AFSWC-TR-6 1-60 DYNAMIC RESPONSE OF THE 6-FOOT DIAMETER SHOCK TUBE TO A CONSTANT VELOCITY PRESSURE FRONT by Sing-Chih Tang University of Michigan August 1961 Research Directorate AIR FORCE SPECIAL WEAPONS CENTER Air Force Systems Command Kirtland Air Force Base New Mexico Approved: ONALD I. PR CKETT Colonel USAF Project No. 1080 Director, Research Directorate Task No. 10802 Contract No. AF 29(601)-2793

This research was administered through The University of Michigan's Office of Research Administration, by the Civil Engineering Department and under the general supervision of Bruce Go Johnston, Project Supervisor.

TR-61-60 ABSTRACT The problem of determining the dynamic effect of an internal nondecaying pressure front which moves with constant velocity parallel to the axis of a circular tube is treated. It was known that the equations for the problem of a beam resting on an elastic foundation are equivalent to those for a circular tube, after a simple replacement of constant terms. Thereafter, the equiva_ lent beam problem is considered with both infinite and finite lengths. The finite beam is investigated with two types of boundary conditions: both ends simply supported, and both ends fixed. Viscous damping is considered for the infinite beam. The effects of shear and rotatory inertia are neglected in all cases. It is shown that the dynamic factors for the infinite and finite beams are nearly identical, and the solution for one circular tube is obtained on the basis of an equivalent infinite beam. PUBLICATION REVIEW This report has been reviewed and is approved. Fo and in the bseve of JOHN J. DISHU K Colonel USAF Deputy Chief of Staff for Operations iii

TABLE OF CONTENTS Page LIST OF MAJOR NOTATIONS vii INTRODUCTION 1 PART I. BEAMS WITH FINITE LENGTH 3 1. Equation of Free Vibrations 3 2. Case for a Beam with Both Ends Simply Supported 4 3. Case for a Beam with Both Ends Clamped 9 PART II. BEAMS WITH INFINITE LENGTH (REF. 6) 13 1. General Equation of Motion 13 2. Case without Damping (P = 0) 15 3. Case with Damping 17 CONCLUSION 23 REFERENCES 25 APPENDIX I 27 APPENDIX II 29 DISTRIBUTION 33

LIST OF MAJOR NOTATIONS E = Modulus of elasticity of the beam I = Moment of inertia of the beam M = Bending moment R = Radius.of the circular tube Po = Magnitude of the concentrated force a = Cross sectional area of the beam c = Viscous damping coefficient per unit length g = Gravitational acceleration h = Wall thickness of the circular tube k = Spring constant per unit length of the beam ~ = Span of the finite beam p = Intensity of the uniformly distributed force t = Time v = Velocity of the moving force y = Transverse deflection of the beam 5 = -Damping factor (defined later) y = Weight of the beam per unit volume a = Fiber stress = Poisson's ratio vii

INTRODUCTION The purpose of this investigation is to determine the dynamic effect of an internal nondecaying pressure front which moves with constant velocity parallel to the axis of a circular tube, The differential equation of equilibrium and compatibility for a circular tube under the action of internal pressure with small deformation is: d4y Eh D 2d + R- Y = p(x) (Ref. 1) (0.1) Ehp3 where D = and p(x) = internal pressure. The differential equation 12(1-l2) of equilibrium and compatibility for a transversely loaded beam resting on elastic foundation, if the transverse deflection is small, has the same form as (0.1), i.e. d4y EI + ky = p(x) (Ref. 2) (0.2) 4dx These two equations —(0.l) and (0,2) -are entirely equivalent, if we set EI = D = Eh and k = (003) 12(1-_2) R2 Thus the solution of the problem of a beam resting on an elastic foundation provides also the related solution for a circular tube, if replacements Eh3 Eh EI = and k = -- are made, Thus this paper deals not only with the 12(l-f2) R2 dynamic effect of a beam resting on elastic foundation under forces moving with constant horizontal velocity, but also provides a solution for an internal pressure front moving down a circular tube. Both the finite beam (Part I) and the infinite beam (Part II) are considered.o Two kinds of boundary conditions —both ends simply supported and both ends clamped —are investigated in Part I. Viscous damping effect is considered in Part II. In all cases, in calculating dynamic load amplification factor, we neglect the effects of shear and rotatory inertia. We also assume that the circular tube vibrates radial-symmetrically. We neglect the effect of the 1

longitudinal stress wave, as its velocity, v = JgE/y = 16,150 ft/sec, is almost ten times the velocity of the transverse wave propagated down the tube. In Timoshenko's paper (Refo 4), "Method of Analysis of Statical and Dynamical Stresses in Rail," he has derived a formula to calculate the deflection of a simply supported beam resting on elastic foundation under a moving force with constant velocity. His formula is identically the same as Formula (l10) in this paper, Using this formula, we can find the maximum bending moment in the beam. One numerical example has been done by taking the summation of one hundred terms of the series. Using the same idea as his, we derived the formula of the deflection curve and bending moment for the beam with both ends clamped, but did not try to do any numerical computation. The series in this case is very slow in its convergence. In Kenney's paper (Ref. 6), "Steady-State Vibrations of Beam on Elastic Foundation for Moving Load," he established a solution for the infinite beam resting on elastic foundation. His formula is used to find maximum bending moment both for concentrated load and uniformly distributed loado Dynamic load amplification factors based upon the derived maximum moment formula are computed with various velocities and wall thickness of the circular tube of 6-foot diameter. 2

PART I BEAMS WITH FINITE LENGTH 1. Equation of Free Vibration. PO Take the origin of the coordinate V at the left end of the beam, then X EI- + _ Y_ + ky = 0 (1.1) - L gx" @;. 9 t1 Fig. 1.1 Assume y = X(x)-T(t) By the method of separation of variables, we have xiv k ay " 2 X EI EIg T Where p is a real arbitrary constant. Now xiv k X k r__.+( _ p) = O X EI Put k2 k. - p2 < 0 and 4 p2 _ EI Where a is a real arbitrary constant too. Then iv x 4 and X = Acos cax + Bsinoax + Ccoshcax + Dsinhax (1.2) where A, B, C, and D are constants and determined by boundary conditions. 3

2. Case for a Beam with Both Ends Simply Supported~ Boundary conditions: X(O) = X( ) = O X"(0) = X"( ) = 0 (1l3) o A = C = D = 0 or X =..B sin ax, sin aG = 0 ol = nT, a = n — where n = 1,2,3,.eo (1) From the above solution, we have the modal shapes of the simply supported beam for free vibration as the following: Xn = Bn sin nrx (1o4) The modes are orthogonal. (2) Norms of X and X" n =' 2 dx = n4B o sin2 n dx = - B 2 (1 6) >Un = T(X9)2 dx = 4 nff 20 Bn n o ~ 0 ~ 2~3 (3) Response due to a moving concentrated force P0. i,o By the method of superposition of normal modes, we may assume the deflection curve of the beam as the following: 00 Y = n n sin nx (Refo 3) (107) n=l where qn is a function of time t only and qn. is known. as the generalized di splacement o LagrangeTs equation will be used to find the response. Total strain energy in'the beam:

E-!,ay) 2 k I E=-j (8a) dx + - y2dx 2 x 2 2 00 00 EI E n4 4 k 2 2 y2 2yj2 n=l n=l EI___ n3 -4 4~ n=l n=l Total kinetic energy for the beam: 00 T= a (ay)2 dx a= 2 Po 2g o 4g n=l Lagrange' s equation:.X d aT;T )V ( = a + a (1.8) Fig. 1.2 where 4n =- and Qn is the generalized force defined as follows: dt We assume the deflection curve of the beam due to a moving force PO to be 00 Y= qnsin n (1.9) n=l At time t = O, Po is at the origin and at t = tl, Po moves a distance c from the origin, thus c = vt. Let 6y be the change of deflection of the beam of c due to change of i-th generalized coordinate qi, then by = bqi sin - = Iqi sin irt The work done by the external force Po on the beam due to by will be PoSqi sin -

The generalized force times the change of i-th generalized displacement qi will be equal to that amount of work. i.e., Qibqi = PoSqi sin i Qi = Po sin it-vt Insert V, T and Qn into Lagrange's equation, we have n+ 2g (EI n4 + k gP in nvt yal 213 2 yal ~ Let 2 2g EIT n4 + k) = 7al 21a 2 The general solution of the differential equation will be qn = An cos wht + Bn sin ut + g ~ o Vtsin n!vtl sin wn(t-ti)dti Suppose that both initial displacement and velocity of the beam are zero, then An = Bn = 0 sin O vt sin sn (t-ti)dt1.....2gP... (wn sin n-vt - sin r it) yaa(-2n( 2c 4-na v ~v The term nrv/~sinwt due to free vibration will be gradually damped out, so we neglect it, and 6

n = -—; 2gP02 Wn- sil nncvt yanr( ~2w-n2Tr2v2) 2 00 Y = I qn sin n n=l 00 o iox nx nrvt sin -N sin, 2P~ ~ 2 4E nln4 + kQ4 n v2ya ~2 (Ref. 4) (1.10) EI Ir n=l n4 + EIEr4 g EIt2 ii. Maximum stress in the beam. Since the bending stress is linearly proportional to bending moment M, the maximum stress, will be determined by the maximum bending moment. Now M = -EI 2 [x2 co njx nn-vt _2_ 2Po P V sin - sin -7~y 2P ~k 2 ~x2 EI 2 M:2 ~ y Y T n=l n + - -- EI4n2 g EI T2 For maximum value of a sin nx sin nvt = 1 axx2' 00.Mmax = - EI(2) - 4 2 2 6x2 max T2 2 kM vya 1, n=l Ei4n2 g EIt iii. Numerical example. Take a circular tube with radius R and wall thickness h as an example. Given data as following: E = 29 x 103 k/ina 1 = 0o3 h = 1,0 in. R = 36 in, 2 = 120 in, o y = 0.284 x 10-3 k/in.3 g = 386 in o/sec2 7

From (0.3) Equivalent k = Eh 22.377 k/in.3 R2 Equivalent I = h = 0.0916 in.3 12( 1, _2) EI = 2655.6 k-in. For the static case, i.e., v + 0, but vt # 0 00 199 M = 2Po~ j 1 2P 1 n=l2 n 1 + n= n2 + EIl 4n2 EIT4n2 1.108 Po, n = 1,3,5,.... For v = 1600 ft/sec 00 M 2Po 2 k2~4 1v2ya n=l n2 + EI3r4n2 g EIjT 199 2P 7 n=l n2 k2Q4 v E_ EIT4n2 g EIn = 1.690 Po, n = 1,3,5,.... 1.690P Dynamic load amplification factor F = 69oo 1.53 1.108Po (4) Deflection curve for uniformly distributed load. Let p be the intensity

pdx = pvdt1 0. nTrx. njvt 2pv3 3 sn) n - dy = _ dt EIn 4 L...k4 n 2V a 2t n=l n4 g —--- 0o n7~nx_ n n-vt1 2pv3t sin nl sin -(1 y p I ~ I EI1T4 n4 + k~4 n2v2a ~2 dt (1.12) EI~r n=l +4 1 kf. n2vya. n=L EIr4 g EITr2 x=vtl dx. Fig. 1.3 3. Case for a Beam with Both Ends Clamped. Boundary conditions: X(O) x = a) xt(O) = Xt() = 0 (1.13) X = Acos o + B sin + C cosh E+ D sinh C C = -A, D = -B (1.14) A = (sin ~ - sinh c~) B (cos cd - cosh co) A = + (cos c~ - cosh c) B (1.15) (sin CZ~ + sinh cll~) 9

(cos ce - cosh c~) (sin al - sinh al) = 0 (-sin ae - sinh el) (cos we - cosh we) or cos Con~ cosh cn~ - 1 = 0 (1.16) (1) Modal Shape Xn = Ancos Qnx + Bnsin anX + Cncosh anx + Dnsinh CnX = An(cos enx - cosh onx) + Bn(sin anx - sinh onx) n = 1,2,3, ooo XnTs are orthogonal. (Ref. 5) An, Bn and cn have the following relations: An = _ (sin ani - sinh an2~) B (cos ani - cosh ani) or (cos an~ - cosh aCn~) An = + Bn (sin aen + sinh an) n cos ~nQ cosh Gnf = 1 a'. ani = 4.7307 7.853, 100996, 140137, 17.279,.(Ref. 3) (2) Norms of Xn and Xn I = W(X2{)2 - A2' sin2anQ sinh2an ( Pn = (X) = A... (2 17) 0~ (cos an.e cosh anl)2 (Ref. 5) sin2 n~ sinh2acn On1"= 1(X~)2dx = Q4Ane (cos cn~ - cosh an~ )2 (1018) (3) Reponse due to P0 Assume the deflection curve of the beam as the following: Y = qL(t)~Xn(x) n=l 10

From Lagrange's equation, we have qn + ya (EI ) + k) qn = Qn where Qn = PoXn(vt) is the generalized forced and defined in Section 2-(3). Let t =~ g — (EI + k) ya ~n Then the solution of the differential equation will be: qn = g 1 Qnsin hn(t-tl)dti yarn 0 0 if we assume the initial displacement and velocity of the beam to be zero. The smallest value of c~ is 4.730, so we can put An - Bn from (1.15) Qn = PoXn(vtl) ) PoAn(cos anvtl - sin anVtl - e_-nvtl) q~gn = PoAn r f o (cos canvt - cos cnt) ga'n n yaa, C~v (consin 1 anvt cnv s in nt nt) + c~ ~Od2V2 UAS;1 n- nvt (-+~v (alcncOs aunt - CnV sin wnt - cane If we neglect the terms due to free vibration, then [ 1 e~envt qn = g PO 1 (cos anvt - sin anvt) e Ya 7n n _ 2cof-v2 n n2 V o(cos cosh n)2 Y = I qn(t)~Xn(x) ya 1nl in s inh2 n cos in nX- + sinh nX+ (1.19) 11

where OCns are roots of the equation cos %n~ cosh cn~ = 1 and = i(EI + k) (EIn + k) ya 00 a2y _ gPo 1 (cos an~ - cosh oan~)2en ax2 ya n= sin2 cn~ sinh2 cn~ 1 __ _nvt K{_- nev (cos anvt - sin cnvt) - e cos Unx + cosh -nx - sin anx + sinh cnx4 (1.20) a~2 M = EI x2 (4) Deflection Curve for Uniformly Distributed Force p. Use the same process as that in Section 2-(4), to get the deflection curve. 12

PART II BEAMS WITH INFINITE LENGTH (REF. 6) 1. General Equation of Motion. E x4 g at2 C + ky= Po(x,.t) (2.1) ax4 g 0t2 We transpose the y-axis so that it is attached to the moving force Po which moves with constant velocity v. Let xl be horizontal coordinate in the new system xl = x - vt PO then (2.1) becomes EI ~y ayv%2y \) /y /// EI a4 x + - 0 v y + ky y ax4 g ax axxl (2.2a) - a2v 2 a +2 aty oxFig. 2.1 +2 + aY Poy yt) - x1 t g t2 Put xl = x in (2.2a), since xi is a dummy variable. As this is an infinite beam of indefinite extent both fore and aft of the moving load, the deflected shape will not change with respect to the moving coordinate axis after the applied load p moves on the beam for a suitable time. All the time derivatives are zero and the equation for steady state response will be: EI d4y av2 d2y _ v + ky = 0 (2.2b) E. + ky dx4 g dx2 dx Put A = (k/4EI)4 (2.3) 13

Ccr = 2(k Lg)2, Vcr 2 (2. 4) G = v/Vcr, I = C/Ccr (2.5) The equation becomes yiV + 4(\)2y,, - 896X3y' + 4y = (2,6) and is subjected to the following boundary conditions. (a) lim [y(O +') - y(O - ) ] = 0 (deflection continuous) C+0 (b) lim [y'(O + c) - y'(O - c)] = 0 (slope continuous) *+0 (2.7) (c) lim [y"(0 + C) - y"(O -,) ] = 0 (curvature continuous) (d) lim [y"' (0 + e) - y'" (0 - )] = (shear discontinuous due to - EI concentrated load). The solution of equation (2,6) with boundary conditions shown in (2.7) is x<O Px.< - Y - 0 -- -1 2k +4 + ( jG)2 + 2 (/1) 2 f ( G/ +12) sin(2eG2+2a-2GS/q) 2?\x + cos(2G2 + 2 ) x>O P0' -1 I e h 1 2k LrL4 + (rG)2 + 2 (/) t(e6/q-.2)sin_(29e+a+2e6/n)LxA + cos(2G + r + 2Ge/r)2x q( 2G2+a2+2G6/~) 2

where q is the positive real root of the equation ~6 + 292rn4 + (4-_1)rj2 _ 9262 = 0 (2.9) if 9 is less than 1. 2. Case Without Damping (5 = 0). From (2.9) = (=1- 2)2 (1) Deflection curve. 1 2k (192)1 1 sin (1+G2) ~ x + cos(l+G2) 2\x~ (2.10) ( 1+92)2 (2) Bending moment. M = - EI 2 dx2 Po e 4- (G2) 2 2_ sin(l+G2)2 |\xI + cos(l+G2) 2- (2.11) Maximum moment occurs at x = 0 and Po. 1 Mmax = 4- (l_eG)9 (.12) (3) Dynamic load amplification factor. For static case Mrnax 4 (Ref. 1) (21) 15

Dynamic load amplification factor F = (2.14) (192)>2 (4) Uniformly distributed load. Let 0 be the point where maximum moment occurs, and xl, x2 be the distances from each end of the uniformly distributed load to point 0. From (2.12) _(1-G2) BX = P A e dM (e l) 4\ ( _92) 2..... sin(l +2) 2\x + cos(l+2)r?\x dx L l+G 2 M-= *. K.. 1 2 sin(l+m2)2?mx + cos(i +2)2 dx ~i~L~c'4_2(1_0=4) xL2 e.. P isin( 1+2)2)Ax - t 21_0~)'2x1~\i z, _(ls@2) 2X2 As x2 becomes large, e approaches zero if G A 1, we neglect it _(l_9p2)x1 P pdx M \2 1 XI O 4 (1-G4) (2.14) X2 I e, sin( 1+Q2) 2Ax1 Fig. 2.2 For maximum moment For the static case (@ = O), maximum moment occurs at tan Axl = 1, or Axl = nax -42e 4 sin - (2.15) 16

Dynamic load amplification factor = I -( l.-G2)AX1i 1+2) tan( 1+2) 2Ax = ( 2 1 L~ 2)(1 2(5) Graphs of dynamic load amplification factors versus velocity. These are shown both for concentrated load and uniformly distributed load on the following pages. Since the dynamic load amplification factor depends on G, or v/vcr where kEg2\ Vcr = (h2y2) we should use vcr as a parameter. For a circular tube with radius R and wall thickness h, equivalent k and I will be: Eh h3 k = I = R2 12(1- [2) vcr depends on h only if E, I, and R are constants, so the graphs are drawn with h as a parameter. We use E = 29 x 103 k/ino2, = 0~3, and R = 36 in. to plot all curveso From Graph 1, for h = 1 ino, v = 1600 ft/sec dynamic load amplification factor F = 1l51. In comparing that with the simply supported beam calculated in Part I, F = 1,53, dynamic load factors both for simple beam and infinite beam are almost the same. 3. Case with Damping. (1) Deflection curve for steady state response due to concentrated moving force Po. From (2o8), we get 17

6 -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~I".. *-'" 0 C-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~, 100 0 — ~-____...0 080 2 VELOCITY 00 ~~~2400 20 VELOCITY OF MOVING FORCE IN FT. / SEC. disraphu Dynamic load amplificati factors for a movin foce distributed radially along a circular tube of radi us R 36 in.

t5 Z = U-A, ~ ~ ~ ~ ~ ~ ~ ~ ~ - 5 C)~~~~~ I -40 (-.. 10 100 160 800 2000 0 26 200 3000 VELOCITY OF MOVING PRESSURE IN FT. i SEC. nGraph 2. Dynam ic c rulo aar amb O IN FT.n SEC.m l 0 t3000 Gr a lp. D n~ l a d'p lification factors for a m oving u n iform ly d istrib u ted internal pressure acting on a circular tube of radius R 36 in. ~~~~'u36 i.

i. x<O 2k 4 + (r1j)2 + ~ ( 2/~) I(G/rl+Q2) sin(202+I2 -2G0/n) x c(2+-20/ - ( 2G2+r2-2Gp/r1) 2 ii. x>O lo__ X>e - (?~/ -~2) sin(2G2+2+2G/ri~) Ax cos2S2n2+@X/) x _r (2G2+-2-2~ /rl) where q is the positive real root of the equation 26 + (24- + (>4.1) 2 = (2) Bending moment. d2y M EI dx2 i, x < P ne e ]3 - 2 + + + G35//2 - G2p2/3 M = - -- ~G+ 22 I -/ —r 1~h G / 4 i x4 + (TG) 2 + 7 2 2 2 /) sin(2G2+2-2G/r) 2"2- x - ( 2+02) cos(2G +% -2G/~)2X|ii x > M O e, 0_~ 3 + 921 + GP+ G3p/2 + G22/A 0 LMA 4 + ()2 (G) (22+2+2 2 1 1 sin(2G+2+2eG/r) 2-x - (12+92) cos(292+12+29e/1))2\x 20

(3) Numerical example with damping ratio 5 = 0.02. i. Dynamic load factor for 0 = 1, the most critical case. x< 0 M = e Po e3 + + /2 _- 2/3. 7A q4 f r'2 + - -' — ~- ( +-'-_t/-2 /')2 sin(2+~2-2/1) 2Ax - (12+1) cos(2+~2-22/1)?xj x>O PO r -e -9 13 + _ + p + p3/112 + p2/r3 M = ~I? 1 47 lh4 -/ + ( (2+~2+2 p/~)sin(2+r2+2~/~)2x - (12+1)cos(2+e2+T25/2 ) 27\x where q is the real positive root of 6 + 2 44 _ 0 0 = 0.02 By Newton's method, we find 1 = 00 1185 Po x < O, M = 5.14 Po x > O, M = -5.44 Dynamic load amplification factor F = 5.44~ iio Dynamic load factor for G < 1. 16 + 2G214 + (G4-1) 2 - G022 = 0 21

If G = 0,5,' 0.866 = (1-. 2)2 Since ~ = (1-.2)2 for p = 0. here r is the same as that without damping. The dynamic load amplification factor does not change from that obtained for the case without damping. For 1 G = o.8, oo6oo = (J-e2)~ 9 = 0.9, rl 0.436 = (1-92)' 9 = 0-95, r 0.323 = (1-02)2 Once q is determined the dynamic load factor can be computed. For 9 < 0o95, P is the same for both cases with and without damping, so that the dynamic load amplification factor for both cases are equal. 22

CONCLUSION From the example of a circular tube with h = 1.0 in., R = 36 in., and v = 1,600 ft/sec, the dynamic load factor for an equivalent beam with length ~ = 120 in. is 1.53 and for an equivalent infinite beam is 1.51. Those two factors are practically the same, The deflection y of an infinite beam resting on an elastic foundation is a function including e-(102) 1Xxl as a multiplier. As Ixl increases, e-(1- 2)1 x decreases very rapidly, so the deflection at a short distance away from the loading point is nearly equal to zero. This is equivalent to the beam's having a support there. For this reason the simply supported, or even the clamped beam (zero end slopes), may be treated as an infinite beam, insofar as effects of end conditions on support interruptions are concerned. Once we are allowed to treat the finite beam as an infinite beam, then we can use the graphs on pages 18-19 to find the closely approximate dynamic load amplification factors for the finite beam. The exact solution for a finite beam derived in Part I is in a form of infinite series which converges very slowly, so we can ignore it. For ordinary structural members 5 varies from 0.005 to 0.03. From the end of this paper, we know that the dynamic amplification factors with these small damping ratios do not deviate much from the case without damping except for G approaching 1. As the velocity of the moving force approaches the critical, the maximum dynamic load factor is 5.15 for B = 0.02. The actual internal pressure is a decaying wave. The calculation of the response of the circular tube due to such a decaying wave would be complexo For an approximate estimation of the maximum bending moment in the circular tube caused by a decaying pressure wave, we can use an equivalent rectangular wave pulse instead of a decaying wave to calculate the static bending moment in the tube and multiply it by the dynamic load amplification factor from Graph II on page 19. If the duration of the shock wave is large in comparison with the time required for the wave front to travel through one segment of the tube, we can use the maximum intensity of the decaying wave pressure as that of the rectangular wave pressure. This is the case for our problem. Since the velocity of the wave front is from 1,000 ft/sec and the maximum length of the circular tube section in the test region is 10 ft it takes less than 1/100 sec to travel over the whole length; while the duration of the shock wave is 1/10 seco Within 1/100 sec the intensity of the decaying wave pressure will not change much. 23

REFERENCES 1. "Theory of Plates and Shells," by Timoshenko and Woinowsky-Krieger, second edition, published by McGraw-Hill Book Co., 1959, ppo 466-481. 2. "Strength of Materials, Part II," by Timoshenko, S. P., third edition, published by D. VanNostrand Co., 1956, pp. 1-11. 3. "Vibration Problems in Engineering," by Timoshenko, S. P., second edition, published by D. VanNostrand Co., 1937, ppo 348-370 and pp. 343. 4. "Method of Analysis of Statical and Dynamical Stresses in Rail," by Timoshenko, S. P., Proceedings of the Second International Congress for Applied Mechanics, 1926, ppo 407-418o 5. "Partial Differential Equations of Mathematical Physics," by Webster, A. G., published by Hafner Publishing Co., 1950, ppo 138-142. 6. "Steady-State Vibrations of Beam on Elastic Foundation for Moving Load," by Kenney, J. T., Jr., Journal of Applied Mechanics, Vol. 21, No. 4, Dec. 1954, ppo 359-364. 25

.APPENDIX I 1. Calculate maximum bending stress for a tube of radius 36 in. under static pressure p = 100 psi M = 6M 6 22 3p h2 h2 h2-A2' T 1 P 100 psi;2 2_ h,(21-R1) X.. = ) = s - 0... ____h __.I _ I 4 v _ O h2B hR = 3p.hR - pR 6, 550 Fig. A-1 h2'5(l-) hh Wall thick- 1/2 3/4 1 1-1/4 1-1/2 1-3/4 2 Max. bending stress o(psi) 13,000 8,730 6,550 5,240 4,370 3,750 3,270 2. Possible'maximum stress in 1 in. wall tube if shock front velocity coincides with the critical velocity. i.e., v = 2145 ft/sec Ass. B = 0.02, p = 100 psi Dynamic load amplification factor F = 5.44 Static stress due to bending, from Section 1 in Appendix I: a = 6,550 psi Dynamic stress 1 = 5.44 x 6,550 = 35,700 psi Londitudinal stress, assume no dynamic amplification factor involved 27

a2 = 1- 62 = 1800 psi 2o5 36.1 Total maximum possible stress a = l1 + a2 = 37,500 psi 28

APPENDIX II Transient Response (Time required to build up 90% of the maximum deflection) 1. For moving coordinate y = mass of the beam per unit length EI a + yv2 a2y Cv ay + ky - 2yv + Y + Cy t) axxt axt -x at2 at p(x,t) Let y = Y(X)*T(t) where Y = P r for x = O 2k [4 + (Ti)2 +1 (@2/)2 y1 = pA 1 1 Te e+~ for x = 0 2k + /)2 EI + yv d2Y - Cv dY Po(xlt), for x = 0 dx4 dx2 dx -2yvY'T' + yYT" + CYtI = 0 T +.CY-2vY' T = 0 c Y+ t 0 7YY C 2v,V 0 V t - 2v-)t= 0 = -, for x = O Boundary conditions: Fig. A-2 T(0) = 0, limT(t) = 1 t2oo 29

Let v + 2v e-rt T7T + VT = 0 Boundary conditions: T(O) = o, lim T(t) = 1 t-oo (a) Complementary function: m2 +vm = 0 m = - vO T = A + Bet when (b) t = 0, T = 0 Fig. A-3 A + B = or A = -B t - o, T = 1. A = 1 -vt.'. T = 1 - e 30

Investigate v v = + 2v Y rT2 C 1 CC C C C Ccr - 2(ky)2 - = Ccr -cr = - = 2P =eg eG, v 2v 2 \ e 7kE 4 2 - IEk T Ry7 *.v =2 + t 2. Numerical computation For h = lin, G = 1 = 0.02, T = 0.1185 0.284 lb-sec2 0.284 K-sec2 386 in.3 386,000 in.3 k = 22.38 K/in.3 51

v =. 0 2238 x 386,000 1+2 ( 0.284 0.1185 = 0,04 x 5520 x 72 = 15,900 - vt T = 1 - e For T = 0~9 0.o = elt ~ t =.2.3 1.45 x 0-4 sec 15,900 Distance travelled = 2,398 x 1.45 x 10-4 = 0. 347 ft 32

TR-61-60 DISTRIBUTION No. Cys HEADQUARTERS USAF 1 Hq USAF (AFOIE), Wash 25, DC 1 Hq USAF (AFOCE), Wash 25, DC 1 Hq USAF (AFTAC), Wash 25, DC 1 AFOAR, Bldg TD, Wash 25, DC 1 AFOSR, Bldg TD, Wash 25, DC MAJOR AIR COMMANDS 1 AFSC (SCR), Andrews AFB, Wash 25,.DC 1 AUL, Maxwell AFB, Ala 1 USAFIT, Wright-Patterson AFB, Ohio AFSC ORGANIZATIONS 1 ASD, Wright-Patterson AFB, Ohio BSD, AF Unit Post Office, Los Angeles 45, Calif 1 (BSL) 1 (B'ST) 1 (BSQ) I (BSR) 1 ESD (ESAT), Hanscom Fld, Bedford, Mass 1 AFCRL (CRZG), L. G. Hanscom Fld, Bedford, Mass KIRTLAND AFB ORGANIZATIONS AFSWC, Kirtland AFB, NMex 1 (SWNH) 75 (SWOI) 2 (SWRS) OTHER AIR FORCE AGENCIES 1 Director, USAF Project RAND, via: Air Force Liaison Office, The RAND Corporation (RAND Library), 700 Main Street, Santa Monica, Calif ARMY ACTIVITIES 1 ARGMA Liaison Office, Bell Telephone Labs, Whippany, NJ 533

TR-61-60 DISTRIBUTION ( cont' d) No. Cys 1 Director, Ballistic Research Laboratories, (Library), Aberdeen Proving Ground, Md 1 Director, Evans Signal Laboratory, (Weapons Effects Section), B elmar, NJ 1 Commanding Officer, US Army Engineers, Research & Development Laboratories, Ft. Belvoir, Va 1 Office of the Chief, Corps of Engineers, US Army (Protective Construction Branch), Wash 25, DC 1 Director, US Army Waterways Experiment Sta (WESRL), P. O. Box 60, Vicksburg, Miss NAVY ACTIVITIES 1 Commanding Officer and Director, Naval Civil Engineering Laboratory, Port Hueneme, Calif OTHER DOD ACTIVITIES 1 Chief, Defense Atomic Support Agency (John E. Lewis), Wash 25, DC 10 ASTIA (TIPDR), Arlington Hall Sta, Arlington 12, Va AEC ACTIVITIES 1 President, Sandia Corporation (Mr. W. Perret), Sandia Base, NMex OTHER 1 Armour Research Foundation (Dr. T. Schiffman), 3422 So. Dearborn St, Chicago, Ill 1 Severdrup and Parcel and Associates, Inc, (Mr. W. Rivers), 933 Olive St, St. Louis, Mo 1 University of Illinois, Talbot Laboratory, Room 207, Urbana, Ill I-4 The University of Michigan, University Research Security Office, Lobby 1, East Engineering Bldg, Ann Arbor, Mich Massachusetts Inst. of Tech, Dept of Civil and Sanitary Engineering, 77 Massachusetts Ave, Cambridge 39, Mass (Dr. Charles H. Norris) 1 (Dr. R. V. Whitman) 1 American Machine and Foundry Company, (Thomas G. Morrison), 1104 So Wabash Ave, Chicago 5, Ill 1 Director, Stanford Rsch Institute, (Dr. Robert Vaile, Jr), Menlo Park, Calif Official Record Copy (SWR, Capt Pantall) 34

UNIVERSITY OF MICHIGAN 3 9015 03527 1991