THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING A PROPOSED METHOD FOR THE STRESS ANALYSIS OF PROPELLER BLADES Y. Lahav September, 1956 IP-178

AC KNOWT FGMENTS The author wishes to express his gratitude to the following people who have helped make this paper possible. 1. Professor L. A. Baier for technical assistance. 2. Professor Leo Legatski for assistance in performance of tests. 3. Professor Harry Benford for encouragement and assistance in publication. 4. Mr. Wesley D. Wheeler whose idea it was to publish the idea in the form of a paper. ii

TABLE OF CONTENTS Page INTRODUCTION 1 Summary of Propeller Stress Formulae 2 1. GENERAL 3 2. THE TORQUE 3 2.1 The Torque Force 3 2.2 Center of Torque 5 2.3 Moment due to Torque 7 3. THRUST 7 3.1 General 7 3.2 The Magnitude of the Thrust 7 3.3 The Center of Thrust 9 3.4 The Moment due to Thrust L1 4. CENTRIFUGAL FORCE 12 4.1 The Equation of the Centrifugal Force 12 4.2 Moment due to Centrifugal Force 14 5. VIBRATION 15 6. PROPERTIES OF BLADE SECTION 16 7. MOMENT DIAGRAM 17 8. STRESSES AT ROOT SECTION 19 8.1 Stress due to Torque 19 8.2 Stress due to Thrust 20 8.3 Stress due to Centrifugal Moment 21 8.4 Stress due to Centrifugal Force 22 8.5 Maximum Total Stress 22 APPENDIX I. STRAIN TEST ON PROPELLER BLADE 24 APPENDIX II. STRESS CALCULATION FOR THE FOLLOWING PROPELLER 32 REFERENCES 34 iii

INTRODUCTION In the early stages of propeller design, the engineer must attempt to estimate stresses in his proposed blade design. This operation is quite troublesome since available techniques have been either too simple to be considered accurate or so complicated as to require completion of design before use. The object of this paper is to present a method of preliminary design stress analysis developed by the writer. It is believed that this system will allow a relatively accurate check to be made on blade stresses in the earliest design stages. Admiral David W. Taylor's stress theory, while relatively simple to use, is based on outmoded ogival sections and is really so overly simplified as to raise doubts as to its accuracy. Later attempts at improvement on Taylor's method have culminated in the vortex theory which is unsuitable for preliminary design purposes. Many designers prefer Taylor's method, even for final analysis, since even the vortex theory is only approximate (mathematics of blade loading being still unknown) and large safety factors are required in any event. The method presented in this paper is somewhat akin to Taylor's method in that both are based on the blade element theory which assumes a straight line relationship between both thrust and torque and the change of blade radius. Both theories neglect the losses due to the free vortex and blade surface friction, these being small enough to be safely overlooked in preliminary work. In both cases stresses are based on the prismatic beam theory. The method presented here is felt to be an improvement on Taylor's, however, in that centrifugal forces have been introduced as a factor and constants used are appropriate for modern airfoil blade sections. The present development has also resulted in equations which are easier to apply than are Taylor's. Experimental results, detailed in the Appendix, tend to confirm the theory presented in this paper. 1

Summary of Propeller Stress Formulae For a propeller made of bronze, the following are the maximum stresses: Stress due to torque BDHP CMQ = 4.18 x 205 x Z ~rtr2 Stress due to thrust DHD x D UMT = 365 x Z Va x r x tr2 Stress due to centrifugal moment = 08 x 107 D4 x N2 x DAR MF = 1.08 x 10r xtr Stress due to centrifugal force D3 x N2 x DAR aF = 0.3 x 10-7 X x The total maximum stress a = "-MQ- C'MT ~'MF + aF DHP = Developed area ratio Developed Area Disc Area N = rpm D = tip diameter of blades (in.) Z = number of blades,r = blade width at 0.2 D tr = maximum blade thickness at 0.2 D

1. GENERAL A propeller rotating behind a ship is driven by the main engines through the shaft, which transfers the torque to the propeller. The main object of the propeller is to convert the torque into thrust with optimum efficiency. This is accomplished by the pitched blades acting like a screw. The forces acting on the blades are the following: a. Torque b. Thrust c. Centrifugal d. Vibration Each of the above forces develop stresses in the blade sections. The maximum stress will, in normally shaped blades, be at the root section where the moment due to the forces is the greatest. 2. THE TORQUE 2.1 THE TORQUE FORCE The torque or transverse force acts normal to the propeller axis (see Fig. 1). Its magnitude can be calculated as follows: The work transferred through shaft: W DHP x 35,000 N DHP = Delivered horsepower at propeller's axis N = rpm. Center of torque Fig. 1.

R The work done by the blades: Wb =Z dQ 2ir Z = number of blades Ro = radius at root section R = radius at tip. dQT Fig. 2. Both works are equal; thus rR DHP,x 33,000 = z dQ 2cr N 0 DHP x 33,000 R Z x N x 2t dr It can be assumed that the torque distribution along the blade radius changes linearly with the change of the radius; thus dQ = constant = K dr dQ = Kdr DHP x 33,000 = rdr K I rdr Z x N x 2t JR DHP x 5250 K (R2 Ro2) NxZ 2 DHP x 5250 2 K = x N x Z xR-R as dQ = Kdr Q = K dr = K(R-Ro) o

Substitute the value of K DHP x 5250 2 [R0 =[ N 5 x 22 2 [R-R~] Q DHP x 10,500 (1) N x Z(R+Ro) at root section Ro = 0.2 R DHP x 10,500 DHP x 10,500 x 2 x 12 NxZ x 1.2 R N x Z x 1.2 D Q = 210,000 DHP (2) Q = torque (lb) N = rpm Z = number of blades D = tip diameter (in.) DHP = delivered horsepower 2.2 CENTER OF TORQUE The summation of the element torque distributed over the blade face can be presented by a single force concentrated at a point called the center of attack or torque center (see Fig. 1). Its distance from the propeller axis is RQ. The magnitude of RQ can be calculated by dividing the moment torque at any section by the torque itself. Fig. 3.

The center of torque about an arbitrary radius r is equal to R dQ(r-rl ) by substituting the value of the elementary torque dQ = Kdr The numerator will be dQ(r-rl) = K (r-r)dr = r- = - rl - r l K[ r lR [(2 ) (R 2r) R2 2 r += (Rr2 K )2 - [ 2 Developing the denominator rR R dQ = K dr = K(R-rl) Jl r! Substitute both values in Equation 3 K/2(R-rl)2 R-rl Y = K(R-rl) 2 y R-r (4) 2 at root section r = Ro = 0.2 R y = - (R-0.2R) = 0.4 R 2 y = 0.2 D The value of RQ RQ = 0.4 R + 0.2 R = 0.6 R RQ = 0.6 R = 0.3 D. (5)

2.3 MOMENT DUE TO TORQUE The torque moment at root section will be MQ = Q x y; by substituting the value of y in Equation DHP M = 210,000 x DHP x 0.2 D NxZ m = 42,OOO x DHP (6) l x Z 3. THRUST 3.1 GENERAL Center of thrust Fig. 4. It was mentioned that the propeller converts the torque into thrust. The thrust has to overcome the resistance or the tow rope horsepower (TRHP) of the ship's hull. By the interaction of the propeller and the hull on augmentation or a thrust decuction is involved, and as a result the stream line pressure behind the ship is reduced, thus only part of the thrust developed by the propeller is transferred to overcome the hull resistance. 3.2 THE MAGNITUDE OF THE THRUST The relation between the hull resistance and thrust is expressed as follows.

R = T(l-t) (7) The thrust being a function of the horsepower and speed of ship is expressed as T = 33,oo000 TRHP (8) 101.33 VK(l-t) R = hull resistance T = thrust developed by the propeller (l-t) = thrust deduction VK = speed of ship in knots The thrust can be expressed in terms of propeller's speed and horsepower, where Va = VK(l-w) Va vK (l-w) TRHP = DHP x P.C. Substitute in Equation 8 T = 33,o00 x DIP x P.C.(l-w) (9) 101.33 Va (1-t) Va = speed of advance of the propeller (knots) w = percentage of wake P.C. = propulsive coefficient DHP = delivered hp at the propeller The following average numerical values are used for vessels with one shaft 1-t = 0.80 For l-t = 0.95 For 1-w = 0.70 low-speed 1-w = 0.80 high-speed P.C. = 0.75 ships P.C. = o.65 3 ships P.C. x (l-w) = 0.75 x 0.70 = o.656 (l-t) 0.80 P.C. x (l-w) - o.65 x 0.80 = 0.547 (l-t) 0.95 33,000 x DHP Z x 101.53 x V x o.66 8

For slow rpm T = 214 DHP (10) Z Va For high rpm T = 178 DHP Z Va 3.3 THE CENTER OF THRUST The following calculations were based on the assumption that the thrust is linearly related to the radius (see Fig. 5). Tr a t, RT -- r Blade radius Fig. 5. The elementary thrust dT = Trdr The total thrust T = Trdr. (11) OR Fig. 6.

Substitute the value of Tr = Kr in Equation 11 rR T = K rdr JR0 K T = 2 (R1-Ro2) (12) K was eliminated from Equation 12 2T R2_Ro2 The thrust from tip to any radius rl can be expressed as follows: R K T1 = K rdr = - (2-r 12) (1) J1 2 Substitute the value of K in Equation 13 T1 = T B2 - r (1_4) The moment thrust at the radius rl will be dM = dT(r-rl) (15) as dT = Trdr = Krdr than dM = Krdr(r-rl ) by substituting the expression for K dM = R2 2 rdr(r-rl) R2-Ro2 The moment from tip to the radius rl M1 B R21 r(r-rl)dr 2T r= r2rB 10

2T f -2r Ml 2T R3 R2rl_ r12 + ri3 1 R2-Ro2 3 2 3 2 r&D3 7D2 I ml T r2R3 - 3R2r1 + rl3] R2-R0 L j M1 = T [2R3 3R2rl + r13] 5 (B2-R2) at root section where rl = Ro T M = 3(+Ro R) (2R2 - RRo - Ro2) (16) The thrust moment about the root section can be expressed as the total thrust concentrated at the center of thrust multiplied by the lever arm to the root section. M = T(RT-Ro). (17) By equating Equations 16 and 17 RT - R Bo - 2 3(R+Ro) 2R2 _RRo - Ro2 RT = Ro + 3T (R+Ro) (18) of root section Ro = 0.2 R 2R2 - 0.2 R2 - 0.04 R2 RT =Ro + 3.6 RT = 0.69 R. (19) 3.4 THE MOMENT DUE TO THRUST M = T(RT - Ro) M = T(0.69 R - 0.2 R) M = Tx 0.49R 11

Substitute the value of T from Equation 10 T = 214 HP x O.49 R Z Va Low rpm MT = 52.4 DHP x D (20) Z X Va High rpm T 36 DHP x D Z x V 4. CENTRIFUGAL FORCE 4.1 THE EQUATION OF THE CENTRIFUGAL FORCE F mv mw2r (21) r For a given speed w is constant; thus the force will change with the change of r Sh " -... dm Fig. 7. The forces at section x-x equal to the centrifugal force due to the mass beyond the section. The elementary mass dm = Z S dr (22) g where Y = specific weight S = area of section. 12

The total force rR F = Sw2 rdr (23) g J The expression of S as a function of r is very complicated and without approximations it can hardly be derived. This is due to the variation of the blade section thickness and shape from tip to root and also the shape of the blade contour. The simplest expression to use will be the ratio of mean thickness tm maximum thickness of hub tr This ratio was experimentally found to be 0.428 for most of the propeller blades. The approximate weight of blade: G = y x 144 x A x 0.428 r (24) Z G = weight of one blade (lb) A = developed area of blade (sq ft) y = specific weight of material (lb per cu in.) For bronze y = 0.315 For cast steel y = 0.284 tr = maximum thickness of blade section at root RG = radius of center of blade weight (in.) VG = the velocity of the center of blade (fpm). By experimental test the approximate distance of center of the blade's mass was found to be RG = 0.485 R The velocity of the center VG 2= RG N 60 VG2x weight of one blade (25) RG g By substituting the values of Vo and G in Equation 25 13

F = IyDN2 Atr (26) 2350 Z D = tip diameter (in.) A = developed area (sq ft) 4.2 MOMENT DUE TO CENTRIFUGAL FORCE F Rake Fig. 8. Most of the new built propellers are raked at a certain angle. It is acceptable to use an angle of 5 degrees for the average type of propeller. For a 5-degree angle the lever arm b (see Fig. 8) will be b = r x tg 50 x 0.485 b = 0.0875 x 0.485 x r b = 0.0212 D. The moment due to centrifugal force M = Fxb; using F from Equation 26 yD2N2 A tr 111,000 xZ 14

Substituting the value of the developed area ratio (DAR) DAR = developed area disc area A DAR icD2/ DAR x oD2 4 x 144 Substitute the value of A. yD2 N tr DAR x iD2 MF x - lll,000xZ 4 x 144 7zD2N x DAR x tr =O O (27) 2.03 x 107 x Z 5. VIBRATION Stresses due to vibration of propeller blades are known to exist. In a well shaped and balanced propeller the vibrations are considerably reduced. The critical vibration will rise when the frequency of the blades is equal to the natural frequency of the hull. At high modes the effect of the resonance is much more destructive. There are two sources of vibration; either by unbalanced blades or by the shock of the blades entering and leaving the wake region. The vibration frequency due to the unbalanced blade equals the number of revolutions per minute. The frequency of the second type is equal to the number of blades multiplied by the rpm and an integer number. The vibration may be either flexural, where the blade vibrations are perpendicular to its surface, or torsional, where the blade twists around the root section. The calculation of the stress due to vibration, which is a fatigue stress, is laborious due to the many factors involved; in general they will not exceed the stresses due to the moment. When a propeller is well designed and perfectly balanced, and its natural frequencies such that they will not be in phase with those of the hull, especially at high modes, no vibration stress problem will exist. 15

6. PROPERTIES OF BLADE SECTION Figure 9 shows three types of blade section of which the aerofoil type is the most commonly used. d m - = o.-k t A m Td L 0.5 T = 0.5 1, (a) Ogival d = 0.5 m = 0.5 7 l (b) Symmetrical d - = 0.44 _m t - = 0.46 (c) Aerofoil Fig. 9. The areas and moment of resistance of cross sections were experimentally tested. The constants tabulated in Table I are average values which are very close to actual conditions. The area of the blade section can be expressed as S = K I t (28) as S = area (in.2) t = thickness of section (in.) = width of section (in.). 16

yI CY xI- it - - -Ymt -x Fig. 10. The moment of inertia about xx axis Ixx = K t3 (29) The moment of inertia about yy axis I = p 13t. (30) The constants Kn and Kp are tabulated in Table I. The xx and yy axes are parallel and normal to the face of the blade. These axes are also called the principal axes. TABLE I Type of Section Ka K Kp Ogival o.67. o46 0. 033 Aerofoil 0.71 o.o46 0.39 7. MOMENT DIAGRAM The three moments acting at different centers of attack are shown in Fig. 11. Figure 12 illustrates the components of the moments on xx and yy axes. M1y is the sum of the moment on yy axis. Mx is the sum of the moment on xx axis. 17

Fig. 11. / 4)4/ Axis i 1 / IW Fig. 12.

The angles can be expressed by the pitch. 2os r COS = -2- -r)2 + p2 at root section Axis cos = 1 1 + [Pr2/(4IRo)2] i + (Pr/IdDo)2 Pr/icDo L sin / = +(r/Do Pitch DO = diameter at root section (in.) Pr = pitch at root section (in.) Fig. 13. The average ratio for Pr/Do = 0.45 will result in a pitch angle at root 0 = 55 degrees. cos 0 = o.57358 sin ~ = 0.81915 8. STRESSES AT ROOT SECTION 8.1 STRESS DUE TO TORQUE Since the moment of resistance of the blade section about yy axis is much greater than that about xx axis, it can be proved that the maximum stress will appear at c (see Fig. 10) which is the point of the greatest distance from the xx axis. The stress equation for a beam can be applied. Mxy Ixx From Fig. 12 the component of the torque along yy axis is MyQ = MQ sin 0 Vqy = 0.81915 MQ The stress at c will be ~Q = My~ x Ymax Ixx 19

From Equation 6 MQ 42,000 DHP NxZ From Equation 30 Ixx = Kn itr3 = 0.046 itr3 From Fig. 9 Ymax = 0.56 tr Thus stress at root section aIQ1 =42 000 x DP_ x 0.81915 x 0.56 tr 3 N x Z 0.046 irtr3 aMQ = 4.18 x 105 x DKP N Z Irtr2 (2r and tr expressed in inches). 8.2 STRESS DUE TO TERUST The moment due to thrust from Equation 20: DHP x D MT = 52.4 Z x Va zx Va The component of this moment on yy axis: MYT = MT cos. ]DHP x D MYT = 52.4 x 0.57358 x MYT = 0 DHP Z x Va The maximum stress at point c will be MyT x Ymax Ixx 30 x DIHP x 0.5s6 tr Z x Va x 0.046 Qrtr3 20

DHP x D Low rpm a = 365 x Z x Vatr (32) DHP x D High rpm MT = 304 V x lrtr (D, Qr, and tr expressed in inches) 8.3 STRESS DUE TO CENTRIFUGAL MOMENT The moment due to the force and rake arm was expressed in Equation 27. yD4N2x DAR x tr 2.03 x 107 x Z The component on yy axis: MYF = C os MyF = 2.82 x 10-l8 D4N2 x DAR x tr z The stress at point c: MF = MYF x Ymax Ixx cr = 3.44 x 10-7 D4N2 DAR (33') Z x Irxtr For bronze y = 0.315 o4N2 I.AR (Low rpm) r = 1.08 x 10-7 D (34) F - Z XXIr x tr D4N2 DAR (High rpm) aF = 7.8 x 10 x x t Z x Ir x tr For cast steel y = 0.284 (Low rpm) = 9.8 x 1- D4N2 DA(35) Z x Lr x tr (High rpm) = 7 x 10-8 D4N2 DAR Z x r x tr 21

8.4 STRESS DUE TO CENTRIFUGAL FORCE The centrifugal force due to the mass of the blade tends to tear off the blade from its root. This force creates a tensile stress over the entire cross-sectional area. F aF = Sr Sr = area of blade section at root. Sr = Ka Lrtr From Table I Ka = 0.71 Sr = 0.71 ~rtr From Equation 26 F = zDN2 A tr 2350 Z or F = 2.32 x 10-6 yDN2 x DAR x tr (36) =.27 x 0-6 ZD3N2 DAR Z x ~r For bronze 7 D3N2 DAR aF = 10.3 x 10- 7 D IA (38) r For cast steel D3N2 DAR CF = 9.3 x 10-7 (3 DA9) 8.5 MAXIMUM TOTAL STRESS The maximum stress at the root section will be at point c which is the furthese point from xx axis. As shown in Figs. 11 and 12, the stresses at point c due to torque thrust and centrifugal moment are compressive stresses while the stress at the same point due to the centrifugal pull on the section area is a tensile stress. The total stress at the point will be 22

2 = & cMQ - MIvT % IF + aF For a bronze propeller and low rpm DHP 365 x DHP x D -7 D4N2 DAR N x Z x 0rtr2 Z x Va x ~rtr~2 Z lrtr (40) + 10.3 x 10-7 D3N2 DAR Z x Ir For a bronze propeller and high rpm DH DHP x D _ DN2 DAR a = -4.18 x 105 304 x.-7.8x.0 N x Z x rtr2 Z x Vax Irtr2 Z x Vartr (40) +10.3 x 10-7 D3N2 DA Z x r 23

APPENDIX I STRAIN TEST ON PROPELLER BLADE Propeller's Characteristics Diameter = 19" = 1.583 ft Disc area = 283.53 sq in. Developed area = 144 sq in. DAR = o. 5o8 Number of blades = 3 Tip pitch = 13.35" Mean pitch = 12.88" Root pitch = 9.81" at 0.2 D Pitch ratio (a) = 0.678 Max thickness of blade at root = 0.58" Width of blade at root = 4.1" A G.M. diesel engine developing 110 hp (max) at 1840 rpm was selected for driving the propeller with no gear reduction. Ting Troost charts for threebladed screw series type B3-50 the following table was arranged. The speed of advance Va was arbitrarily chosen. TABLE II Va Va2 5 8 Bp a E 8.3 200 350 96.3 0.62 43.6 = 11 400 265 48.2 O. 675 52.2 Va 12 500 243 38.5 0.695 55 3 BD 14.55 800 200 24.1 0.74 61.2 BP 2 Va2.5 From the curves on Fig. 14, the efficiency and speed corresponding to the propeller pitch ratio,a = 0.678 are: = 47.8% Va = 11.15 knots 24

0 z w o64 ljo I 60 i 58 0.76 56 0.74- 54 0.721 52 0.70 50 / E= 47.8% a=0.678 0.68 48- ___ o.66 46/ 0.64 44 -.. Ile 0.62 42 1 0.60 40. I.. 8 9 10 1 1 12 13 14 15 16 - - Va (Knots) Fig. 14. Efficiency, pitch ratio versus speed of propeller. THE LOADS APPLIED IN TEST 1. Thrust. —From Equation 10, T = 178 DHP/ZVa for high rpm. As the propeller is driven directly by the engine shaft without a gear reduction, we can neglect friction losses in shaft bearings and use the shaft horsepower as the delivered horsepower. 178 x 110 l 3 x 11:.15 5 The center of thrust Rt = 0.69 x R = 6.55". 2. Torque. — Q = 210000 DHP Nx 2 xD QA 2210,000 x 110 220 lb 1840 x 3 x 19 - The center of torque RQ = 0.6 R = 5.7". 25

3. Centrifugal Force. — zDN2 Atr 0.315 x 19 x 18402 1 x 0.58 F 2350 Z 2350 x 3 F = 1665 lb Center of force RF = 0.485 R = 4.61". TEST PROCEDURE The propeller was secured to a 12" x 12" x 1/2" plate. The plate was fastened to a built-up frame by clamps (Fig. 15). Holes of 1/4" were drilled at the centers of the forces where a bolt was secured so that the weight basket could be suspended. Bars of approximately 50 lb were used as loads. The maximum load which could be handled in the test was 715.5 lb. The centrifugal load was applied as shown in Figs. 15 and 16. The position of the loaded blade in the torque loading is shown in Fig. 17. The blade was secured horizontally and the load was applied vertically. Fig. 15. Fig. 16. For the thrust loading the propeller was turned and secured as shown in Fig. 1]8. In each test three strain gages of 1/4" and gage factor of 1.91 were used. The gages were tied to the propeller one at each side of the blade at root section at the maximum thickness of blade (center of blade). The third gage, was a dummy gage tied to a free blade for temperature compensation. The wiring, gages, and strain indicator are shown in Fig. 15.

Fig. 17. Fig. 18. The results of the strain indicator readings are tabulated in the following tables. TORQUE TEST ON PROPELTLER BLADE Gage No. 1 0ont Gage No. 2 Back Indicator Increase'Indicator Increase Test Load Ref 4Reading Difee inStrain Ref Reading in Strain NoI. nlb ~ pin./in. lbkin. /in. I n./in. uin./in. in./in. 1 0 3 1335 0 0 6 1210 0 0 2 83 3 1170 -165 -165 6 1360 150 150 3 132 3 1080 - 9C -225 6 1445 85 235 4 L82.5 3 975 -105 -360 6 1535 90 325 5 220 3 900 - 75 -435 6 1605 70 395 6 283.5 3 775 -125 -560 6 1715 110 505 7 331 3 680 - 95 -655 6 1800 85 590 Note: uin. means microinch. 27

THRUST TEST ON PROPELLER BLADE Gage No. 1 (Front) Gage No. 2 (Back)! Indicator Increase Indicator Increase TestLoad Ref Readig Difference Difference toa nRef/Readino. in Strain Reading in./in. in Strain l NI jion./in.'''n.I' I,n.in.in./in. 1 0 6 1320 0 0 3 1420 0 0 2 112.5 6 1130 -190 -190 3 1577 157 157 3 218 6 935 -195 -385 3 1770 193 350 4 343 6 730 -205 -590 3 1950 180 530 5 440 6 570 -160 -70 *4 1080 150 680 6 541 6 390 -180 -930 4 1240 160 840 7 630 6 245 -145 -1075 4 1375 135 975 *The difference between ref 3 and 4 is 980 kin./in. CENTRIFUGAL FORCE TEST Gage No. 1 (Front) Gage No. 2 (Back) Indicator L Increase Indicator Increase Test lLoad Difference Difference Sti Refeading in./in. in Strain Ref Reading in. /in. in Strain Re Reaing tin./in in./in. l in./in. |uin./in. No. lb k~in./in. 1 0 8 890 0 0 7 155 0 0 2 214.5 8 850 -40 -40 7 16o 5 5 3 311.5 8 835 -15 -55 7 165 5 10 4 412 8 815 -20 -75 7 170 5 15 5 513 8 800 -15 -90 7 175 5 20 6 612 8 785 -15 -105 7 180 5 25 7 715.5 8 765 -20 -125 7 185 5 30 The tabulated values were plotted on Figs. 19, 20, and 21 and the strain corresponding to the required forces are as follows: Load Strain lb kin./in. Torque 220 435 Thrust 586 1010 Centrifugal force 1665 290 THE STRESS CALCULATED FROM STRAIN For biaxial strain the stress equation is a E(E1 + kE2) 1 - 2 28

600 / 500 435 Micro-in.per in. / Co 400 / Z 300 / I- Compression / 200 // // ~l 0 50 100 150 200 250 300 LOAD (LBS.) Fig. 19. Strain-load diagram for torque. 1200 0 1100 / /? 1/ 0 50 100 150 200 250 300 00700'.J Compressn Tension /;z 7300 2500 /9 I 400 7oo 200 0 100 200 300 400 500 600

300 290 Micro- in.per in. / 250 I.s200 7 Compression / o I>7 C ID 100/ 50 7 U) 7 100 I 50 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 LOAD (LBS.) Fig. 21. Strain-load diagram for centrifugal force. E = strain (in./in.) stress (psi) E = modulus of elasticity = 15 x 106for bronze = Poissons ratio = 0.35 for bronze y I - jy

For the blade section of the tested propeller which is symmetric about the yy axis no tension or compression stress will exist at point c due to the moment Mx along xx axis. The center of attack at 0 is on the same yy axis as c. Due to the moment My point c will be under the maximum tensile stress which will be uniaxial. The uniaxial stress will be expressed as a = EC Applying the results of the strain test, the following stresses resulted. Torque stress aQ = 15 x 106 x 435 x 10- = 6520 psi Thrust stress aT = 15 x 106 x 1010 x 10-6 - 15,150 psi Centrifugal stress oF = 15 x 106 x 290 x 10-6 = 4350 psi Total stress = 6520 + 15,150 + 4350 = 26,020 psi. STRESS CALCULATION BY PROPOSED EQUATIONS 4.18 x 105 x DHP 4.18 x 105 x 110 6,o4o psi Torque rMQ =,840 3 41 0psi N x z x Ir x tr 2 1840 x 3 x 4.1 x 0.58 Thrust M 304 x DHP x D 304 x 110 x 19 Z x Va x l' t2 3 x 11.15 x 4.1 x 0.582 Centrifugal 78 x 10-8 4 N2 8 4 2 Centrifugal 7.8 x 10 x D x N2 x DAR 7.8 x 10-8 x 194 x 1840 x 0.52 Moment NMF Z x r x tr 3 x 4.1 x.58 2,490 psi Centrifugal 10.3 x 10-7 x D3 x N2 x DAR 10.3x10-7 x 193x18402x0.52 Force oF Z x ~r 3 x 4.1 1010 psi Total stress a = 6040 + 13,900 + 2490 + 1010 = 23,440 psi 31

APPENDIX II STRESS CALCULATION FOR THE FOLLOWING PROPELLER The following propeller data is required w~hen calculating stress by the proposed equations: Diameter D = 210 in. Shaft horsepower SHP = 4000 Number of blades Z = 4 Revolutions per minute rpm = 87 Speed of ship VK = 14.5 knots Developed area D.A. = *110 sq ft Thickness of blade at root tr = 7.12 in. Width of blade at root r = 38.3 in. CALCULATION The developed horsepower will be 3850 hp considering about 5% bearing losses. Speed of advance of propeller Va = VK(1 - w) = 14.5 x 0.7 Va = 10.15 knots Developed area ratio 110 x 4 DAR - 10 2 = 0.458 x ( -) 12 4.18 x l0O x 3805o Stress due to torque aM = 87 x 4 x 38.3 x 7.12 = -2390 psi Stress due to thrust 36 x 3850 x 210 GMT = 4 x 10.15 x 38.3 x 7.122 Stress due to centrifugal moment 1.08 x 10-7 x 2104 x 872 x 0.458 4 x 38.3 x 7.12 32

Stress due to centrifugal force 10.3 x 10-7 x 2103 x 872 x 0.458 +160 psi'F = 4 x 38.3 Sum of stress = - 2390 - 3750 - 660 + 160 = - 6640 psi CONCLUSIONS FROM TESTING The strain test as described in Appendix I does not show a complete picture of the stress distribution in the blade. Such a test involves a great number of strain gages. The main object of the test was to estimate the maximum stress by assuming the location and direction of the maximum strain, based on the theory and conditions described in the paper. The results of the strain Zage test compared with those calculated by the proposed equations clearly indicate that a stress concentration factor has not been considered in the calculations. This concentration factor is a result of both the single load applied at the center of attack, and the change from the blade section to the root section. These factors explain why the stress measured by strain gages are higher than those calculated. There is no reason to expect the stress results, obtained by the two different methods, to be very close. To the writer they seem too close. From the two examples in Appendices I and II, it can be stated that the proposed equation lead to a fairly accurate prediction of the stress in the blade. In case of high speed vessels, the constants in the equations have been changed. It should also be remembered that these constants change wshen a twin screw vessel is considered. The difference will be due to valuations in the thrust load, the wake thrust deduction and propulsion coefficients with resulting variations in loadings. For high speed vessels the factors mentioned above will change, particularly the centrifugal force which will have considerable effect on the stress. To decrease the centrifugal stresses, in most high speed wheels, the rake of the blades is decreased. These factors have been considered and the constant in the equation for calculating the stress due to centrifugal force has been reduced in proportion to the decrease in the rake. The torque stress constants are independent of rpm. 33

UNIVERSITY OF MICHIGAN 11111111111111111 11111111111 1111 3 9015 03466 3917 REFERENCES 1. W. B. Morgan, An Approximate Method of Obtaining Stress in a Propeller Blade, David W. Taylor Model Basin, Report 919, Washington, D. C., 1954. 2. W. Mackle, "Stress in Propeller Blades," Shipbuilder and Marine Engine Builder, Vol. 38, 1941. 3. S. F. Dorey, "Stresses in Propellers," Transactions of the Institution of Naval Architects, 1951. 4. G. C. I. Gardiner, "Aspect of Propellers' Fatigue," Aircraft Engineering, 1949. 5. G. S. Baker, "Vibration Patterns of Propeller Blades," Shipbuilder and Marine Engine Builder, Vol. 38, 1941. 6. L. A. Baier, "Propellers of Propulsion," Paper presented to the Society of Naval Architects and Marine Engineers, Great Lakes Section, Feb. 3, 1956. 34