PM
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3
Torsion
Torsional Deformations
Problem 3.2-1 A copper rod of length L 18.0 in. is to be twisted by torques T (see figure) until the angle of rotation between the ends of the rod is 3.0. If the allowable shear strain in the copper is 0.0006 rad, what is the maximum permissible diameter of the rod?
d T T
L
Probs. 3.2-1 and 3.2-2
Solution 3.2-1
Copper rod in torsion
d T T
L
L f
18.0 in. 3.0 p (3.0)a 180 b rad
From Eq. (3-3):
max
rf L 2L
df 2L
allow
0.05236 rad
allow
dmax dmax
0.0006 rad
f 0.413 in.
(2)(18.0 in.)(0.0006 rad) 0.05236 rad
Find dmax
;
Problem 3.2-2 A plastic bar of diameter d 56 mm is to be twisted by torques T (see figure) until the angle of rotation between the ends of the bar is 4.0. If the allowable shear strain in the plastic is 0.012 rad, what is the minimum permissible length of the bar?
249
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Solution 3.2-2
NUMERICAL DATA d
a
56 mm 0.012 radians 4a p b radians 180 Lmin df 2a
f
solution based on Equ. (3-3): Lmin 162.9 mm
;
Problem 3.2-3 A circular aluminum tube subjected to pure torsion by torques T (see figure) has an outer radius r2 equal to 1.5 times the inner radius r1.
(a) If the maximum shear strain in the tube is measured as 400 10 6 rad, what is the shear strain 1 at the inner surface? (b) If the maximum allowable rate of twist is 0.125 degrees per foot and the maximum shear strain is to be kept at 400 10 6 rad by adjusting the torque T, what is the minimum required outer radius (r2)min?
T
T
L r2 r1
Probs. 3.2-3, 3.2-4, and 3.2-5
Solution 3.2-3 NUMERICAL DATA
r2 u 1.5r1 0.125a 1.818
max
(b) MIN. REQUIRED OUTER RADIUS 400 (10 6) radians r2min r2min
max
u
r2min
max
u
p 1 ba b 180 12 10
4
2.2 inches
;
rad /m.
(a) SHEAR STRAIN AT INNER SURFACE AT RADIUS r1
1
1
r1 r2 267
max
1
6
1 1.5
max
10
radians
;
Problem 3.2-4 A circular steel tube of length L
1.0 m is loaded in torsion by torques T (see figure).
(a) If the inner radius of the tube is r1 45 mm and the measured angle of twist between the ends is 0.5, what is the shear strain 1 (in radians) at the inner surface? (b) If the maximum allowable shear strain is 0.0004 rad and the angle of twist is to be kept at 0.45 by adjusting the torque T, what is the maximum permissible outer radius (r2)max?
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Torsional Deformations
251
Solution 3.2-4
NUMERICAL DATA L r1 f 1000 mm 45 mm 0.5 a p b radians 180 (b) MAX. PERMISSIBLE OUTER RADIUS f 0.45 a
max
p b radians 180
max
r2
f L L f
0.0004 radians 50.9 mm
r2max
max
(a) SHEAR STRAIN AT INNER SURFACE f r1 393 10 6 radians 1 1 L
r2max
;
;
Problem 3.2-5 Solve the preceding problem if the length L 0.5, and the allowable shear strain is 0.0004 rad.
56 in., the inner radius r1
1.25 in., the angle of twist is
Solution 3.2-5 NUMERICAL DATA
L f
a
(b) MAXIMUM PERMISSIBLE OUTER RADIUS (r2)max 1.25 inches f
max
a
56 inches r1 0.5 a
p b radians 180
0.5 a
p b radians 180 f r2 L
0.0004 radians (a) SHEAR STRAIN g1 (IN RADIANS) AT THE INNER
SURFACE
0.0004 radians L r2max a f r2max 2.57 inches
1
r1
f L
;
1
195
10
6
radians
;
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Torsion
Circular Bars and Tubes
P
Problem 3.3-1 A prospector uses a hand-powered winch (see figure) to raise a bucket of ore in his mine shaft. The axle of the winch is a steel rod of diameter d 0.625 in. Also, the distance from the center of the axle to the center of the lifting rope is b 4.0 in. If the weight of the loaded bucket is W 100 lb, what is the maximum shear stress in the axle due to torsion?
d W b W
Solution 3.3-1
Hand-powered winch
d b W 0.625 in. 4.0 in. 100 lb MAXIMUM SHEAR STRESS IN THE AXLE From Eq. (3-12): tmax 400 lb-in. tmax tmax 16T pd3 (16)(400 lb-in.) p(0.625 in.)3 8,340 psi ;
Torque T applied to the axle: T Wb
Problem 3.3-2 When drilling a hole in a table leg, a furniture maker uses a hand-operated drill (see figure) with a bit of diameter d 4.0 mm.
(a) If the resisting torque supplied by the table leg is equal to 0.3 N m, what is the maximum shear stress in the drill bit? (b) If the shear modulus of elasticity of the steel is G 75 GPa, what is the rate of twist of the drill bit (degrees per meter)?
d
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Circular Bars and Tubes
253
Solution 3.3-2
Torsion of a drill bit
(b) RATE OF TWIST From Eq. (3-14): u T GIP 0.3 N # m p (75 GPa)a b (4.0 mm)4 32 0.1592 rad/m 9.12/m ;
d
4.0 mm
T
0.3 N m G
75 GPa u
(a) MAXIMUM SHEAR STRESS From Eq. (3-12): tmax 16T pd3 16(0.3 N # m) p(4.0 mm)3 23.8 MPa ;
u
tmax tmax
Problem 3.3-3 While removing a wheel to change a tire, a driver applies forces P 25 lb at the ends of two of the arms of a lug wrench (see figure). The wrench is made of steel with shear modulus of elasticity G 11.4 106 psi. Each arm of the wrench is 9.0 in. long and has a solid circular cross section of diameter d 0.5 in.
(a) Determine the maximum shear stress in the arm that is turning the lug nut (arm A). (b) Determine the angle of twist (in degrees) of this same arm.
P
9.0
in.
A
9.0
in.
d = 0.5 in. P = 25 lb
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Solution 3.3-3
Lug wrench
P L d G 25 lb 9.0 in. 0.5 in. 11.4 106 psi (a) MAXIMUM SHEAR STRESS From Eq. (3-12): (16)(450 lb - in.) 16T tmax 3 pd p(0.5 in.)3 tmax 18,300 psi ; (b) ANGLE OF TWIST From Eq. (3-15): (450 lb-in.)(9.0 in.) TL f GIP p (11.4 * 106 psi)a b (0.5 in.)4 32 f 0.05790 rad 3.32 ;
T torque acting on arm A T P(2L) 2(25 lb) (9.0 in.) 450 lb-in.
Problem 3.3-4 An aluminum bar of solid circular cross section is twisted by torques T acting at the ends (see figure). The dimensions and shear modulus of elasticity are as follows: L 1.4 m, d 32 mm, and G 28 GPa.
(a) Determine the torsional stiffness of the bar. (b) If the angle of twist of the bar is 5, what is the maximum shear stress? What is the maximum shear strain (in radians)?
d T T
L
Solution 3.3-4 (a) TORSIONAL STIFFNESS OF BAR
d kT Ip 32 mm GI p L 1.029 G Ip 28 GPa p4 d 32
(b) MAX SHEAR STRESS AND STRAIN f 5a p b radians 180 d Ta b 2 Ip ;
105 mm4
T
max
kT f
tmax
kT kT
p 2811092a 0.0324 b 32 1.4 2059 N # m ;
27.9 MPa tmax G 997 10
6
gmax
max
radians
;
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Circular Bars and Tubes
255
Problem 3.3-5 A high-strength steel drill rod used for boring a hole in the earth has a diameter of 0.5 in. (see figure). The allowable shear stress in the steel is 40 ksi and the shear modulus of elasticity is 11,600 ksi. What is the minimum required length of the rod so that one end of the rod can be twisted 30 with respect to the other end without exceeding the allowable stress?
T
d = 0.5 in.
T L
Solution 3.3-5
T
Steel drill rod
d = 0.5 in. T L
From Eq. (3-15): f T tmax
TL GIP
32TL Gpd4
Gpd 4f , substitute Tinto Eq. (1): 32L a 16 pd ba 3 Gpd4f b 32L Gdf 2L
G d f
11,600 psi 0.5 in. 30 (30)a 40 ksi 16T pd3 p b rad 180 0.52360 rad
Lmin
Gdf 2t allow (11,600 ksi)(0.5 in.)(0.52360 rad) 2(40 ksi)
allow
MINIMUM LENGTH From Eq. (3-12): tmax (1)
Lmin
38.0 in.
;
Problem 3.3-6 The steel shaft of a socket wrench has a diameter of 8.0 mm. and a length of 200 mm (see figure). If the allowable stress in shear is 60 MPa, what is the maximum permissible torque Tmax that may be exerted with the wrench? Through what angle (in degrees) will the shaft twist under the action of the maximum torque? (Assume G 78 GPa and disregard any bending of the shaft.)
d = 8.0 mm T L = 200 mm
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Torsion
Solution 3.3-6
Socket wrench
ANGLE OF TWIST From Eq. (3-15): f TmaxL GIP pd3tmax 16 IP pd4 32
From Eq. (3-12): Tmax d 8.0 mm 60 MPa L 200 mm G 78 GPa f a pd3t max L ba b 16 GIP
allow
MAXIMUM PERMISSIBLE TORQUE 16T From Eq. (3-12): tmax pd3 3 pd tmax Tmax 16 Tmax Tmax p(8.0 mm) (60 MPa) 16 6.03 N # m ;
3
f
pd3tmaxL(32) 16G(pd4)
2tmaxL Gd 0.03846 rad
f
2(60 MPa)(200 mm) (78 GPa)(8.0 mm) 10.03846 rad2a
f
180 deg/radb p
2.20
;
Problem 3.3-7 A circular tube of aluminum is subjected to torsion by torques T applied at the ends (see figure). The bar is 24 in. long, and the inside and outside diameters are 1.25 in. and 1.75 in., respectively. It is determined by measurement that the angle of twist is 4 when the torque is 6200 lb-in. Calculate the maximum shear stress max in the tube, the shear modulus of elasticity G, and the maximum shear strain max (in radians).
T
T
24 in.
1.25 in. 1.75 in.
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Circular Bars and Tubes
257
Solution 3.3-7 NUMERICAL DATA
L f 24 in. 4a r2 1.75 in. 2 T r1 1.25 in. 2
MAX. SHEAR STRAIN
max
gmax ; G
r2 f L tmax gmax
0.00255 radians
p b radians 180
6200 lb-in. Tr2 Ip 0.681 in.
4
SHEAR MODULUS OF ELASTICITY
G
G MAX. SHEAR STRESS Ip p4 1 r2 2 Tr2 Ip r142
max
3.129 G
106 psi TL fIp G 3.13 106 psi
tmax Ip
or
;
tmax
7965 psi
;
Problem 3.3-8 A propeller shaft for a small yacht is made of a solid steel bar 104 mm in diameter. The allowable stress in shear is 48 MPa, and the allowable rate of twist is 2.0 in 3.5 meters. Assuming that the shear modulus of elasticity is G 80 GPa, determine the maximum torque Tmax that can be applied to the shaft.
d T T
L
Solution 3.3-8
NUMERICAL DATA d 104 mm p 2a b 180 rad 3.5 m p4 d 32 Ip
a
FIND MAX. TORQUE BASED ON ALLOWABLE RATE OF TWIST 48 MPa u f L Tmax Tmax G 1.149 80 GPa 107 mm4 GIpf L Tmax GIp ;
u Ip
9164 N # m ^ governs
FIND MAX. TORQUE BASED ON ALLOWABLE SHEAR STRESS taIp Tmax 10,602 N # m Tmax d 2
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CHAPTER 3
Torsion
Problem 3.3-9 Three identical circular disks A, B, and C are welded to the ends of three identical solid circular bars (see figure). The bars lie in a common plane and the disks lie in planes perpendicular to the axes of the bars. The bars are welded at their intersection D to form a rigid connection. Each bar has diameter d1 0.5 in. and each disk has diameter d2 3.0 in. Forces P1, P2, and P3 act on disks A, B, and C, respectively, thus subjecting the bars to torsion. If P1 28 lb, what is the maximum shear stress max in any of the three bars?
P3
C
P1 A
135 d1 D 135 P1 90
P3
d2
P2 P2 B
Solution 3.3-9
Three circular bars
THE THREE TORQUES MUST BE IN EQUILIBRIUM
T3 is the largest torque T3
MAXIMUM SHEAR STRESS (Eq. 3-12) 16T3 16P1d2 12 16T tmax 3 3 pd pd1 pd3 1 tmax 16(28 lb)(3.0 in.) 12 p(0.5 in.)3 4840 psi ;
T1 12
P1d2 12
d1 d2 P1 T1
diameter of bars 0.5 in. diameter of disks 3.0 in. 28 lb P1d2 T2 P2d2 T3 P3d2
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Circular Bars and Tubes
259
Problem 3.3-10 The steel axle of a large winch on an ocean liner is subjected to a torque of 1.65 kN m (see figure). What is the minimum required diameter dmin if the allowable shear stress is 48 MPa and the allowable rate of twist is 0.75/m? (Assume that the shear modulus of elasticity is 80 GPa.)
T
d T
Solution 3.3-10
NUMERICAL DATA T ua 1.65 kN # m 0.75 a p rad b 180 m
a
dmin 48 MPa G 80 GPa
0.063 m
dmin
63.3 mm ^ governs
;
MIN. REQUIRED DIAMETER OF SHAFT BASED ON ALLOWABLE
SHEAR STRESS
MIN. REQUIRED DIAMETER OF SHAFT BASED ON ALLOWABLE RATE OF TWIST
t
Td 2Ip
t
u
T GIp
4
Ip
T Gu
p4 d 32 32T a b pGua
1 4
T Gu dmin
Td p4 2a db 32
1
d
32T pGua
dmin
16T 3 c d pta
dmin dmin
0.056 m 55.9 mm
Problem 3.3-11 A hollow steel shaft used in a construction auger has outer diameter d2 6.0 in. and inner diameter d1 4.5 in. (see figure). The steel has shear modulus of elasticity G 11.0 106 psi. For an applied torque of 150 k-in., determine the following quantities:
(a) shear stress (b) shear stress (c) rate of twist
2 1
at the outer surface of the shaft, at the inner surface, and (degrees per unit of length).
d2
Also, draw a diagram showing how the shear stresses vary in magnitude along a radial line in the cross section.
d1 d2
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Torsion
Solution 3.3-11
Construction auger
d2 d1 G T IP 6.0 in. 4.5 in. 11 150 k-in. p4 (d 32 2
4 d1 )
r2 r1
3.0 in. 2.25 in.
(c) RATE OF TWIST u T GIP (150 k-in.) (11 * 106 psi)(86.98 in.)4
6
106 psi u 86.98 in.4
157 * 10
rad/ in.
0.00898/ in.
;
(d) SHEAR STRESS DIAGRAM
(a) SHEAR STRESS AT OUTER SURFACE t2 Tr2 IP (150 k-in.)(3.0 in.) 86.98 in.4 5170 psi ;
(b) SHEAR STRESS AT INNER SURFACE t1 Tr1 IP r1 t r2 2 3880 psi ;
Problem 3.3-12 Solve the preceding problem if the shaft has outer diameter d2 150 mm and inner diameter d1 Also, the steel has shear modulus of elasticity G 75 GPa and the applied torque is 16 kN m.
100 mm.
Solution 3.3-12 d2 150 mm
d1 G T IP 100 mm 75 GPa
Construction auger r2 75 mm
r1 50 mm
(b) SHEAR STRESS AT INNER SURFACE t1 Tr1 IP r1 t r2 2 20.1 MPa ;
16 kN # m p4 (d2 d14) 32 Tr2 IP
39.88 * 10 mm
6
4
(c) RATE OF TWIST T 16 kN # m u GIP (75 GPa)(39.88 * 106 mm4) 0.005349 rad/m (d) SHEAR STRESS DIAGRAM 0.306/m ;
(a) SHEAR STRESS AT OUTER SURFACE t2 (16 kN # m)(75 mm) 39.88 * 106 mm4 30.1 MPa ;
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Circular Bars and Tubes
261
P B
Problem 3.3-13 A vertical pole of solid circular cross section is twisted by horizontal forces P 1100 lb acting at the ends of a horizontal arm AB (see figure). The distance from the outside of the pole to the line of action of each force is c 5.0 in. If the allowable shear stress in the pole is 4500 psi, what is the minimum required diameter dmin of the pole?
c A P d
c
Solution 3.3-13
Vertical pole
tmax ( P c 1100 lb 5.0 in. 4500 psi P(2c + d)d pd /16 (16P)d
4
16P(2c + d) pd3 32Pc 0
3 max)d
SUBSTITUTE NUMERICAL VALUES: UNITS: Pounds, Inches ( )(4500)d3 or d3 1.24495d 12.4495 d dmin 0 2.496 in. 2.50 in. ; Solve numerically: (16)(1100)d 32(1100)(5.0) 0
allow
Find dmin
TORSION FORMULA Tr Td tmax IP 2IP T P12c + d2 IP pd 4 32
Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P and the allowable shear stress is 30 MPa.
5.0 kN, the distance c
125 mm,
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Torsion
Solution 3.3-14
Vertical pole
TORSION FORMULA tmax Tr IP Td 2IP IP pd 4 32 16P (2c + d) pd 3 32Pc 0
T
P(2c + d) P(2c + d)d pd /16
3
tmax (
4
max)d
(16P)d
SUBSTITUTE NUMERICAL VALUES: P c
allow
5.0 kN 125 mm 30 MPa
UNITS: Newtons, Meters ( )(30 or d3 848.826 10 6d d dmin 212.207 64.4 mm 10
6
106)d3
(16)(5000)d
32(5000)(0.125) 0
0
Find dmin
Solve numerically:
0.06438 m ;
Problem 3.3-15 A solid brass bar of diameter d 1.25 in. is subjected to torques T1, as shown in part (a) of the figure. The allowable shear stress in the brass is 12 ksi.
(a) What is the maximum permissible value of the torques T1? (b) If a hole of diameter 0.625 in. is drilled longitudinally through the bar, as shown in part (b) of the figure, what is the maximum permissible value of the torques T2? (c) What is the percent decrease in torque and the percent decrease in weight due to the hole?
T1
d
T1
(a) d T2 T2
(b)
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Circular Bars and Tubes
263
Solution 3.3-15
d2
1.25 in. d1 ta
0.625 in.
a
12 ksi
(a)
MAX. PERMISSIBILE VALUE OF TORQUE
T1 SOLID BAR
T2max
T1max
T1max T1max T1max
T1max d 2 1 tapd3 16 1 (12)p (1.25)3 16 ; 4.60 in.-k
taIp
ta
p4 d 32 d 2
T2max T2max (c) PERCENT
p 1 d24 d142 32 d2 2 d24 d12 1 tp 16 a d2 ; &
PERCENT DECREASE
4.31 in.-k
DECREASE IN TORQUE
IN WEIGHT DUE TO HOLE IN
(b)
(b) MAX. PERMISSIBILE VALUE OF TORQUE T2
HOLLOW BAR
percent decrease in torque T2max T1max (100) 6.25% T1max
;
percent decrease in weight (weight is proportional to x-sec area) A1 A1 A1 p2 d 42 A2 (100) A2 p2 1d 42 ; d122
d1 d2
25 %
Problem 3.3-16 A hollow aluminum tube used in a roof structure has an outside diameter d2 104 mm and an inside diameter d1 82 mm (see figure). The tube is 2.75 m long, and the aluminum has shear modulus G 28 GPa.
(a) If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist (in degrees) when the maximum shear stress is 48 MPa? (b) What diameter d is required for a solid shaft (see figure) to resist the same torque with the same maximum stress? (c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft?
d1 d2 d
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Solution 3.3-16
NUMERICAL DATA d2 d1 L G Ip Ip 104 mm 82 mm 2.75 28 GPa (p/32)(d 2 4 7.046 d 14 ) d3 dreqd 106 mm4 103 mm 32Td2 p ad24 d24 d2 a d24 d2 d14 b dreqd 88.4 mm
1 3
set tmax expression equal to
Td2 2 p a d4 2 32 d4 b 1
d14 b d14
then solve for d
;
(c) RATIO OF WEIGHTS OF HOLLOW & SOLID SHAFTS (a) FIND ANGLE OF TWIST f f TL GIp (tmax) f 2L Gd2 a
max
48 MPa
WEIGHT IS PROPORTIONAL TO CROSS SECTIONAL AREA
Td2 2L b 2Ip Gd2
Ah As
p2 Ad 42 p 2 d 4 reqd
d12 B Ah As 0.524 ;
0.091 radians 5.19
DIAMETER
So the weight of the tube is 52% of the solid shaft, but they resist the same torque.
;
(b) REPLACE HOLLOW SHAFT WITH SOLID SHAFT - FIND d 2 p
T tmax
tmax
16T d3p
32d 4
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Circular Bars and Tubes
265
Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is subjected to a torque produced by forces P 900 lb (see figure). The forces have their lines of action at a distance b 5.5 in. from the outside of the tube.
If the allowable shear stress in the tube is 6300 psi and the inner radius r1 1.2 in., what is the minimum permissible outer radius r2?
P P
P
r2 r1 P b 2r2 b
Solution 3.3-17
Circular tube in torsion
SOLUTION OF EQUATION UNITS: Pounds, Inches Substitute numerical values: 6300 psi or 4(900 lb)(5.5 in. + r2)(r2) p[(r4) 2 (1.2 in.)4]
P b
allow
900 lb 5.5 in. 6300 psi 1.2 in.
r4 2.07360 2 r2(r2 + 5.5) or r4 2 r2 r2 0.181891 r2 2 1.3988 in.
0.181891
0
r1
1.000402 r2
2.07360
0
Find minimum permissible radius r2 TORSION FORMULA T IP 2P(b p4 (r 22 Tr2 IP r2) r4) 1
Solve numerically:
MINIMUM PERMISSIBLE RADIUS 1.40 in. ;
2P(b + r2)r2 4P(b + r2)r2 p4 p (r4 r4) 2 1 (r2 r4) 1 2 All terms in this equation are known except r2. tmax
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Nonuniform Torsion
T1
Problem 3.4-1 A stepped shaft ABC consisting of two solid circular segments is subjected to torques T1 and T2 acting in opposite directions, as shown in the figure. The larger segment of the shaft has diameter d1 2.25 in. and length L1 30 in.; the smaller segment has diameter d2 1.75 in. and length L2 20 in. The material is steel with shear modulus G 11 106 psi, and the torques are T1 20,000 lb-in. and T2 8,000 lb-in. Calculate the following quantities: (a) the maximum shear stress max in the shaft, and (b) the angle of twist C (in degrees) at end C.
d1
d2 B
T2
A L1
C L2
Solution 3.4-1
Stepped shaft
SEGMENT BC TBC tBC fBC T2 16 TBC pd3 2 TBCL2 G(Ip)BC 8,000 lb-in. 16(8,000 lb-in.) p(1.75 in.)3 7602 psi
(8,000 lb-in.)(20 in.) (11 * 106 psi)a p b(1.75 in.)4 32
d1 d2 G T1 T2
2.25 in. 1.75 in. 11
6
L1 L2
30 in. 20 in.
0.015797 rad (a) MAXIMUM SHEAR STRESS Segment BC has the maximum stress
max
10 psi
20,000 lb-in. 8,000 lb-in.
7600 psi
;
SEGMENT AB TAB tAB fAB T2 T1 pd3 1 12,000 lb-in.
(b) ANGLE OF TWIST AT END C
`
16 TAB
`
16(12,000 lb-in.) p(2.25 in.)3 (
C
AB
BC
( 0.013007 0.16 ;
0.015797) rad
5365 psi
C
0.002790 rad
TABL1 G(Ip)AB
12,000 lb-in.)(30 in.) p b(2.25 in.)4 32
(11 * 106 psi)a
0.013007 rad
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Nonuniform Torsion
267
Problem 3.4-2 A circular tube of outer diameter d3 70 mm and inner diameter d2 60 mm is welded at the right-hand end to a fixed plate and at the left-hand end to a rigid end plate (see figure). A solid circular bar of diameter d1 40 mm is inside of, and concentric with, the tube. The bar passes through a hole in the fixed plate and is welded to the rigid end plate. The bar is 1.0 m long and the tube is half as long as the bar. A torque T 1000 N m acts at end A of the bar. Also, both the bar and tube are made of an aluminum alloy with shear modulus of elasticity G 27 GPa.
(a) Determine the maximum shear stresses in both the bar and tube. (b) Determine the angle of twist (in degrees) at end A of the bar.
Tube Fixed plate End plate Bar T A
Tube Bar
d1 d2 d3
Solution 3.4-2
Bar and tube
TORQUE T 1000 N m
(a) MAXIMUM SHEAR STRESSES Bar: t bar Tube: t tube TUBE d3 Ltube (Ip) tube 70 mm 0.5 m p4 (d 32 3 d2 G 60 mm 27 GPa
4 d2)
16T
79.6 MPa ; pd3 1 T(d3/2) 32.3 MPa (Ip) tube
;
(b) ANGLE OF TWIST AT END A Bar: fbar Tube: ftube
A bar
TL bar G(Ip) bar TL tube G(Ip) tube
tube
0.1474 rad 0.0171 rad 0.0171 0.1645 rad
1.0848 * 106 mm4 BAR d1 40 mm
4 pd1
A
0.1474
9.43
;
Lbar
1.0 m
G
27 GPa
(Ip) bar
32
251.3 * 103 mm4
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Problem 3.4-3 A stepped shaft ABCD consisting of solid circular segments is subjected to three torques, as shown in the figure. The torques have magnitudes 12.5 k-in., 9.8 k-in., and 9.2 k-in. The length of each segment is 25 in. and the diameters of the segments are 3.5 in., 2.75 in., and 2.5 in. The material is steel with shear modulus of elasticity G 11.6 103 ksi.
(a) Calculate the maximum shear stress max in the shaft. (b) Calculate the angle of twist D (in degrees) at end D.
12.5 k-in. 3.5 in.
9.8 k-in. 9.2 k-in. 2.75 in. 2.5 in. C D 25 in.
A 25 in.
B 25 in.
Solution 3.4-3 NUMERICAL DATA (INCHES, KIPS)
TB TD dAB dCD 12.5 k-in. 9.2 k-in. 3.5 in. 2.5 in. TC L dBC G 9.8k-in. 25 in. 2.75 in. 11.6 (103) ksi T1 (TB TC TD) 31.5 in.-kip IP1 IP3 fD TD Check CD: tCD dCD 2
CD BC
Check BC:
tBC
1TC + TD2 p d4 32 BC 4.65 ksi
dBC 2
; controls
(b) ANGLE OF TWIST AT END D (a) MAX. SHEAR STRESS IN SHAFT torque reaction at A: RA RA |RA| tAB dAB 2
max
|RA|
T2
TC IP2
TD
T3
TD
p d4 32 AB
3.742 ksi
p d4 32 AB p d4 32 CD TiLi a GI
p d4 32 BC
fD
pi
T2 T3 L T1 a + + b G IP1 IP2 IP3 fD 0.978 degrees ;
p d4 32 CD
2.999 ksi
D
0.017 radians
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Problem 3.4-4 A solid circular bar ABC consists of two segments, as shown in the figure. One segment has diameter d1 56 mm and length L1 1.45 m; the other segment has diameter d2 48 mm and length L2 1.2 m. What is the allowable torque Tallow if the shear stress is not to exceed 30 MPa and the angle of twist between the ends of the bar is not to exceed 1.25? (Assume G 80 GPa.)
d1 T A L1 B L2
d2 C
T
Solution 3.4-4 NUMERICAL DATA
d1 L1
a
Tallow based on angle of twist d2 L2 fa 48 mm 1200 mm 1.25 a G 80 GPa p b radians 180 Tallow tapd23 16 a Tallow fmax T G
56 mm 1450 mm 30 MPa
J 32
a p L1
L1 d14 b
+
L2 a p4 db 32 2 K
Gf a + a
Allowable torque Tallow based on shear stress tmax Tallow 16T d3p 2 Tallow
L2 p4 d2 b 32 governs
p4 d1 b 32
459 N # m
;
651.441 N # m
Problem 3.4-5 A hollow tube ABCDE constructed of monel metal is subjected to five torques acting in the directions shown in the figure. The magnitudes of the torques are T1 1000 lb-in., T2 T4 500 lb-in., and T3 T5 800 lb-in. The tube has an outside diameter d2 1.0 in. The allowable shear stress is 12,000 psi and the allowable rate of twist is 2.0/ft. Determine the maximum permissible inside diameter d1 of the tube.
T1 = T2 = 1000 lb-in. 500 lb-in.
T3 = T4 = 800 lb-in. 500 lb-in.
T5 = 800 lb-in.
A
B
C
D d2 = 1.0 in.
E
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Solution 3.4-5
Hollow tube of monel metal
REQUIRED POLAR MOMENT OF INERTIA BASED UPON
ALLOWABLE SHEAR STRESS
tmax d2
allow
1.0 in. 2/ft
allow
12,000 psi
Tmaxr Ip
IP
T max(d2/2) tallow
0.05417 in.4
0.16667/in.
REQUIRED Tmax GIP
POLAR MOMENT OF INERTIA BASED UPON
0.002909 rad/in. From Table H-2, Appendix H: G TORQUES 9500 ksi
ALLOWABLE ANGLE OF TWIST
u
IP
Tmax Gu allow
0.04704 in.4
SHEAR STRESS GOVERNS Required IP IP p4 (d 32 2 d4 3 0.05417 in.4 d4) 1 (1.0 in.)4 32(0.05417 in.4) p
T1 T4
1000 lb-in. 500 lb-in.
T2 T5
500 lb-in. 800 lb-in.
T3
800 lb-in. d4 1
32IP p
INTERNAL TORQUES TAB TBC TCD TDE Tmax T1 T1 T1 T1 T2 T2 T2 1000 lb-in. 500 lb-in. T3 T3 T4 1300 lb-in. 800 lb-in. d1
0.4482 in.4 0.818 in. ;
(Maximum permissible inside diameter)
Largest torque (absolute value only): 1300 lb-in.
Problem 3.4-6 A shaft of solid circular cross section consisting of two segments is shown in the first part of the figure. The left-hand segment has diameter 80 mm and length 1.2 m; the right-hand segment has diameter 60 mm and length 0.9 m. Shown in the second part of the figure is a hollow shaft made of the same material and having the same length. The thickness t of the hollow shaft is d/10, where d is the outer diameter. Both shafts are subjected to the same torque. If the hollow shaft is to have the same torsional stiffness as the solid shaft, what should be its outer diameter d?
80 mm
60 mm
1.2 m
0.9 m d d t= 10
2.1 m
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Solution 3.4-6 Solid and hollow shafts SOLID SHAFT CONSISTING OF TWO SEGMENTS
TORSIONAL STIFFNESS kT T f Torque T is the same for both shafts.
1 2
For equal stiffnesses, 98,741 m f1 TLi g GIPi T(1.2 m) p Ga b(80 mm)4 32
3 3
3.5569 m d4 36.023 * 10 77.5 mm
6
T(0.9 m) + p Ga b (60 mm)4 32 d4 d
3.5569 98,741 0.0775 m
m4
;
32T (29,297 m pG 32T (98,741 m pG HOLLOW SHAFT
+ 69,444 m
3
)
3
)
d0 f2
inner diameter TL GIp
0.8d T(2.1 m)
Ga
p b[d4 32
(0.8d)4] 32T 3.5569 m a b pG d4
2.1 m 32T a b pG 0.5904 d4 UNITS: d meters
Problem 3.4-7 Four gears are attached to a circular shaft and transmit the torques shown in the figure. The allowable shear stress in the shaft is 10,000 psi. 8,000 lb-in.
(a) What is the required diameter d of the shaft if it has a solid cross section? (b) What is the required outside diameter d if the shaft is hollow with an inside diameter of 1.0 in.?
A B C D 19,000 lb-in. 4,000 lb-in. 7,000 lb-in.
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Solution 3.4-7
Shaft with four gears
(b) HOLLOW SHAFT Inside diameter d0 Tr Ip 1.0 in. d Tmax a b 2 Ip
allow
10,000 psi 8000 lb-in.
TBC TCD
11,000 lb-in. 7000 lb-in.
tmax
t allow
TAB
(a) SOLID SHAFT tmax d3 16T pd3 16Tmax pt allow 16(11,000 lb-in.) p(10,000 psi) 1.78 in. ; 5.602 in.3
10,000 psi
d (11,000 lb-in.) a b 2 a p b[d4 32 (1.0 in.)4]
UNITS: d 10,000 or d4
inches 56,023 d d4 1
Required d
5.6023 d
1
0 ;
Solving, d Required d
1.832 1.83 in.
Problem 3.4-8 A tapered bar AB of solid circular cross section is twisted by torques T (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. For what ratio dB/dA will the angle of twist of the tapered bar be one-half the angle of twist of a prismatic bar of diameter dA? (The prismatic bar is made of the same material, has the same length, and is subjected to the same torque as the tapered bar.) Hint: Use the results of Example 3-5.
T
A
B T
L dA dB
Problems 3.4-8, 3.4-9 and 3.4-10
Solution 3.4-8
Tapered bar AB
TAPERED BAR (From Eq. 3-27) f1 b2 + b + 1 TL a b G(IP)A 3b 3 b dB dA
ANGLE OF TWIST f1 or 1 f 22 3
3
b2 + b + 1 3b 3 2
2
1 2 2 0
2
PRISMATIC BAR TL f2 G(IP)A
SOLVE NUMERICALLY: b dB dA 1.45 ;
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273
Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T 36,000 lb-in. (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. The bar has length L 4.0 ft and is made of an aluminum alloy having shear modulus of elasticity G 3.9 106 psi. The allowable shear stress in the bar is 15,000 psi and the allowable angle of twist is 3.0. If the diameter at end B is 1.5 times the diameter at end A, what is the minimum required diameter dA at end A? (Hint: Use the results of Example 35).
Solution 3.4-9
Tapered bar
dB T L G
allow allow
1.5 dA 36,000 lb-in. 4.0 ft 3.9 3.0 0.0523599 rad 48 in. 106 psi
TWIST
MINIMUM DIAMETER BASED (From Eq. 3-27) b f dB/dA 1.5
UPON ALLOWABLE ANGLE OF
15,000 psi
b2 + b + 1 TL a b G(IP)A 3b 3 (36,000 lb-in.)(48 in.) (3.9 * 106 psi)a 2.11728 in.4 d4 A 2.11728 in.4 fallow 40.4370 in.4 2.52 in. p4 bdA 32
TL (0.469136) G(IP)A (0.469136)
MINIMUM
STRESS
DIAMETER BASED UPON ALLOWABLE SHEAR
tmax
16T pd3 A
d3 A
16 T ptallow 2.30 in.
16(36,000 lb-in.) p(15,000 psi)
d4 A
2.11728 in.4 0.0523599 rad
12.2231 in.3 dA dA
ANGLE OF TWIST GOVERNS Min. dA 2.52 in. ;
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Problem 3.4-10 The bar shown in the figure is tapered linearly from end A to end B and has a solid circular cross section. The diameter at the smaller end of the bar is dA 25 mm and the length is L 300 mm. The bar is made of steel with shear modulus of elasticity G 82 GPa. If the torque T 180 N m and the allowable angle of twist is 0.3, what is the minimum allowable diameter dB at the larger end of the bar? (Hint: Use the results of Example 3-5.)
Solution 3.4-10
Tapered bar
dA L G T
allow
25 mm 300 mm 82 GPa 180 N m 0.3
p (0.3)a 180
rad b degrees (180 N # m)(0.3 m) a b2 + b + 1 3b 3 b
p (82 GPa)a b (25 mm)4 32 0.304915 0.914745
3
b2 + b + 1 3b 3
2
Find dB DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (From Eq. 3-27) b f dB dA b2 + b + 1 TL a b(IP)A G(IP)A 3b 3 p4 d 32 A
1
0
SOLVE NUMERICALLY: 1.94452 Min. dB dA 48.6 mm ;
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Nonuniform Torsion
275
Problem 3.4-11 The nonprismatic cantilever circular bar shown has an internal cylindrical hole from 0 to x, so the net polar moment of inertia of the cross section for segment 1 is (7/8)Ip. Torque T is applied at x and torque T/2 is applied at x L. Assume that G is constant.
(a) (b) (c) (d) (e) Find reaction moment R1. Find internal torsional moments Ti in segments 1 & 2. Find x required to obtain twist at joint 3 of 3 TL/GIp What is the rotation at joint 2, 2? Draw the torsional moment (TMD: T(x), 0 x L) and displacement (TDD: (x), 0 x L) diagrams.
R1 1
Segment 1 x 7 Ip 8
Segment 2
Ip T 2
T 2 3
x T1
Lx
T2 TMD 0 2 0 3
TDD 0
0
Solution 3.4-11 (a) REACTION TORQUE R1
a Mx 0 R1 aT+ T b R1 2 3 2 T;
L x
17 1 x+ L 14 2 14 L ab 17 2 x 7 L 17 ;
(b) INTERNAL MOMENTS IN SEGMENTS 1 & 2 T1 R1 T1 1.5 T T2 T 2
(d) ROTATION AT JOINT 2 FOR X VALUE IN (C) T1x 7 Ga Ip b 8 12TL 17GIP 3 7 a Tb a Lb 2 17 7 Ga Ip b 8
(c) FIND X REQUIRED TO OBTAIN TRWIST AT JT 3 f3 TL GIP TiLi a GI Pi T1x 7 Ga IP b 8 3 a Tbx 2 7 Ga IP b 8 3 a bx 2 7 ab 8 + + + x) T2(L GIP T a b(L 2 GIP
f2
f2
f2
;
(e) TMD & TDD SEE PLOTS ABOVE x) TMD is constant - T1 for 0 to x & T2 for x to L; hence TDD is linear - zero at jt 1, 2 at jt 2 & 3 at jt 3
TL GIP
L
1 (L 2
x)
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Problem 3.4-12 A uniformly tapered tube AB of hollow circular cross section is shown in the figure. The tube has constant wall thickness t and 2dA. The polar length L. The average diameters at the ends are dA and dB moment of inertia may be represented by the approximate formula IP L d3t/4 (see Eq. 3-18). Derive a formula for the angle of twist of the tube when it is subjected to torques T acting at the ends.
A T
B T
L t t
dA dB = 2dA
Solution 3.4-12
Tapered tube
t dA, dB dB 2dA thickness (constant) average diameters at the ends Ip pd3t (approximate formula) 4
ANGLE OF TWIST
Take the origin of coordinates at point O. d(x) Lp(x) x (d ) 2L B p[d(x)]3t 4 x d LA ptd3 A 4L
3
For element of length dx: df Tdx GIP(x)
2L
Tdx ptd3 A G a 3 bx3 4L 4TL3 pGtd3 A
2L
4TL3dx
3 pGtdA x3
x3 f L L
df
dx
3TL 2pGtd3 A
L x3 L
;
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Nonuniform Torsion
277
Problem 3.4-13 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is shown in the figure. The outside diameters at the ends are dA and dB 2dA. A hollow section of length L/2 and constant thickness t dA/10 is cast into the tube and extends from B halfway toward A.
(a) Find the angle of twist of the tube when it is subjected to torques T acting at the ends. Use numerical values as follows: dA 2.5 in., L 48 in., G 3.9 106 psi, and T 40,000 in.-lb. (b) Repeat (a) if the hollow section has constant diameter dA. (See figure part b.)
dA
T
A
L 2
t constant dB 2t B T
L (a) L 2 T A
dB
B T
dA
dA
L (b)
dB
Solution 3.4-13
PART (A) - CONSTANT THICKNESS use x as integration variable measured from B toward A from B to centerline outer and inner diameters as function of x 0x d0(x) di (x) di (x) L 2 d0(x) xdA L 2t) dA [(2dA 2t) (dA L 2t)] x dB a dB L dA bx
2d A (dB 1 5
9L + 5x L x dA L
solid from centerline to A L x L 2 f d0(x)
L
2dA
L T 32 1 1 2 ab dx + L 4 dx 4 4 G p P L d0 di L d0 Q 0 2
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f
T 32 2 ab G p L 0 T a Gp
L
1 a 2dA xdA 4 b L a 1 5 dA 9L + 5x 4 b L 125 L 2
L
dx +
1 a2dA + TL
4 GdA
L 2
L
xdA 4 b L 19
dx
f
32
125 3ln(2) + 2ln(7) L 2 dA 4 16TL
4 81GpdA
ln(197)
2ln(19) + ln(181) dA 4 3.868
81dA 4
Lb
Simplifying: f
a38 + 10125 lna
71117 bb 70952 G
or fa 3.9
Use numerical properties as follows L
a
48 in. ;
106 psi
dA
2.5 in.
t
dA 10
T
40000 in.-lb
0.049 radians
fa
2.79
PART (B) - CONSTANT HOLE DIAMETER 0 x L 2 d0( x) d0(x)
4 0 P L d0
L L 2
dB 2dA 1 di
a
dB L xdA L
L
dA
bx
d0(x)
2 dA
xdA L
di (x)
dA
L x L 2 f T 32 ab Gp
dx + 4
1
L d0 4
L 2
dx
f
2 T 32 ab Gp L 0
fb
3 ln(5) + 2atan a b 2 T1 L 32 Gp 4 dA 4 TL
4 GdA
J
a2dA
1 xdA 4 b L
Q
1 L 2
L
dx + dA
4
1 a2dA xdA 4 b L
dx
K
1 ln(3) + 2atan(2) 19 L + L 4 dA 4 81dA 4
Simplifying, fb
3.057
Use numerical properties given above
b
0.039 radians 1.265
fb
2.21
;
fa fb
so tube (a) is more flexible than tube (b)
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Nonuniform Torsion
279
Problem 3.4-14 For the thin nonprismatic steel pipe of constant thickness t and variable diameter d shown with applied torques at joints 2 and 3, determine the following.
(a) Find reaction moment R1. (b) Find an expression for twist rotation 3 at joint 3. Assume that G is constant. (c) Draw the torsional moment diagram (TMD: T(x), 0 x L).
R1 1
2d
t T/2
d 2
t
d T, f 3 3 L 2 T
L 2 x
T 2 0
TMD
Solution 3.4-14
(a) REACTION TORQUE R1 statics: R1 T 0 0 R1 T 2 + (b) ROTATION AT JOINT 3 d12( x) d23(x) 2d a1 d
L 2
L 2 2T Gpt L 0
f3 ;
1 c2da1
L
T +T 2
x3 bd L
dx
4T
L Gpd3t L 2 L
dx
x b L
0 x
L 2
f3
2 2T Gpt L 0
1 c2d a 1 x3 bd L
dx
L x L 2 T 2 p 3 Ga d12(x) tb 4
L
2TL + Gpd3t 3TL 8Gp d t 19TL 8Gpd3t
3
f3
L 0
dx
f3 f3 (c) TMD
2TL + Gpd3t ;
T
+
p Ga d23(x)3tb 4 use IP expression for thin walled tubes
L L 2
dx
TMD is piecewise constant: T(x) T/2 for segment 1-2 & T(x) T for segment 2-3 (see plot above)
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Problem 3.4-15 A mountain-bike rider going uphill applies torque T Fd (F 15 lb, d 4 in.) to the end of the handlebars ABCD (by pulling on the handlebar extenders DE). Consider the right half of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths L1 2 in. and L3 8.5 in., and with outer diameters and thicknesses d01 1.25 in., t01 0.125 in., and d03 0.87 in., t03 0.115 in., respectively as shown. Segment BC of length L2 1.2 in., however, is tapered, and outer diameter and thickness vary linearly between dimensions at B and C.
Consider torsion effects only. Assume G 4000 ksi is constant. Derive an integral expression for the angle of twist D of half of the handlebar tube when it is subjected to torque T Fd acting at the end. Evaluate D for the given numerical values.
Handlebar extension d01, t01 B A L1 L2 C L3 d03, t03
E T = Fd D
d 45 D Handlebar
Handlebar extension F
Solution 3.4-15 ASSUME THIN WALLED TUBES
fD Segments AB & CD p3 IP1 dt IP3 4 01 01 Segment BC d02(x) d01 a1 d01L2 t01 a1 t01L2 0 x p3 dt 4 03 03 L2
L1 4Fd c Gp d01 3t01 L3 d03 3t03 d
L2
L2 4
3 L (d01L2 0 d01x + d03x) t01x + t03x) * (t01L2
dx
+
;
x x b + d03 a b L2 L2 d01x + d03x L2 x x b + t03 a b L2 L2 t01x + t03x L2
NUMERICAL DATA L1 2 in. L2 1.2 in. L3 t03 0.115 in. t01 0.125 in. F 15 lb d d03 0.87 in. G 4 (106) psi fD 0.142 ; 8.5 in. d01 1.25 in. 4 in.
d02(x) t02(x)
t02(x)
fD
L2 L3 1 Fd L1 + dx + p G IP1 IP3 L 0 P Q d (x)3t02(x) 4 02
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281
Problem 3.4-16 A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant intensity t per unit distance (see figure).
(a) Determine the maximum shear stress max in the bar. (b) Determine the angle of twist between the ends of the bar.
A
t
L
B
Solution 3.4-16
Bar with distributed torque
(a) MAXIMUM SHEAR STRESS Tmax tL tmax 16Tmax pd3 pd4 32 32 tx dx pGd4 32t pGd4 L 0
L
16tL pd3
;
(b) ANGLE OF TWIST T(x) df t d G intensity of distributed torque diameter shear modulus of elasticity f L 0 tx IP
T(x)dx GIP
L
df
x dx
16tL2 pGd4
;
Problem 3.4-17 A prismatic bar AB of solid circular cross section (diameter d) is loaded by a distributed torque (see figure). The intensity of the torque, that is, the torque per unit distance, is denoted t(x) and varies linearly from a maximum value tA at end A to zero at end B. Also, the length of the bar is L and the shear modulus of elasticity of the material is G.
(a) Determine the maximum shear stress max in the bar. (b) Determine the angle of twist between the ends of the bar.
t(x) A
L
B
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Solution 3.4-17
Bar with linearly varying torque
(a) Maximum shear stress tmax 16Tmax pd
3
16TA pd
3
8tAL pd3
;
(b) ANGLE OF TWIST T(x) T(x) df f torque at distance x from end B t(x)x 2 T(x) dx GIP
L
tAx2 2L
IP
pd4 32
16tAx2 dx pGLd4 16tA
L
L 0
df
pGLd4 L 0
x2 dx
16tA L2 3pGd4
;
t(x) tA d G TA
intensity of distributed torque maximum intensity of torque diameter shear modulus maximum torque 1 2 tAL
Problem 3.4-18 A nonprismatic bar ABC of solid circular cross section is loaded by distributed torques (see figure). The intensity of the torques, that is, the torque per unit distance, is denoted t(x) and varies linearly from zero at A to a maximum value T0/L at B. Segment BC has linearly distributed torque of intensity t(x) T0/3L of opposite sign to that applied along AB. Also, the polar moment of inertia of AB is twice that of BC, and the shear modulus of elasticity of the material is G.
(a) Find reaction torque RA. (b) Find internal torsional moments T(x) in segments AB and BC. (c) Find rotation C. (d) Find the maximum shear stress tmax and its location along the bar. (e) Draw the torsional moment diagram (TMD: T(x), 0 x L).
T0 L
RA A
2Ip B L 2 2 2 0 T0 12 T0 3L L 2
IP C
F c
T0 6
TMD
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283
Solution 3.4-18
(a) TORQUE REACTION RA STATICS: RA + RA + T 0 1 T0 L a ba b 2 3L 2 RA T0 6 ; 0 (d) MAXIMUM SHEAR STRESS ALONG BAR For AB 2IP For BC IP
1
1 T0 L a ba b 2L 2 1 T 60 0
p d4 32 AB p d4 32 BC
dBC
14 a b dAB 2 T0 dAB 62 p d4 32 AB
(b) INTERNAL TORSIONAL MOMENTS IN AB & BC TAB (x) T0 6 a T0 6 (L L 2 ca x L ; tmax T0 x x ab L L2 PQ 2 x2 L2 T0 b 0 x At A, T T0/6 tmax
TAB (x)
L 2
;
tmax
8T0 3pdAB3 T0 dBC a b 12 2 p d4 32 BC
; controls T0/12
TBC (x)
x) T0 (L x) ab 3L 2 L b
2T 0
Just to right of B, T
TBC (x)
3
d
tmax
L x L 2 (c) ROTATION AT C fC
T0 0.841dAB a b 12 2 p (0.841dAB)4 32 2.243T0 pdAB 3
L TBC(x) TAB(x) dx + dx L L G(2IP) 0 L GIP 2
L 2
tmax
L 2
fC
T0 6
2 x T0
L 0
L
3L G(2IP) x L
2
dx L b
2T 0
(e) TMD two 2nd degree curves: from T0/6 at A, to T0/12 at B, to zero at C (with zero slopes at A & C since slope on TMD is proportional to ordinate on torsional loading) see plot of T(x) above d
ca
3
+
L L 2
GIP T0L 72GIP ;
dx
fC fC
T0L 48GIP T0L 144GIP
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Problem 3.4-19 A magnesium-alloy wire of diameter d 4 mm and length L rotates inside a flexible tube in order to open or close a switch from a remote location (see figure). A torque T is applied manually (either clockwise or counterclockwise) at end B, thus twisting the wire inside the tube. At the other end A, the rotation of the wire operates a handle that opens or closes the switch. A torque T0 0.2 N m is required to operate the switch. The torsional stiffness of the tube, combined with friction between the tube and the wire, induces a distributed torque of constant intensity t 0.04 N m/m (torque per unit distance) acting along the entire length of the wire.
(a) If the allowable shear stress in the wire is allow 30 MPa, what is the longest permissible length Lmax of the wire? (b) If the wire has length L 4.0 m and the shear modulus of elasticity for the wire is G 15 GPa, what is the angle of twist (in degrees) between the ends of the wire?
T0 = torque
Flexible tube B
A t
d
T
Solution 3.4-19 Wire inside a flexible tube
d 4 mm T0 0.2 N m t 0.04 N m/m (a) MAXIMUM LENGTH Lmax 30 MPa allow Equilibrium: T tL T0 16T From Eq. (3-12): tmax pd3 tL + T0 L Lmax pd 3tmax 16 1 (pd3tmax 16t 1 (pd3tallow 16t 16T0) 16T0) ; 4.42 m ;
(b) ANGLE OF TWIST L
1
4 m G 15 GPa angle of twist due to distributed torque t 16tL2 pGd4 (from problem 3.4-16)
T
pd3tmax 16
2
angle of twist due to torque T0 T0 L GIP
1
32 T0 L pGd4
2
(from Eq. 3 -15)
total angle of twist 16L pGd 4
f
(tL + 2T0) ;
Substitute numerical values: Lmax
Substitute numerical values: 2.971 rad 170 ;
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Nonuniform Torsion
285
Problem 3.4-20 Two hollow tubes are connected by a pin at B which is inserted into a hole drilled through both tubes at B (see cross-section view at B). Tube BC fits snugly into tube AB but neglect any friction on the interface. Tube inner and outer diameters di (i 1, 2, 3) and pin diameter dp are labeled in the figure. Torque T0 is applied at joint C. The shear modulus of elasticity of the material is G. Find expressions for the maximum torque T0,max which can be applied at C for each of the following conditions.
(a) The shear in the connecting pin is less than some allowable value ( pin p,allow). (b) The shear in tube AB or BC is less than some allowable value ( tube t,allow). (c) What is the maximum rotation C for each of cases (a) and (b) above?
TA A
d3 d2 LA
B
d2 d1 Pin dp LB C
T0, Fc
Cross-section at B
Solution 3.4-20 (a) T0,max BASED ON ALLOWABLE SHEAR IN PIN AT B
Pin at B is in shear at interface between the two tubes; force couple V # d2 T0 V T0 d2 T0 d2 a pdp 2 4 b tpin V AS ttubeAB
T0 a p (d 4 32 3
d3 b 2 d2 4)
ttubeAB 4T0 pd2dp 2
16T0d3 p(d3 4 tt,allow c d2 4) p(d3 4 d2 4 ) d 16d3
so based on tube AB: T0,max ;
tpin
tpin
T0,max
tp,allow a
p d2 d2 p 4
b
and based on tube BC: ; ttubeBC T0 a p (d 4 32 2 16T0d2 p(d2 4 tt,allow d1 4) p(d2 4 d1 4) 16d2 d2 b 2 d1 4)
(b) T0,max BASED ON ALLOWABLE SHEAR IN TUBES AB & BC IPAB IPAC p ad3 4 32 p ad 4 32 2 d3 T0 a b 2 IPAB d2 4 b d1 4 b
ttubeBC
T0,max
ttubeAB
J
K
;
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(c)
CHAPTER 3
Torsion
MAX. ROTATION AT SHEAR IN PIN AT TUBES
C
BASED ON EITHER ALLOWABLE
MAX. fCmax
ROTATION BASED ON ALLOWABLE SHEAR STRESS
B
OR ALLOWABLE SHEAR STRESS IN
IN TUBE
AB tt, allow c 2(d3 4 d2 4) d Gd3 LB (d2
4
PIN AT
MAX. ROTATION BASED ON ALLOWABLE SHEAR IN B fC T0,max G a LA IPAB + LB IPBC b LB p (d 4 32 2 b
fCmax
t p,allow a
p d2d2 p 4
MAX. fCmax d1 4) K
ROTATION BASED ON ALLOWABLE SHEAR STRESS
J
LA (d3
4
d2 )
4
+
d1 4)
K
;
IN TUBE
BC tt, allow c 2(d2 4 d1 4) d Gd2 + (d2 LB
4
G
J 32 (d3
p fCmax tp, allow a
LA
4
+ d2 4) b + (d2
8d2dp 2 G LA
J
LA (d3
4
d2 )
4
d1 4)
K
;
J
LB
4
(d3
4
d2 )
4
d1 4)
K
;
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287
Pure Shear
Problem 3.5-1 A hollow aluminum shaft (see figure) has outside diameter d2 4.0 in. and inside diameter d1 2.0 in. When twisted by torques T, the shaft has an angle of twist per unit distance equal to 0.54/ft. The shear modulus of elasticity of the aluminum is G 4.0 106 psi.
(a) Determine the maximum tensile stress max in the shaft. (b) Determine the magnitude of the applied torques T.
d2
T
T
L
d1 d2
Probs. 3.5-1, 3.5-2, and 3.5-3
Solution 3.5-1
Hollow aluminum shaft
d2 G
4.0 in. 4.0
d1
2.0 in.
0.54/ft
(a) MAXIMUM TENSILE STRESS
max max
106 psi
occurs on a 45 plane and is equal to
max
max.
MAXIMUM SHEAR STRESS
max
6280 psi
;
Gr (from Eq. 3-7a) d2 /2 2.0 in. 1 ft prad (0.54/ft)a ba b 12 in. 180 degree 785.40 10
6
(b) APPLIED TORQUE Use the torsion formula tmax T 10
6
r u
Tr IP (2.0 in.)4]
rad/in. rad/in.)
tmaxIP r
IP
p [(4.0 in.)4 32
max
(4.0
106 psi)(2.0 in.)(785.40
23.562 in.4 T (6283.2 psi) (23.562 in.4) 2.0 in. 74,000 lb-in. ;
6283.2 psi
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Problem 3.5-2 A hollow steel bar (G 80 GPa) is twisted by torques T (see figure). The twisting of the bar produces a maximum shear strain max 640 10 6 rad. The bar has outside and inside diameters of 150 mm and 120 mm, respectively.
(a) Determine the maximum tensile strain in the bar. (b) Determine the maximum tensile stress in the bar. (c) What is the magnitude of the applied torques T?
Solution 3.5-2
Hollow steel bar
G d2 IP
80 GPa 150 mm p4 (d 32 2
max
640
10
6
rad
(b) MAXIMUM TENSILE STRESS
max
d1 d 4) 1
120 mm
G
max
(80 GPa)(640 51.2 MPa ;
10 6)
51.2 MPa
max max
p [(150 mm)4 32
(120 mm)4]
(c) APPLIED TORQUES Torsion formula: tmax T 2IPtmax d2 Tr IP Td2 2IP
29.343 * 106 mm4 (a) MAXIMUM TENSILE STRAIN max gmax 2 320 * 10
6
;
2(29.343 * 106 mm4)(51.2 MPa) 150 mm
20,030 N # m 20.0 kN # m ;
Problem 3.5-3 A tubular bar with outside diameter d2 4.0 in. is twisted by torques T action of these torques, the maximum tensile stress in the bar is found to be 6400 psi.
(a) Determine the inside diameter d1 of the bar. (b) If the bar has length L 48.0 in. and is made of aluminum with shear modulus G twist (in degrees) between the ends of the bar? (c) Determine the maximum shear strain max (in radians)?
70.0 k-in. (see figure). Under the
4.0
106 psi, what is the angle of
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Pure Shear
289
Solution 3.5-3
Tubular bar
d2
max
4.0 in.
T
70.0 k-in.
max max
70,000 lb-in. 6400 psi
f
6400 psi
TL GIp 2IP tmax d2 2Ltmax Gd2 0.03840 rad
(a) INSIDE DIAMETER d1 Torsion formula: tmax IP Td2 2tmax Tr IP Td2 2IP
From torsion formula, T f 2IPtmax L a b d2 GIP
(70.0 k-in.)(4.0 in.) 2(6400 psi) f
4 d1)
2(48 in.)(6400 psi) (4.0 * 106 psi)(4.0 in.) 2.20 ;
21.875 in.4 Also, Ip p4 (d 32 2 p [(4.0 in.)4 32
4 d1 ]
(c) MAXIMUM SHEAR STRAIN gmax tmax G 6400 psi 4.0 * 106 psi
6
Equate formulas: p [256 in.4 32 Solve for d1: d1 (b) ANGLE OF TWIST L 48 in. G 4.0 106 psi
4 d1 ]
21.875 in.4 ;
1600 * 10
rad
;
2.40 in.
Problem 3.5-4 A solid circular bar of diameter d 50 mm (see figure) is twisted in a testing machine until the applied torque reaches the value T 500 N m. At this value of torque, a strain gage oriented at 45 to the axis of the bar gives a reading P 339 10 6. What is the shear modulus G of the material?
d = 50 mm T
Strain gage
T = 500 Nm
45
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Solution 3.5-4
Bar in a testing machine
Strain gage at 45:
max
SHEAR STRESS (FROM EQ. 3-12) tmax 16T pd tmax gmax
3
339 50 mm
10
6
16(500 N # m) p(0.050 m)3 20.372 MPa 678 * 10
6
20.372 MPa
d T
500 N m
6
SHEAR MODULUS G 30.0 GPa ;
SHEAR STRAIN (FROM EQ. 3-33)
max
2
max
678
10
Problem 3.5-5 A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 When twisted by a torque T, the tube develops a maximum normal strain of 170 10 6. What is the magnitude of the applied torque T?
1.5 in.
Solution 3.5-5
Steel tube
G
max
11.5 170 p2 1d 32 2
106 psi d2 10
6
2.0 in.
d1
1.5 in.
Equate expressions: Td2 2IP Ggmax
IP
4 d1 2 4
p [(2.0 in.)4 32
(1.5 in.)4]
SOLVE FOR TORQUE T 2GIPgmax d2 2(11.5 * 106 psi)(1.07379 in.4)(340 * 10 2.0 in. 4200 lb-in. ;
6
1.07379 in.
SHEAR STRAIN (FROM EQ. 3-33)
max
2
max
340
10
6
)
SHEAR STRESS (FROM TORSION FORMULA) tmax Also, Tr IP
max
Td2 2IP G
max
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291
Problem 3.5-6 A solid circular bar of steel (G 78 GPa) transmits a torque T 360 N m. The allowable stresses in tension, compression, and shear are 90 MPa, 70 MPa, and 40 MPa, respectively. Also, the allowable tensile strain is 220 10 6. Determine the minimum required diameter d of the bar.
Solution 3.5-6
T
Solid circular bar of steel
78 GPa DIAMETER BASED UPON ALLOWABLE TENSILE STRAIN
max
360 N m G
ALLOWABLE STRESSES Tension: 90 MPa Compression: 70 MPa Shear: 40 MPa Allowable tensile strain: max 220 10
2
max;
max
G
max
2G
max
tmax
6
16T pd
3
d3
16T ptmax
16T 2pGmax
6
DIAMETER BASED UPON ALLOWABLE STRESS The maximum tensile, compressive, and shear stresses in a bar in pure torsion are numerically equal. Therefore, the lowest allowable stress (shear stress) governs.
allow
d3
16(360 N # m) 2p(78 GPa)(220 * 10 53.423 * 10
6
)
m3
d
0.0377 m
37.7 mm
40 MPa 16T pd
3
TENSILE STRAIN GOVERNS 16T pt allow 10
6
tmax d
3
d3
16(360 N # m) p(40 MPa)
dmin
37.7 mm
;
45.837 0.0358 m
m3
d
35.8 mm
Problem 3.5-7 The normal strain in the 45 direction on the surface of a circular tube (see figure) is 880 10 6 when the torque T 750 lb-in. The tube is made of copper alloy with G 6.2 106 psi.
If the outside diameter d2 of the tube is 0.8 in., what is the inside diameter d1?
Strain gage T d 2 = 0.8 in. 45 T = 750 lb-in.
Solution 3.5-7
Circular tube with strain gage
d2
0.80 in. T
750 lb-in.
max
G 10
6.2
6
106 psi
MAXIMUM SHEAR STRAIN
max
Strain gage at 45:
880
2
max
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MAXIMUM SHEAR STRESS tmax tmax IP
4 d2
INSIDE DIAMETER Substitute numerical values: Td2 2tmax Td2 4Gmax Td2 4Gmax d4 2 (0.8 in.)4 0.4096 in.4 d1 8Td2 pGmax 0.60 in. ; 8(750 lb-in.) (0.80 in.) p (6.2 * 106 psi) (880 * 10 0.2800 in.4 0.12956 in.4
6
Ggmax T(d2/2) IP p4 (d 32 2
4 d1
2Gmax IP
4 d1 )
)
8Td2 pGmax
4 d1
4 d2
Problem 3.5-8 An aluminium tube has inside diameter d1 50 mm, shear modulus of elasticity G 27 GPa, and torque T 4.0 kN m. The allowable shear stress in the aluminum is 50 MPa and the allowable normal strain is 900 10 6. Determine the required outside diameter d2.
Solution 3.5-8
Aluminum tube
d1 T
50 mm
G
27 GPa
allow
NORMAL STRAIN GOVERNS
allow
4.0 kN m
50 MPa
900
10
6
allow
48.60 MPa
Determine the required diameter d2. ALLOWABLE SHEAR STRESS (
allow)1
REQUIRED DIAMETER t Tr IP 48.6 MPa (4000 N # m)(d2/2) p4 [d2 (0.050 m)4] 32
50 MPa
ALLOWABLE SHEAR STRESS BASED ON NORMAL STRAIN max (
allow)2
g 2 2G
t 2G
allow
Rearrange and simplify: t 2Gmax 10 )
6
d4 2
(419.174 * 10
6
)d2
6.25 * 10
6
0
2(27 GPa)(900
Solve numerically: d2 d2 0.07927 m 79.3 mm ;
48.6 MPa
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Pure Shear
293
Problem 3.5-9 A solid steel bar (G 11.8 106 psi) of diameter d 2.0 in. is subjected to torques T 8.0 k-in. acting in the directions shown in the figure.
(a) Determine the maximum shear, tensile, and compressive stresses in the bar and show these stresses on sketches of properly oriented stress elements. (b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these strains on sketches of the deformed elements.
T d = 2.0 in. T = 8.0 k-in.
Solution 3.5-9
Solid steel bar
T G
8.0 k-in. 11.8 10 psi
6
(b) MAXIMUM STRAINS gmax gmax G 5093 psi 11.8 * 106 psi
6
(a) MAXIMUM STRESSES tmax 16T pd
3
16(8000 lb-in.) p(2.0 in.) ;
c
432 * 10 max gmax 2 216
rad
;
6
3
5093 psi
t
216 * 10 10
6 c
5090 psi
5090 psi
;
t
216
10
6
;
Problem 3.5-10 A solid aluminum bar (G 27 GPa) of diameter d 40 mm is subjected to torques T 300 N m acting in the directions shown in the figure.
(a) Determine the maximum shear, tensile, and compressive stresses in the bar and show these stresses on sketches of properly oriented stress elements. (b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these strains on sketches of the deformed elements.
d = 40 mm T
T = 300 Nm
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Torsion
Solution 3.5-10
Solid aluminum bar
(a) MAXIMUM STRESSES tmax 16T pd
3
(b) MAXIMUM STRAINS m)
3
16(300 N # ;
c
p(0.040 m)
gmax
tmax G
23.87 MPa 27 GPa
6
23.87 MPa
t
884 * 10 23.9 MPa ; max
t
rad
;
6
6
23.9 MPa
gmax 2 442
442 * 10 10
6 c
442
10
;
Transmission of Power
Problem 3.7-1 A generator shaft in a small hydroelectric plant turns at 120 rpm and delivers 50 hp (see figure).
(a) If the diameter of the shaft is d 3.0 in., what is the maximum shear stress max in the shaft? (b) If the shear stress is limited to 4000 psi, what is the minimum permissible diameter dmin of the shaft?
120 rpm d
50 hp
Solution 3.7-1
Generator shaft
50 hp d diameter tmax tmax 16T pd
3
n 120 rpm H TORQUE H T 2pnT H 33,000 33,000 H 2pn 2188 1b-ft d 3.0 in.
16(26,260 1b-in.) p (3.0 in.)3 ;
hp n
rpm T
1b-ft
4950 psi
(33,000)(50 hp) 2p(120 rpm) 26,260 1b-in.
max
(b) MINIMUM DIAMETER dmin
allow
4000 psi 16T ptallow 3.22 in. 16(26,260 1b-in.) p (4000 psi) ; 33.44 in.3
d3 dmin
(a) MAXIMUM SHEAR STRESS
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SECTION 3.7
Transmission of Power
295
Problem 3.7-2 A motor drives a shaft at 12 Hz and delivers 20 kW of power (see figure).
(a) If the shaft has a diameter of 30 mm, what is the maximum shear stress max in the shaft? (b) If the maximum allowable shear stress is 40 MPa, what is the minimum permissible diameter dmin of the shaft?
d
12 Hz
20 kW
Solution 3.7-2 f 12 Hz P
TORQUE P T T 2 fT P
Motor-driven shaft 20 kW 20,000 N m/s
1
tmax
16T pd3
16(265.3 N # m) p(0.030 m)3 ;
watts f
Hz
s
50.0 MPa
Newton meters P 2pf 20,000 W 2p(12 Hz) 265.3 N # m
max
(b) MINIMUM DIAMETER dmin
allow
40 MPa 16T pt allow 33.78 10 16(265.3 N # m) p(40 MPa)
6
d3
(a) MAXIMUM SHEAR STRESS d 30 mm
m3 ;
dmin
0.0323 m
32.3 mm
Problem 3.7-3 The propeller shaft of a large ship has outside diameter 18 in. and inside diameter 12 in., as shown in the figure. The shaft is rated for a maximum shear stress of 4500 psi.
(a) If the shaft is turning at 100 rpm, what is the maximum horsepower that can be transmitted without exceeding the allowable stress? (b) If the rotational speed of the shaft is doubled but the power requirements remain unchanged, what happens to the shear stress in the shaft?
18 in.
100 rpm
12 in. 18 in.
Solution 3.7-3
d2
Hollow propeller shaft
(a) HORSEPOWER n n H 100 rpm rpm T H 2pnT 33,000 hp
18 in. d1 12 in. allow 4500 psi p4 IP (d d4) 8270.2 in.4 2 32 2 TORQUE tmax T T(d2/2) IP T 2t allowIP d2
lb-ft H
2(4500 psi)(8270.2 in.4) 18 in. 4.1351 * 106 1b-in. 344,590 1b-ft.
2p(100 rpm)(344,590 lb-ft) 33,000 6560 hp ;
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(b) ROTATIONAL SPEED IS DOUBLED H 2pnT 33,000
If n is doubled but H remains the same, then T is halved. If T is halved, so is the maximum shear stress. Shear stress is halved ;
Problem 3.7-4 The drive shaft for a truck (outer diameter 60 mm and inner diameter 40 mm) is running at 2500 rpm (see figure).
(a) If the shaft transmits 150 kW, what is the maximum shear stress in the shaft? (b) If the allowable shear stress is 30 MPa, what is the maximum power that can be transmitted?
2500 rpm 60 mm
40 mm 60 mm
Solution 3.7-4 Drive shaft for a truck d2 60 mm d1 40 mm n 2500 rpm
IP p4 (d 32 2
4 d1 )
tmax
Td2 2 IP
(572.96 N # m)(0.060 m) 2(1.0210 * 10
6
1.0210 * 10
max
6
m
4
m4)
16.835 MPa tmax 150,000 W 16.8 MPa ;
(a) MAXIMUM SHEAR STRESS P T P T power (watts) P
150 kW
(b) MAXIMUM POWER Pmax
allow
torque (newton meters) n 2pnT 60 T 60P 2pn
rpm
30 MPa P tallow tmax (150 kW) a ; 30 MPa b 16.835 MPa
Pmax 572.96 N # m
60(150,000 W) 2p(2500 rpm)
267 kW
Problem 3.7-5 A hollow circular shaft for use in a pumping station is being designed with an inside diameter equal to 0.75 times the outside diameter. The shaft must transmit 400 hp at 400 rpm without exceeding the allowable shear stress of 6000 psi. Determine the minimum required outside diameter d.
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SECTION PM
Page 3.7
Transmission of Power
297
Solution 3.7-5
d
Hollow shaft
H T hp n rpm T lb-ft
outside diameter d0 H inside diameter 0.75 d 400 hp 6000 psi p4 [d 32 (0.75 d)4] 0.067112 d 4 n 400 rpm
33,000 H 2pn 5252.1 lb-ft
(33,000)(400 hp) 2p(400 rpm) 63,025 lb-in.
allow
MINIMUM OUTSIDE DIAMETER tmax Td 2IP IP Td 2tmax Td 2t allow
IP TORQUE H
2pnT 33,000
0.067112 d 4 d3
(63,025 lb-in.)(d) 2(6000 psi) dmin 4.28 in. ;
78.259 in.3
Problem 3.7-6 A tubular shaft being designed for use on a construction site must transmit 120 kW at 1.75 Hz. The inside diameter of the shaft is to be one-half of the outside diameter. If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d?
Solution 3.7-6 Tubular shaft d outside diameter
d0 P
allow
T
inside diameter 0.5 d 120 kW 45 MPa p4 [d 32 (0.5 d)4] 0.092039 d 4 120,000 W f 1.75 Hz
P 2pf
120,000 W 2p(1.75 Hz)
10,913.5 N # m
MINIMUM OUTSIDE DIAMETER tmax Td 2IP IP Td 2tmax Td 2t allow
IP
0.092039 d4 d3
(10,913.5 N # m)(d) 2(45 MPa) d ; 0.1096 m
TORQUE P T 2 fT P watts f Hz
0.0013175 m3 110 mm
dmin
newton meters
Problem 3.7-7 A propeller shaft of solid circular cross section and diameter d is spliced by a collar of the same material (see figure). The collar is securely bonded to both parts of the shaft. What should be the minimum outer diameter d1 of the collar in order that the splice can transmit the same power as the solid shaft?
d1
d
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Torsion
Solution 3.7-7
Splice in a propeller shaft
SOLID SHAFT tmax 16 T1 pd3 T1 pd3tmax 16
EQUATE TORQUES For the same power, the torques must be the same. For the same material, both parts can be stressed to the same maximum stress. T1 or a T2 pd3tmax 16 d1 d 1 ptmax 4 (d 16d1 1 0 d 4)
HOLLOW COLLAR IP p4 (d1 32 2tmaxIP d1 ptmax 4 (d1 16 d1 d ) tmax
4
T2r IP
T2(d1/2) IP d)
4
T2
2tmax p 4 a b(d1 d1 32 d 4)
d1 4 b d
(Eq. 1)
MINIMUM OUTER DIAMETER Solve Eq. (1) numerically: Min. d1 1.221 d ;
Problem 3.7-8 What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter 50 mm, inside diameter 40 mm, and shear modulus of elasticity 80 GPa) turning at 600 rpm if the allowable shear stress is 100 MPa and the allowable rate of twist is 3.0/m?
Solution 3.7-8
d2 G 50 mm
Hollow propeller shaft
d1 40 mm 600 rpm
allow
80 GPa n 100 MPa p4 (d 32 2 d4 ) 1
allow
3.0/m
9
BASED UPON ALLOWABLE RATE OF TWIST T2 u T2 GIP allow GIP m4 T 2 (80 GPa) (362.3 * 10 *a T2 1517 N m 94 m )(3.0/m)
IP
362.3 * 10
BASED UPON ALLOWABLE SHEAR STRESS tmax T1 T1(d2/2) IP T1 2t allowIP d2
9
p rad /degreeb 180
2(100 MPa)(362.3 * 10 0.050 m 1449 N # m
m4)
SHEAR STRESS GOVERNS Tallow T1 1449 N m MAXIMUM POWER 2p(600 rpm)(1449 N # m) 2pnT P 60 60 P Pmax 91,047 W 91.0 kW ;
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299
Problem 3.7-9 A motor delivers 275 hp at 1000 rpm to the end of a shaft (see figure). The gears at B and C take out 125 and 150 hp, respectively. Determine the required diameter d of the shaft if the allowable shear stress is 7500 psi and the angle of twist between the motor and gear C is limited to 1.5. (Assume G 11.5 106 psi, L1 6 ft, and L2 4 ft.)
Motor A d B C
L1
L2
PROBS. 3.7-9 and 3.7-10
Solution 3.7-9
Motor-driven shaft
FREE-BODY DIAGRAM
L1 L2 d n (
6 ft 4 ft diameter 1000 rpm 7500 psi 1.5 11.5 0.02618 rad 106 psi G
AC)allow
TA TC d TB
17,332 lb-in. 9454 lb-in. diameter 7878 lb-in.
allow
INTERNAL TORQUES TAB TBC lb-ft 17,332 lb-in. 9454 lb-in.
TORQUES ACTING ON THE SHAFT H 2pnT 33,000 33,000 H 2pn 33,000(275 hp) 2p(1000 rpm) 1444 lb-ft 17,332 lb-in. At point B: TB At point C: TC 125 275 150 275 TA TA 7878 lb-in. H hp n rpm T
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS The larger torque occurs in segment AB tmax 16TAB pd
3
T
d3
At point A: TA
16TAB pt allow 11.77 in.3
16(17,332 lb-in.) p(7500 psi) d 2.27 in.
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST IP 9454 lb-in. pd4 32 f TL GIP 32TL pGd4
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Segment AB: fAB 32TAB LAB pGd 4 32(17,330 lb 1.1052 d
4
fBC
0.4018 d4 fAB + fBC 1.5070 d4
From A to C: fAC in.)(6 ft)(12 in./ft) (
AC)allow
p(11.5 * 106 psi)d 4 fAB
0.02618 rad 1.5070 d4 and d 2.75 in.
0.02618
Segment BC: fBC 32 TBCLBC pGd
4
Angle of twist governs d 2.75 in. ;
32(9450 lb-in.)(4 ft)(12 in./ft) p(11.5 * 106 psi)d 4
Problem 3.7-10 The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of 32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1 1.5 m and L2 0.9 m. Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist between points A and C is 4.0, and G 75 GPa.
Solution 3.7-10
Motor-driven shaft
At point A: TA At point B: TB At point C: TC 300,000 W 2p(32 Hz) 120 T 300 A 180 T 300 A 1492 N # m
L1 L2 d f G (
1.5 m 0.9 m diameter 32 Hz 50 MPa 4 0.06981 rad
596.8 N # m 895.3 N # m
FREE-BODY DIAGRAM
allow
75 GPa
AC)allow
TORQUES ACTING ON THE SHAFT P T T 2 fT P watts f Hz TA TB TC d 1492 N m 596.8 N m 895.3 N m diameter
newton meters P 2pf
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301
INTERNAL TORQUES TAB TBC 1492 N m 895.3 N m
Segment BC: fBC fBC 32 TBCLBC pGd 4 0.1094 * 10 d4 fAB + fBC 0.4133 * 10 d
4 6 6
32(895.3 N # m)(0.9 m) p(75 GPa)d 4
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS The larger torque occurs in segment AB tmax d3 16 TAB pd 3 d
3
16 TAB pt allow d
16(1492 N # m) p(50 MPa) 53.4 mm
From A to C: fAC (
AC)allow
0.0001520 m3
0.0534 m
0.06981 rad 0.1094 * 10 d
4 6
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST IP pd 32
4
0.06981 and d
f
TL GIP
32TL pGd 4
0.04933 m 49.3mm
Segment AB: fAB fAB 32 TABLAB pGd
4 6
SHEAR STRESS GOVERNS 32(1492 N # m)(1.5 m)
4
d
53.4mm
;
p(75 GPa)d
0.3039 * 10 d4
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Statically Indeterminate Torsional Members
T0 2T0 C D TD
Problem 3.8-1 A solid circular bar ABCD with fixed supports is acted upon by torques T0 and 2T0 at the locations shown in the figure. Obtain a formula for the maximum angle of twist max of the bar. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)
TA
A
B
3L 10
3L 10 L
4L 10
Solution 3.8-1
Circular bar with fixed ends
ANGLE OF TWIST AT SECTION B
From Eqs. (3-46a and b): TA TB T0 LB L T0 LA L T0 a 7 4 b + 2T0 a b 10 10 fB fAB TA(3L/10) GIP 9T0 L 20GIP
ANGLE OF TWIST AT SECTION C fC fCD TD(4L/10) GIP 3T0 L 5GIP 3T0 L 5GIP
APPLY THE ABOVE FORMULAS TO THE GIVEN BAR: TA TD 15T0 10 15T0 10
MAXIMUM ANGLE OF TWIST fmax fC
3 6 T0 a b + 2T0 a b 10 10
;
Problem 3.8-2 A solid circular bar ABCD with fixed supports at ends A and D is acted upon by two equal and oppositely directed torques T0, as shown in the figure. The torques are applied at points B and C, each of which is located at distance x from one end of the bar. (The distance x may vary from zero to L/2.)
(a) For what distance x will the angle of twist at points B and C be a maximum? (b) What is the corresponding angle of twist max? (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)
T0 TA A B
T0 C D TD
x L
x
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303
Solution 3.8-2
Circular bar with fixed ends
(a) ANGLE OF TWIST AT SECTIONS B AND C
B max
0
L/A
L/2
x
From Eqs. (3-46a and b): TA TB T0 LB L T0 LA L
fB dfB dx dfB dx or x
fAB
TAx GIP
T0 (L GIPL 4x)
2x)(x)
T0 (L GIPL 0; L L 4 4x
0
APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:
;
(b) MAXIMUM ANGLE OF TWIST fmax (fB)max (fB)x
L 4
T0L 8GIP
;
TA
T0(L L
x)
T0x L
T0 (L L
2x) TD
TA
Problem 3.8-3 A solid circular shaft AB of diameter d is fixed against rotation at both ends (see figure). A circular disk is attached to the shaft at the location shown. What is the largest permissible angle of rotation max of the disk if the allowable shear stress in the shaft is allow? (Assume that a b. Also, use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)
Disk A d B
a
b
Solution 3.8-3
Shaft fixed at both ends
Assume that a torque T0 acts at the disk. The reactive torques can be obtained from Eqs. (3-46a and b): T0b T0a TB L L Since a b, the larger torque (and hence the larger stress) is in the right hand segment. TA
L a
a b
b
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tmax T0
TB(d/2) IP 2LIPtmax ad
T0 ad 2LIP (T0)max 2L IPt allow ad
ANGLE OF ROTATION OF THE DISK (FROM Eq. 3-49) f fmax T0 ab GLIP (T0)maxab GLIp 2bt allow Gd
;
Problem 3.8-4 A hollow steel shaft ACB of outside diameter 50 mm and inside diameter 40 mm is held against rotation at ends A and B (see figure). Horizontal forces P are applied at the ends of a vertical arm that is welded to the shaft at point C. Determine the allowable value of the forces P if the maximum permissible shear stress in the shaft is 45 MPa. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.)
200 mm A P C B P 600 mm 400 mm 200 mm
Solution 3.8-4 Hollow shaft with fixed ends GENERAL FORMULAS:
T0 LB LA L d2
allow
P(400 mm) 400 mm 600 mm LA LB d1 1000 mm 40 mm 50 mm T0 LB L T0 LA L
45 MPa P(0.4 m)(400 mm) 1000 mm P(0.4 m)(600 mm) 1000 mm 0.16 P 0.24 P
TA APPLY THE ABOVE FORMULAS TO THE GIVEN SHAFT TB
UNITS: P
Newtons T
Newton meters
From Eqs. (3-46a and b): TA TB T0 LB L
T0 LA L The larger torque, and hence the larger shear stress, occurs in part CB of the shaft.
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Statically Indeterminate Torsional Members
305
Tmax
TB
0.24 P 2tmaxIP d
Substitute numerical values into (Eq. 1): 0.24P (Eq. 1) 2(45 * 106 N/m2)(362.26 * 10 0.05 m 652.07 N # m P
9 9
SHEAR STRESS IN PART CB tmax Tmax(d/2) IP
6
m4)
Tmax
UNITS: Newtons and meters
max
45
10 N/m
2
652.07 N # m 0.24 m 2720 N
2717 N
Ip d
p4 (d d 4) 32 2 1 d2 0.05 mm
362.26 * 10
m4
Pallow
;
Problem 3.8-5 A stepped shaft ACB having solid circular cross sections with two different diameters is held against rotation at the ends (see figure). If the allowable shear stress in the shaft is 6000 psi, what is the maximum torque (T0)max that may be applied at section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.)
0.75 in. A C
1.50 in. B
T0 6.0 in. 15.0 in.
Solution 3.8-5
Stepped shaft ACB
dA LA
allow
0.75 in. dB 6.0 in. LB 6000 psi
1.50 in. 15.0 in.
Combine Eqs. (1) and (3) and solve for T0: (T0)AC LAIPB 1 pd3 t a1 + b 16 A allow LBIPA
4 LAdB 1 pd3 t allow a1 + b 4 16 A LBdA
Find (T0)max REACTIVE TORQUES (from Eqs. 3-45a and b) LBIPA b TA T0 a LBIPA + LAIPB TB T0 a LAIPB b LBIPA + LAIPB
(4)
(1) (2)
Substitute numerical values: (T0)AC 3678 lb-in.
IN SEGMENT IN SEGMENT
ALLOWABLE TORQUE BASED UPON SHEAR STRESS CB tCB TB 16TB
3 pdB
ALLOWABLE TORQUE BASED UPON SHEAR STRESS AC tAC TA 16TA
3 pdA 1 pd3 t 16 A AC
1 pd3 t 16 A allow
1 pd 3 t 16 B CB
1 pd 3 t 16 B allow
(5)
(3)
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Combine Eqs. (2) and (5) and solve for T0: (T0)CB LBIPA 1 pd3 t a1 + b 16 B allow LAIPB (6)
SEGMENT AC GOVERNS (T0)max (T0)AC (T0)CB 3680 lb-in. a LA dB ba b LB dA
;
NOTE: From Eqs. (4) and (6) we find that
4 LBdA 1 pd3 t allow a1 + b 4 16 B LAdB Substitute numerical values: (T0)CB 4597 lb-in.
which can be used as a partial check on the results.
Problem 3.8-6 A stepped shaft ACB having solid circular cross sections with two different diameters is held against rotation at the ends (see figure). If the allowable shear stress in the shaft is 43 MPa, what is the maximum torque (T0)max that may be applied at section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.)
20 mm A C T0 225 mm
25 mm B
450 mm
Solution 3.8-6
Stepped shaft ACB
dA dB LA LB
allow
20 mm 25 mm 225 mm 450 mm 43 MPa
TA
1 pd 3 t 16 A AC
1 pd 3 t 16 A allow
(3)
Combine Eqs. (1) and (3) and solve for T0: (T0)AC LAIPB 1 3 pdAt allow a 1 + b 16 LBIPA (4)
Find (T0)max REACTIVE TORQUES (from Eqs. 3-45a and b) TA TB LBIPA T0 a b LBIPA + LAIPB T0 a LAIPB b LBIPA + LAIPB (1) (2)
4 LAdB 1 3 pdAt allow a 1 + b 4 16 LBdA
Substitute numerical values: (T0)AC 150.0 N m
IN SEGMENT
ALLOWABLE TORQUE BASED UPON SHEAR STRESS CB tCB TB 16TB pd3 B 1 pd3 t 16 B CB 1 pd3 t 16 B allow (5)
ALLOWABLE TORQUE BASED UPON SHEAR STRESS IN SEGMENT AC tAC 16TA pd3 A
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307
Combine Eqs. (2) and (5) and solve for T0: (T0)CB LBIPA 1 pd 3 t a1 + b 16 B allow LAIPB (6)
SEGMENT AC GOVERNS (T0)max (T0)AC (T0)CB 150 N m a LA dB ba b LB dA ;
NOTE: From Eqs. (4) and (6) we find that
4 LBdA 1 3 pdBt allow a 1 + b 4 16 LAdB
Substitute numerical values: (T0)CB 240.0 N m
which can be used as a partial check on the results.
Problem 3.8-7 A stepped shaft ACB is held against rotation at ends A and B and subjected to a torque T0 acting at section C (see figure). The two segments of the shaft (AC and CB) have diameters dA and dB, respectively, and polar moments of inertia IPA and IPB, respectively. The shaft has length L and segment AC has length a.
(a) For what ratio a/L will the maximum shear stresses be the same in both segments of the shaft? (b) For what ratio a/L will the internal torques be the same in both segments of the shaft? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.)
dA A
IPA
dB C T0
IPB
B
a L
Solution 3.8-7
Stepped shaft
SEGMENT AC: dA, IPA LA
a
or
(L
a)dA
adB a L dA dA + dB LAIPB IPA IPA + IPB
SEGMENT CB: dB, IPB LB L a REACTIVE TORQUES (from Eqs. 3-45a and b) TA T0 a LBIPA b ; TB LBIPA + LAIPB T0 a LAIPB b LBIPA + LAIPB
Solve for a/L:
;
(b) EQUAL TORQUES TA TB or LBIPA or (L a)IPA a L aIPB
(a) EQUAL SHEAR STRESSES TA(dA/2) tCB tAC IPA
AC CB
TB(dB/2) IPB TBdB IPB (Eq. 1)
Solve for a/L: or a L
4 dA
or
TAdA IPA
4 4 dA + dB
;
Substitute TA and TB into Eq. (1): LBIPAdA IPA LAIPBdB IPB or LBdA LAdB
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Problem 3.8-8 A circular bar AB of length L is fixed against rotation at the ends and loaded by a distributed torque t(x) that varies linearly in intensity from zero at end A to t0 at end B (see figure). Obtain formulas for the fixed-end torques TA and TB.
t0 t(x) TA A x L B TB
Solution 3.8-8
Fixed-end bar with triangular load
ELEMENT OF DISTRIBUTED LOAD
dTA t(x) T0 T0 t0x L Resultant of distributed torque
L
Elemental reactive torque Elemental reactive torque
dTB
From Eqs. (3-46a and b): t0 x dx LL 0 t0L 2
L
L 0
t(x)dx
t0 L 2
dTA
t(x)dxa
L L
L
x
b
dTB
x t(x)dxa b L
REACTIVE TORQUES (FIXED-END TORQUES) TA TB dTA L dTB L t0L x L x a t0 b a bdx L L 6 L 0 L t0 L x x a t0 b a bdx ; LL 3 L 0 t0L (check) 2
EQUILIBRIUM TA + TB T0
;
NOTE: TA + TB
Problem 3.8-9 A circular bar AB with ends fixed against rotation has a hole extending for half of its length (see figure). The outer diameter of the bar is d2 3.0 in. and the diameter of the hole is d1 2.4 in. The total length of the bar is L 50 in. At what distance x from the left-hand end of the bar should a torque T0 be applied so that the reactive torques at the supports will be equal?
25 in. A 3.0 in.
25 in. T0 B
x
2.4 in.
3.0 in.
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SECTION 3.8
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309
Solution 3.8-9
Bar with a hole
IPA IPB fB Polar moment of inertia at left-hand end Polar moment of inertia at right-hand end TB(L/2) TB(L/2) + GIPB GIPA T0(L/2) GIPA Substitute Eq. (1) into Eq. (2) and simplify: fB T0 L L c + G 4IPB 4IPA
B
T0(x L/2) GIPB (2)
L L/2 d2 d1 T0
50 in. 25 in. outer diameter 3.0 in. diameter of hole 2.4 in. Torque applied at distance x TB T0 2
x L + IPB 2IPB
L d 2IPA
COMPATIBILITY x IPB 3L 4IPB
0 L 4IPA
Find x so that TA EQUILIBRIUM TA TB
soLVE FOR x: TB (1) x IPB IPA x L a3 4
4 d2 4 d2
T0 TA
IPB b IPA
4 d1
REMOVE THE SUPPORT AT END B
1
a
d1 4 b d2
d1 4 L c2 + a b d 4 d2
;
SUBSTITUTE NUMERICAL VALUES: x
B
50 in. 2.4 in. 4 c2 + a bd 4 3.0 in.
30.12 in.
;
Angle of twist at B
Problem 3.8-10 A solid steel bar of diameter d1 25.0 mm is enclosed by a steel tube of outer diameter d3 37.5 mm and inner diameter d2 30.0 mm (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has a length L 550 mm, is twisted by a torque T 400 N m acting on the end plate.
(a) Determine the maximum shear stresses 1 and 2 in the bar and tube, respectively. (b) Determine the angle of rotation (in degrees) of the end plate, assuming that the shear modulus of the steel is G 80 GPa. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.)
Tube A Bar B T
L
End plate
d1 d2 d3
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Solution 3.8-10
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) Bar: T1 Tube: T2 Ta IP1 b IP1 + IP2 IP2 b IP1 + IP2 100.2783 N # m 299.7217 N # m
Ta
(a) MAXIMUM SHEAR STRESSES Bar: t1 Tube: t2 d1 G 25.0 mm 80 GPa d2 30.0 mm d3 37.5 mm T1(d1/2) IP1 T2(d3/2) IP2 32.7 MPa 49.0 MPa
; ;
(b) ANGLE OF ROTATION OF END PLATE f T1L GIP1 1.03 T2L GIP2 ; 0.017977 rad
POLAR MOMENTS OF INERTIA Bar: IP1 Tube: IP2 p4 d 32 1 p 1d 4 32 3 38.3495 * 10
4 d2 2 9
m
4
(c) TORSIONAL STIFFNESS
9
114.6229 * 10
m4
kT
T f
22.3 kN # m
;
Problem 3.8-11 A solid steel bar of diameter d1 1.50 in. is enclosed by a steel tube of outer diameter d3 2.25 in. and inner diameter d2 1.75 in. (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has length L 30.0 in., is twisted by a torque T 5000 lb-in. acting on the end plate.
(a) Determine the maximum shear stresses 1 and 2 in the bar and tube, respectively. (b) Determine the angle of rotation (in degrees) of the end plate, assuming that the shear modulus of the steel is G 11.6 106 psi. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.)
Tube A Bar B T
L
End plate
d1 d2 d3
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SECTION 3.8
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311
Solution 3.8-11
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) Bar: T1 Tube: T2 Ta IPI b IPI + IPI IP2 b IP1 + IP2 1187.68 lb-in. 3812.32 lb-in.
Ta
(a) MAXIMUM SHEAR STRESSES Bar: t1 Tube: t2 T1(d1/2) IP1 T2(d3/2) IP2 1790 psi 2690 psi
; ;
(b) ANGLE OF ROTATION OF END PLATE d1 G 1.50 in. d2 11.6 10 psi
6
1.75 in. d3
2.25 in. f
T1L GIP1 0.354
T2L GIP2 ;
0.006180015 rad
POLAR MOMENTS OF INERTIA Bar: IP1 Tube: IP2 p4 d 32 1 p4 1d 32 3 0.497010 in.4
4 d2 2
(c) TORSIONAL STIFFNESS
4
kT
1.595340 in.
T f
809 k- in.
;
Problem 3.8-12 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 40 mm for the brass core and d2 50 mm for the steel sleeve. The shear moduli of elasticity are Gb 36 GPa for the brass and Gs 80 GPa for the steel.
Assuming that the allowable shear stresses in the brass and steel are 48 MPa and s 80 MPa, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.)
b
T
Steel sleeve Brass core T
d1 d2
Probs. 3.8-12 and 3.8-13
Solution 3.8-12
Composite shaft shrink fit
d1 d2 GB 40 mm 50 mm 36 GPa GS
80 GPa
Allowable stresses: 48 MPa S 80 MPa B
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BRASS CORE (ONLY)
Eq. (3-44b): TS
Ta
GSIPS b GBIPB + GSIPS 0.762082 T
T
TB
TS (CHECK)
ALLOWABLE TORQUE T BASED UPON BRASS CORE IPB GBIPB p4 d 32 1 251.327 * 10
9
m4
9047.79 N m2
TB(d1/2) 2tBIPB TB IPB d1 Substitute numerical values: tB TB 0.237918 T 2(48 MPa)(251.327 * 10 40 mm T 2535 N m
9
STEEL SLEEVE (ONLY)
m4)
ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE p4 (d 32 2
4 d1 ) 9
IPS GSIPS
362.265 * 10
m4
TS(d2/2) 2tSIPS TS IPS d2 SUBSTITUTE NUMERICAL VALUES: tS TS 0.762082 T 2(80 MPa)(362.265 * 10 50 mm T 1521 N m Tmax 1520 N m ; STEEL SLEEVE GOVERNS
9
28,981.2 N m2
TORQUES Total torque: T Eq. (3-44a): TB TB Ta TS GBIPB b GBIPB + GS IPS 0.237918 T
m4)
Problem 3.8-13 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 1.6 in. for the brass core and d2 2.0 in. for the steel sleeve. The shear moduli of elasticity are Gb 5400 ksi for the brass and Gs 12,000 ksi for the steel.
Assuming that the allowable shear stresses in the brass and steel are b 4500 psi and s 7500 psi, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.)
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SECTION 3.8
Statically Indeterminate Torsional Members
313
Solution 3.8-13
Composite shaft shrink fit
TORQUES Total torque: T Eq. (3-44 a): TB TB Ta Ta TS GBIPB b GBIPB + GSIPS GSIPS b GBIPB + GSIPS
0.237918 T d1 d2 GB 1.6 in. 2.0 in. 5,400 psi GS 12,000 psi T 7500 psi TB Eq. (3-44 b): TS
0.762082 T TS (CHECK)
Allowable stresses:
B
4500 psi
S
ALLOWABLE TORQUE T BASED UPON BRASS CORE TB(d1/2) 2tBIPB TB IPB d1 Substitute numerical values: tB TB 0.237918 T 2(4500 psi)(0.643398 in.4) 1.6 in.
BRASS CORE (ONLY)
IPB GBIPB
p4 d1 32
0.643398 in.4 106 lb-in.2
T
15.21 k-in.
ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE TS(d2/2) 2tSIPS TS IPS d2 Substitute numerical values: tS tS T 0.762082 T 9.13 k-in. Tmax 9.13 k-in. ; 2(7500 psi)(0.927398 in.4) 2.0 in.
3.47435
STEEL SLEEVE (ONLY)
STEEL SLEEVE GOVERNS IPS GSIPS p (d 4 32 2 11.1288 d14) 0.927398 in.4
106 lb-in.2
Problem 3.8-14 A steel shaft (Gs 80 GPa) of total length L 3.0 m is encased for one-third of its length by a brass sleeve (Gb 40 GPa) that is securely bonded to the steel (see figure). The outer diameters of the shaft and sleeve are d1 70 mm and d2 90 mm. respectively.
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(a) Determine the allowable torque T1 that may be applied to the ends of the shaft if the angle of twist between the ends is limited to 8.0. (b) Determine the allowable torque T2 if the shear stress in the brass is limited to b 70 MPa. (c) Determine the allowable torque T3 if the shear stress in the steel is limited to s 110 MPa. (d) What is the maximum allowable torque Tmax if all three of the preceding conditions must be satisfied?
Brass sleeve T A
d2 = 90 mm
Steel shaft
d1 = 70 mm T
1.0 m L = 2.0 m 2 d1 Brass sleeve d1
B L = 2.0 m 2 d1 Steel shaft d2
C
d2
Solution 3.8-14
(a) ALLOWABLE TORQUE T1 BASED ON TWIST AT ENDS OF 8 DEGREES First find torques in steel (Ts) & brass (Tb) in segment in which they are joined - 1 degree stat-indet; use Ts as the internal redundant; see equ. 3-44a in text example Ts Gs IPs T1 a b Gs IPs + Gb IPb statics Gb IPb b Gs IPs + GbIPb now find twist of 3 segments: Tb T1 Ts Tb L L L Ts T1 4 4 2 + + Gb IPb Gs IPs Gs IPs T1 T1 a
For middle term, brass sleeve & steel shaft twist the same so could use Tb(L/4)/(Gb IPb) instead Let a = allow; substitute expression for Ts then simplifiy; finally, solve for T1, allow Gs IPs L L L T1 a b T1 Gs IPs + Gb IPb 4 4 2 + + Gb IPb Gs IPs Gs IPs T1 L L L T1 T1 4 4 2 + + Gb IPb Gs IPs + Gb IPb Gs IPs T1 1 L 1 2 + T1 a + b 4 Gb IPb GsIPs + Gb IPb Gs IPs 4a Gb IPb(GsIPs + Gb IPb)Gs IPs c22 d 2 L Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb
fa
fa fa
T1, allow
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315
NUMERICAL VALUES Gs d1 IPs 80 GPa Gb 70 mm p4 d1 32 d2 IPs
fa
8a
p b rad 180 3.0 m
T2, allow
13.69 kN m
so T2 for hollow segment controls 40 GPa L 90 mm 2.357 10
6
(c) ALLOWABLE
s
TORQUE
s
T3
BASED ON ALLOWABLE SHEAR
STRESS IN STEEL,
m4
110 MPa First check segment 2 with brass sleeve over steel shaft 10
6
p Ad 4 d14 B 32 2 T1, allow 9.51 kN m IPb (b) ALLOWABLE
TORQUE
b
IPb ;
4.084
m4 t
Ts
d1 2
IPS T3 a
where from stat-indet analysis above GsIPS b Gs IPS + Gb IPb 2ts (Gs IPs + Gb IPb) d1Gs 13.83 kN m
T2
BASED ON ALLOWABLE SHEAR
STRESS IN BRASS,
Ts
70 MPa b First check hollow segment 1 (brass sleeve only) T2 t d2 2 2tbIPb d2 ;
T3, allow T3, allow
IPb
T2, allow
also check segment 3 with steel shaft alone T3 t d1 2 IPs 2ts IPs d1 ; controls over T3 above
T2, allow
6.35 kN m
controls over T2 below also check segment 2 with brass sleeve over steel shaft Tb t d2 2
T3, allow 7.41 kN m
T3, allow where from stat-indet analysis above
IPb T2 a
(d) Tmax IF ALL PRECEDING CONDITIONS MUST BE
CONSIDERED
Tb
T2, allow
GbIPb b Gs IPS + Gb IPb 2tb(Gs IPS + Gb IPb) d2 Gb
from (b) above Tmax 6.35 kN m ; max. shear stress in hollow brass sleeve in segment 1 controls overall
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Problem 3.8-15 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is fixed against rotation at A and B, as shown in the figure. The outside diameters at the ends are dA and dB 2dA. A hollow section of length L/2 and constant thickness t dA/10 is cast into the tube and extends from B halfway toward A. Torque To is applied at L/2.
(a) Find the reactive torques at the supports, TA and TB. Use numerical values as follows: dA G 3.9 106 psi, T0 40,000 in.-lb. (b) Repeat (a) if the hollow section has constant diameter dA.
Fixed against rotation TA dA L 2 A T0 L (a) Fixed against rotation T0 dA L 2 B Fixed against rotation TB dB (b) dA Fixed against rotation TB dB
2.5 in., L
48 in.,
B
d(x) t constant
TA
A
L
Solution 3.8-15
Solution approach-superposition: select TB as the redundant (1 SI ) L 2 TA1 A T0 f1 L
B
1(same results for parts a & b) 608T0L 81GpdA
4
f1
2. 389
T0L GdA 4
+
TA2 A
L 2
B 2 TB
See Prob. 3.4-13 for results for w2 for Parts a & b
f2a f2a 3.868 3.057 T0 L Gd A 4 T0L Gd A 4
L
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317
(a) REACTIVE TORQUES, TA & TB, FOR CASE OF
CONSTANT THICKNESS OF HOLLOW SECTION OF TUBE
(b) REACTIVE TORQUES, TA & TB, FOR CASE OF
CONSTANT DIAMETER OF HOLE
compatibility equation: TB redundant T0 40000 in.-lb TB TB TA TA 608T0 L
1
2
0 TB TB TA ; ; TA
a
608T0 L
81GpdA
ba 4
GdA4 b 3.05676L 31266 in.-lb 8734 in.-lb ; ;
GdA4 a ba b 81G pdA4 3.86804L T0 1.94056 p T0 TB TB TA TB 24708 in-lb 15292 in-1b
2.45560 T0 TB
T0 p
TB
TB TA
40,000 in.-1b (check)
40,000 in.-lb (check)
Problem 3.8-16 A hollow circular tube A (outer diameter dA, wall thickness tA) fits over the end of a circular tube B (dB, tB), as shown in the figure. The far ends of both tubes are fixed. Initially, a hole through tube B makes an angle with a line through two holes in tube A. Then tube B is twisted until the holes are aligned, and a pin (diameter dp) is placed through the holes. When tube B is released, the system returns to equilibrium. Assume that G is constant.
(a) Use superposition to find the reactive torques TA and TB at the supports. (b) Find an expression for the maximum value of if the shear stress in the pin, p, cannot exceed p,allow. (c) Find an expression for the maximum value of if the shear stress in the tubes, t, cannot exceed t,allow. (d) Find an expression for the maximum value of if the bearing stress in the pin at C cannot exceed b,allow.
IPA TA A Tube A C L b Pin at C
IPB Tube B
TB B
L
Tube A Tube B
Cross-section at C
Solution 3.8-16
(a) SUPERPOSITION compatibility:
B1
TO FIND TORQUE REACTIONS
-
USE
TB
(b) ALLOWABLE SHEAR IN PIN RESTRICTS MAGNITUDE OF
TORQUE PIN AT
AS THE REDUNDANT
B1 B2
0
joint tubes by pin then release end B TBL 1 1 a + b G IPA IPB TBL IPB + IPA a b G IPAIPB Gb IPAIPB a b L IPA IPB TB ; statics ;
V
fB2 fB2 TB TA
TB FORCE COUPLE VdB WITH V C TB V tp dB As TB dB p2 d 4P tp, allow L 4G ;
SHEAR IN
tp, allow
tp, allow
Gb IPAIPB a b L IPA IPB p2 dB dp 4
b max ca
IPB + IPA b dBpdP 2 d IPAIPB
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(c) ALLOWABLE SHEAR IN TUBES RESTRICTS MAGNITUDE
OF
Bearing stresses from tubes A & B are: sbA FA s dPtA bB TB dA tA dPtA FB dPtB TB dB tB dP tB
tmax
dA TB 2 IPA
or
tmax
dB TB 2 IPB
tmax or
Gb IPAIPB dA a b L IPA IPB 2 IPA Gb IPAIPB dB a b L IPA IPB 2 IPB
sbA
sbB
substitute TB expression from part (a), then simplify solve for b Gb IPAIPB a b L IPA + IPB sbA dA tA dPtA Gb IPAIPB a b L IPA + IPB dB tB dPtB b b L(IPB GIPAIPB + IPA)(dA tA) dP tA tB)dPtB
tmax
simplifying these two equ., then solving for gives: tmax or tmax b max or b max IPA + IPB 2L tt, allow a ba b GdB IPA controls ; b max Gb dB IPA a b L IPA + IPB 2 IPA + IPB 2L tt, allow a ba b GdA IPB ; Gb IPB dA a b L IPA + IPB 2 sbB sbA sbB b max
IPAIPB G L (IPB + IPA)(dB L G tA)dPtA
sb, allow c
(IPB + IPA)(dA IPAIPB L G
d
;
where lesser value of
s b, allow c
(d) ALLOWABLE BEARING STRESS IN PIN RESTRICTS
MAGNITUDE OF
(IPB + IPA)(dB IPAIPB
tB)dPtB
d
;
Torque TB force couple FB (dB tB) or FA(dA tA), with F ave. bearing force on pin at C
where lesser value controls
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Strain Energy in Torsion
Problem 3.9-1 A solid circular bar of steel (G 11.4 106 psi) with length L 30 in. and diameter d 1.75 in. is subjected to pure torsion by torques T acting at the ends (see figure).
(a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 4500 psi. (b) From the strain energy, calculate the angle of twist (in degrees).
T d T
L
Solution 3.9-1
Steel bar
pd2Lt2 max 16G Substitute numerical values: U 32.0 in.-lb ; (Eq. 2)
G L d
max
11.4 30 in. 1.75 in.
106 psi
(b) ANGLE OF TWIST U Tf 2 f 2U T
4500 psi 16 T pd3 pd 4 32 T pd3tmax 16 (Eq. 1) Substitute for T and U from Eqs. (1) and (2): f 2Ltmax Gd (Eq. 3)
tmax
IP
Substitute numerical values: 0.013534 rad 0.775 ;
(a) STRAIN ENERGY U T2L 2GIP a pd3tmax 2 L 32 ba b a 4b 16 2G pd
Problem 3.9-2 A solid circular bar of copper (G 45 GPa) with length L to pure torsion by torques T acting at the ends (see figure).
0.75 m and diameter d
40 mm is subjected
(a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 32 MPa. (b) From the strain energy, calculate the angle of twist (in degrees)
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Solution 3.9-2
Copper bar
(a) STRAIN ENERGY U T2L 2GIP a pd3tmax 2 L 32 ba ba b 16 2G pd 4 (Eq. 2)
G L d
max
45 GPa 0.75 m 40 mm 32 MPa 16T pd
3 3
pd2Lt2 max 16G Substitute numerical values: U pd tmax 16 (Eq. 1) 5.36 J ;
(b) ANGLE OF TWIST Tf 2U f 2 T Substitute for T and U from Eqs. (1) and (2): 2Ltmax (Eq. 3) f Gd Substitute numerical values: U 0.026667 rad 1.53 ;
tmax IP
T
pd4 32
Problem 3.9-3 A stepped shaft of solid circular cross sections (see figure) has length L 45 in., diameter d2 1.2 in., and diameter d1 1.0 in. The material is brass with G 5.6 106 psi. Determine the strain energy U of the shaft if the angle of twist is 3.0.
d2 T
d1 T
L 2
L 2
Solution 3.9-3
Stepped shaft
8T2L 1 1 a 4 + 4b pG d2 d1 Also, U Tf 2 (Eq. 1) (Eq. 2)
Equate U from Eqs. (1) and (2) and solve for T: d1 d2 L G 1.0 in. 1.2 in. 45 in. 5.6 3.0 10 psi (brass) 0.0523599 rad
6
T U
44 pGd1 d2 f
16L(d4 + d4) 1 2 Tf 2
44 pGf2 d1 d2 a4 b 32L d1 + d4 2
f
radians
SUBSTITUTE NUMERICAL VALUES: U 22.6 in.-lb ; 16 T2(L/2) + pGd4 1
STRAIN ENERGY U T2L a 2GI 16 T2(L/2) pGd4 2
P
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321
Problem 3.9-4 A stepped shaft of solid circular cross sections (see figure) has length L and diameter d1 30 mm. The material is steel with G 80 GPa. Determine the strain energy U of the shaft if the angle of twist is 1.0.
0.80 m, diameter d2
40 mm,
Solution 3.9-4
Stepped shaft
Also, U Tf 2 (Eq. 2)
Equate U from Eqs. (1) and (2) and solve for T: T d1 L 30 mm 0.80 m 1.0 d2 G 40 mm 80 GPa (steel) U
44 pG d1 d2 f 4 4 16L(d1 + d2 )
Tf 2
44 pGf2 d1 d2 a4 b 4 32L d1 + d2
f
radians
0.0174533 rad SUBSTITUTE NUMERICAL VALUES: U 16T2(L/2)
4 pGd2
STRAIN ENERGY U TL a 2GI
2
1.84 J
;
16T2(L/2) +
4 pGd1
P
8T2L 1 1 a 4 + 4b pG d2 d1
(Eq. 1)
Problem 3.9-5 A cantilever bar of circular cross section and length L is fixed at one end and free at the other (see figure). The bar is loaded by a torque T at the free end and by a distributed torque of constant intensity t per unit distance along the length of the bar.
(a) What is the strain energy U1 of the bar when the load T acts alone? (b) What is the strain energy U2 when the load t acts alone? (c) What is the strain energy U3 when both loads act simultaneously?
t
L
T
Solution 3.9-5
Cantilever bar with distributed torque
G IP T t shear modulus polar moment of inertia torque acting at free end torque per unit distance
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(a) LOAD T ACTS ALONE (Eq. 3-51a) U1 T2L 2GIP ;
(c) BOTH LOADS ACT SIMULTANEOUSLY
(b) LOAD t ACTS ALONE From Eq. (3-56) of Example 3-11: U2 t2L3 6GIP At distance x from the free end: ; T(x) U3 T + tx
L
L [T(x)]2 1 dx (T + tx)2dx 2GIP 2GIP L L 0 0 T2L TtL2 t2L3 + + ; 2GIP 2GIP 6GIP
NOTE: U3 is not the sum of U1 and U2.
Problem 3.9-6 Obtain a formula for the strain energy U of the statically indeterminate circular bar shown in the figure. The bar has fixed supports at ends A and B and is loaded by torques 2T0 and T0 at points C and D, respectively. Hint: Use Eqs. 3-46a and b of Example 3-9, Section 3.8, to obtain the reactive torques.
2T0 A C L 4 L 2
T0 B D L 4
Solution 3.9-6
Statically indeterminate bar
REACTIVE TORQUES From Eq. (3-46a): (2T0)a TA TB 3T0 L TA 3L b 4 + 5T0 4 L T0 a b 4 L 7T0 4
STRAIN ENERGY (from Eq. 3-53) U Ti2Li a i 1 2GiIPi
n
L L L 1 2 2 cT 2 a b + TCD a b + TDB a b d 2GIp AC 4 2 4 1 ca 2GIP +a U 7T0 2 L ba b 4 4
INTERNAL TORQUES TAC 7T0 4 TCD T0 4 TDB 5T0 4
T0 2 L 5T0 2 L ba b+a b a bd 4 2 4 4 ;
2 19T0 L 32GIP
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323
Problem 3.9-7 A statically indeterminate stepped shaft ACB is fixed at ends A and B and loaded by a torque T0 at point C (see figure). The two segments of the bar are made of the same material, have lengths LA and LB, and have polar moments of inertia IPA and IPB. Determine the angle of rotation of the cross section at C by using strain energy. Hint: Use Eq. 3-51b to determine the strain energy U in terms of the angle . Then equate the strain energy to the work done by the torque T0. Compare your result with Eq. 3-48 of Example 3-9, Section 3.8.
A
IPA
T0 C IPB
LA LB
B
Solution 3.9-7
Statically indeterminate bar
WORK DONE BY THE TORQUE T0 W T0f 2
EQUATE U AND W AND SOLVE FOR Gf2 IPA IPB a + b 2 LA LB f STRAIN ENERGY (FROM EQ. 3-51B) U GIPif2 i a 2L i i1
n
T0f 2 ;
T0LALB G(LBIPA + LAIPB)
GIPAf2 GIPBf2 + 2LA 2LB
(This result agrees with Eq. (3-48) of Example 3-9, Section 3.8.)
Gf2 IPA IPB a + b 2 LA LB
Problem 3.9-8 Derive a formula for the strain energy U of the cantilever bar shown in the figure. The bar has circular cross sections and length L. It is subjected to a distributed torque of intensity t per unit distance. The intensity varies linearly from t 0 at the free end to a maximum value t t0 at the support.
t0
t
L
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Solution 3.9-8
Cantilever bar with distributed torque
x distance from right-hand end of the bar
ELEMENT d Consider a differential element dj at distance j from the right-hand end.
STRAIN ENERGY OF ELEMENT dx dU [T(x)]2dx 2GIP t0 2 1 a b x4dx 2GIP 2L t2 0 8L2GIP
L
x4 dx
STRAIN ENERGY OF ENTIRE BAR U dT dT external torque acting on this element t(j)dj j t0 a bdj L ELEMENT dx AT DISTANCE x U t2L3 0 40GIP L 0 dU x4 dx 8L2GIP L 0 t2 L5 0 ab 8L2GIP 5 ; t2 0
L
T(x) T(x) T(x)
internal torque acting on this element total torque from x
x x
0 to x
x
L 0 t0x2 2L
dT
L 0
t0 a
j bdj L
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325
Problem 3.9-9 A thin-walled hollow tube AB of conical shape has constant thickness t and average diameters dA and dB at the ends (see figure).
(a) Determine the strain energy U of the tube when it is subjected to pure torsion by torques T. (b) Determine the angle of twist of the tube. Note: Use the approximate formula IP ring; see Case 22 of Appendix D. d3t/4 for a thin circular
T
A
B T
L t t
dA
dB
Solution 3.9-9
Thin-walled, hollow tube
Therefore,
L
dx cdA + a dB L dA bx d
3
L 0
1 t dA dB d(x) d(x) thickness average diameter at end A average diameter at end B average diameter at distance x from end A dA + a pd3t 4 p[d(x)]3t 4
L
L
2(dB L
dA) L
cdA + a
2
dB L
dA L
bx d
2
0
+ 2(dB
2(dB
22 2dA dB
dA)(dB)
dA)(dA)2
dB L
dA
bx
L(dA + dB) Substitute this expression for the integral into the equation for U (Eq. 1): U 2T2 L(dA + dB) 22 pGt 2dAdB T2L dA + dB a 22b pGt dA dB Tf 2 ;
POLAR MOMENT OF INERTIA IP IP(x)
pt cd + a 4A
dB L
dA
bx d
3
(b) ANGLE OF TWIST Work of the torque T: W W (Eq. 1) U Tf 2
(a) STRAIN ENERGY (FROM EQ. 3-54) U T2dx 0 L 2GIP(x) L dx 2T2 3 pGt L dA dB 0 cdA + a bx d L
T2L(dA + dB) pGt d2 d2 AB
Solve for : f 2TL(dA + dB) pGt d2 d2 AB ;
From Appendix C: dx L (a + bx)3 1 2b(a + bx)2
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Problem 3.9-10 A hollow circular tube A fits over the end of a solid circular bar B, as shown in the figure. The far ends of both bars are fixed. Initially, a hole through bar B makes an angle with a line through two holes in tube A. Then bar B is twisted until the holes are aligned, and a pin is placed through the holes. When bar B is released and the system returns to equilibrium, what is the total strain energy U of the two bars? (Let IPA and IPB represent the polar moments of inertia of bars A and B, respectively. The length L and shear modulus of elasticity G are the same for both bars.)
IPA Tube A
IPB Bar B
L b
L
Tube A Bar B
Solution 3.9-10
Circular tube and bar
TUBE A
COMPATIBILITY
A B
FORCE-DISPLACEMENT RELATIONS fA TL GIPA fB TL GIPB
T
A
torque acting on the tube angle of twist
Substitute into the equation of compatibility and solve for T: T bG IPAIPB a b L IPA + IPB T 2L 2GIP T 2L T 2L + 2GIPA 2GIPB
BAR B
STRAIN ENERGY U g
T2L 1 1 a + b 2G IPA IPB Substitute for T and simplify: U T
B
b 2G IPA IPB a b 2L IPA + IPB
;
torque acting on the bar angle of twist
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327
Problem 3.9-11 A heavy flywheel rotating at n revolutions per minute is rigidly attached to the end of a shaft of diameter d (see figure). If the bearing at A suddenly freezes, what will be the maximum angle of twist of the shaft? What is the corresponding maximum shear stress in the shaft? (Let L length of the shaft, G shear modulus of elasticity, and Im mass moment of inertia of the flywheel about the axis of the shaft. Also, disregard friction in the bearings at B and C and disregard the mass of the shaft.)
Hint: Equate the kinetic energy of the rotating flywheel to the strain energy of the shaft.
A
d B C
n (rpm)
Solution 3.9-11
Rotating flywheel
IP d U p4 d 32 diameter of shaft pGd 4f2 64L
UNITS: G d n diameter rpm L U IP (force)/(length)2 (length)4 radians length (length)(force) p2n2Im 1800 pGd 4f2 64 L
KINETIC ENERGY OF FLYWHEEL K.E. v n K.E. 1 Imv2 2 2pn 60 rpm 2pn 2 1 Im a b 2 60 p n Im 1800 UNITS: Im K.E. (force)(length)(second)2 radians per second (length)(force)
22
EQUATE KINETIC ENERGY AND STRAIN ENERGY K.E. U
2A
Solve for : f 2n 15d 2pImL G ;
MAXIMUM SHEAR STRESS t T(d/2) IP f TL GIP
Eliminate T: t tmax tmax 2pImL 2L15d A G 2pGIm n 15dA L Gdf 2L Gd2n
2
STRAIN ENERGY OF SHAFT (FROM EQ. 3-51b) U GIPf 2L
2
;
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Thin-Walled Tubes
Problem 3.10-1 A hollow circular tube having an inside diameter of 10.0 in. and a wall thickness of 1.0 in. (see figure) is subjected to a torque T 1200 k-in. Determine the maximum shear stress in the tube using (a) the approximate theory of thin-walled tubes, and (b) the exact torsion theory. Does the approximate theory give conservative or nonconservative results?
1.0 in.
10.0 in.
Solution 3.10-1
Hollow circular tube
APPROXIMATE THEORY (EQ. 3-63) t1 T 2pr t
2
1200 k-in. 2p(5.5 in.)2(1.0 in.) ; 6310 psi
6314 psi
approx
EXACT THEORY (EQ. 3-11) T t r r d2 d1 1200 k-in. 1.0 in. radius to median line 5.5 in. outside diameter inside diameter 12.0 in. 10.0 in. t exact t2 T(d2/2) IP Td2 p 4 2a b 1d2 32
4 d1 2
16(1200k-in.)(12.0 in.) p[(12.0 in.)4 6831 psi 6830 psi ; (10.0 in.)4]
Because the approximate theory gives stresses that are too low, it is nonconservative. Therefore, the approximate theory should only be used for very thin tubes.
Problem 3.10-2 A solid circular bar having diameter d is to be replaced by a rectangular tube having cross-sectional dimensions d 2d to the median line of the cross section (see figure). Determine the required thickness tmin of the tube so that the maximum shear stress in the tube will not exceed the maximum shear stress in the solid bar.
t t d d
2d
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Thin-Walled Tubes
329
Solution 3.10-2 SOLID BAR
Bar and tube
Am tmax 16T pd3 (Eq. 3-12) tmax (d)(2d) T 2tAm 2d2 T 4td2 (Eq. 3-64) (Eq. 3-61)
EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t RECTANGULAR TUBE 16T pd3 tmin T 4td2 pd 64 ;
If t tmin, the shear stress in the tube is less than the shear stress in the bar.
Problem 3.10-3 A thin-walled aluminum tube of rectangular cross section (see figure) has a centerline dimensions b 6.0 in. and h 4.0 in. The wall thickness t is constant and equal to 0.25 in.
(a) Determine the shear stress in the tube due to a torque T 15 k-in. (b) Determine the angle of twist (in degrees) if the length L of the tube is 50 in. and the shear modulus G is 4.0 106 psi.
t h
b
Probs. 3.10-3 and 3.10-4
Solution 3.10-3
Thin-walled tube
Eq. (3-64): Am Eq. (3-71) with t1 J 28.8 in.4 bh t2 24.0 in.2 t: J 2b2h2t b+h
(a) SHEAR STRESS (EQ. 3-61) t b h t T L G 6.0 in. 4.0 in. 0.25 in. 15 k-in. 50 in. 4.0 106 psi f T 2tAm TL GJ 0.373 1250 psi ;
(b) ANGLE OF TWIST (EQ. 3-72) 0.0065104 rad ;
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Problem 3.10-4 A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions b and h 100 mm. The wall thickness t is constant and equal to 6.0 mm.
150 mm
(a) Determine the shear stress in the tube due to a torque T 1650 N m. (b) Determine the angle of twist (in degrees) if the length L of the tube is 1.2 m and the shear modulus G is 75 GPa.
Solution 3.10-4
Thin-walled tube
b h t T L G 150 mm 100 mm 6.0 mm 1650 N m 1.2 m 75 GPa (a) SHEAR STRESS (Eq. 3-61) t T 2tAm 9.17 MPa ;
(b) ANGLE OF TWIST (Eq. 3-72) f TL GJ 0.140 0.002444 rad ;
Eq. (3-64): Am Eq. (3-71) with t1 J 10.8 10
6
bh t2
0.015 m2 t: J 2b2h2t b+h
m4
Tube (1) Bar (2)
Problem 3.10-5 A thin-walled circular tube and a solid circular bar of the same material (see figure) are subjected to torsion. The tube and bar have the same cross-sectional area and the same length. What is the ratio of the strain energy U1 in the tube to the strain energy U2 in the solid bar if the maximum shear stresses are the same in both cases? (For the tube, use the approximate theory for thin-walled bars.)
Solution 3.10-5 THIN-WALLED TUBE (1)
Am tmax T U1 T2L 2GJ r2 T 2tAm 2 rt
2
J
2 r3t A T 2pr t
2
2 rt
SOLID BAR (2) A tmax U2 T2L 2GIP
2 pr2
IP 2T
p4 r 22 T
3 pr2 tmax 2
max
12pr2ttmax22L 2G(2pr3t)
prtt2 L max G A But rt 2p U1 At2 L max 2G
3 pr2 3 22 (pr2 tmax)2L pr2 tmaxL 4G p4 8Ga r2 b 2
Tr2 IP
2 But pr2
A
U2
2 Atmax L 4G
RATIO U1 2 U2
;
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Problem 3.10-6 Calculate the shear stress and the angle of twist (in degrees) for a steel tube (G 76 GPa) having the cross section shown in the figure. The tube has length L 1.5 m and is subjected to a torque T 10 kN m.
t = 8 mm r = 50 mm r = 50 mm
b = 100 mm
Solution 3.10-6
Steel tube
SHEAR STRESS G L T 76 GPa. 1.5 m 10 kN m t T 2tAm 10 kN # m 2(8 mm)(17,850 mm2) 35.0 MPa ANGLE OF TWIST f 2(100 mm)(50 mm) TL GJ (10 kN # m)(1.5 m) (76 GPa)(19.83 * 106 mm4) 0.00995 rad 0.570 2 (50 mm) ; ;
Am Am Lm
r2
2br
(50 mm)2 17,850 mm2 2b 2r 2(100 mm) 514.2 mm 4tA2 m Lm
J
4(8 mm)(17,850 mm2)2 514.2 mm
19.83 * 106 mm4
Problem 3.10-7 A thin-walled steel tube having an elliptical cross section with constant thickness t (see figure) is subjected to a torque T 18 k-in. Determine the shear stress and the rate of twist (in degrees per inch) if G 12 106 psi, t 0.2 in., a 3 in., and b 2 in. (Note: See Appendix D, Case 16, for the properties of an ellipse.)
t
2b
2a
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Solution 3.10-7
Elliptical tube
FROM APPENDIX D, CASE 16: Am pab Lm L p[1.5(a + b) p[1.5(5.0 in.) J 4tA2 m Lm 26.0 in.2] 1ab]
p(3.0 in.)(2.0 in.)
18.850 in.2 15.867 in.
4(0.2 in.)(18.850 in.2)2 15.867 in.
17.92 in.4 T G t t 18 k-in. 12 0.2 in 106 psi a 3.0 in. b 2.0 in. constant SHEAR STRESS t T 2tAm 2390 psi 18 k-in. 2(0.2 in.)(18.850 in.2) ;
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST) u u f L T GJ 18 k-in. (12 * 106 psi)(17.92 in.)4
6
83.73 * 10
rad/in.
0.0048/in.
;
Problem 3.10-8 A torque T is applied to a thin-walled tube having a cross section in the shape of a regular hexagon with constant wall thickness t and side length b (see figure). Obtain formulas for the shear stress and the rate of twist .
t
b
Solution 3.10-8
Regular hexagon
b t Lm Length of side Thickness 6b
FROM APPENDIX D, CASE 25: 3 13b2 2
2
60 n
6 6b2 cot 30 4
Am
b nb cot 4 2
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333
SHEAR STRESS t T 2tAm 4A2 t m
Lm
u 9b2t 4A2 t m Lm ;
T13
T GJ
2T G(9b t)
3
2T 9Gb3t
;
(radians per unit length) 9b3t 2
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST) J ds t
L 0
Problem 3.10-9 Compare the angle of twist 1 for a thin-walled circular tube (see figure) calculated from the approximate theory for thin-walled bars with the angle of twist 2 calculated from the exact theory of torsion for circular bars.
(a) Express the ratio 1/ 2 in terms of the nondimensional ratio r/t. 5, 10, and 20. What conclusion (b) Calculate the ratio of angles of twist for about the accuracy of the approximate theory do you draw from these results?
t r C
Solution 3.10-9
Thin-walled tube
(a) RATIO f1 f2 Let b 4r 2 + t 2 4r 2 r t f1 f2 1+ t2 4r 2 1 4b 2
2
1+
1/
;
APPROXIMATE THEORY f1 TL GJ J 2 rt
3
(b) f1 TL 2pGr3t 5 10 20
EXACT THEORY f2 f2 TL GIP TL GIP From Eq. (3-17): Ip 2TL pGrt(4r 2 + t 2) prt (4r 2 + t 2) 2
1.0100 1.0025 1.0006
As the tube becomes thinner and becomes larger, the ratio 1/ 2 approaches unity. Thus, the thinner the tube, the more accurate the approximate theory becomes.
Problem 3.10-10 A thin-walled rectangular tube has uniform thickness t and dimensions a b to the median line of the cross section (see figure). a/b if How does the shear stress in the tube vary with the ratio the total length Lm of the median line of the cross section and the torque T remain constant? From your results, show that the shear stress is smallest when the 1). tube is square (
t
b
a
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Solution 3.10-10
Rectangular tube
T, t, and Lm are constants. Let k 2T tL2 m constant t k (1 + b)2 b
t a, b b Lm T
thickness (constant) dimensions of the tube a b 2(a b) constant constant t ab k min 4 tmin 8T tL2 m is minimum when 1
SHEAR STRESS t Lm b Am t T 2tAm 2b(1 Am ) ab b2
From the graph, we see that and the tube is square. ALTERNATE SOLUTION t constant Am
2 Lm bc d 2(1 + b)
2T (1 + b)2 c d b tL2 m (1 + b)2(1) 2T b(2)(1 + b) c d tL2 b2 m ) (1 )2 0 1 0
Lm 2(1 + b) bL2 m
2
dt db
or 2 (1 2T(1 + b)2 tL2 b m
4(1 + b) T(4)(1 + b)2 T 2tAm 2tbL2 m
;
Thus, the tube is square and is either a minimum or a maximum. From the graph, we see that is a minimum.
Problem 3.10-11 A tubular aluminum bar (G 4 106 psi) of square cross section (see figure) with outer dimensions 2 in. 2 in. must resist a torque T 3000 lb-in. Calculate the minimum required wall thickness tmin if the allowable shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft.
t 2 in.
2 in.
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335
Solution 3.10-11
Square aluminum tube
THICKNESS t BASED UPON SHEAR STRESS t T 2tAm tAm in. t)2 1 b T 2t in. t(b T t)2 lb-in. 13 in. 3 T 2t psi
UNITS: t t(2.0 in. 3t(2 Outer dimensions: 2.0 in. G T 4 2.0 in. 106 psi 4500 psi 0.01 rad/ft 0.01 rad/in. 12 t)2
3000 lb-in. 2(4500 psi) 0 0.0915 in.
Solve for t: t
THICKNESS t BASED UPON RATE OF TWIST u T GJ T Gt(b in. t)3 G t)
3
3000 lb-in.
t(b
t)3 rad/in.
T Gu
allow
UNITS: t t(2.0 in.
psi
u allow Let b
3000 lb-in (4 * 106 psi)(0.01/12 rad/in.) 9 10 9 0
outer dimension 2.0 in. b 4(b t) t)
4
Centerline dimension Am J (b 4tA2 m Lm t)
2
t t) t(b t)3
10t(2 t
t)3
Lm
Solve for t: 0.140 in.
4t(b 4(b
ANGLE OF TWIST GOVERNS tmin 0.140 in. ;
Problem 3.10-12 A thin tubular shaft of circular cross section (see figure) with inside diameter 100 mm is subjected to a torque of 5000 N m. If the allowable shear stress is 42 MPa, determine the required wall thickness t by using (a) the approximate theory for a thin-walled tube, and (b) the exact torsion theory for a circular bar.
100 mm t
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Solution 3.10-12
Thin tube
(b) EXACT THEORY t Ip Tr2 Ip p4 (r 22 r4) 1 p [(50 + t)4 2 (50)4]
T
5,000 N m d1 42 MPa
inner diameter
100 mm
42 MPa
allow
(5,000 N # m)(50 + t) p [(50 + t)4 (50)4] 2 (5000 N # m)(2) (p)(42 MPa) 5 * 106 mm3 21p
t is in millimeters. r Average radius 50 mm + r1 r2 t 2
(50 + t)4 (50)4 50 + t
Inner radius 50 mm Outer radius 50 mm t Am r2
Solve for t: t 7.02 mm ;
(a) APPROXIMATE THEORY t T 2tAm T T 2t(pr 2) 2pr 2 t 5,000 N # m 2pa50 + or ta50 + t2 b 2 5,000 N # m 2p(42 MPa) 5 * 106 mm3 84p t2 bt 2
The approximate result is 5% less than the exact result. Thus, the approximate theory is nonconservative and should only be used for thin-walled tubes.
42 MPa
Solve for t: t 6.66 mm ;
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Problem 3.10-13 A long, thin-walled tapered tube AB of circular cross section (see figure) is subjected to a torque T. The tube has length L and constant wall thickness t. The diameter to the median lines of the cross sections at the ends A and B are dA and dB, respectively. Derive the following formula for the angle of twist of the tube:
f 2TL dA + dB a 22 b pGt dAdB
T
A
B T
L t t
Hint: If the angle of taper is small, we may obtain approximate results by applying the formulas for a thin-walled prismatic tube to a differential element of the tapered tube and then integrating along the axis of the tube.
dA
dB
Solution 3.10-13
Thin-walled tapered tube
For entire tube: f 4T pGT L 0
L
dx 3 dA dB cdA + a bx d L
From table of integrals (see Appendix C): t dA dB T d(x) d(x) J J(x) thickness average diameter at end A average diameter at end B torque average diameter at distance x from end A. dA + a 2pr t
3 L
1
dx (a + bx)3
1 2b(a + bx)2
f dB L
3
4T pGt
dA
bx
J
1 2a dB L L 2(dB dA)d2 B ; + 2(dB dA b adA + dB L L dA)d2 A dA
2
# xb
d
K0
pd t 4
3 dA dB pt cdA + a bx d 4 L
4T c pGt f
pt [d(x)]3 4
2TL dA + dB a 22 b pGt dAdB
For element of length dx: df Tdx GJ(x) 4Tdx dB GptcdA + a dA L bx d
3
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CHAPTER 3
Torsion
Stress Concentrations in Torsion
The problems for Section 3.11 are to be solved by considering the stress-concentration factors.
D2 T R D1 T
Problem 3.11-1 A stepped shaft consisting of solid circular segments having diameters D1 2.0 in. and D2 2.4 in. (see figure) is subjected to torques T. The radius of the fillet is R 0.1 in. If the allowable shear stress at the stress concentration is 6000 psi, what is the maximum permissible torque Tmax? Solution 3.11-1 Stepped shaft in torsion
Probs. 3.11-1 through 3.11-5
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR R D1 K D1 D2 R
allow
0.1 in. 2.0 in. 1.52 tmax
0.05
D2 D1
2.4 in. 2.0 in. Ka pD3 1
1.2 b
Kt nom
16 Tmax
2.0 in. 2.4 in. 0.1 in. 6000 psi
Tmax
pD3tmax 1 16K p(2.0 in.)3(6000 psi) 16(1.52) 6200 lb-in. ;
6200 lb-in.
Tmax
Problem 3.11-2 A stepped shaft with diameters D1 40 mm and D2 60 mm is loaded by torques T 1100 N m (see figure). If the allowable shear stress at the stress concentration is 120 MPa, what is the smallest radius Rmin that may be used for the fillet? Solution 3.11-2 Stepped shaft in torsion
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax K D1 D2 T
allow
Kt nom
Ka
16T pD3 1
b 1.37
40 mm 60 mm 1100 N m 120 MPa
D2 D1
pD3tmax 1 16 T 60 mm 40 mm
p(40 mm)3(120 MPa) 16(1100 N # m) 1.5 D2 D1 1.5 and K 1.37,
From Fig. (3-48) with we get Rmin R L 0.10 D1
0.10(40 mm)
4.0 mm
;
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Problem 3.11-3 A full quarter-circular fillet is used at the shoulder of a stepped shaft having diameter D2 1.0 in. (see figure). A torque T 500 lb-in. acts on the shaft. Determine the shear stress max at the stress concentration for values as follows: D1 5 0.7, 0.8, and 0.9 in. Plot a graph showing max versus D1.
Solution 3.11-3
Stepped shaft in torsion
D1 (in.) 0.7 0.8 0.9
D2/D1 1.43 1.25 1.11
R(in.) 0.15 0.10 0.05
R/D1 0.214 0.125 0.056
K 1.20 1.29 1.41
max(psi)
8900 6400 4900
D2 T D1
1.0 in. 500 lb-in. 0.7, 0.8, and 0.9 in. D1 2R)
Full quarter-circular fillet (D2 R D2 2 D1 0.5 in. D1 2
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax Kt nom K Ka 16 T pD3 1 b 2546 K D3 1 NOTE that max gets smaller as D1 gets larger, even though K is increasing.
16(500 lb-in.) pD3 1
Problem 3.11-4 The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm. The shaft has a full quarter-circular fillet, and the smaller diameter D1 100 mm. If the allowable shear stress at the stress concentration is 100 MPa, at what diameter D2 will this stress be reached? Is this diameter an upper or a lower limit on the value of D2?
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Solution 3.11-4
Stepped shaft in torsion
P n
600 kW 400 rpm
D1
allow
100 mm 100 MPa
Use the dashed line for a full quarter-circular fillet. R L 0.075 D1 D2 D2 D1 R 0.075 D1 7.5 mm 2R 100 mm ; ; 2(7.5 mm) 115 mm 115 mm 0.075 (100 mm)
Full quarter-circular fillet POWER P P T 2pnT ( Eq. 3-42 of Section 3.7) 60 rpm T Newton meters 14,320 N # m 60(600 * 103 W) 2p(400 rpm)
watts n 60P 2pn
This value of D2 is a lower limit
(If D2 is less than 115 mm, R/D1 is smaller, K is larger, and max is larger, which means that the allowable stress is exceeded.)
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax K Kt nom tmax(pD3) 1 16T (100 MPa)(p)(100 mm)3 16(14,320 N # m) 1.37 Ka 16T pD3 1 b
Problem 3.11-5 A stepped shaft (see figure) has diameter D2 stress is 15,000 psi and the load T 4800 lb-in. What is the smallest permissible diameter D1?
1.5 in. and a full quarter-circular fillet. The allowable shear
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Stress Concentrations in Torsion
341
Solution 3.11-5
Stepped shaft in torsion
D2
allow
1.5 in. 15,000 psi D1 D1 2 2R 4800 lb-in. D2 2 D1
Use trial-and-error. Select trial values of D1 D1 (in.) 1.30 1.35 1.40 R (in.) 0.100 0.075 0.050 R/D1 0.077 0.056 0.036 K 1.38 1.41 1.46
max(psi)
T
Full quarter-circular fillet D2 R 0.75 in.
15,400 14,000 13,000
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax pD3 1 K 16(4800 lb-in.) c d p D3 1 K D3 1 Kt nom Ka 16T b
24,450
From the graph, minimum D1
1.31 in.
;
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