quadrature. How many Gauss points (abscissa) will be needed, at the minimum, for exact integration? (a) none - won't work. (b) one (c) two (d) three (e) four. 20. Suppose one uses Radau integration on the line element in problem 19. How many Gauss points are needed for an exact integration? (a) none - still won't work (b) one (c) two (d) three (e) four. 21. A Turner triangle is heated 100~F. What are a possible set of equivalent nodal loads due to thermal strain? (a) -1000 +1000 +1000 -1000 1414 / \ (b) 1000 -1000 1000. 1000 I c) neither of the above are remotely possible. -1414 (c) neither of the above are remotely possible.

5 22. Bending rigidity of a plate of thickness h, compared to a rectangular cross-section beam of thickness h, will be (a) 10% smaller (for a similar amount of material) (b) 10% greater (for a similar amount of material) (c) the same (for a similar amount of material) 23. Successively refined, nested finite element grids, for the displacement method, yield convergence: (a) from below on potential energy (monotonically increasing) (b) from above on strain energy (monotonically decreasing) (c) from below on stress (monotonically increasing) (d) all of the above. 24. Conservative force fields (a) possess a potential (b) can do no work in a closed cycle of motion (c) can be used in the potential energy method (d) include forces of fixed direction and magnitude. (e) all of the above. 25. The serendipity linear mapping functions (a) map a line element domain onto a double unit line segment (b) map a triangle onto a double unit square (c) map a tetrahedron onto a double unit cube.

Open Book AE 510/ ME 557/ AM 510 50 Minutes Finite Elements in Mechanics I 10/10/80 Honor Code Hour Exam W.J.A. Problem 1 (40 points). (a) Propose a displacement function u(x, y) for a 5-sided plane stress element as shown: u(x, y) = 2 e (b) Propose a shape function N (x) for a 4-noded -X line element as shown: N (x) 1 21 z 3 4 (c) Give the relation between displacement and shear strain in the quadrilateral shown: YvY(,y)= ]{ } y \ (d) The tetrahedron shown has 4 sides and is imbedded in 3-D space. If the body is unconstrained, how many rigid body 1 modes will it have? Number R.. Modes = Problem 2 (20 points) A constant strain, plane stress triangle is shown. If the 3rd node (at the top) is moved horizontally one unit to the right, and if the other degrees of freedom are constrained, 3 how much force is required to keep node 3 from moving vertically? / x< 2

-2Problem 3 (30 points) A beam has a load over half its span, as shown. Find the vertical compo- 2c0 Ib/C nent of the equivalent nodal load on A 4i 4 the left end. I A = 5 in2 --- E =107 psi = 0.3:-< 5'3 ---- Problem 4 (10 points) A 200-lb. man is poised (motionless) at the end of a thin, flexible diving board. If the end of the board is given a virtual displacement downward of 1", how much virtual work is done on the board by gravity (the man' s weight)? AW = ext How much work is done by the internal elastic forces during the same 1" displacement? i t. =1 internal

2 Hours AERO 5100, 197 Honor Code ME 557 D W. J. A. Open Book AMES 510 FINAL EXAMINATION Please place all answers on the special computer-graded form provided. Your final score will be the total number of right answers. Wrong answers will not count against you, so you should guess answers to any unsolved problems. Choose the answer supplied which is closest to the correct answer. It is not intended that the list always contains the exact answer. Each answer will count 2 points. 1. The triangle studied by Turner, Clough, Martin and Topp was originally developed using: a) energy ideas ) equilibrium ideas c) set theory d) integral equations 2. The Turner triangle is important because of its a) accuracy b) linear strain field ( historical significance d) use in beam theory 3. An attempt to separate the shearing and direct stress effects in the Turner triangle stiffness matrix 0) is successful b) fails because shear and direct stress effects are always coupled c ) is not useful d) can only be done in nonorthotropic cases 4. Virtual work is the work done: a) during a displacement from deformed to undeformed states b) by external forces c) by internal forces O() during a virtual displacement

2 A line element has EA/L = 10 Ib/in. The element is loaded with 2000 lb. A virtual displacement of -0. 002" is then given at the right node. 5. What is the strain energy in the element at equilibrium??,0 a a) 20 in.lb ql b) 100ft.lb ( 200 in lb d) 400 in. lb 6. What is the total virtual work done during the virtual displacement? (S 0 in. lb b) - 20 in. lb c) + 20 in. lb d) 32 ft. lb 7. What is the value of the work potential of the line element at equilibrium? a) 400 in* lb b1 200 in. lb c) 0 in-lb - 400 in- lb What is the potential energy of the line element at equilibrium? a) 400 in-lb b) 10 in.lb c) 0 in. lb @ - 200 in-lb 9. A line element with three. nodes as shown can have a shape function a) N(x) = x-x 3. b) N(x) 2 +x 40 N(x)= - (x 3L) 2 43L d) N(x) 1 + 2x + 3x 10. An equivalent nodal load vector for a two-noded line element under uniform load could be: -1001 ___ a) -100 b) 23 c) [^ o 1(jo)

3 11. An Euler-Bernoulli beam in 2-dimensions, neglecting column effects, has how many degrees of freedom per node? a) 1 c) 3 d) 4 12. A beam-column in three dimensions has how many degrees of freedom per node? a) 2 b) 4 0)) 6 d) 8 13. The purpose of the decomposition of a real symmetric matrix before solution is to a ) reduce storage requirenents b ) allow iteration 0) organize a sequential solution d ) improve positive definiteness 14. Solving "in place" - ) increases the number of equations required to find a total solution b ) prevents finding of stresses c ) is not practical d) can be done only on real, symmetric matrices 15. For two-dimensional finite elements, there are hco many rigid body motions? a ) One b ) Two 0) Three d) Six 16. A beam -column in 3 dimlensions has how many rigid body n-odes? a ) Two b) Four c ) Five: ) Six 17. Bandwidth Q describes the width. of the region along the main diagonal of a matrix, containing all nonzero, terml s b ) is defined for rectangular' r matrices as it is for square matrices c ) has nro meaning for mllass man.trices d ) hlas no n-eaning for dinlping mlatrices

4 18k What is the half bandwidth of the stiffness matrix composed of 6 beams connected end to end in 3-D? 0( 4;t~jl Bi *ttu) a) Two b) Six c) Ten (G Twelve 19. NASTRAN uses a matrix storage method: a) based on bandwidth ideas b) based on solving-in-place ideas 3O based on sparse matrix concepts, using a pointer system d) based on ninirizing input data 20. Natural modes for an element always include: a) shape functions b) transcendental functions c) cubics (D rigid body modes 21. The stiffness matrix for a beam D is invariant with respect to translations of a coordinate system b) is invariant with respect to rotations of a coordinate system c) depends on the mass distribution of the beam d) has an odd number of rows and columns 22. Skewed boundary conditions often occur: a) because of a mistake in boundary conditions b)!in problems solved by SAP6 c) because of angled forces (d because of planes of symmetry at angles 23. Nonlinear problems in a network often arise because: a) Ohm' s law is nonlinear (Q) the pressure drop across a length of pipe varies as a noninteger power of velocity c) the method of assembly uses a law such as Kirchhoff' s current law at a node, which is linear d) the sign convention requires flow into a node to be positive 24. If one wishes to interpolate a field variable in one dimension, and wishes to use four nodes: Q a cubic polynomial is needed b) a quartic polynomial is needed c) natural coordinates must be used d) derivatives of the function must be used

5 25. Area coordinates for a triangular region a) are difficult to integrate 0( add up to unity c) are four in number d) are five in number 26. If one uses Gauss integration with 4 sample points on a one-dimensional region: a) there are 4 corresponding deriving equations in the integration scheme () a 7th degree polynomial can be exactly integrated on the region c) a 5th degree polynomial can be exactly integrated on the region d) there are 8 corresponding abscissae 27. Gauss quadrature is particularly useful in finite elements because: a) the domain of the problem is complicated the degree of polynomial being integrated is known c) rigid body modes are automatically included and simplify the problem d) stress problems don' t have to be accurate 28. Equivalent nodal loads: (_p can be derived using virtual work b) are simplier than lumped load ideas c) allow shorter computer solution times d) do not work for distributed loads 29. Consistent mass matrices: a) are simpler than lumped mass matrices _ are often "full" matrices c) lead to less expensive solutions d) uncouple the degrees of freedom more than lumped masses Consider the 3 noded, heat conduction 3 element shown. It has thickness of 2 inches in z direction. The conduction coefficients BTU k = k =.001 (o t) x y in-sec F. (IjJ 30. The shape functions include: -. 31. a) N (x,y) = -x b) N2(x,y) - Z-x c) N3(x,y) y-x+l Q) N(x,y) = 2-y

6 32. The value for k is: 33. a).001 BTU/(sec-in) b).0005 BTU/(sec~F) c).002 BTU/sec ~.001 BTU/(sec~F) A bar bell (for weight-lifters) is W L -' L -- L-4 modelled as a beam using 3 twodimensional Euler Bernoulli beam elements as shown. The points where the weight-lifter' s hands are applied are considered pinned. No axial dis - constant area and placements are considered. mechanical properties 34. What characterizes the term k in the assembled stiffness matrix? (This term relates the vertical displacement at node 1 to the moment created at node 4.) a) it is negative Q Pit is zero c) it is positive d) one cannot tell from the data given 35. How many of the 8 degrees of freedom are constrained due to boundary conditions? a two b) three c) four d) six 36. From the sketch shown, try to relate k and k58 -D k 14 k58 ) k14 =k58 b) k k 8 4) k4 58 - d) one cannot tell from the data given 37. What is the relation between k14 and k? 14 76b a) k14 k76 C) 4 - - 0 one cannot tell f the data gien d) one cannot tell fromn the data given

9 50. Isoparametric elements are different from "straight-sided" (common) elements in that: a) the number of nodes is smaller with isoparametric elements b) isoparametric elements don' t use shape functions c) isoparametric elements must be smaller d) straight-sided elements are more accurate G) isoparametric elements almost always use numerical integration

Two Hours FINITE ELEMENTS IN MECHANICS I. Dec. 18, 1978 Open Book AERO 510 M. E. 557 AM/ES 10 WJA Honor Code FINAL EXAM Problem 1. (25 points) A) A constant strain (Turner) triangle is given as shown. The second node is moved 0.050" to the right while the other nodes are fixed. If the element is made of steel and is 0.100" thick, how much strain energy is stored -- in the triangle due to this deformation? 6 /3 E 30x10 psi V 0~3 6 4 — a G 11. xl0 psi B) If such a triangle were actually cut from steel and loaded at one corner to produce the same -.-.X nodal displacement, would the stored strain energy be less than, equal to, or greater than the value found for the finite element? Problem 2. (25 points) A plane stress problem has the mesh shown. T A) What is the semi-bandwidth of the assembled set 2 of equations for the nodal numbering given? Use detailed notation. -5 -- B) What is the optimum semi-bandwidth which could be obtained by renumbering the nodes? Again use 4 detailed notation. C) For the given element numbering, what is the q maximum wavefront for the problem, in terms of number of active equations? a- --. II 10 _~ 13 Problem 3. (25 points) A two-dimensional, constant flux triangle (,z) (comparable to a constant strain Turner triangle) 13 (2) is shown. Find the term hlt in the internal energy (heat conduction) matrix. Reduce the expression as much as possible for the given triangle. | C- ) 164

165 Problem 4. (25 points) 3 An electrical network of resistance RI R+ is shown. $s (a) Write out the matrix of electrical conductivities for this network by --- inspection. z 4 6 (b) What is the semi-bandwidth for the matrix.? Problem 5. (25 points) A rectangular, thin elastic plate is suspended from its center by a rigid rod as shown. The physical problem involved is to find the deflection of the plate due to gravity. It is desired here, however, to ask only questions about boundary conditions and symmetry. t Plate-rod system. Quarter-plate element. Degrees of freedom. (enlarged) Suppose the plate is modeled by 4 rectangular finite elements as shown, and by symmetry, only 1/4 of the problem (one element) need be solved. Provide the conditions to be placed on the following variables due to symmetry and boundary conditions. Reduce the number of constraints to the minimum number needed to run a typical finite element program such as SAP6. w 0 x0, z u2, v2, w2, x2 e, F F F M M M F F F M M M xl Y1 Z1 x1 Y1 Z1 X2 Y2 z2 X2 Y2 Z21 Problem 6. (2 5 points) Two thermal line elements are connected thermally as shown. An external heat source at 200~F. insulated is provided at node 2. What heat flux into surface. rode 2 at 2000F. is required to maintain the temperature at node 1 at 100~F.? I H. k = k2 5x 103 BTU/(sec. F. in) It t1 A1 = 1 in 5i ~~~~~~~~1.l 2'i <-10 — f — t15"......

7 A triangular sheet of metal. has a hole punched as shown, centered on the centroid of the triangle. / The sheet is put in compression as shown. / 38. What is the smallest portion of the problem that can be solved, using.. t.. symmetry? t -t t a) 1/12 l X 1/8 A) 1/6 d) 1/3 e) 1/2 39. If you solved one half of the problem, using symmetry, would you have to constrain any rigid body modes, in addition to applying symmetry conditions? a) no yes, constrain one rigid body mode c) yes, constrain two rigid body modes d) yes, constrain three rigid body modes e) one can' t tell - it depends on which way the body is divided 40. If the body is clamped at the lowver boundary and subjected to gravity in the vertical direction, what is the smallest portion that can be solved, using symmetry? a) 1/8 b) 1/6 c) 1/3 Q 1/2 e)' the entire body 41. Active joints in robots: Q require special consideration b) can be studied using conventional programs such as SAP6 42. Nunmerical integration in time: a) can be explicit, inmplicit or Gauissian quadrature b) is used in static stress analysis o) only works for beatn elements d) is a method peculiar to finite element theory Q) can have stability problelns

8 43. SAP6 has a problenl in that it: a) can' t find static stresses b) doesn' t have any dynamics capability c) doesn' t have a plate element d) doesn' t have automatic node generation < cannot apply symmetry conditions on a skewed boundary 44. The boundary element in SAP6 is used to: a) impose specific forces at a point c( impose nonzero displacements at a point c) apply dynamic loads d) provide an artificial node for beams A plane stress problem is shown: 7 45. What is the maximum wavefront, f 46. in detailed notation?? 4 8 b) 5 e) 10 47. What is the half bandwidth of the stiffness I 48. matrix, using compact notation? a) 4 ) 6 c) 8 d) 10 e) 12 49. The dotted line connecting the origin to a point on the stress-strain law represents an n-dimensional surface with "slope" we call a) elastic stiffness b) secant stiffness c) tangent stiffness (~ secant modulus e) tangent modulus

166 IMPORTANT INSTRUCTION: SOLVE ONLY ONE OF PROBLEMS 7 AND 8. ---------------- (IF YOU DO TWO, ONLY THE FIRST W ILL BE GRADED L) Problem 7. (50 points) A beam is clamped at both ends. A rod extending from above helps to support it. Find the deflection of the beam at the center when a 1000 lb. load is applied as shown. 9 2 N\ % EI = 2. x0l9 lb. in. for the beam T EA = 6.5xl6 lb for the rod 50 | o (This problem appeared on the...._____. national Professional Engineer's f Exam three years ago. It required / one hour to solve by conventional analysis.) ~ _ t.. 7.j, St 0ooo 1 Problem 8. (50 points) Alternate problem. One step in the generation of (z ) ) two-dimensional isoparametric elements is the mapping between a quadrilateral and the square shown. Suppose a Gauss integration point lies at a point (O.5,0.5) in (2) the $,' plane. To what point in I the x, _ plane does this correspond? ( ) (-,,,((s,,) ) %'(i)' -----— ^5~(.s.5

167 50 minutes HOUR EXAM December 6, 1978 Honor Code W.J.A. Open Book &. Notes AERO 510 ME 557 AtM,. S1() Problem 1 (25 points) A) Integrate the function () 6 2^ 42 over the interval (-1,1) using Gaussian integration and 3 example points. B) Is your answer exact? Why'? Problem 2 (25 points) J ( 4) Calculate the consistent mass matrix term m33 3 for the constant strain triangle shown. The triangle is of constant thickness t and I density Co -— X Fig, 1. Constant strain triangle Problem 3 (25 points),t. A truss element is imbedded in 2-D space as shown. 9 p If the stiffness matrix in global coordinates T EA I has a term 41 -.5J10) what is the basic stiffness A for the truss member? X Fig. 2. Truss element Problem 4 (25 points) An Euler-Bernoulli beam element in a two dimensional space must have two rigid body motions -- translation and rotation. If O(t and ~( are the two generalized coordinates in rigid body translation and rotation respectively, write out a suggested [Al matrix for the beam element.

168 HOUR EXAM Open Book. Aero 510 M. E. 557 AM/ES 510 October 13, 1978 Honor Code. WJA 50 Minutes. Problem 1. (20 points) A Turner (constant strain) triangle is shown. Find the shape functions for this triangle. 1, (0,0) (,0) Problem 2. (40 points) A Turner (constant strain) triangle is used is a vertical orientation as shown so that gravity loads the triangle at 50 lb/in2. Find the equivalent nodal loads on node 2, only. The shape functions are: X I 0- ((-x- ^ 0: o3 3 0( 0 4 44

Problem 3. (20 points) A hypothetical element has -o -o 0 [A — 1 o I Lo - es->el E The element volume is V. A) What is the stiffness matrix for the element? B) Is there a rigid body mode present in this description of the element? Prove this. Problem 4. (20 points) Two problems as shown are to be solved by finite element methods. A Gauss elimination equation solver is to be used* Prove which problem will take more CPU time for the equation solver. - - - -, -Ii A) Three-dimensional brick B) Two-dimensional quadrielements. 8 nodes per lateral elements. 4 nodes brick, per element.

170 2 HOURS AERO 510 M E 557 Am/ES 510 17 Dec 1977 Honor Code FINAL EXAM WJA Open Book FALL 1977 Problem 1. (20 points) Given the equations r, + r2 r, +2r =,13 carry out a triangular decomposition of the stiffness matrix. Find only the matrices in the form L. >-3r[LHT and do not solve for the unknown displacements. Problem 2. (40 points) A constant-strain triangle is shown. It Af~ | uses linear shape functions, of course. Find the_ equivalent nodal loads for the element due to the " " S: " ) 0 distributed load shown, which extends from x = -2" to x = +2" on the top edge. \ Hint: Because both u(x,y) and v(x,y) are interpolated in the same way, the form of the shape function matrix is: \ n:9ou) m Ot wrk(,t) O p ro0 e) o si1 #tr. oI Comment: You must work this problem out using theory. No intuitive solutions will be accepted. Problem 3. (20 points) Two hexagonal bars have been pierced by triangular holes as shown and are loaded on three sides by uniform distributed stresses. * r' ~ -- The polygons are regular and are concentrically \ located such that their centroids coincide. Each represents a two-dimensional, plane strain / \S problem, with the only difference being the \ \.. orientation of the triangle. / A) Which one(s),if any,of the problems 1 l can be solved by modeling a smaller portion of the problem using symmetry? Sketch. r B) If either or both can be solved using a smaller portion, give the type of boundary conditions to be imposed on the cut edges to preserve symmetry,

171 Problem 4. (30 points) An Euler-Bernoulli beam has the following properties: A= In o,,-. —200 -- J X^ 15;- -- a s 30( I /;^ x M = 0.3 L 200 no How much strain energy is absorbed by the beam if the right end is rotated -10~ and no other end deflections or rotations occur? Problem 5. (50 points) A heat conduction line element with three nodes is to be developed. The nodes are to be equally spaced. A) Develop the three shape functions from first principles, explaining the geometrical requirements. B) Using the variational expressions developed in the course, find the the term h!, using the proper terminology for the heat conduction problem. L Problem 6. (40 points) A) A Radau integration is to be used with a total of 5 sample points, including the fixed one. What degree polynomial, at most, can be exactly integrated with this procedure? Explain your reasoning. 3) What is the half-bandwidth, in detailed notation, of the 2-D structure shown? You may use compact notation for intermediate calculations, of course. z 13 1+43 __ 412 5 bC 8 e 1~ G) What is the maximum:lavefront, in detailed notation, of the same 2-D structure shown above?

172 D) If the set of equations shown is to be solved by first modifying them with the Payne and Irons' method for solving "in place", how many equations must be solved simultaneously after the modification? I 3 -r 9 Z~ nr, zo 3 1o 5 8 4 r, 30 7 57 40 Z ^ t 10 q 8 2. loo 0 o (2. 4 4 o 400 r5 (5

173 50 minutes ME 557/Aero 510/AMES 510 Open Book Honor Code Second Hour Exam 21 Nov. 1977 Honor Code Problem 1. (30 points) A slender elastic beam is clamped at both ends by rigid walls as shown. This support constrains both the deflections and rotations at the ends. A lateral force of magnitude P~ acts at the center of the beam. Using / _t known stiffness matrices, calculate' the deflection at the center of the beam due to P. Use two EulerBernoulli beam elements. The beam has stiffness EI and length L. Problem 2. (10 points) A plane stress problem has been 3Z posed where all stress boundary conditions have been given. A 12-element model has been proposed V as shown, where only degrees of -- - freedom 1 and 2 have been constrained. (a) How many additional degrees of freedom must be constrained. -- to properly remove rigid body modes? (b) Name all of the candidate de- - -- grees of freedom shown (9, 10, Xl D 31 and 32) that are acceptable choices to remove the remaining degree(s) of freedom. Problem 3. (10 points) A plane strain problem is modeled as shown. What is the half band-. width of the given system, in de- _10 tailed notation? How many zero terms will there be in the stiffness 3 matrix, in detailed notation for the entire matrix? I _5 81 1 7

174 Problem 4. (40 points) Two line elements are connected as shown and their outer ends rigidly pinned. The elements are then cooled 20 F. Find the displacement of the center connection. E1A1/L = 10 lb/in. L = 10 in. / j_>-_ L... 1 [k -- L, -— 4 O —-- L LIZ_: a1 = 1.2 x 105 /~F. v = 0.3 EzA2/L2 = 2 x 10 lb/in. L = 20 in. 2= -5 a2 0. 5x 10 1/F. v = 0.3 Problem 5. (10 points) A 4-node line element has been proposed as shown. The nodes are equally spaced. The element extends between xl and x2. Natural length coordinates L and L2 are defined for the element. Write down the shape function N (L, L2) which gives unit displacement at the first node. ~~1 1 ^ ~~ ~ MX, 2.

175 Aero 510 Me 557 Ames 510 Honor Code 50 minutes First Hour Exam Oct 17, 1977 Open Book W. J. Anderson Prob 1 (33 points) A plane strain quadrilateral has been developed using the following displacement 1 functions:. 2 u(x,y:) al + x x + X y + xy + x22 ) 2 2 v(xy) = a6 + 7 x + a 8y + Q9xy + a10x y A) Write down the l-I(x, y) matrix, which relates internal displacements to generalized coordinates. B) Write down the jA] matrix, which relates generalized coordinates and nodal coordinates, for the specific element shown. Prob 2 (33 points) A plane-stress, constant strain triangle is given as shown. If the shape functions Lod 3 0 1/;^ for the triangle are - l u1 [N11(xi,) 0 N13(x,Y) ql - \);, lH J0L N z(xy) *. j 2 \ q5 q6 Fill in any missing notation using the obvious generalization of the sketchy version given. Then calculate the equivalent nodal loads on the element due to the given vertical line load on the top edge. Evaluate as completely as possible without finding the explicit polynomials for N..(x, y), which is a harder problem.

176 Page 2 10/17/77 Prob 3 (34 points) Given a line element with two nodes and stiffness [k = 10.n1 1 In -I I Ai The left end of the pin is fixed and the right end is loaded so that VI = (1000 lb A) Find the displacement of the right end of the line element using our matrix format (i. e. don't use freshman physics). B) If the line element is 2mpressed 10- in as a virtual displacement (from equilibrium) calculate the virtual work done by the external forces during this virtual displacement.

177 2 Hours AERO 510 23 Dec. 1976 OPEN BOOK FINAL EXAM Honor Code Problem 1 (25 points) A constant area line element is to be developed. It will have four equally.spaced nodes. L (a) Write out a suitable displacement function for u(x), using the [%] matrix. (b) Write out the [A] matrix, relating nodal displacements and generalized coordinates. Use the local coordinate system shown. Problem 2 (25 points) Calculate the potential energy at the equilibrium position for a line ele- t ment with!, E = 30 X 106 psi A 2 in.2 L = 6 in. -a 0 Q0i = 1000 lb Qc2z where Qa1 and QaQ are the generalized forces corresponding to the natural displacements {q} 1 q -zIL$L V I, Problem 3 (25 points) A uniform beam is fixed at both ends. e.,M Using finite element theory, and using a two Euler-Bernoulli beam elements, find the rotation 0 2 at the center of the beam due to a moment Ma = 5000 in. lb. E = 10 X 106 psi I = 400 in.4 ---- L - > v = 0.3 L = 100 in

178 Problem 4 (25 points) Wind pressure is acting on the left face of a concrete wall as shown. The wall is modelled with constant strain, plane strain triangles (Turner's). (a) How many elements are there? (b) How many nodes, including constrained nodes at the base? __ (c) What is the minimum half bandwidth possible, using detailed notation? /'77 (d) How many nodes will be given nonzero equivalent nodal loads due to the wind, including any constrained nodes? Problem 5 (25 points) R, — Three identical line elements are joined as shown. The left node rl is constrained and the third node r3 is moved 1" to the right. -> Set up the equations of equilibrium for the 4X4 matrix problem, using the artifice of Payne and Irons to modify the equations for solution "in place." Let k E1A1/L, etc. Kll Ki2 K13 K14 [K41 K44 [ Express the equations in the simplest modified form, without solving for any of the unknowns. Problem 6 (50 points) Three identical segments are assembled in a Y. For an individual pipe element, where jQe[ is the fluid flow 1 1e IQ.il,iaQel ^IQ'I P [ _-/J0 ^JQ6e ^T

179 9 K, Problem 6 (continued) in the element in question. \Ib (a) Assemble the equations for the "Y", using a node 4 at the junction. (b) If l' i = 100 psf'z = 200 psf 3 = 150 psf'4 = 125 psf p / solve for the fluid flows using the j "variable stiffness " iteration (secant stiffness approach). Carry the solution out to 3 iterations. It is suggested that you use the notation 1/f1 I/N_/QfI, etc., for the assembly process. This will simplify your writing and also grading. Problem 7 (25 points) (a) Does SAPIV require careful numbering of nodes or of elements to minimize high speed storage requirements? (b) Which takes less CPU time to run on SAPIV: problem 1 with 1000 equations and a bandwidth of 200 or problem 2 with 600 equations and a bandwidth of 300? (c) A two-dimensional problem involves a square cut out in a round tube. The tube is to be internally pressurized and free on the outer surface. Using symmetry arguments, what is the minimum fraction of the body you would have to model for a solution? _ (d) A plate is fixed at a wall and supported by a roller as shown, with two vertical tip loads. If this problem is modelled with two plate elements as shown: - How many forces and moments 1000 lb have been prescribed by the problem statement? ___? - How many displacements and I00 \ rotations have been prescribed?? 77 / 77

180 50 Min. AERO 510 Honor Code SECOND HOUR EXAM December 3, 19 Open Book ( 0) PROB 1. (25 pts.) A plane strain square is shown. If this elementhas 5 nodes, at corners c t and centroid, write out the relation (-) {q} =[A] {a} using the natural mode approach. Include numerical values in the A matrix to yield the proper rigid body modes, and include numerical values for one straining mode. Use detailed notation, as suggested in the sketch. PROB 2. 114 laso 91 ge 8* "5to 23 0 --'oL0 0..._. coo ~.. _.. -- -4 —-- 8

181 A certain professor has numbered the nodes for a F.E. study of an octagonal crossarm as shown. Plate elements with 6 D. O. F. per node are used. A) What is the current half bandwidth, in detailed notation? B) Show a suggested nodal numbering scheme that cuts the bandwidth to below 130. You may use pencil on the existing grid pattern. C) What is the half bandwidth of your suggested numbering system, in detailed notation? Show the critical element which causes the maximum difference in node numbers. PROB 3 (25 points) A three-point Labotto integration is to be used to integrate a quartic polynomial (4th degree) on a symmetric domain [ -1, 1]. The endpoints are taken to be two integration points. Assume you are asked to develop this 3 point Labotto integration from "scratch". A) How many defining equations are there? B) What degree polynomials can be exactly integrated? Is the integration of the quartic going to be exact? C) Wiite out the first defining equation, and reduce it as much as possible to the case at hand. PROB 4 (2 5 points) A constant temperature gradient triangle has been developed for steady heat conduction. It is the thermal equivalent of a Turner triangle as seen in stress problems. For 6 simplicity, a right triangle with oo i local coordinate system is considered. Sb The shape functions are N1 =1 -x/a - y/b N =x/a (oZo)s 2 ea.- 15," N3 = y/b ~b " (oi Find the equivalent he equivalent due to a flux of intensity 100 BTU/SEC /IN at the left boundary as shown.

HoU0r E6Xo s Oc C7 o I H ^ono Coe n E. 5-5 A, 0 t 5'o A-H/f-S Si 01-eo 6too - -- -—. t 0 5', 50 ^M^ci S PFeo6 ~.Bs,,; ~, —;AKZl tor (..0.-) rot-^tzarhs 0^ c^t- ejow v t^ t ezci i-F 8q ~~;~(e ovt Ca-2r~ Vertex 0 —-— ===== w, ut O i A_ V e- v 5 < A 3^ Pbzte> (-. \ ~a*Yc^J EL, I o 3 -1%)8 ~tl. p qt cqC,. w.te xa.s S.OA,. / 3 ^.; Wr;,ctcY ot o rod t oCA 2l Oe 0 b),Cs,@). @. sn v' Jt,~') 7;t l, 182

183 PfreB 3. tW.ru V CJ,~? rsu$4> (C30 %) a Av,1o.o VirtxcL c, Q+'(.'a t ~l.=t4 tA4 3t \ t pI * -OsX *L I I vIt L 4cV;f 7,,, l'*<.$orcoi? rect L.'~l = -i o 7 IR +. HSS>5LD,-S I hT<i (10%5 0- A plAk 5+tw^ PYOW\ cs:lows Us;i. co~c_ t IsI\+ ness fa^;L rc -cro? Bt! ^orc tn covnt ibo; k^ od I;; 7 e < ^<*.Y^^,;n ^ YO^r Qo)i^Uev^t

184 f4 50 Minutes AERO 510 ME 557 AMES 510 October 24, 1975 Open Book Honor Code MIDTERM EXAM W. J. Anderson All questions refer to the displacement method, and use detailed (as opposed to compact) notation. Answer on these pages. TRUE-FALSE SECTION (30 points - 2 points per problem) T F 1. The bandwidth of an assembled stiffness matrix depends on the jj j way the elements are numbered. 2. Higher order plane strain triangles (those that use higher degree polynomials for displacement functions) are less stiff than the L _ constant strain Triangle of Turner, et. al. 3. The virtual work theorem can be proven from the potential energy 1 ] theo rem. 4. Usual finite element problems involve more displacements which are specified than forces which are specified. I 5. Prestrain can occur because of temperature effects. II 6. Temperature effects cannot be studied in the same solution as external, prescribed loads because two different stiffness matrices are required. 7. A useful beam element can be made with as few as 4 d. o. f. l 3 8. The stress field in an Euler-Bernoulli beam is uniaxial. [3 1 9. A column inthe shape function matrix [ N] represents internal displacement fields due to a unit nodal displacement at a specific d. o. f. 10. The strain energy in the body will be lower for an approximate 13 solution than for the exact. 11. Stiffness matrices for a single element should be singular (have Pj 1 zero determinant) in order to allow convergence. 12. The global shape function matrix [ N] is useful for creating assembled [ 1 equivalent nodal loads in problems where many different local coordinate systems are used. 13. The work potential as it appears in the potential energy theorem i1 is equal numerically to the work done by the external loads. 14. The natural mode method of Argyris is useful because the rigid v, j body modes are explicitly seen and because the natural stiffness a L matrix is simpler than the physically defined stiffness matrix. 15. The first variation of potential energy yields the equilibrium E [ equations.

jB5 Section No. 2 (f5 points) 3. i IA plane stress study uses rectangular I I elements (see sketch). The nodes are shown. 5. A) How many nodal degrees of freedom are there? 30 1 B) What is the half bandwidth? 12 C) If you renumbered, what is the Cu T J MCrAL optimum bandwidth? J( D) If nodes 1 and 3 are restrained in the horizontal direction and node 2 is constrained in the vertical direction, are rigid body motions possible? O -t6 4Section No. 3 (15 points) (0,4) i A plane stress quadrilateral uses /(4 )t di.splacement functions u(x,y) = a+ + x + +3Y + 2k +J / v(x.y)= a5 + a6x + a txy + ao)y 1- Find the matrix A relating the nodal coordinates to the generalized coordinates. r +i'. O 7 \ D a g ~ o3c oo I S 4- zo o o o o A10 o o I 4-0 zo 0 0 4 0 o ~ 0 O o o o, o - I -Z O o o o o o 0 0 0 0 -2 00 el), C0 0) 0 0

186 Section 4 (10 points) A beam element has stiffness matrix 12 6L -12 6L [K] = 4L2 -6L 2L2 [ K]: L' 12 -6L 4L2 Write out the assembled equations of equilibrium for a two-element representation of a clamped-clamped 1000 1b beam, using the numerical value for loads and proper end conditions. Use the symbols for element lengths, L and L2. - W:l\ need e GK^ vic<;~ t 2 -L I _ -I- _ 1 _ I 0 0 0,Q La' <^ - - (S,^ i - -I- I -- - I o k., -.:I i-ll -7-,. -— ^'t a-\-. ~.4L $ *^: ^-^. I?2.

187 Section 5 (10 points) A natural mode solution is to be done for a plane problem. Write down the rigid body modes in terms of nodal coordinates for the given square element with unit sides. f L0 0 It)O I~~~~~~~~~~~(, ~,tcrmi p)$l 1M0(38I~ hl~YI~-i~S j 4' 1 ~s'. ", "' S^ 0t / I 0 ^ I J t.^'t I 9l 2,:'*;l) Q * ***' OI^ cvi 0 9 0(1-,' _0 ~I 0 JV^'

188 Section 6 _(20 points) A two-node, constant area, line -- L element has a displacement function. —:-..-___ u(x)= + x = +2 x A line load is F x/L O<x<L/2 F 0 L/2<x<L Calculate the equivalent nodal loads. for this distributed load.', /^ L.L L - IQbY S A "A O FO I o 5O 1t-l/o cscck 4 loR~c - \ F, 2-. ( [E,,, F~F~(v ~ i~l ^J

189 ME 557 AMES 510 AERO 51U FINAL EXAM 2 HOURS DECEMBER 20,1974 CLOSED BOOK HONOR CODE Problem 1 25 points Carry out a formal solution of the equations 6 +26 =5 26 + 66 = 14 by the Choleski decomposition. (Decompose the coefficient matrix into the product of a lower triauglar matrix and its transpose. ) Show all of your steps. Describe the prQcedure with proper terminology. Problem 2 25 points Attack the same set of equations as in Problem 1 with Gauss-Seidel iteration. Assume an initial vector{,and find the next vector iterate. Problem 3 25 points 10o 1b +4-oo b 3%0 1b Two prismatic links are connected axially as shown. The three external forces are prescribed. Can you solve for the displacements? Show exactly your calculations and reasoning.

190 Problem 4 (Z5 points) A plane stress problem is solved by finite elements. A total of 8 elements G with 10 nodes are used to model the 2 structure. ( ) \ 3 b The half bandwidth of the assembled stiffness matrix (in compact notation) is The number of zero trgcs in the assembled stiffness matrix (in compact notation) is The maximum wavefront (in compact notation) is

191 MULTIPLE CHOICE (5 points per problem) a b c d 1. Turner, et al developed the constant strain triangle and rectangle using a) Energy arguments b) Galerkin' s method c) equilibrium methods d) Lagrange multipliers. D D D D 2. Energy approaches in F.E. have advantages in a) development of equivalent nodal loads b) creation of consistant mass matrices c) development of elements with complicated shape functions d) all of the above. ~] D D ~ 3. Vector force equilibrium at a node is described by a) a single equilibrium equation in "detailed" notation b) a single equilibrium equation in "compact" notation c) a column of the stiffness matrix d) summation of external forces at the node.D L D L 4. Fluid and electrical networks a) consist of 1 dimensional elements imbedded in a two or three dimensional space b) must be studied by energy methods since equilibrium doesn' t apply to flow c) have a strict sign convention that all flows are positive when to the right d) cannot be a linear problem. D I O 5. Numbering of the nodes affects a) the bandwidth b) the wavefront size c) the number of nonzero terms in the assembled stiffness matrix d) all of the above. 6. Prestrain a) can be handled as an equivalent nodal load b) can be caused by temperature change c) can be caused by shrinkage, such as drying of wood d) all of the above. D L L L

192 7. For an element to provide a solution which converges as more elements are used a) linear strain must be possible when nodal movement allows it b) rigid body motion must result for any combination of nodal movements c) constant strain must result if the nodal displacements are computable with that condition d) all of the above. [D D W [ 8. In the basic theory of finite elements, an element with n nodes always has a) n generalized coordinates in detailed notation b) 2n shape functions in detailed notation c) three displacement fields: u (x,y), v (x,y), w (x,y) d) n nodal dis- D D CD placements in compact notation. 9. Lagrange multipliers are used in F.E. theory to a) form equivalent nodal loads b) calculate residual error c) constrain -displacements d) impose material conditions. D D D 10. A method for discretizing a field problem directly from the differential field equations is a) Potential energy theorem b) Galerkin' s method c) virtual work theorem d) all of the above. II III 11. If a material is orthotropic, one must account for this in the a) stress-strain law b) strain-displacement law c) equivalent nodal loads due to volume loads d) all of the above. I D | 12. The size of the element chosen is determined by a) the gradient of the stress field (how rapidly stress changes across the body) b) the accuracy desired c) the amount of

193 money available for computation d) all of the above. D D D 13. The main reason that 3-D elements (solids) cost so much to run is a) a solid body requires so many equations for its description b) solid elements such as the Tetrahedron must have complicated shape functions c) computers can handle matrices K(I, J) but have more trouble with matrices K(I, J, L) d) the bandwidth (or wavefront) is larger for 3-D bodies. E l E I 14. Why should displacement functions be continuous across element faces in plane stress problems? a) they don't have to be! b) to avoid infinite contributions to the potential energy functional at the interface c) so that rigid body modes of the assembled structure are possible d) so that strains (the derivatives of displacements) can be continuous across the interfaces. DD0 0 15. An acceptable shape function for an EulerBcrnoulli beam in the x, y plane is a) X(X-L)2 b) X (X-L)2 c) X (X-L) d) all of the above. - C - 16. The virtual work theorem requires a) an infinitesimal, virtual displacement b) an energy conserving system c) that one account for work done by external forces on the body in question d) the first and third answers above. I -D [

194 17. The potential energy theorem a) uses a strain energy measured from some initial reference position b) uses a work function defined from some initial reference c) requires a variation of displacement quantities d) all of the above. D D 18. The natural mode method a) allows solution in terms of generalized coordinates instead of nodal coordinates b) is a version of the Lagrange multiplier method c) creates a natural stiffness matrix which is nonsymmetric d) all of the above. - [D D 19. Gaussian elimination is an efficient way to solve: b) sets of linear, symmetric differential equations c) eigenvalue problems d) sets of linear algebraic equations. 20. Wavefront methods are better than banded decomposition methods because they a) use far less CPU time b) use far less core storage c) don' t require efficient numbering of nodes or elements d) all of the above. [-D - -

\400 lk ~ o- l. - -- REP Rl#TS S IApx 1 vam MEMO

Appendix Stiffness and Deflection Analysis of Complex Structures M. J. TURNER,* R. W. CLOUGH,t H. C. MARTIN, AND L. J. TOPP** ABSTRACT tion on static air loads, and theoretical analysis of aeroA method is developed for calculating stiffness influence co- elastic effects on stability and control. This is a probefficients of complex shell-type structures. The object is to pro- lem of exceptional difficulty when thin wings and tail vide a method that will yield structural data of sufficient accuracy surfaces of low aspect ratio, either swept or unswept, to be adequate for subsequent dynamic and aeroelastic analyses. are involved Stiffness of the complete structure is obtained by summing stiffnesses of individual units. Stiffnesses of typical structural It is recognized that camber bending (or rb bending) components are derived in the paper. Basic colnitions of con- is a significant feature of the vibration modes of the tinuity and equilibrium are established at selected points (nodes) newer configurations, even of the low-order modes; in the structure. Increasing the number of nodes increases the in order to encompass these characteristics it seems accuracy of results. Any physically possible support conditions likely that the load-deflection relations of a practical can be taken into account. Details in setting up the analysis can be performed by none.igineering trained personnel; calculations structure must be expressed i the form of either deare conveniently carried out on automatic digital computing flection or stiffness influence coefficients. One apequipment. proach is to employ structural models and to determine Method is illustrated by application to a simple truss, a flat the influence coefficients experimentally; it is anticiplate, and a box beam. Due to shear lag and spar web deflection, pated that the experimental method will be employed the box beam has a 25 per cent greater deflection than predicted from beam theory. It is s wn that the proposed method cor-.extensively in the future, either in lieu of or as a final rectly accounts for these effects. check on the result of analysis. However, elaborate Considerable extension of the material presented in the paper models are expensive, they take a long time to build, is possible. and tend to become obsolete because of design changes; for these reasons it is considered essential that a con~(I) INTRODUCTION ~tinuing research effort should be applied to the develTRESENT CONFIGURATION TRENDS in the design of opment of analytical methods. It is to be expected high-speed; aircraft have created a number of that modern developments in high-speed digital comdifficult, fundamental structural problems for the puting machines will make possible a more fundamental worker in aeroelasticity and structural dynamics. The approach to the problems of structural analysis; we chief problem in this category is to predict, for a given shall expect to base our analysis on a more realistic elastic structure, a comprehensive set of load-deflection and detailed conceptual model of the real structure relations which can serve as structural basis for dynalmic than has been used in the past. As indicated by the load calculations, theoretical vibration and flutter title, the present paper is exclusively concerned with analyses, estimation of the effects of structural deflec- methods of theoretical analysis; also it is our object to outline the development of a method that is well Received June 2., 1955. This paper is based on a paper adapted to the use of high-speed digital computing presented at the Aeroelasticity Session, Twenty-Second Annual machinery Meeting, IAS, New York, January 25-29, 1954. * Structural D)ylnluics Unit Chief, Boeing Airplane Company. Seattle Division. (11) REvIEw oF EXISTING METIIODS OF STRUCTURAL t Associate Professor of Civil Engineering. University of Cali- ANALYSIS forniu, Berkeley. $ Professor of Aeronautical Engineering, University of Wash- (1) Elementary Theories of Flexure and Torsion ington. Seattle. * Structures Engineer, Structural Dynamics Unit, Boeing Air- The limitations of these venerable theories are too plane Company, Wichita Division. well known to justify extensive comment. They are Reprinted with permission from the Journal of Aeronautical Sciences, Vol. 23, September 1956, pp. 805-823, 854 (with corrections by authors). 13

JOURNAL OF TH E AERONAUTICAL SCIENCES-SEPTEMBER, 19 5 adequate only for low-order Inodes of elongated struc- (5) Direct Stiffness Calculation: Levy, Schuerch' " tures. When the loading is complex (as in the case In a recent paper Levy has presented a method of of inertia loading associated with a mode of high order) analysis for highly redundant structures which is parrefinements are required to account for secondary ticularly suited to the use of high-speed digital comeffects such as shear lag and torsion-bending. puting machines. The structure is regarded as an assemblage of beams (ribs and spars) and interspar (2) Wide Beam Theory: Schuerch' torque cells.'The stiffness matrix for the entire strucSchuerch has devised a generalized theo of co- ture is computed by simple summation of the stiffSchuerch has devised a generalized theory of com-....h ness matrices of the elements of the structure. Fibined flexure and torsion which is applicable to multinally, the matrix of deflection influence coefficients is spar wide beams having essentially rigid ribs. Torsion- nay the matx of d ction influence coefficients is bobtatined by inversion of the stiffness matrix. Schuerch benllcling effects are included but inot shear lug. It is has also presented a discussion of the problem from the expected that wide beam theory will be used extensively has also a discussion of the problen from the poinrt of view oF determr1ininllg'tle stifflcss cocllicients. in the solution of static acroclastic problems (effect of air-frame flexibility on steady air loads, stability, etc.). However, the rigid rib assumption appears to limit its (III) SOME UNSOLVED PROIULMS utility rather severely for vibration and flutter anal-, At the present tine, it is believed that the greatest ysis of thin low aspect ratio wings. need is to derive a numerical method of analysis for a class of structures intermediate between the thin (3) Method of Redundant Forces: Levy, Bispllnghofftiffened shell and the solid plate. These are hollow -Lang, Langefors, Rand, Wehle and Lansings-' structures having a rather large share of the bending These writers have contributed the basic papers material located in the skin, which is relatively thick leading to the present widespread use of energy prin- but still thin enough so that we may safely neglect cipltes, tmatrix algebra, and influence coellicieuts in the its plate bending stiffness. In order to cope with this solution of structural deflection problems. Redundant class of structures successfully, we must base our internal loads are determined by the principle of least analysis upon a structural idealization that is suffiwork, and deflections are obtained by application of ciently realistic to encompass a fairly general twoCastigliano's theorem. The method is, of course, dimensional stress distribution in the cover plates; perfectly general. However, the computational diffi- and our method of analysis must yield the load-deflccculties become severe if the structure is highly re- tion relations associated with such stresses. It is chardundant, and the method is not particularly well acteristic of these problems that the directions of prinadapted to the use of high-sI'peed comllputing tiachinlles. ip)al stresses in certain critical parts of the structure Rand has suggested a method of solution for stresses cannot be determined by inspection. Hence, the in highly redundant structures which might also be familiar methods of structural analysis based upon the used for calculating deflections. Instead of using concepts of axial load carrying members, joined by inember loads as redundalts, lie plroposes to enlmlloy membranes carrying pure shear, are not satisfactory, systems of setf-equilibrating internal stresses. These even if we employ effective width concepts to account redundant stresses may be regarded as perturbations for the bending resistance of the skin. We should like of a primary stress distribution that is in equilibrium to include shear lag, torsion-bending, and Poisson's with the external loads (but does hot generally satisfy ratio effects to a sufficient approximation for reliable colll)atibility conditions).'The lnumbler of properly prediction of vibration modes and natural frequencies chosen redundants required to obtain a satisfactory of moderate order. Also, we should like to avoid any solution may be considerably less tlan the "degree of assumptions of closely spaced rigid diaphragms or of: redundancy." Successful application of this method orthotropic cover plates, which have been introduced requires a high degree of engineering judgment, and in many papers on advanced structural analysis. The the accuracy of the results is very difficult to evaluate. actual rib spacing and finite rib stiffnesses should be accounted for in a realistic fashion. In summary, what (4) Plate.Methods: Fung, Reissner, Benscoter, and is required is an approximate numerical method of MacNeaP-* ~M~~~ac ~~~NeaV~ ~analysis which avoids drastic modification of the As the trend toward thinner sections approaches the geometry of the structure or artificial constraints of its ultimalte limit, we enter first a regime of very thick elastic elements. This is indeed a very large order. walled hollow structures, such that the flexural and However, modern developments in high-speed digital torsional rigidities of the individual walls make a computing machines offer considerable hope that significant contribution to the overall stiffness of the these objectives can be attained. entire wing. Finally we come to the solid plate of variable thickness. During the past few years a sub(IV) METHIOD OF DIRECT STIFFNESS CALCULATION stantial research effort has been devoted to the development of methods of deflection analysis for these struc- For a given idealized structure, the analysis of tural types, and important contributions have been stresses and deflections due to a given system of loads made by all of the aforementioned authors. is a purely mathematical problem. Two conditions 14

STIFFNESS AND DEFLECTION ANALYSIS ullst be satisfied in tle analysis: (1) the forces developed in tle members must be in equilibrium and (2) L- LENGTH the deformations of the members must be compatible — 2 U2 M i.e., consistent with each other and with the boundary col;ditions. In addition, the forces and deflections in L each member must be related in accordance with the\/ v /> stress-strain relationship assumed for the material. The analysis may be approached from two different 0 GOS e - points of view. In one case, the forces acting in the 1 GCOS Oy"' members of the structure are considered as unknown (uantities. In a statically indeterminate structure, (b.) (b) an infinite number of such force systems exist which...~,....,~~..,. _ FIGo. 1. Typical pin-cndcd truss member. will satisfy the equations of equilibriuml. The correct force system is then selected by satisfying the condiiotnis of lcoi)patible (leform:,llations inl thle Imemll)ers. limatrix notationl. Iqs. (I) and (2) becomie, respectively, This approach has been widely used for the analysis of F -= [K] (a! (3) all types of indeterminate structures but is, as already noted, particularly advantageous for structures that { K} = [K]-' {IF = [C] JF} (4) are not highly rendundant. I-Here [K] is the mIatrix of stiffness influence coefficients. In the other approach, the displacements of the A typical element of [K] is kt = force required at i joints in the structure are considered as unknown in the {-direction, to support a unit displacement at j quantities. An infinite number of systems of mutuallydirection. If and always refer to the same colpatible ldefonrlationls ill thle me~berl s Iarepossible n the ensl-direction. If and the always refer to the sarme eonil)atil)le (tleforimmtiols in time memi)ers are I)ossible; the correct pattern of displacements is the one for which se the simpler fori k In eithe case an element of [K], and also of [C], must obey the the equations of equilibrium are satisfied. The con- well-known reciprocal relations. In other words, the well-known reciprocal relations. In other words, the cept of static determinateness or indetenninateness i [K] and [C] atrices are symetric, provided they irrelevant when the analysis is considered from this re d to oth al r ate ste i are referred to orthogonal coordinate systems. As will viewpoint. This approach is the basis for many re- be seen later, the symmetry codition does not apply be seen later, the symmetry condition does not apply laxation type analyses (such as moment distribution) if oblique coordinates arc used. and has been applied to the analysis of complex aircraft structures by Levy in the aforementioned paper. This (3) Truss Member. will be calle(l the method of direct stiffiness calculation. hercaftcr. IFig. l(a) shows a typical pin ended truss member. hereafter. After revieing the various metods available to the - We wish to determine its matrix of stiffness influence After re'YWviIIg tlie vario(us me1Pthod(]s avaIilable to the I., * COC'l*i~lltS, Lords may be applsd at ncoefficients. (nodes) dynamics engineer for computing load-deflection rela- i1 and 2. Each node can experience two components tionis of elastic structures, it is concluded( that tie most 2. E no can xperienc two components of displacementt. Therefore, prior to introducing promising approach to our present dilliculties is to cx- tel(1 furter tlle etil of (lirct stiffless clcltatiot. boundary conditions (supports), [K] for this member t',nd further the mnethod of direct stiffness calculation. wo The remainder of this paper is concerned with methods l b orer y w that extsimi y* e a is To develop one column of [K], subject the member by whichI tlat extensio(n m1ay be acco(nplisheli d. S. to U2 0, UI = vi = v1 t = 0. The11 (V) SIMPI',1: EXAMIP'I.tS I )F STIFFNI SS INFLUIlINCI: AL, = Uc2 COS O = IX2 COEFFICIENTS The axial force needed to produce AL is (1) Elastic Spring P = (AE/L)AL - (AE/L)A u2 If an elastic spring deflects an amount 5 under axial load F, Hooke's Law applies and The components of P at node 2 are F = k6 (1) Fr, = P cos O = (AE/L) X2 u2 Here k can be regarded as the force required to produce Fv, = P cos Oy = (AE/L) XAS U2;. iinil h'* (i'il, ); wcn' it r Ic', s't to be a..c.,,.i......(.I. 1(, r;. a 1 de1cCiE he, iols.idt- to hbe a Plquiilibriumi gives the forces at lno(d I as stiffness influence coellicient. Eq.. (1) can also be written as Fr = - F1, = (l/k)F = c (2) Fv, = -Fv, where c is the.deflection due to a unit force (deflection Eq. (3) for this member then takes the form influenlce coeflicient). F. \2 (2) Two-Dimensional Elastic Body [z, A, k A2.. U Extending the above relations to the two-dimensional Fy, L - X. v bo(ly is mmiost convemniently acconilllished by iintroduecing Fy, X.. V2 15

JOURNAL OF THE AERONAUTICAL SCI ENCES-SEPTEMBER, 1 9 5 The other elements in [K J arc found in a sillilar manner. (VI) STIFFNESS ANALYSIS OF SIMPLE TRUSS We get ~We~~ ~geA~t OIICCOnce stiffness matrices for the various component units of a structure have been determined, the next Ul U2 VI v2 Ui U2 Vl VI2 step of finding the stiffness of the composite structure [K] AE 2 2 1 may be taken. The procedure for doing this is essentrusls L _ x (6) tially independent of the complexity of the structure. mnctnbcr [ -A X 2 2: As a result, it will be illustrated for a simple truss as _-X, X~ -~ _J shown in Fig. 2. The stiffness of any one member of the truss is given As given in Eq. (6), [K] is singular —that is, its deter- by Eq. (G). Since length varies for the truss members, mtilnalnt vanishes and its inverse does not exist. This this terml should be brought inside the matrix. It is is overcome by supplying boundary conditions or sup- then convenient to call the elements of the stiffness ports for the bar suflicient to prevelt it froml mloving Imatrix X2 = X2/length, etc. Then X2;, P, and Xa repreas a rigid body. For example, we may choose ul = sent the essential terms defining the stiffness of the vI = u2 = 0, v2 d 0. Node 1 is then fixed, while node selparate truss members. These are conveniently cal2 is provided with a roller in the y-direction. The only culated by setting up Table 1. force component now capable of straining the bar is Fron the last three columnis of Table 1 the truss Fy,. The force in the bar and the reactions are given stiffness matrix can be written directly. This is best' by Eqs. (5) and (6). seen by fonling the truss equation [Eq. (7a) J analogous Any other physically correct boundary conditions to Eq. (5) for the single member. can be imposed. In other words, once [K] has been The formation of all columns in Eq. (7a) can be exdetermined, a solution can be found for any set of sup- plained by considering any one of them as an example. port conditions. The only requirement is that the'I'le second columl will be chosen. It represents the structure be fixed against rigid body displacement. case for which v\ 0 0, all other node displacements = 0. |F_ V 1 - 1 I 2v2L 2V\2L 2/2VL 2V22L,0 0 ~F^ I,0 0 1 1 0 U2 IL - L F. = AE L (7a) 1 I 2V'2L 2V/2L L L 2V/ L 2 2-L. J L 2 2 2L L ~ 2V/L 2V2L tv or {F} = [K] {f} (7b) signs follow from the basic stiffness matrix given in Eq. (6). Since equilibriunm must hold, the sum of these In this second column the y-components of' force y-components of force nust vanish. are given by the j32 terms in Table 1; the x-comn- Similarly, F/, is the sum of the Xi, terms for members ponents of force are given by the Xj; terms. Thus 1-2 and 1-3. Likewise, F,, is the negative value of F,, is the sum of,a for members 1-2 and 1-3 since these Xa for member 1-2. Finally F., is - Xi for member are strained due to displacement vi. Also F,, is -:A2 1-3. These forces imust also sunt to zero if equilibriui for member 1-.2, and I/, is -/2 for lmemtber 1-3. The is to hold. TABLE 1 Member x y Length X X' ^I X XA'' ) 1-2 0 -L L 0 -1 0 1' 0 0 O 1-3 L L X/I 1 1 1 1 1 1 1 1 3 L L V2L V2 2 2 2viZL 2v"~L 2V.L 2-3 L 0 L 1 0 1 0 0 0 0 16

STIFFNESS AND DEFLECTION ANALYSIS This process is repeated for all columns. In this way all possible node displacement components are taken into account. In each case the displacements are compatible ones for all members of the truss. A structure having various kinds of structural corn- po)iel1ts beatIls as well as axially loaded memlbers, for example-would be treated in the same manner. A (SAME FOR AL. MEMBERS) However, the basic stiffness matrix for each type of member would have to be known. Deriving these for units of interest in aircraft design represents a L \?L major part of this paper. The nlatrix of Eq. (7a) is singular. This is altered by providing supports for the truss sufficient to prevent it from displacing as a rigid body when loads are applied. Any sufficient set of supports may be imposed; here 2 3 we choose to put FIG. 2. Simple truss. uI = VI = u2 - V: = 0 It is now convenient to rewrite Eq. (7a) and simulIn other words, nodes 1 and 2 are fixed, while 3 is left taneously partition it as shown by the broken lines in free. Eq. (7c). | For0 20,20|,2^ 21 -1 0 u1 1 i 1 1 ________ 2f/ V 0 0 V3 2V2 2V2 2V2N _2 2 I =2V2 2V 2i 2 1 I0 01 U l= (7c) F, - 0 0 0 1 0 U2 0 0 0 -1 0 1 V2 O If the partitioned square (stiffness) matrix is designated FZ hy A l -' 1 F-,,} (9b) [ii'zxz I Ih4XJI A2 4X2 1 /94X4 In dynamic analyses of aircraft structures it is ordiexpand(ing Eiq. (7c) leads to the following two sets of narily sufficient to detennine [A]-'. This is the equations: flexibility matrix. It is interesting to note that [A] Fr, t h can be found from the complete [K] matrix by merely \pv ( =[A 1J i j (Xa) striking out columns and rows corresponding to zero displacements as prescribed by the support conditions. and FZ, A complete stress analysis leading to the truss memFy 5F, =u b er forces can also be carried out. It is merely neces/',, tva fb) sary to know the force-deflection relations for the Ft, individual members, or components, of the structure. This is a straightforward problem for the truss and, Eq. (Sa) gives unknown node displacements in teherefore, will not be discussed further in this paper. applied forces, It is worth while to notice that once the stiffness }/ \ ( matrix has been written, the solution follows by a V = [A]-' F' (9a) series of routine matrix calculations. These are rapidly carried out on automatic digital computing while Eq. (8b), together with Eq. (9a), gives unknown equipment. Changes in design are taken care of by reactions in terms of applied forces, properly modifying the stiffness matrix. This cuts 17

JOURNAL OF THE AERONAUTICAL SC IENCES-SEPTEMBER, 1956 12 10~1 l~0o (6) Forces in the internal members can be found by applying the appropriate force-deflection relations. The primary functions of the engineer will be to D/// A / w/ I \ p)rovide tle information required in steps (I) and (2) I 2! ii Sf/i o /1' l above and to provide the individual member forcedeflection relations if a stress analysis is to be carried K9 8 s (f - / out. Steps (3) through (6) can be performed by nonengineering trained personnel. Changes ill design can be taken into account by correcting local stiffness con//Z///9 D/ A~/ ///\ ftributions to K. Node densities can be increased in RIB// D / /D Rregions of maximum complexity and importance. If ~__L6 {D. - h- AcELEMENTS vertical deflections only arc required, as in the case of the aircraft wing problem, the 3n X 3n mlatrix for K can be reduced to order n X n by a sequence of matrix D64 ll5! 7/fj t ^SPAR Calculations. Physically, continuity of displacements ELEMENTS in three directions at each node will still be maintained. M! ~ nID A IL COVER PLATE(VIII) STIFFENED SHIELL STRUCTURES COVER PLATE ELEMENTS In carrying the above procedure over to stiffened shell structures, it is first necessary to perform steps FLANGE AREAS (1) and (2) of the previous outline. FIG. 3. Wing structure breakdown. For a wing structure the idealization will be ialla(l by replacing the actual structure by an assemblage of spar segmlents, rib segments, stiffeners, and cover analysis time to a minimum, since development of the plate elements, joined together at selected nlodes. stiffness matrix is also a routine procedure. In fact, Fig. 3 shows the proposed idealized structure. The it may also be programmed for the digital computing decomposition of the structure can be carried further machine. with some increase in accuracy (for example, by decomposing spar segments into spar caps and shear (VII) SUMMARY-MEiTIIOD OiF DIRKICT STIFFN.rSS webs), or it canl be simlll)lified( by treating tle structure CALCULATION as an assemblage of spars and torque boxes. Tie (1) A complex structure must first be replaced by an degree of breakdown should be consistent withlthe equivalent idealized structure consisting of basic struc- coml)lexity of structural deformations required by the tural parts that are connected to each other at selected problem at hand. (In a vibration analysis the order of node points the highest mode is a determining factor.) In light (2) Stiffness Imatrices tmnust be cither knowll or (Ic- of the proposed idealization, it is necessary that stiffness tennined for each basic structural unit appearing in the matrices be develoed for tie following comollnelts: idealized structure. bcan segments consisting of flanges joined by thin (3) While all other nodes are held fixed, a given webs, and l)lap t clmlllels of arbitrary shpel. In node is displaced in one of the chosen coordinate direc- addition, provision must be made for taking stiffeners tions. The forces required to do this and the reactions in accounl t and possily for including the effect of set up at neighboring nodes are then known from the sandwich type skin panels. various individual member stiffness matrices. These In the general case, spars will be swept, lnoinparallel, forces and reactions detentine one column in the overall and not necessarily orthogonal to ribs. It will generally stiffness matrix. When all components of lisplacemlenlt be convenient to transfer stiffness values for any given at all nodes have been.considered in this manner, the member to a fixed set of reference axes. These refercomplete stiffness matrix will have been developed. ence axes will be chosen as rectangular Cartesian In the general case, this matrix will be of order 3n X 3n, (x, y, z) in order to preserve symmetry in the total where n equals tlhe number of notles. T'lle stiffness K-lmatrix. matrix so developed will.be singular. An outline of the determination of member stiffness (4) Desired support conditions can be imposed by for simple structural elements is given in the paper. striking out columns and corresponding rows, in the Further details are presented in Appendixes. Derivastiffness matrix, for which zero displacements have tion of stiffness matrices for more complex elements been specified. This reduces the order of the stiffness can be accomplished in a straightforward manner. matrix and renders it nonsingular. However, in tile analysis of an actual structure, it will (5) For any given set of external forces at the nodes, be necessary to weigh the relative advantages of ernmatrix calculations applied to the stiffness matrix then ploying a small number of large complex elements yield all components of node displacement plus the against the advantages of using a larger number of cxternal reactions. small elements for wllich silmple stiffness coefflicients 18

STIFFNESS AND DEFLECTION ANALYSIS may be employed. The main criterion to be observed in resolving this issue is that the problem must be programmed so that as much as possible of the data processing is performed automatically by the computer < and not by human operators substituting in complex ionnulas. x (IX) SPARS AND RIBS \ First we consider the untapered beam segment of uniform cross section shown in Fig. 4. Its stiffness X matrix will be determined by application of beam 0 theory, which is extended, however, to include shear 3'' web flexibility. Nodes, 1, 1', 2, and 2' are established as shown in Fig. 4. The following notation is used: I = moment of inertia of beam section about FIG. 5. Rectangular Cartesian axes systems. neutral (y) axis;4 = t = thickness of shear web WI = wI,, w = W'. (1 I =- modulus of elasticity of flange material = - -U, U2 = -U2 G =- modulus of rigidity of shear web material G = modulus of rigidity of shear web material Stiffness in the y-direction is assumed negligible. An outline of the derivation of the stiffness matrix Displacements are assumed such as to be compatible for the above beam segment is given in Appendix (A). with elementary beam theory. In other words, It is slhown to be of the form U1 VI WI U2 V2 W2 (4/3) (1 + n) 0 0 6EI - (WL) 0 h2/L2 [K] Lh2(l + 4n) (Ila) (2/3) (1 - 2n) 0 -(h/L) (4/3) (1 + n) 0 0 0 0 0 h/L 0 - (h2/VL:2) h /L 0 h2/L2. where n = 3(I/G) [I/(htL2)] (lib) 4 L2 LI Contrilbutini (,f shear web deforlnation to the above 3d ~ I U2I / stiffness nmatrix is indicated by values of n > 0; for a -F = L I (12) rigid shear web n 0. /L As a simple example of the use of the beam stiffness imatrix, we consider a cantilever of length L and loaded Eq. (12) may be inverted to yield tip displacements by force P at the free end (nodes 2. and a ). Putting MLand ucin terms of applied load P (PF, 0, Fp n - 0 and applying Eq. (1 Ia) gives: P/2). The results are -~ ~^ - = --.(pI'2/2p!) (1/:2), w:2 =7/ 3EI 0 NODES i p4k f,AF which agree with known results. _.2_ In an actual wing structure, spar and rib segments x"l i xl~ f jwill be more or less randomly oriented with respect --- x,u'h )~1 ~c to a set of standard reference axes. As a result, transF., _C ~formation of stiffness matrices for these members to the F r7- - - - -( standard set of axes will generally be necessary. The p F~~ ~~~-L~~~~ ~basis for such transformations is given below. Let the direction cosines of x, y, z-axes with respect FIG'. 4. Beam (spar or rib) segment. to standard 2, 9, f-axcs, Fig. 5, be 19

JOURNAL OF T H A RONAUTICAL SCIENCES-SE PTEMB ER, 1956 (X) STIFFENED PLATES C/OL>f? SKy^ _ ~(1) Stiffeners } ^ Jv- _^f RSPARI A plan view of a typical portion of stiffened' cover Np D~E ~/ - G ^ f skin structure is shown in Fig. 6. Nodes are initially I ~"~ —-/ -_ -I- -. ~established at points 1, 2, 3, and 4. The included structure then consists of spar segments (1-2 and 3-4), /- J~~ FTENg^ I / rib segments (1-3 and 2-4), and stiffened plate element 1-2-3-4. Stiffeners may be conveniently lumped with / -- - i { sl;spar caps and, if desired, into one orr iore equ(ivalelit )~ / — 7s"~ — /(~ ~ ~stiffeners located between spars. In this latter event ) _ __ SPAR _ _ additional nodes must be established, as at the interf,6-{~~~ 3Sr~ j4 ( sections of these equivalent stiffeners with the ribs. The stiffness matrix for a lumped stiffener of constanlt area A, length L, anid o(dulus 1~ is FIG. 6. Stiffened cover skin element. [ t1 [K] = 1- (1() stiffener L L1 1 x y z x XI A X Derivation of a similar matrix for a tapered imember is Y A # A.t - Lstraightforward; the area A is replaced by a suitable SIz X, mean value. The influence of shear lag effects on load-deflection relations for the panel and stiffeners Simple geometrical considerations then give the follow- can only be included if nodes are established at intering equation for relating forces in the x, y, z system to mediate points on the ribs, between spars. forces in the x, y, 2 system: (2) Plate Stiffness Fw1 y y X. 0 0 0 F I The quadrilateral plate element 1-2-3-4 of Fig. 6 I, = v, v, 0 0 FO tis assumed to possess in-l)lane stiffness only. Since Ir. = n oV V 0 0 0 \ F' (l3a) two independent displacement components can occur F,, 0 0 0 A, X, X, ) 1ZI., 0 0 0'Z j at each node,(" tle or(lert of tlhe K'-matrix for this plate Fi0P L 0 0 v. Pv;' J FZ Jelement will be 8 X 8. The problem of calculating K L0"/ 00 3is not an easy one, an11 the solution offered here is felt or, I{} - [] tF} (13b) to have potential usefulness for finding approximate solutions to many two(-dillensional l)robllems in elasDisplacements are vectors similarly related to the ticity. coordinate systems as forces and hence transform under Before proceeding with the method developed for a rotation of axes in the same manner. Consequently, calculating K of the plate element, it is pointed out that a so-called framework tanalolgy'2 exists, wllicl p)rts} = j 14)] 6} (14) mnits one to replace tile elastic plate witl a lattice of elastic bars. Under certain conditions the franmework 4 l\ ICthen deforms as does the plate and hence can, be used vI to calculate the plate stiffness. The determination of where e~~ =.)etc. a lattice representation for a rectangular plate is relawhere' ~. I ~tively straightforward; however, plate elements of nonrectangular formn present basic difficulties. For ex~tllp&9~2 Iample, if one attempts to apply the rectangular gridwork to a nonrectangular plate, difficulties arise in From the above and Eq. (3) it follows that, attempting to satisfy boundary conditions. On the ()ther hand, if (ne goes to tnontrectalgular lattice formls, R[] = (14] [K] [(]-1 = [()] [K] [4)]' (15) difficulties arise when attempting to satisfy the stressstrain relations ill the interior of the plate. Considerwhere [K] is the stiffness matrix referred to the stand- ations such as these led to eventual abandonment of ard 2, y, i set of axes. Beam segments encountered in this approach. the analysis of real structures will be tapered in depth, The concept finlally employed for determining plate and flange areas will be variable; generally the segments stiffness is based on approximating actual plate strains will be taken short enough so that the variation in by a restricted strain representation. In other words, depth may be assumed linear. Derivation of stiffness no matter what the actual strains in the plate may be, matrices for elements of this kind is straightforward, these will be approximated by a superposition of and details will not be included in the present paper. several simple strain states.'lIhe mlethod for doing 20

STIFFN' ESS AND DEFLECTION ANALYSIS this and the accuracy of results based on such a representation form an important portion of this paper. 3 (y, To give an initial illustration, the actual strain distribution in a rectangular plate element can be alproxilmated by superimnosing the strains that correspond to each of the simple external load states shown in Fig. 7. These load states are seen to rclre- sent uniform and linearly varying stresses plus constant, x shear, along the plate edges. Later it will be seen 1 x,1) 2(txz that the number of load states must be 2n - 3, where that the numberof load states musbe 2n FIG. 8. Node designation for triangular plate clement. n = number of nodes. Before commenting further on the scheme suggested here for analyzing plate elements, the method will be angle. Hence the triangle can dislplace as a rigid body applied to the triangular plate of Fig. 8. The triangle in its own plane and undergo uniform straining accordis not onily siil)pler to handle than the rectangle but ingto Eq. (17a). later it will be used as the basic "building block" for Displacements at the nodes can be determined by calculating stiffness matrices for plates of arbitrary inserting applicable node coordinates into Eq. (17b). shape. In this way six equations occur which are just sufficient We start by assuming constant strains, or for uniquely determining the six constants of Eq. (17b). As a result the constants become known in terms of = -a = (/E) ( - v ) = u/x node displacements and coordinates. It is this part v =b (1/E) (a, - vo) = v/by (17a) of the solution which determines the number of terms Trv c C-= (l/G)r^7 - (u/by) 4+ (6v/x)J which must be chosen in the strain expressions or alterLater it will be pointed out why we are restricted in the natively the number of applied edge stress states which choice of strain expressions. Integrating we find the must be used. The number is always twice the number displacements to be of nodes minus three. Hence, for the triangle we reu = ax + Ay + B )quire three terms and five for the rectangle (or quadri^r by + (C - A)x + ^r\ (17b) lateral). v -y c x + To proceed with the solution, we solve directly for where, A, B, and C are constants of integration which stresses in terms of node displacements ui, vl, U2, etc. define rigid body translation and rotation of the tri- If x = xi - xj and X = (1 - v)/2, this gives I vXa^ 1 vx O X2 XLY3 X2 X2Y3 Y3~ Z; k. P Xjm v 1 U2 i (18a) J 1 - x 2 x 2y X X2 xYys V2 XlX32 Xl Xx3 X1 X1 u3 0 Va _ Xy3 x'3.ry3 x' Ya or {oa [S] ({} (18b) Y~~~~~ S ~The next step is to obtain the concentrated forces at t the nodes which are statically equivalent to the applied L ^~~~x - constant edge stresses. The procedure for doing this will be briefly illustrated for the ease of the shear stress. Fig. 9(a) shows the shear stresses on the circumQg,, J t t __ scribed rectangular element, and Fig. 9(b) shows the _ o.) A^. h>0 S, e e 4 corresponding edge shear forces on the triangle. As (- ) (b) h ( b) before xa, v, refer to coordinates of node points. Forces on any edge are equally distributed between \ffi 14\~~ ~nodes lying on that edge. For the forces as given in Fig. 9(b), this leads to ^17~ i~ _~ Ft,^~(aF), = - -(X2 - X3) (t/2) rty (d) 4 <l 0 F,,c3' = — +ya(t/2) (19) ~ a t'b f F,, (= -x:y(t/2) T,, v fZ,(3) - +aX2(t/2) TX FIG. 7. Applied loads on edges of rectangular plate element. F~,(s) = 0 21

JOURNAL OF THE AERONAUTICAL SCIENCES-SEPTEM IER, 1956 where the superscript refers to case 3 (that of shear or stress). This procedure is repeated for the two normal stresses. Superimposing results for these three cases F} = [r] { r} (20b) then leads to the following systenm f equations for node Sustitutig Eq. (lbb) into Eq. (20b), forces in tensis of applied edge stresses: r-y3 0 -(X - X = [T] S] {} (20c) IV 0 1 (' 2- X3) -X I | arx Comparing this last equation with Eq. (3) shows that F t Ya 0 — x3 Y 21 0 -X 3 Y3 [K] = [T] [S] (21) F., 0 0 x: 7,i 1, -, 0( x2 l Carrying out the indicated matrix multiplication and (20a) puttitng A: = (1 + v)/2 gives y3 + Xx2321 X2 X2Y3 X2X32 ~ X232 1XY3 X2 X2Y3 x2 _ + XIX3X23 VX32 X13 y3 1X32 [K] = l x2 X2Y3 X2 X2 X2 X2Y3 I,,; 2(1 - v2) X AX32 X23 XIY3 X3 X32 Xy (triauglc). a a x-+ — y X2 X2 X X23 X2 X2 X2Y3 X2 XIX23 X1xs AXX2 -- --- — h --- X hY3 Y3 Y3 X23 xa X2 —, -- - - - 0-' Y3 Y3 Y3 An alternative approach to the above miethod for to spar, rib, etc., stiffilesses which are also given for calculating the plate stiffness mIatrix is to calculate the specilield nodal poilts. IIowever, til ) plate 1node strain energy in the plate due to the assumed strain forces are statically equivalent to certain plate edge distribultionl au(l to telie alpply Castigliau's'Theoreli stresses. Furtlhermore, these edge stresses will tend to for finding the node forces. This procedure can also approach actual edge stresses, even of a complex nature, be conveniently carried out in terims of mlatrix oper- if sullicielt sublelelicnts are used. A result of these ations; details will not be included here, however, since equivalent edge stresses is that continuity will tend to tlhe result is the s:iae as that alreadly ()btain(ed. be alproximately Ilailltailled alonlg Ceoiimll ed(ges of Stiffness matrices for plates having four and more subelements, between nodes. In other words, we are nodes have been derived and studied. The advantage assuming that a plate utnder complex strains will deform in introducing additional nodes lies in the fact that a in a manner that can be approximated by relatively more general strain expression may then be emlployed — simple strains acting on subeleiments into which the or equivalently additional load states as illustrated by larger plate has been divided. The accuracy of this Fig. 7 may be used for the plate. As a result a choice representation should increase as thle nlnber of subbetween two points of view may be adopted; first, the elements increases simplest or triangular plate stiffness lmatrix imay he usel and the desired accuracy obtained by using a sufficient (3) Quadrilateral Plates number of subelements, or second, a more general plate In tile analysis of wings and tail surfaces it is generally stiffness matrix may be used with fewer subelements. convenient to emnploy a subdivision of cover plates Experience to date indicates that satisfactory results such that most elements are of quadrilateral shape. can be obtained using the triangular plate stiffness'he stiffness matrix for such elements can then be dematrix. rived in one of two ways: (a) the previous solution Some additional plate stiffness matrices are given demonstrated for the triangle can be extended to inin Appendix (B). elude the quadrilateral and (b) the quadrilateral can'o summarize briefly the meaning and significance be subdivided into triangles and its stiffness matrix of the plate stiffness matrix, it is first pointed, out that determined by superposition of the stiffnesses of the this matrix relates node forces to node displacements. individual triangles. In this section the latter proAs a result the plate stiffness can be imlmediately addedl cedure will be adopted. 22 (?

STIFFNESS AND DEFLECTION ANALYSIS Two simplle subdivisionls of the quadrilateral into. triangles are shown in Figs. 10(a) and 10(b). These 3 r _it 3j t lead to different stiffness matrices for the quadrilateral. ^ -xt A tnique result is obtained by using the subelements / < shown in Fig. 10(c). The interior node will be located at the centroid, although any other choice could be used. I rx 2 Xzt For the general quadrilateral plate it has proved to be preferable to program the calculation of the stiffness (a) (b) matrix'for higl-speed comiputing equillnent. In the FIG. 9. Shear loading on triangular plate element. case of the rectangle, however, an explicit derivation call be readily carried out.'he necessary calculations, included below, are given here, since the end result is useful and since these calculations serve to illustrate a step of some importance in carrying out the analysis 2 2 l, 2 of a more complete structure — for exapllle, a wing or / tail surface. / The rectangle and its four triangular subelenents, 3 with interior node nutnber 5 at the centroid, is shown 4 4 4 in Fig. 11. Stiffness mt.itrices for the triangles can be calculated from Eq. (22), or more conveniently from (Q) (b) () Eq. (B-3) of Appelndix (B). In dleterniining K of the;. 10. l)ecomiposition of (liuarihttr;tl plate int( trialigultr rectangle, superposition in the following form is used: subelements. K = K, + K1 +l Kil t+ Kiv rectangle Since five nodes have been established, K for the rectangle will initially be of order 10 X 10. This will later be reduced to order 8 X 8 to give a result consistent with the choice of four external nodes; only at 44,Y4).3... s) these external nodes is contact implied with adjoining / m structure.'T'le immllediate toilnt is, however, that K for each triangle lmust be increased to order 10 X 10 S before superposition is carried out. This is accomn- [ plished in the usual way-that is, by introducing appro- I priate rows andl columns of zero elemclt!ts. X u 1n order to silnmlify the expressions for elements l(XliY) 2(2Jy appearing in the stiffness matrices the derivation of K tG. 11. Triangular subclements for rectangular llate. for the rectangle will be restricted to v = 1/3. On superimposing stiffnesses for the component triangles of Fig. 11 it becomes possible to express Eq. (3) in the form (- 100"< F., U2 —1., u[4 140" —— x.u --'X (UNIFORM) FI [ 8x8 |Bx2 vi (23) (23)\ FY, VI 1', Utu F; J t V6 Since forces are to be applied to the rectangle by stresses t = 0.050 IN. equivalent to forces acting at nlodes 1, 2, 3, and 4, the E = 10.5 X 10 PSI condition t) 1/3 F7,t P= F, = 0 TOTAL LOAD - 2 LBS. con be applied to Eq. (23). Doing this results in the 12. Clamped rectangular plate subjected to uniform tensile two sets of equations written below: loading. 23

JOURNAL OF THE AERONAUTICAL SCIENCES-SEPTEMBER, 1956 Solving Eq. (24b) for displacements at node 5 and sub] ~~~~Fz, j4 | ut~ |stituting the result into Eq. (24a), (FvU3 [1 Us (24a) 1 [B [C] 1 [B[')v } (25) [A I + [B \ (24a) FI;,V2 P where i = 1, 2, 3, 4. Comparing Eq. (25) with 1Eq1. (3) 1,Y~ V3 gives F,, V4 I /P [ K] = [A - [B] [C]-t [B]' (26) rectangle UI u2 Carrying out the calculations required by Eq. (26) reU3 suits in the following rectangular plate stiffness matrix: 0J U4 V )' where, when m = (x2 - xl)/(y - y), Ul U2 U3 U4 U, U2 U3 UI 9 1 1 3m + C-1 9 9 1 m — 3m +- 1 1 -1 K, = 4 m-m +m + 1 (27b) 3 3 9 3m + - m --- - 3m + 3m. + m -1 1-1 m in m,3 3 9 9- I - h -:m + -- -m- "I-:3Mr+ - 1 - n 4- win m - J V1 V2 V3 V4 VI V2 V3 V4 m + 1 1 m -!)Oi + 1 1 -1 m in + 4 -3 -- - - m - m — 1 1 - 1 -9m +- -3m -- 3m — )n + m m n mn VI V2 V3 V4 placements are to be retained in a wing analysis. In 11 ] " this latter problem it then becomes necessary to eliilliKhtz = 0 0 -1 nuate all u and v components of displacement. The -1 0) 1 (27d) procedure for doing this is the saime as that used inl -K29,= K' (27) gi rectangular plate. A;LL == kh^'I (7c-) If the order of v-terms in the above equations are re- (4) Example arranged from vl, v2, v3, V4 to VI, V4 V3, V2, it will be dis- It is of interest to carry out calculations on a simple covered that K2l equals Kt provided we replace m in example and compare results obtained by applying the K'i everywhere by 1/m. The corresponding form for plate stiffness matrix with values that can be regarded Ku2 may be written without difficulty. It is again as correct. pointed out that the above plate stiffness matrix ii For this purpose the plate of Fig. 12 is analyzed using based on v = 1/3. several different methods. Deflections at several points Tlie process of eliminating displacements at node 5 due to the indicated loading will be calculated. Since is simllilar to the situation tlhat arises when only w dis- azl exact solution is not available, correct (disl)lacemelluets 24 \

S TIFFNESS AND D E FL E C T ON ANALYSIS TABLE 2 Solution u& u2 u U4 us5 VI VU V4 No. Method Fig. Multiply all values by 10-6 RRelaxation 13 2.703 2.607 2.703 1.391 1.218 0.0 686 -0. 85 0.562 Simple theory 13 2.721 2.721 2.721 1.300 1.360 (0.35 -0.635 Plate K-matrix 13a 2.595 2.595 0.740 -0.740 Plate K-matrix 13b 2.692 2.578 2.692 1.355 1.199 0.680 -0.680 0.568 $ Plate K-matrix 13c 2.71 2 697 0.68(i -0.717 i6 P late K-matrix 1id 271.1 2 712 0.688 -(.691 will be taken as those calculated by applying the re- Each subquadrilateral was considered as consisting of laxation method to the fun(damental (equations govern- four triangles in a manner analogous to t he treatment ing! this problem. Although details of these calcula- described previously for the rectangle of Fig. 11. In tions arc not presenlted, results are listedl in Table 2. Solution No. 5 we note that Ui anld u3 are not equal, a The problem is interesting for at least two reasons. consequence of the random nature of orientation of the First, the accuracy obtainable using various numbers subelemcnts. By increasing the number of random of subelements can be observed, and second, the effect subelemcnts as in Solution No. 6, this lack of symmetry of using random orientation of subelements-with in results is virtually removed. Comparison with respect to the plate edges-can be observed. relaxation values is seen to be very good for both SoluResults of all calculations are sulmmarized in Table 2. tions 5 and 6. Node locations and subelements are illustrated in Fig. A more comprehensive example is given in the next 1.3. section of the l)ap)er. In'able 2 the solution based on simple theory was obtained from u = PL/AE and e; = -v E,. It is observedl that on this basis both tu alnd vt agree quite (X1) ANALYSIS o Box BAM well with the relaxation solution. well- wt the relaxatin l n. As a final example, the box beam of Fig. 14 will be The crudest plate matrix solution is listed in Table 2 analyzed for deflections, using the stiffness matrices as Solution No. 3. It was obtained by considering the previously derived. plate as a single element whose stiffness is given by Eq. (27). The results for u, and vl are seen to be The box is'uniform in section, unswept, and contains reasonably good. Solution No. 4 considers the pllte a rib at tle ulsu)lorted en(. The following dillnas consisting of four rectangular subelements as shown sionsapply: /b =, 2b/h 10 t = 3 t- = 0.05 in Fig. 13(b). Again the stiffness natrix was obtainedl i., A, = b/2, a 400 i. hy using Eq. (27), this time for each subelement. As the simplest possible breakdown, we consider the Agreement with relaxation results is seen to be satis- box to consist of two spars, one rib, and two cover factory, particularly in regard to ul. Also the dif- skins. The nodes are then as shown in Fig. 15. Forces ferences between tin and u1 are approximated accu- lay be applied at the nodes at the free end. Two rately by this solution. It is to be remembered that cases will be investigated: (1) up loads at each spar the actual straitl distribution ill the plate is compllex (bending) and (2) ul) loadl 1 one spar and a down in nature. load at the other spar (twisting). Solutions 5 and 6 in Table 2 were carried out ill it The spar lmatrix is given by E1. (I Ia). Calculation matter of minutes on a high-speed digital computer. shows it to be uI or U2 wi or w'2 Ut or U4 W3 or w4 t1.13903 l = Altj 0.:052.27 0.0:333 (28) spar 2 0.50303 (.05227 1.13903 -0.05227 -) )0( 3: -0.05227 0.00333 Cover plate stiffness is given by Eq. (27a1) and for this clse beoiies Ul Vl U2' Z'. 1U3 V3 14 Vt4 0.90878 -0.37500 1.39778 -0.19329 0 0.90879 [,K] | O - 1.15928 0.3 7.500 1.3()77 S (29) cover 2 -0.31916 0 -0.39634 0.37500 0.90879 |latc 0 0.37109 -0.37500 - 0.60959. 3750 1.39778 -0.39634 0.37500 -0.31916 0 -0.19329 0 0.90879 L 0.:7500 -04.(09.}59 0 0.37109 0 -1. 15928 -0.37500 1.39778 25

JOURNAL OF THE AERONAUTICAL SCIENCES-SE PTEMBER, 19 5 (; The rib has not been defined as yet. Two possible rib configurations will be analyzed in this paper. In the first case, the rib is considered as a beam identical in section to the spar, This leads to the following stiffness matrix for the rib: /ess fOanges. Vi Wl V1 W2 0.13086 [K] = t -0.00976 0.00098 rib 2 0.06413 -0.00976 0.13086 (0a) 0.00976 -0.00098 0.00976 0.00098. In the second case, the rib is treated as a flat plate. The general stiffness matrix which has been derived for a rectangular flat plate is of order 8 X 8. However, in the present instance, the following conditions must be introduced ti insure compatibility with tlle other portions of tlle structure (see Fig. 15 for subscript locations): wlW = Wt v1 = -vrs Using the same technique as described for the simple W2 s W2, 2 = -v2, truss, it is now a straightforward matter to form the stiffness matrix for the complete box. Advantage can an, likewise, for tile forces be taken of the following: (1) structural symmetry F,, =F F,, = -Fy, that exists for the box with resl)ect to the xy-mlli(llale and and (2) restriction in this problem to loads that act FZ, = F,,, F, = -,,' nlormal to this plane. Inder these conditions each l 0.00 i) ad a y- pair of upper and lower surface nlodes will experience,''Treatinlg the rib as a flat plate (t = 0.0(50 in.) and apply-.. t. i.ng... a latplin addition to equal vertical deflections, equal but ing the above conditions leads to the following rib stiffngessthe maboverix: c i e topposite displacements with respect to the xy-midplane. stiffness matrix: In other words, the box will deflect in the sense of a Vt W1 V W2 conventional beam. The spar and rib stiffness ma~5 65.650~88 1~ ~trices already provide for such elastic behavior. The E -0.37500 0.03754 )plte stiffness matrices mlake no distinction, other tlan [K] =2 1 84181 -0 37500 5 6508 in the sign of the node forces, for a reversal in direction 0.37500 -0.03754 0.37500 0.0 4 of nlode displacemnlt. Consequently, if tile normal loading is carried equally by upper and lower nodes, (30b) only the upper set will need be considered when forming It is anticipated that the choice of rib will have little the box stiffness matrix. Due to the division of loadeffect on deflections due to tile bending-type loading ing, correct deflections will result. In this mannler and a more pronounced effect on the twisting-type the stiffness matrix for the box is found to be [Eq. loading. (30a) used for rib stiffness ] Uj vI WI U2 V2 W2 2.0.17S82 -0.37500 1.52864 Et -t0.05227 -0.00976 0.00430 [K Lj l (31) 0 -1.09515 -0.00976 0.37500 1.52864 0 0.00976 -0.00098 -+0.05227 0.00976 0.00430 The inverse of this matrix is the flexibility matrix. F,, F,, F,, F,, F,, 0.81646 0.22705 1.66224 2 -10.47344 2.72965 409.39998 [K]-' ='C] E -=.- (32) box = Et 0.20384 -0.08123 -5.55027 0.8i646 ( 0.08123: 1.26026 5.01982 -0.22705 1.66224 -5. 55027 -5.01982 142.67751 -10.4.7344 -2.72965 409.39998 From the flexibility matrix, deflections due to applied loads can be found at once. For the two cases of applied loadings we find the following (rib treated as beam). 26

STIFFNESS AND DEFLECTION ANALYSIS Case 1 (bending): Forces of 1 lb. acting upward at each spar (nodes 1 and 2). W= - 11,041.55/E ul = -:320.47/E vi = -45.80/1 w2 = 11,041.55/1 u = — 320.47/E v2 = 45. 80/F Case 2 (twisting): Force of 1 lb. upward at node 1 and 1 lb. downward at node 2. wt = 5,334.45/FE u = -98.46/ Vi - 154.99/E W2 -5,334.45/E U2 = 98.46/E V2 - 154.99/E Similar results may be calculated for the case when plates. It can therefore be felt that this node pattern the rib is assumed as a plate. Complete details are will give final results which represent convergence of not given. In bending we get w, =- 10,88. 12/I/, the method. As Illentione(ld previously, this is substallu\ = -310.56/E, and vi = -18.25/E. Twisting tiated by comparison with values obtained from Fig. results are wl = 3615.72/E, ul = -25.84/E, and vi = 16(b). 349.52/E. There remains the question as to what is the correct,,...., value for w, for this problem. IElemelltary betlla It is now advisable to select additional nodes and v e fr w fr t ro I theory gives w =- 6,900/E, and, if extended to include recalculate the previous deflection data. When added, ad, i e t shear distortion of spar webs, gives w, = 7,740/E. nodes have little effect on results, the process can be consde t v. W e Using Reissner's shear lag theory,13 the tip deflection is considered to have converged. Whether convergence.,. i obtained as w, = 7,900/1. Finially if Reissner's shear be to the correct values requires additional information.. lag theory is modified to include spar shear web deThese questions are now examined. Ts q n ae nw formation, the result is w, = 8,740/E. This is the First, solutions are found for the node patterns. - Fw ir st,. suo ae -.found,. fote node pa tt ern ll most accurate theory available. It agrees to approxishown in Fig. 16. Vertical deflections at node 1 for m ately 2 per cent with the numerical solution based on bending-type loading are as follows: e stiffness matrices. Fig. 16(a) w, = 8558.0/E The pronounced shear lag effect in this problem and its marked influence on the vertical tip deflection are Fig. 16(b) wi = 8$591.2/E significant. It is precisely this effect that produces a Fig. 16(c) w, = 8548.4/E very complex stress distribution in the cover skins. Nevertheless the plate stiffness matrix developed in It is seen that the change in w, in goiilg from the node Eq. (27a) and based on triangular subelements reprepattern of Fig. 16(b) to 16(c) is about 1/2 per cent. sents this stress pattern with gratifying effectiveness. Consequently convergence canl be::ssumllle to have'The solution for tile node pattern of Fig. 16(c) was been attained with the solution found from Fig. 16(b). obtained in a few minutes by utilizing a program for a Obviously the first solution, based on Fig. 15, is in high-speed digital computer that computed individual considerable error. This is due to the poor tie between plate and spar stiffnesses and then combined these spars andc cover plate. Fig. 16(a() introdtrucs ai.atdd(i- ilto th e stiffness matrix for the complete box. tional tie between these two compo nents. The decreased value of w1 for this case therefore reflects the (XII) RRDnDUCTION IN ORDIKR OF ST'rIlPtNSS MATRIX added stiffness due to illcluding tile two nodes at the mid-span location. (1) Eliminating Components of Node Displacement An unexpected result is the close agreement between In an actual problem-as a wing analysis-the numthe solutions based on Figs. 16(a) and 16(h). In fact l)er of nol(ls1 to be lused( cant bcoJII (llite large. If, for it would seenm reasonablle to exlpect Fig. 1(i(b) to lead purposes of discussion, 50 nodes are assumed, the stiffto a smaller value for wl than that given by Fig. 16(a). ness matrix becomes of order 150 X 150. By elimiCareful scrutiny, howeve, indicates that these results nating i and v components of displacement at each node, are quite reasonable. Whereas the node pattern of the stiffness lmatrix can be reduced to order 50 X 50. Fig. 16(b) accounts for slear lag in the cover plate, this However, this reduction process [see treatment of Eq. is not the case with Fig. 16(a). As a result, the added (23), for example can require the calculation of tle stiffness ill Fig. 16(b), (ue to tole additional todes inverse of a 100 X 100 matrix. Such calculations are connecting spars and cover skins, is offset by the best avoided at present. added flexibility introduced by sliear lag in cover skins. The problem that arises in eliminating the u and v The results indicate these factors to be nearly equal; components can be handled satisfactorily in any one herce the reason for the nearly correct values given by of several ways. First, the calculation of the inverse Fig. 16(a). of a large-order matrix can be avoided by eliminating a Fig. 16(c) allows for shear lag and, at the same time, single component at a time. This is a practical exprovides for adequate tie between spars and cover lIedient when autolmatic dligital copll)utinlg equipmllent 27.4

JOURNAL OF THE AERONAUTICAL SCIENCES-SEPTEMBER, 195 g^~50_ method leads to the highest frequency and correspond-50| ing mode. If the order of the stiffness matrix is high ll..1 1.. (say, 50 X 50), it becomes impractical to eliminate Tsuccessively the higher mllodes and so eventually obtain 70 I 1I. the lowest modes. Inversion of the stiffness matrix leads to the flexi2' rbility matrix. This matrix used in the matrix iteration m[[ II procedure yields results for the lowest mode. There-,>df i|> ___~ ~fore, it is ordinarily preferable to know the flexibility...3.....6 3 matrix. If tile stiffness matrix is of high order (say, 50 X (aQ (b) 50), inverting it becomes a major problem in itself. This can be overcome to some extent by employing the ^~ ~65-~ 1~~ 41 ccapabilities of present-day digital computing equipT- ^x > 0 -rlc r >-, 9_, meint. H-owever, in many instances an alternative 56. 1 / 11^. /I procedure may either be useful or necessary. ConseL5 j76./2 / It... quently, a possible approach to overconing this difli^ / | 1 ^- g / I culty will be outlined here. t —--- > — ^ V -t 1lThe prolosed metlhod colsists of converting the m]l \ IT -, - ^ original stiffness matrix K into a lower order stiffness,y I { >. V tl iatrix K*. This is accomplished by introducing a set 75-| 3 C 3 of generalized coordinates which are related to the original displacellents ((on which K is based) tllrough a set of appropriately chosen functions. The accuracy (C) (d) inherent in K will have a direct influence on K*. FIG. 13. Nodes and supports for clamped rectangular plate. Suppose K is known for the cantilever beam of Fig. 17. The order of K is 10 X 10. Now assume a set of polynomials of the form is available. Second, in some cases it may be feasible to eliminate "blocks" of u and v components at a time, - thereby reducing the order of mlatrices to Ie' inverted at any one time to a reasonable size (say 20 X 20).'Third, the analysis can be carried out for sections of the structure, taken one by one. For each section, as / a spanwisi portion of the wing, the complete stiffness matrix can be determined. Elimination of u and v COVER components can then be carried out at any selected *,X SKIN-tC 0.O5" nodes, except those commont to two distinct sections of the structure. Each section can he treated in this manner. Bl y lprolerly adlding tile indlividltual sectionl //h,SPAR WEBt 0.05" stiffness matrices, the total stiffness nlatrix can be ob- 25.4 RIB tained. lFi;nally u 1and v displacelents at nodes where t A -0 the sections join together can be eliminated. The stiff- [- 2bu254.6" —- A. ness matrix that remains will apply to w deflections 6.365 SIN. op.iy. From a practical standloint, the method just de- F1i. 14. C;tltilevercd boxx beam. scribed has several worth-while features. For example all components of dlisplacemlellt at a given ode 4 may be eliminated. This can be useful when additiolal Iod(les are felt to, be necess;ry in or(ler to account / properly for regions of maxilnum structural com- 4 plexity. Even though eventuallly eliminated, these / / nodes will have contributed to the elements retained / in the stiffness matrix. / / (2) Inversion of Stiffness Matrix / Ordinarily, only the first few low-order vibration 2 nodes and frequencies are required for the purpose of h_ carrying out subsequent dynamic analyses. Using 1 2 x the stiffness mlatrix directly in the matrix iteration FIG. 15. Simplest node pattern for box beam. 28 3)

STIFFNESS AND DEFLECTION ANALYSIS Pil(x) = (s1X2 + b1X3 + CIx4 24 3 4 5.. 4 P2(x) = a2x2 + b2x' + C2X n P6(x) = aEx2 + b x + cs"xj - - Each of these will be made to satisfy the boundary conditions of the cantilever which are, 1 2 1 2 1I 2 P,(O) = P/'(0) = PJ(L) = P,"'(L) = 0 (a) (b) (c) Applying these conditions results in FIG. 16. Additional node patterns for box beam. Pi(x) = 6(x/L)2 - 4(x/L)3 + (x/L)4.l P2(x) - 20(x/L)2 - 10(x/L)3 + (x/L)5 (34) P,(x) = 140(x/L)2 - 56(x/L)3 + (x/L)8 _ A -I We now introduce generalized coordinates qj which are related to the (lisplactlelnts y, througll the above polynoinials. This relationship is established through the _ 10 EQUAL PARTS - (,llatios @ L/j0 Vy r P,(xi) P2(x,).. Ps(xi) F I;. 17. Statioa seklctiions,11 cantilever ba.in:. yI /'I (x) /'z(.X ).. I/b(X,) q1 determined, starting with the highest. This is feasible q2 if K* is of sufficiently low order (say, 10 X 10). _ = Q4 (35T)) This process can be modified in several respects, and q4 the purpose here is not to give an exhaustive treatment q but rather to simply point out a possible approach to Io*~~~~~ I I the problem. Preliminary calculations indicate that 310 J LPi(XI) I1(xQ). 0) J the idea Inay possess practical value. Extension to a two-dimenlsional grid caln Ib made by generalizing the It is seen that the ten displacements yl, y,..., yo are lrocedure suggested above. to be replaced by the five coordinates ql, q2, ~., qs.'I'le free vilrationl iroblenl forr the catiilever cam I) APPENDIX (A) set up in terms of kinetic and potential energies. In termts of original (lisplacelnents y,, y2,...y,, these DERIVATION OF SPAR STIFFNESS MATRIX energies are, resplectively,''The structure and notation are described in Section l7 (1/2)' A v' \ All tj l )al (IX) anId Fig. 1. V = (1/2) {y}' K y} ([ ) Flantges are assumi(ed to carry axial stresses, while the web carries shear stresses. Cover plate material is not where [A.lJ is thie i!!ertia;t (Imass) umatfri.x ad(l [Aj tw included as;lart of spar flanges. J)erivation below is original 1. X 1o stiffness matrix. based on conventional beam theory. Writing (Eq. (135) as ty} -I [P] } Casel and substituting into Eqs. (36), U1 = - ul' ~ 0; al other collents of ode s7' = 1/2) j' [1' IAI.t placement for the beamni 0. 1 -= (1/2) {}' [1\' [AIJ [Pi [\'} d c'llThe lellecteld beaml ald neccssary forcesand reacV - (1/2) ( P I' [P1' [K] [PJ {</ ttions are shown In Fig. A-i. Due to forces F, at the left ell!,- tl cl)eaII (Icl'ct'Cs Ul)pwar.''lTh /, forces fromll which we define cause a downward deflection. Beaml theory, including I[*] = [P [PP' [K [PI effects of uniformly distributed shear in web, gives [Ka* = [P]'i [K ] r[PI' (37) [M*] = [P]' (M] [P](37) F,,hL2 2F,,L: w= + - (l + n) (A-l) If K is of order of 10 X 10 and P of order 10 X 5, K* 2EL 3E1' will be of order 5 X 5. The vibration analysis is now hL 2 performed using K* and M*. By inverting A'* the 0 = + (A-2) lower modes can be calculated directly. Or alternatively, K* can be used and all modes and frequencies where w and 0 are deflection and slope at the left end of 29.^0

JOURNAL OF T'1 I 1A A RON U 1 AUTICAL S CI C S - S E P T E M B E R, 195( U Y, il a similar manner. When w, = wl, I 0, while all!1F-^ F other nodes are held fixed, the forces of Fig. A-2 apply. F -__iji_* _ ___1_~ Fz L_ Forces due to displacements imposed on the rightFxX!~ 2 y "' --:-... 27 hand end of the beam may be written froil the above -" ~ XJ U xF results by analogy.'he final spar stiffness matrix f'Xrc=, - I FI isgiven as Eq. ( ll). F^ F}' APPENDIX (B) F;G. A-i. First beam displacement required in developing beam stiffness matrix. PLATE STIFFNESS MATRICES Several plate stiffness matrices are given here withtFel F7)z out derivation. F -F................2 (1) Triangle-Arbitrary Node Locations F.. —.,_. —-- 1_. VS )2' X3) Y3 FIG. A-2. Second beam displacement required in developing beam stiffness mat rix. the beam, respectively, and n is given by Eq. (1lb). D)ue I' I1)tIIlI(l;iy conl(itioI)ns, w -.); ails(, fro(n tihe geometry of the deflected beam, 0 2= 2u/h. Using these relations in Eqs. (A-l) and (A-2) and solving for I1' Y1 forces gives 8EI I + n 6E I 4 = /2J, + = 2(j + ) 1T (l + n) u ~ (A-3) FI''. l-I. T1- riangular plate element with arbitrary node locations. oh!/ *L.~1 =- - -- -'I'A4) hec stiffness matrix will be defined with respect to hL2 + - 4n =/(l + (A-4)l the erti;ltioll Forces at node 2 follow from equilibrium considerations. F|, u, They are - vs 4EI 1'- 2n 6EI 2 FI, = [K] u2 (B-1). 4= t - (,I -- 22n F 2, + 4n uL (i-+ 1n),( - 2n)? I,) I F,,, v:3 (A-5) 1 % v; F,, - F, (A-6) Again adopting the notation'I'le above forces relpreselt the first column of the re- x^i = - x s, X = (1 - v)/2, X2 = (1 + v)/2 quired stiffness matrix. The other columns are found (B-2) we get XIX2~32 + Y23J X2X32y23 X232 4- Xly232 Et XlX23X31 + 23Y3aY' XX13Y23 + VX32Y31 XlX312 + Y312 [K] =. 0 2 XIX3h2y3l + VX13y23 X2X31 +,tIY23Sy31 XIXIyI X312 + XiY3l2 XiXl + Y2 XY22X233 + V2X32Y12 IX12X^31 + YISVY XIX2131 + VX13Y12 1X122 2 Y1i2 L Xlx32Y12 + vX21y23 X12Xa2 + 2,yl1223 XIXy2 12+ XYI X2X1+ X X 2y22 y22 + X2X2y22 X122 + X (B-3) rere,= 1/(1 - 2) Yhere + x13y2 + xy:YI1 30

STIFFNESS AND DEFLECTION ANALYSIS (2) Rectangle The stiffness matrix given below for the rectangle is l v. Vbased on the load states shown in Fig. 7. As a result this matrix is more general than that given in Eq. (27) due to the inclusion of linear terms in the strain expres^~~~~~~~~~~~I | ssions. __ _ __ 3 Again the stiffness matrix is arranged to agree with the equation 4 - ^Tn fl = al | b =tr F:YI~Fz, jU2 ( _______X.oU 1 F;' ^ = [K] U2 (B-4) --— _ ^*^ F*, Us t' b 2'. FrG. B-2., Node locations for rectangular plate element. |Fr U4V in which [K] is given by U1 VI U2 V2 Us8 V3 U4 V4 ai + bi I + v a2 + b2 al - bl 1 - 3v al + bl K 1E 3 ^ I c2 — a2 - — 1 - - a2 a2 (B-5 8(1 2) -a - -1 - v - a 1 - 3v a, + i5 -1 - - — a2 - c2 3v - 1 a2-b2 1 + a2+ b2 c1 - a1 3v - 1 -ai - ci 1 + a, - b 1 - 3v a + bi 1 - 3v a2 - b2 1 + V -a2- C2 3v - I c - a2 -1 - a2 + b where, in the above matrix, a,= (i - v), bi = (2/3m) (4 - a), ci (2/3m) (2 + v2) ( a2 = (1 - )/, b = (2m/3) (4 -'), C2 = (2m/3) (2 + v) ( m = I/h (see Fig. 7) (-B-7) Eq. (B-5) simplifies to the following if v = 1/3: U1 Vl U'2 V2 U3 V3 U4 V4 -p,(m) 18 Vl(l//n) 402(m) 0 I,(m) f Et |0,03(1/m) -18,o,(1/rm) [K] 96 o4() -18!3(m) 0 ( (m8) -18 4(I/m) 0 o02(1/m) 18 pI0(1/m) V3(m) 0 W4r(m) 18 I2(m) 0 iPl(m) 0 P2(1l/m) 18 o40(/) 0 3(1/m) -18 V,0(1/m) lhere i,(m) = 9m + (35/t), 41i(l/m) = (9/m) + 35m 02(m) = 9m - (35/m), P2(1/m) = (9/m) - 35m v43(m) = — 9m + (19/m), fa3(l/1n) = (-9/rn) + 19n Vo4(m) = -gm - (19/m), 04(1/m) = (-9/m)- 19n (3) Other Shapes tcrmined by following the basic ideas developed in this Although the parallelogram and arbitrary quadri- paper. lateral can be treated in a manner similar to that used for the rectangle, the individual elements in [K] tend to become unwieldy. For that reason use of automatic EFBRENCBS digstal computing equipment is considered to offer the I Schuerch, H. U., Structural Analysis of Swept, Low Aspect practical means for obtaining stiffnesses of such plates. Ratio, Multispar Aircraft Wings, Aeronautical Engineering RePrograms for carrying out such calculations can be de- view, Vol. 11, No. 11, p. 34, November, 1952. 31

Levy, S., Computation of Influence Coefficients for Aircraft Reissner, E., and Stein, M., Torsion and Transverse Bending Structures with Discontinuities and Sweepback, Journal of the of Cantilever Plates, NACA TN 2369, 1951. Aeronautical Sciences, Vol. 14, No. 10, p. 547, October, 1947. 9 Benscoter, S., and MacNeal, R., Equivalent Plate Theoryfor 8 Lang, A. L., and Bisplinghoff, R. L., Some Results of Swept- a Straight Multicell Wing, NACA TN 2786 1952. back Wing Structural Studies, Journal of the Aeronautical Sciences, 10 Levy, S., Structural Analysis and Influence Coefficients for Vol. 18, No. 11, p. 705, November, 1951. Delta Wings, Journal of the Aeronautical Sciences, Vol. 20, No. 4 Langefors, B., Analysis of Elastic Structures by Matrix Trans- 7, p. 449, July, 1953. formation with Special Regard to Semimonocoque Structures, Jour- Schuerch, H. U., Delta Wing Design Analysis, Paper prenal of the Aeronautical Sciences, Vol. 19, No. 8, p. 451, July, 1952. sented at SAB National Aeronautic Meeting, Los Angeles, 9Rand, T., An Approximate Method for the Calculation of September 29-October 3, 1953, Preprint No. 141. Stresses in Sweptback Wings, Journal of the Aeronautical Sciences, 1i Hrennikoff, A., Solution of Problems of Elasticity by the Vol. 18, No. 1, p. 61, January, 1951. Framework Method, Journal of Applied Mechanics, Vol. 8, No. 4, " Welle, I. B., uadl Lansing, W., A Method for Reducing the December, 1941. Analysis of Complex Redundant Structures to a Routine Procedure, " Hemp, W. S., On the Application of Oblique Coordinates to Journal of the Aeronautical Sciences, Vol. 19, No. 10, p. 677, Problems of Plane Elasticity and Swept Back Wings, Report No. October, 1952. 31, The College of Aeronautics, Cranfield, England, 1950. Fung, Y. C., Bending of Thin Elastic Plates of Variable l "Reissner, E., Analysis of Shear Lag in Box Beams by the Thickness, Journal of the Aeronautical Sciences, Vol. 20, No. 7, Principle of Minimum Potential Energy, Quart. Appl. Math., p. 455, July, 1953. Vol. IV, No. 3, October, 1946. 32

INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN ENGINEERING, VOL. 14, 1643-1651 (1979) DIFFERENCE SCHEMES OR ELEMENT SCHEMES?t JOHN H. CUSHMANt Department of Agronomy, Purdue University, W. Lafayette, Indiana, U.S.A. SUMMARY Several examples are presented to illustrate how standard finite difference schemes for the wave equation (e.g. Lax-Wendroff, Leapfrog, etc.) can be developed from finite element analysis. The development of the difference schemes from the element schemes is made possible by using Galerkin's method on both the spacial and temporal dimensions. INTRODUCTION There is general disagreement about the advantages or disadvantages of using a finite difference technique in preference to a finite element method and vice versa. It is generally felt, however, that for irregular domains finite element analysis is often easier to use, and that for regular domains finite difference methods are more easily programmed. The purpose of this paper is to illustrate that when regular meshes are used for the wave equation (we consider only the one-dimensional problems, although multi-dimensional problems can be handled similarly), it is possible to generate the standard finite difference schemes as special cases of finite element schemes. The tool that makes the analysis possible was developed in Reference 7. The idea is simply to apply the finite element discretization process to time as well as to space. This technique has rarely been used because of the lack of theoretical results on stability and convergence. There is also a general belief that there is little to be gained by the method. Another reason people rarely use the technique is that it appears that the storage and computational requirements are increased. These reasons for not using finite elements in time will be dispelled in this paper. In the literature one can find very little concerning the use of finite elements on the temporal dimension, although there are exceptions.1"'9'10 However, the manner in which we subdivide time here will be quite novel. ANALYSIS Since the mathematical details of this method are straightforward we will, in general, overlook them. The finite element method is justified in any finite dimensional topological space and hence is valid when used on the temporal dimension.7 In this paper, time and space will be partitioned in some very unusual ways. Then our residue vector (coming from a trial solution) will be made orthogonal in the L2 inner product to a set of trial functions by the use of Galerkin's method.8 This minimization technique, when used properly on the temporal dimension, will produce the desired finite difference schemes. t Contribution from the Purdue Agri. Exp. Stn, West Lafayette, IN 47907, Journal Paper Number 7332. t Assistant Professor of Soil Physics. 0029-5981/79/11 14-1643$01.00 Received 12 December 1978 ~ 1979 by John Wiley & Sons, Ltd. Revised 13 February 1979 1643

1644 J. H. CUSHMAN The equation which we will examine is simply the one-dimensional wave equation u,+ cu = 0, Ox L (1) Let H((l) be the Sobolev space of order one and let i = X x T be such that X ={x:O<x<L} and T ={t: 0< t <' < o} Then if ueHl(fl) and N is our vector of linear interpolation functions (shape functions),8 Galerkin's method may be stated as f NT(u +cu) dA = (2) where I' c f. The stronger requirement of u e H2(fl) is sufficient if u is to satisfy the following form of the wave equation: Ut, = C2,X (3) After partitioning l' into elements and letting u = Ni (i being the vector of known (unknown) values of u at the nodes), equation (2) may be written as N, dAa=O (4) aA. ax) where the sum is over the number of elements and the elements may cover only a subset f' of 0. In the past, when Galerkin's method was used on time x space, the temporal dimension had been partitioned into layers of uniform height; 6'9'~ the solution then proceeded layer by layer (Figure 1). i -- < -- -- < -- -- ~ -- -- -- In+2 tt -- - - - - - - - n xi Z I i 1 i i x_ i- j++ Figure 1. Marching of elements in time In two instances6'9 the layers were allowed to overlap, producing what have been called lag elements. We will generalize and extend this technique much further. The most concise and easiest method to describe how we can construct duplicates of finite difference schemes via finite element analysis is by examples. Hence the bulk of the remainder of this paper will be so oriented.

DIFFERENCE SCHEMES OR ELEMENT SCHEMES? 1645 EXAM.PLES. EXPLICIT C.ASES ~oauupk.l;: unLstae Eulrscdiem The t: drc i7:e Z tZ -2 ~ W e z- taio k r- Erl- s e:rz 7T-e ihod is U U,;* -- U< u;_,-;U) r5b where j denotes the x-!ocation, n denotes the time step, and the Courant number v =* (cAr)/a x. We will next show how we can derive equation (5) via finite elements in time x space. Figure 2 represents the time-stepping scheme we will use. Note that only the shaded regions are used and that the solution proceeds level by level. The novelty to this approach is that we are using (for any one time level) only half of fl,\+ in our analysis. n+i t k l -*i n. o j-i. j j+t Ax-4 L x — Figure 2. The Euler method As can be seen, the elements we are using are linear and triangular. To derive the equations for the unknown nodal values (fi at the (n + 1st level) it is only necessary to consider one element since, for example, u +l (Figure 2) is only.dependent on u, and ui+1. Had we been using a procedure outlined in Reference 7 this would not be the case. We will thus confine our attention to one element. Integration of the element equation (see any finite element text) and neglecting known values of u (i.e. at nodes (j, n) and (j + 1, n)) the element stiffness matrix can be shown to have the form 0 0 0 0 0 0 k (c+c (cbi + c) (b ) (cbk+ Ck) where i, j, k denote the associated nodes (Figure 2), bi - TI- Tk, bi = Tk - T, bk= T m- 7T, ci = X - Xi, ci =X -Xk, Ck =X, -Xi, and (X, Ti) denotes the (x, t)-co-ordinate of node i. Note that there is only one non-zero row since there is only one unknown per element, and this is at the (n + l)st time level. With this in mind and substitution for the {hi} and {c,} one gets the following equation: (-cAt-Ax)u +cAtu'"+ +Axu7+' =0 (6) or, after rearranging, u+ =u?- v(ui"W l- u) (7) which is the desired result. Note that although it appears that equation (6) was derived from the element stiffness matrix, this is in fact equivalent to the global stiffness matrix equation (assuming that an interior element was used).

1646 J. H. CUSHMAN The reason we were able to obtain equation (7) is that a node at the (n + I)st level of the eth element was only directly dependent on the two nodes in the eth element at the nth level. Example 2: Upstream differencing The finite difference form is u+r =u P(u -U., ) (8) If one proceeds as in example 2 but with the elements in Figure 3, the required equation will be obtained. 4 ^<<- n+2 t <....... n_ x -- j- j j+l I- kx-4 Figure 3. Upstream differencing Example 3: Two-step Lax-Wendroff Step 1..,s+1/2 = +1 +, U, 9 U,+1/2 =- 2 —-—.(uI+ - ) (9) Step 2. U I n ( Ut+/ n+l/2 - ) (1 0 Uj 3 U VU, 41/2 -1/2 Figures 4, 5 and 6 represent the required finite element scheme. Figure 4 corresponds to the first required step. The elements have width Ax and height At/2. ~J +2 T1.i j+;.. Figure 4. Lax-Wcndroff, step 1 Again, as in the previous examples, it is only necessary to examine one element (e) and in particular the node located at (i +, n + ). The integrations will not be carried out here; they are, however, straightforward and can be verified by the reader. Step 2 is more interesting and is depicted in Figure 5. After completion of step 1 all nodes on the nth and (n + 2)th level have known u values. The only unknowns are at (n - 1 )st level. Let us examine node (j, n + ). This nodal value is dependent on both elements 1 and 2 (Figure 5) and thus both elements enter into the local stiffness matrix necessary to calculate u'. Integration of

DIFFERENCE SCHEMES OR ELEMENT SCHEMES? 1647 n+l i 1) (2) "n+~ i4_ i )+? Figure 5. Lax-Wendroff, step 2 the two element equations results in 1 2 4 Element 1: 1 0 0 0 2 0 0 /-Ax cAt Ax cAt 4 \-ct (-+ -Ct) A-x + Ct and 2 3 4 Element 2: 2 O 0 0 3 0 0 0 xcA\ Ax cAJ 4 ) Ct 2 + 2 2/ or upon combining elements 1 and 2 and dividing by Ax we get 1 2 3 4 2 O o o o 2 0 0 0 0 3 0 O O O0 4 -v -1 v I which, when multiplied by u, gives the required result. Figure 6 represents the way the scheme is marched through time. 2v A -,Zs. nf+Z n+I j-, j j +3 Figure 6. Lax-Wendroff marching sequence

1648 1. H. CUSHMAN Example 4: Leap frog The finite difference method for this equation is i"?+c -=0 (11) -l Ul +2C i+a = ( 2Ar 2Ax where initially two time levels must be specified. Figure 7 represents the required discretization to obtain equation (11). n+l n+il n+ n n J-I J j+I j-2 j- j j+u j+2 (a) (b) j-2 j-l j j+l j+2 (c) /,?\,, t,4,,, + Jo~ i ill~~n+ ( j j+d (d) Figure 7. Leap frog: (a) primary elements; (b) first sweep; (c) second sweep; (d) marching in time The primary elements are the same as in step 2 of example 3, but they are used in a different fashion and are larger-Figure 7(a). Since initially two time levels must have known nodal values there isonly one unknown associated with a pair of elements. Noticing that the elements have a width of 2Ax, we see that the method must be broken into two steps, of which both are at the same time level. Step 1(a) is depicted in Figure 7(b) and step 1(b) is depicted in Figure 7(c

DIFFERENCE SCHEMES OR ELEMENT SCHEMES? 1649 Note that the only difference in steps l(a) and 1(b) is the shift required to cover all nodes. The reader can verify that the required results are obtained. Figure 7(d) illustrates how the method proceeds in time. In using the elements in this fashion we obtain a more sophisticated form of lag elements than depicted in Reference 6, and thus are able to maintain an explicit scheme. Example 5: MacCormack method The MacCormack difference scheme can be written as follows: Predictor: u'( = u a - v(u+, -u) Euler scheme (12) Corrector: u -— [u +uT-v(uj — u;s)] (13) where the bar denotes a temporary solution to u. Example 1 illustrates how the predictor step is obtained from an element point of view. The corrector step is somewhat complicated and involves a special interpretation of u. Step 2 must be broken into two steps, steps 2(a) and 2(b), to obtain the required result. Step 2(a). This step should actually be carried out before step 1. Figure 8 illustrates step 2(a). Here we are using two layers of one-dimensional linear vertically oriented elements of length n+il n+~, — i —---,^ n j-( j j+@ j+2 Figure 8. MacCormack, step 2(a) At/2. Again, since at any particular node at the (n + )th level, the node is a function of two elements only (thus the node immediately below and above it), we must consider both associated elements. Integration of equation (4) over these two elements produces (upon combining the element matrices) the equation 1 2 3 I - o\u' o\ 1 2 -2 0 n +1 |0 2 -~ 1 i - u0l/2 O (14) 3 2 2 ^ \ol and thus n+1/2 =..... (r 1- 2 (15) Step 2(b). This step, illustrated in Figure 9, requires special interpretation, as will now be outlined. If u is defined as u -Nu,+12 + Niu,7 +Nku7

1650 J. H. CUSHMAN mOFA n+i and n+i n+~ n i-I ^ —------------ _ "_ - Figure 9. MacCormack, step 2(b), (n + 1) and (n + 1) overlay (where u+1/2 and u7-" are given by steps 2(b) and 1, respectively) for the integral NJT aN-NdA, ax and u is defined by u = Nu+"1/2 +Nju7 + NkU — for INTaNdA at then the elements in step 2(b) are independent. Performing the element integrations and after simplifying one gets the corrector step in the MacCormack schemes. This was possible only because we have interpreted u in a different and new fashion. IMPLICIT EXAMPLE Example 6: Time-centred implicit The finite difference scheme is given by, n + i =^ 7 ~.. ~ ( u 7 + I n U 7 + 1 + U U ) -n+l - - Us =U 2P(Ui- +'1 il + U.' -1 Although this is a one-step finite difference scheme it takes essentially two steps in finite elements. The first step is the same as step 2(a) in example 3, and is illustrated in Figure 10(b). It is necessary to use all elements (1 and 2 of Figure 10(b)) to determine u,, since u7 is known initially and u?"+l+l are known as functions of u, and u'0L. Integration over all elements will produce the required implicit scheme. Note, however, that step 2 needs to be carried out twice before the system is solved. This is necessary since the pairs of elements in step 2 have a combined width of 2Ax. CONCLUSIONS In the previous examples, for brevity, we have only sketched the procedures for developing the various difference schemes; the details are available upon request. We have not encountered a difference equation representing the wave equation for which we were unable to develop a corresponding finite element scheme. The examples presented are only representative of those we tried, and the only difficulties we encountered were on interpreting the interpolated function u for predictor-corrector schemes.

DIFFERENCE SCHEMES OR ELEMENT SCHEMES? 1651 n+l J-I J J+l (a) n+ j-2 j-i j j l j+2 (b) Figure 10. Time-centred implicit: (a) two layers of one-dimensional elements; (b) this layer of elements must be used twice-once shifted to get all nodes The examples presented show the power of using Galerkin's method on the temporal as well as the spatial dimensions in finite element analysis, and should dispel most of the negative attitudes toward using it. The examples presented serve to tie together the stability theory of finite differences to that of the finite elements. This implies a better understanding of finite element schemes. With the presented results one can go in many directions, and ask many questions. For example, can all difference schemes be interpreted as finite element schemes? If not, which cannot and why? What effect would using a curved temporal element have?, etc. REFERENCES 1. J, H. Argyris and D. W. Scharpf,'Finite elements in time and space', Nucl. Eng. Des. 10, 456-469 (1969). 2. J. C. Bruch and G. Zyvoloski,'A finite element weighted residual solution to one-dimensional field problems', Int. J. num. Meth. Engng, 6, 577-587 (1973). 3. J. C. Bruch and G. Zyvoloski,'Finite element solution of unsteady and unsaturated flow in porous media', in The Mathematics of Finite Elements and Applications (Ed. J. R. Whiteman), Academic Press, New York, 1973, pp. 201-211. 4. J. H. Cushman and D. Kirkham,'A two-dimensional linearized view of one-dimensional unsaturated-saturated flow', Water Resources Res. 14(2), 319-329 (1978). 5. 1. Fried,'Finite element and analysis of time-dependent phenomena', AIAA J. 7, 1170-1173 (1969). 6. W. G. Gray and G. F. Pinder,'Galerkin approximation of the time derivative in finite element analysis of groundwater flow', Water Resources Res. 10, 821-828 (1974). 7. J. T, Oden,'A general theory of finite elements. II. Application', Int. J. num. Meth. Engng, 1, 247-259 (1969). 8. J. T. Oden and J. N. Rcddy,'An introduction to the mathematical theory of finite elements', In Pure and Applied Mathematcs, Wiley-lnterscience, New York, 1976. 9. 0. C. Zienkiewicz, The Finite Element Method, 3rd edn, McGraw-Hill, London, 1977, Chap. 21. 10. 0. C. Zienkiewicz and C. S. Parekh,'Transient field problems-two and three dimensional analysis by isoparametric finite elements'. Int. J. num. Meth. Engng, 2, 61-71 (1970).

MATRIC~SES LS A 4 UvI bl App~blxa

MATRICES AND MATRIX EQUATIONS A. 1 Introduction Elementary matrix theory has become an essential tool for all engineers. To read this text, only an elementary knowledge of matrix definitions and operations is necessary. These definitions are summarized here to serve as a ready reference. A reader totally unfamiliar with matrices should consult one of the many excellent texts available, a few of which are listed at the end of this appendix. In addition, many books on solid mechanics and other engineering subjects have included chapters on matrix theory. A.,2 Matrix Definitions and Notation A matrix of order m x n is an array of quantities in m rows and n columns as follows: a11 12 In* a21 2... a2n [a] =: (A-l) a a... a _ ml m2 mn An element in the i-th row and j-th column is represented by the notation a... A square bracket will denote a matrix of any order. For convenience, we shall define some special cases of the general

matrix and use some special notation. A column matrix has m rows and 1 column, or order m x 1, and is denoted by {a}. For example, if al = 1, al = 3 and a3 = Z and the matrix is of order 3 x 1 and {a} 3 A row matrix has 1 row and n columns or order 1 x n, and is denoted by [a]. For example, [a] = [1 3 2] A square matrix has the same number of rows and columns, or m = n. For example, 1 3 2 [a] 2 13 7 6 4 where a matrix of order 3x 3 is shown. A diagonal matrix is a square matrix in which all the elements are zero except those on the principal diagonal. It is denoted by [ a. Another way to say this is a.. = 0 if i i j. For example, 1 0 0 0 0 6 0 0 ae = 0 0 2 0 0 0 0 -5 The identityr matrix is a special case of the diagonal matrix

and is denoted by r I. For example,'1 0 0.I = o 1 o 0 0 1 It is also frequently called the unit matrix. The null or zero matrix is one for which all elements are zero or a.. = 0. It may be of any order. ij A single element matrix is a scalar and may be written without brackets. The determinant la | formed from the elements of a square matrix is known as the determinant of [a]. A symmetric matrix is a square matrix for which a.. = a. 1j j1 A diagonal matrix is always symmetric but so may others be. The transpose of a matrix is found by interchanging rows and T columns and is denoted by [a]. The elements of the transpose of a matrix are found by setting T a.. = a. aij aji For example, 1 21 1 7 6 [a] = 7 9 [a]T _6 3 12 9 3

The transpose of a column matrix is a row matrix and vise-versa. A symmetric matrix is identical to its transpose. A. 3 Matrix Algebra We now define certain rules which make matrices useful. 1. Equality. Two matrices are equal if they are of the same order and all corresponding elements are equal. That is, [a] = [b] if a.. b.. (A-2) 2. Addition and subtraction. Addition and subtraction is defined only for matrices of the same order. Addition is performed by adding corresponding elements, subtraction by subtracting corresponding elements. Thus, [a] + [b] = [c] if a.. + b.. = c.. ij3 J3 ij3 For example 2 1 7- 1 6 7 3 7 14 3 6 9 + 1-2 = 4 9 1 -4 0 00 1 1-41 3. Multiplication by a scalar. Any matrix may be multiplied by a scalar. The result is that each element of the matrix is multiplied by the scalar, a[b] = [c] if ab.. = c.. 1j ij 4. Multiplication. We denote multiplication of two matrices, say [a] and [b], by

[a] [a] [I] (A-8) Thus the matrix equation [a]{x} = {c} becomes [a-l[a]{x} = [I]{x} = x} = [a]- {c} (A-9) A. 4 Partitioned Matrices A useful property of matrices is their ability to be partitioned into submatrices. These submatrices may then be treated as elements of the parent matrix and manipulated by the rules just reviewed. For example I - all a2 a13 I a14 a15 [Al]1 [A]l2 a2 1 a22 a23; a24 a25 [a]=. = [A], [A]22 (A-10) a31 a 3 a33 i a34 a3 where all a1 a133 Fa14 a15 [A]11 - [A]1 a21 a22 a23 a24 a25 [A]21 = [a31 a3 a33] [A]22 = [a34 a35

[a][b] = [c] (A-5) provided certain conditions exist. The number of columns in [a] must equal the number of rows in [b]. Each element in [ c] is obtained by multiplying the elements of the corresponding row in [a] by the elements of the corresponding column in [b] and adding the results according to the rule n Cik = a..bjk (A-6) j=l For example, ]3 2 (3xl+ 2x6)(3x2 - 2x3)(3xl+2xl)1 15 0 5 21 1 11 _ 2 = (lxl + lx6)(1x2 - lx3)(lx1+ xl) 1 7 -12 i > -3 1 7 -1] (7xl - I:x6)(7x2+ 2x3)(7xl - lxl 1 20 Note that if the order of [a] is mxn and the order of [b] is nxr, the order of [c] is mxr. Matrix multiplication is associative, distributive, but, in general, not commutative. That is, [a] ([b][c]) = ([a][b])[c] [a] ([b] + [c) = [a][b] + [a] [c] (A-7) [a][b] f [b][a] 5. Inversion. Division, as such, is not defined for matrices but is replaced by something called inversion. We define an inverse matrix [a] - such that

A. 5 Differentiating a Matrix To differentiate a matrix we differentiate each element in a conventional manner. For example, if x x2 3x [a] = x2 x4 2x 3x 2x x3 then -da] = 2x 4x3 2 dx _3 2 3X2 Partial derivatives are as simple. For example, xy y 2 y O O - 3 Ix xy2 y = L y 0 2 y x2 yZ 0 20 xy In structural theory we often encounter an expression of the following form U = a[ a j[c]{a} (A-1) wnvere [c] is symmetric and we wish to find its derivative 3U/3a.. By expanding the above form, differentiating and reassembling to form a new matrix, the results we get are u = [c]{a} (A-12) l c]\

For example, if c11 C 12 a U= [a, a2] C 12 C22J aj Z (cjla + 2cl2ala2 + c22 a) 2 1 Differentiation yields Cllal + c12a2 a 1 cl2a1 + c22a2 which, in matrix form, is We note further that the second derivatives 2u __ _ __ u al ~c c a a2 = c12 =2 c22 or, in general ~Zu 12 aU = c.. (A-13)'a. ai 1 ij 1 J A. 6 Integrating a Matrix Matrix integration is defined to be consistent with matrix differentiation. To integrate a matrix we integrate term by term.

I 2x 3 [a] = 2x 4x3 2 _3D 2 3x2 then x xz 3x fX [a]dx = 1x] x4 0 3x 2x x A. 7 Summary of Useful Matrix Relations [a]] I = [I]i[a] = [a] a([b] + [c]) = a[b] + a[c] [a]([b] + [c]) = [a][b] + [a][c] (A-14) [a] + [b] + [c] = [a] + ([b] + [c])= ([a] + [b])+ [c] [a][b][c] = [a] ([b] [c]) = ([a] [b])[c] [a] + [b] = [b] + [a] [a][b]: [b] [a] ([a] [b])T = [b]T a]T ([a][b])-1 = [b] l[a]([a] T)-1 = ([a] -)T

AppC'Ld; 3

Q + ajU,, 1f I* ^t; 1tv. 3 C(f^ro-J^') C<Ca 4tA 4' (A -t4yt ~> ) e r 7ZeCCY4OA_, L*t?_$iS ~e..(-e,, G,.' (o-;~e o$ W,ijo(^ Pt6 c~ AJ^,.. c gjD S) b.0..,......................,, _........... _..,.., gOUltlxt Cy C (~,Ay (, o | 10% 06" I g^PP^ SO - O l e U k"+'Q WPA S t 0 o secr'+ > * t.................,, dsl.. r( J s,,+. ~. f tsxt'L V-dL',.~ ~" 4t " wLU. T'-t<U.1 7. 6+ >- &tt~) t S "rr\CbtL fV %z.{>t ffi_ 4 A > 2 < L U*~~LL~r~I Ir.4~~ k~~n.P WC ~GC- Ur-~ $~

AgtO 1+4- LEcTQrC 15 20 Oc 16 |K (QIRCK^F: - Love. PLA-TC'ntILoV - Fws-t- p\ric (ets wvc^!rIvYid;, FraT,.e In trs^e A pr; l. ote, ireJ Ov r>e~<_ g %W<~'e&,tw. -- 9oP e 5t. 6icyV 6XV wuo XL prie lar I wL, lQ t,5 pirOV4-l Wvdfd, -;spcs oic. t'ost>".A 4is [ (il 4r dec~tcs - KircAio $ s cA GCtpt A +k< fpol& 0 Ny 1-' —- ^ -r.j<A 4Ub-r'") E' X (tt. Y pc dsKS 5 b ^ tyo ^r J^ vQ, 3C QxfJO+c +Lt 0flA4Jf'1-* ~25 5 - - p(L 5 N, w N - 2hl - ( (4') a~t 11dT.

fr: 4AGr,C cLc t', #0 [ P^A-IE FL~^RE fC Ta Afeo 41 JL( leCtVUP 4 J 16 17 G aJ I aL.>?-~^(cVuE- qol t- <` <u<Q" s e ~A ~-<~.X c t LUjfc.'U^ l r.L't k~, ~ 4Q, /~~ ~ /',L E~ ~,4'r' Ji-&.jJ/. t.J Q W+ ~.,' frLx Cr^U A,t Q1. 5tl.saQ.E 4, ^' e 1 i~~~;~ S^ & < w26 ^t^ - ^ ^?

t^UL wr<'. kV' /' ^2 ^ 4AA_._ o Stc,,A X i t5oY AA..,$ (3\ ret >ID5F ~-.L [Cl _ _ ( -\, ( o( D 0 pCC - D D o |(6 t5 0 [2] (ITz —

a di s a, (8f )x --- + p, -O ~,'- re_ _,,VtL W-IL ~., 3 -^ ) 7\L0 4 tPv rwto oro De PCjIt', t Jh roc ~ + a' +. 4) i{cjx [QCK(\ <w)? M e t* +3 o 53r Dw). @ ++ 2(,D +F r>.e: p, ( v% ~~a~ J B T r) ~ ^} CW

4 u ir t c> Q.%RAD&J AA 4 ("r (ra ^IgQ ^ e~ QrA^yJ) HL b<^. Eor LuQ h+ ty*,^t.) a<J^^^^N-, W< *>XD D^te r I yDV,'r' A': - o; \,,,.' r'AL cj n + t + - A ~oT fe-a m"c<^c^ I >^ <^^v~~ua

At.o 6(1 lItt.vL 76 (^. Jt 3laft fI"c 5+vai^ Jl^-A*^ t~,.f l4MLfc 7 aA *' aclu kaA~ I' T ~ 5. 16 a c 1, I. (Curt" vy Va V)y( ( "' - - r,<~v..Q / C-Ci: + s A-. J iL - h4.' W 9uL Q. <4< *- ^5'/+ _~f e 4+4 4 T. $XC8 i.),A4p yi s. AA ^< ^, -_ V r>zi c:O'V >, / W jaM33, js C^zZ ct^ <~i ^^<U ^^^ n(<Jt. ~ 8~-<^m. C-ttjC~GP >^U^OoJLV^ J^./l~AA~~t (^fL^ A Ltlc</ <~^ <XA- <~. Awi jA~~~~ji*-x. t~~tj~jj~z~^\ i^<- ~J~,1

vWj VZrtZt4~ Yfi IAA~tc 4 (J $ t - T 1J Acr, (4, ) l - T-[ tSl7u tL- To< L, ( 0 0 <fr.hsL~ 44 44r; A 1r4 Qj4 fr-~S ) i' + ('a. *-(*I^J )tA ) xr4t (' 1 \ 4 h { beA - ^ 4 i frii X Q^ EL, To ux~n ^ ^ <'A cL<x jxtc a., 4 AQ^ tC^J~L JSo <^~ OL \^^, oAA mo tt, k~;4~~,sT

jta, ta,ose A > v A I dW(xlo) 4=o r A Q r I E, t(4V L,'~ - ttt v -#Xt % 4 t 1 O lS (5 $ k^^oG c A^Z AA~ 4L SL 4+ oC 4A1<~-4 4^ (^AX^A A^. O&A^a<gA1 f = L*=M +4 + I, 2.j ^ V^AA. I *^A 3. <.L~ <^J^ ^ I 3 "A'A-t: fe }'~a-L 2p cE~ dQ~ 5 c t=; CkY'<a@ {si22 t vfg~~~~~~~~~~~~t i(/~~4;1 )(X5+ tA IJ3'< ~~~7

.J Uc j t',Ij' IH JL < dAwy. />^ \ -'. - ") / L *A CJ. I-2 L ja to Z/4dAC a_ Ha, 4J b^'4 yret'tolA plite ecle ) C -'' XL 1 46. Am }I f Ie 2; amp c toxi <;/ua>. t Cto, Cv-4~A P-iyt^h tA. -VUJ L<F J k. See STez^(c;w;c.^ ALUte PrLdC 6SJ eC r^ t>& +Ai^ C~k <|L, 4 MoJ^ 4^ A (^t ablltZ^ (^ *~ Xa. S m < <cey ee-n ~cruu g ~0' d""d(j 7-P~ 1/t~~

,t-' ",',,..... /a., \ D C~~ r( IXO - VyC L (XY 4- a, o kXA -J Vt t j-1 Lt) U,,Sck,.iK;\' 59AKC rtI; l * no S n'~ a %~ - %.0 I P11cL. T __L. ~4 z,,,L ~~ ~ 7 x X i t,..,.. a^ r..L9 -XWJL J i"t'u *fr^ s-1$ ci2 rR s P, (tA. (^y "tZZ-r i - o AL f,

-1-ts $ c k<s 1 d ^4J __ = G&, t 2%r x t+ 5-.7 I 'SC a^2^ -^ ^ Q: Ya4X- _ 2%_ _ - s6ic +i^ rUA, G m ^J ^ T)L ^ W(,,o) s = ll ct+B,+ X2,.7 X../>- _ t3-L + a xc+Yt f o7,,':W (~.,% - sZ.t - 3 t a. ox wz, +Xt $ o, + t ca-, X!if ~ % e 3= -._t' ~ L a x ph. n Y * 5y r, Q QX v - 4)L 2. ~ = ac + ^ k +A -t )( ^ ((X',o)( p..~ L~ ~~.~-~~.-t. j," lT~'.,._a A.~,t AA. J^^J,, t-A I"X(. At, m r..

T/^^A- < J? ^- o- A-c At;) j — &1I..'... 1 R Q vuL A-L CL 4AA #4JS4J ~7li 4 ear;, 4 Vr Jf qadnrati At. w A Mixed formulation linearM h, w (Re f. 1 2.22) 1 2-term Eq. (12.32)1 Mixed formulation E: quadratic M w (R ef. 12.22) 16-term polynomial " - - I Eq.. 2,, -- sh s (Ref. 12.6)14) Fig. 12.7 Numrical comparisons: quadrilateral plate element formulat"fnas.

L CC TV R 2-7 Acwo g51o \>?-TUL WOR% B l-o-mainAL ~~S I O~ 5VAc 2,L Tt^ vltJl wofl (-A- v w Lec)L 5 4 u 7 A SIO ah vtij) woelk a04. ^k vs ICvto IlC. e*t;brty^ ev Is,,y 0\4. vMI. r -L L AW1;c vs.;, tA.L AlIs 4l ( - +4 iA^ cft4s W^.'~ Qt & oU~l e.' s, tbL ofs a_ wy^> * u's;^ v; -At;S Isp^^t b1s.ill *S.^'B!. / _I West, d^ t t^4l -t Aft \Ai K TI 4 I f ) C1 5C O; an elslo w' Jll (n- L o vCCSs*e;S (I CL) to s* t t) - f 4. 4^x^c\ = be~tvc r t $o t Sv ae^ ^ sl S *x r t 15 wt nt<,<.) rC ^;~ eS ^C < 5~Ar ^ ^cest... (.ki <o ut 0b +40 c jt C t w tt 3) nowu' r -6\Oj ^ VIA %S cfees * _ tZ> CrtSU2t, (J 4 -~ Sr l Co TcrZA<At Strai ti, t c^^^x <^4^ AV ^ ^^ ^ U^ > -~v ^ ^^^^ <^U^ ^-ole c^.^ ^^^ -<L;~~~~~AA. CA. o^ V~~~~~~~~~~~~~~v'<~ ^^.r pio^. ^ ~^-<( 5'K/^^ ^ ^^r~~~~fc 4 L

(AuJ ) < 4YVl~) tS Ct 5Tra S: ~ 5 de C At c ~; o; <4 C;,ceW hJ w (Ste ^ov} as *h L'. o LoI' w; <v4 r dy^ c (c 4o tf\fr~d~e (^ ~(( (l~nlAC5^^^t.., ve C0oAo dc4 e. 4&; u^JiS < 1~ ^.k1 O^-^^ueis Or Qt. coUlI CY't;4 VSI - 1^ I-Cl-l. V \AJe I K <~ 8^I plos LAW;s 4 i woAs vt a dJY; orC^J5.e, wag 4o e' — t: < *A^.A-(f- C ( ysAacA, <A ) - W^(LV- ^otJL' rS c t ^ CAW~~A 4 VCAAO Qfo=~ vd~ ~6 v %Tr o7L RU AN-A U< al^ T&A U^w1~SAC ~~dr~~~R dif~~ Q~u^'LT <^JL W ^<. ^^^ ^4^^ ^ (N^ —^-*^^

'^ --- ----- -O,53-s I 1 -v CJI: Ve* 7: tJ r k CTT\ - dr4 P7 - \Al -"nCts \4 ^ r 1C. H;Y c-4 of?o.4+ e4 V t v dcAo) eC o5fc tt. ^virrtv;rtA 1 \ir to dor i;^etr, eo ettCr, Ils0 V,S~iLi;. ii. - ^ftq^ A W - - (^r tl - C.^-I^llb iI. Li;;;s -^ Tlo-3Q sLWy>;^ew*>os:t, -J -ti;y ( iA*, a sSe ^ t. *&< St**' *~~<>^;w t IS os s*Y. 9+ st ki^ L t I ) L M-~~< I~l~rnrR 0 at.~\ LatlA Z4l I ~IS~ Inol,)

0> c -^ one.,, i iot, vOqcrw dr. O t.a -t Aet3 C ^^MJL hr wrk ( po544tc);pUA >va ri U i~<u^ (li 5 + ~, W~ cJi" 4 jl (<.^h' 5. sjt -;r f CS\ o-ur a(;. Airr -n'(l^^i~r} ) -t ( )#;Ti JL <71 (XNu~I-tR3) - [~~(W 4Av< - tL —'t ~Vl t -R- uI -^prf ]hy! ^<~cik} ezje\rI AlpA (I. tst~rT T E7 - rQILA dk( E-XQW\PJ{I0R! ~ ~ 2

LtCTU-c Z8 G Z 10 cf cC (T( A-4 SIO AERO s1 ME 5s7 361c^7:. f, )\$CUO(A O, TVWfM-$ UI PD%-D AL 6,u -t^A <t yhuj ^^t y-. t suw IL J xI ),'.?..... -....,, * t, 4r IT d\, +;d^, - ~A;,l ru A-;4 O.L Do, 4\. v af >1,;4e~l C S I o'<, /\ "'Sue DolwsJl - rt) TT 4. ^^^^M / Mbl, -t:X PL S,C;5l^ S; ^I / A WlSPRMLQ pt^<Jc y^T L ujPor j *r f St4J lS~0 W<> O^^ UXM T^W t r, 7,

-T P ~ a, tA }l easi, l: rfeeA' ^ ieves cS Tr -- 2te%[ktg - Bung} A r- V (^H)[k3^-5 a t^TrklC- - (?t A H-N C ft I * *s (; ^S - l^ ^'' <T0 ( (-~~~rR;m9 1g:32jo T^ ~^.<^ ~CS^c T^ ^XU-.-y v Tn. j'4"?enk\eel C d,owif^ -^ M --- ~~ -------------- Lr~~~~ru ^ \. \ V (/ * s~~,,,,^ +w Fl~ l~dy-<j+aVor4; audle h ~~~~~5FR~ ~~~~~~~~~6

"r' ~)10) L k^wo,~;i01 All ~ I'd T^kLPo, em T-v9 vi +0 2 nt-o'' I*Ae mS ~1 Q uc Cay' Cw k; lto l I tt t (' TO- - t^: l —, t'' T ) /(',' ^'? ~4fe ^-~ -<. pl w+CL'' t ____._.. r d'Ir-,40 Wl~f.St~P1\ -n;, J TiV. -V * iT ~c~,,te -tG~Lx), d;1 0(

4~10 i ~ Cstoc) L0 J0 seac ^ sI 0 ^0 0 0:ti: c~, ~ VLUC~4Q IcI bj Tr, ~ I T hy..~ ~,. ~~,~_ I~LU 27, #ntfA A (ew;s e~ vlsT ok'L A YK 0r elo a'e ir B 2 1 t4A io e |O<I<O.T~ ~~<oo~l~ ~ y / "\^^_^y/TT V..j~r^^^^

(P~i^~. bi S^ ^; X2 F...\. 1Sa ID4 pta. SL C 4 S4h:, (A~of <I(L 4 eKj} oLL) = (5thy^ f- 5^^ n bAJM ) I1 7lJ zr u -U = -i^ -- o.7 Ilew 4 5t l;.,,,,J vv, I t+t iot, vva n-.~ -. —,o2 -U Tr~ - V^t - V ^ -2 ^ =- V ot re4, r " joJ* MJLJ S r3~ ~ 1,~ au QrC~ Y,~~ t- A /C~;~ JyC~\ ~~*RL

sA ~ vv~r Z > - ~ I1 TI A'('tuu OVt A 6lA k~ t L 51. CfiL,v tdIS;nu Oio, i X 6 r1 / y // 5 LkCO/isALk SeGs~at Ewov l 3G~ r &X44; t X^ha c/uuv eoy J a. dr, — ~"v/'AI _ 4,,W/ f1 c"4^4Z t LA: _. _^V x 2~ ~O;6 4rfujA inb 3 & Al%4 G5ip cl sk I F =-V W f),) 7114 cLOt4L -W, \4 j, b u

} ~- _ }_G&e4 04A. riva 5)9 l (Hv) f S!~rc L pl-tRw.- C cO.s S c,; A 5,a k (s ~l x,C ).. z r^,,^ M - C-w:(5 3, er I < 7 t_(t; C:j ^);Ig."~^ 1 - ~^b,!^ 7 i-.i'..;'"i" r^ ^^ ^ ^.^o^UJ T7^ ^-;^s^,,s c~~~~~~~;qCj^

3 9015 02499 5790 3 9015 02499 5790 THE UNIVERSITY OF MICHIGAN DATE DUE \\/\ b