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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 ⫽ 18.0 in.
From Eq. (3-3):
p
f ⫽ 3.0° ⫽ (3.0)a 180 b rad
␥max ⫽
⫽ 0.05236 rad
␥allow ⫽ 0.0006 rad
Find dmax
dmax ⫽
rf
df
⫽
L
2L
2L␥ allow
f
dmax ⫽ 0.413 in.
⫽
(2)(18.0 in.)(0.0006 rad)
0.05236 rad
;
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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CHAPTER 3
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Torsion
Solution 3.2-2
NUMERICAL DATA
d ⫽ 56 mm
␥a ⫽ 0.012 radians
f ⫽ 4a
p
b radians
180
solution based on Equ. (3-3):
Lmin ⫽ 162.9 mm
Lmin ⫽
df
2␥a
;
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.
T
T
L
(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?
r2
r1
Probs. 3.2-3, 3.2-4, and 3.2-5
Solution 3.2-3
NUMERICAL DATA
(b) MIN. REQUIRED OUTER RADIUS
r2 ⫽ 1.5r1 ␥max ⫽ 400 ⫻ (10⫺6) radians
u ⫽ 0.125a
p
1
ba b
180 12
r2min ⫽
␥max
u
r2min ⫽
r2min ⫽ 2.2 inches
␥max
u
;
␪ ⫽ 1.818 ⫻ 10⫺4 rad /m.
(a) SHEAR STRAIN AT INNER SURFACE AT RADIUS r1
␥1 ⫽
r1
␥
r2 max
␥1 ⫽
1
␥
1.5 max
␥1 ⫽ 267 ⫻ 10⫺6 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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SECTION 3.2
251
Torsional Deformations
Solution 3.2-4
(b) MAX. PERMISSIBLE OUTER RADIUS
NUMERICAL DATA
L ⫽ 1000 mm
f ⫽ 0.45 a
r1 ⫽ 45 mm
f ⫽ 0.5 a
p
b radians
180
(a) SHEAR STRAIN AT INNER SURFACE
f
␥1 ⫽ r1
␥1 ⫽ 393 ⫻ 10⫺6 radians
L
p
b radians
180
␥max ⫽ 0.0004 radians
r2max ⫽ 50.9 mm
␥max ⫽ r2
f
L
r2max ⫽ ␥max
L
f
;
;
Problem 3.2-5 Solve the preceding problem if the length L ⫽ 56 in., the inner radius r1 ⫽ 1.25 in., the angle of twist is
0.5°, and the allowable shear strain is 0.0004 rad.
Solution 3.2-5
(b) MAXIMUM PERMISSIBLE OUTER RADIUS (r2)max
NUMERICAL DATA
L ⫽ 56 inches r1 ⫽ 1.25 inches
f ⫽ 0.5 a
p
b radians
180
␥max
␥a ⫽ 0.0004 radians
(a) SHEAR STRAIN g1 (IN RADIANS) AT THE INNER
SURFACE
␥1 ⫽ r1
f
L
␥1 ⫽ 195 ⫻ 10⫺6 radians
p
b radians
180
f
⫽ r2
L
f ⫽ 0.5 a
;
␥a ⫽ 0.0004 radians
L
r2max ⫽ ␥ a
f
r2max ⫽ 2.57 inches
;
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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 ⫽ 0.625 in.
MAXIMUM SHEAR STRESS IN THE AXLE
b ⫽ 4.0 in.
From Eq. (3-12):
W ⫽ 100 lb
tmax ⫽
Torque T applied to the axle:
T ⫽ Wb ⫽ 400 lb-in.
tmax ⫽
16T
pd3
(16)(400 lb-in.)
p(0.625 in.)3
tmax ⫽ 8,340 psi
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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SECTION 3.3
Solution 3.3-2
253
Circular Bars and Tubes
Torsion of a drill bit
(b) RATE OF TWIST
From Eq. (3-14):
u⫽
d ⫽ 4.0 mm
T ⫽ 0.3 N⭈m G ⫽ 75 GPa
(a) MAXIMUM SHEAR STRESS
From Eq. (3-12):
tmax ⫽
tmax ⫽
16T
u⫽
T
GIP
0.3 N # m
p
(75 GPa)a
b (4.0 mm)4
32
u ⫽ 0.1592 rad/m ⫽ 9.12°/m
pd3
;
16(0.3 N # m)
p(4.0 mm)3
tmax ⫽ 23.8 MPa
;
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
Page 254
Torsion
Lug wrench
P ⫽ 25 lb
(a) MAXIMUM SHEAR STRESS
From Eq. (3-12):
(16)(450 lb - in.)
16T
⫽
tmax ⫽
3
pd
p(0.5 in.)3
L ⫽ 9.0 in.
d ⫽ 0.5 in.
G ⫽ 11.4 ⫻ 106 psi
T ⫽ torque acting on
arm A
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
T ⫽ P(2L) ⫽ 2(25 lb)
(9.0 in.)
⫽ 450 lb-in.
f ⫽ 0.05790 rad ⫽ 3.32°
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.
;
d
T
T
L
(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)?
Solution 3.3-4
(a) TORSIONAL STIFFNESS OF BAR
d ⫽ 32 mm
GI p
kT ⫽
L
G ⫽ 28 GPa
Ip ⫽
p 4
d
32
Ip ⫽ 1.029 ⫻ 105 mm4
kT ⫽
p
2811092a 0.0324 b
32
1.4
kT ⫽ 2059 N # m
(b) MAX SHEAR STRESS AND STRAIN
f ⫽ 5a
T ⫽ kT f
tmax ⫽
␶max ⫽ 27.9 MPa
gmax ⫽
;
p
b radians
180
d
Ta b
2
Ip
;
tmax
G
␥max ⫽ 997 ⫻ 10⫺6 radians
;
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SECTION 3.3
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?
Solution 3.3-5
Circular Bars and Tubes
d = 0.5 in.
T
T
L
Steel drill rod
T
d = 0.5 in.
T
T⫽
L
G ⫽ 11,600 psi
p
b rad ⫽ 0.52360 rad
180
Lmin ⫽
␶allow ⫽ 40 ksi
⫽
MINIMUM LENGTH
From Eq. (3-12): tmax ⫽
16T
pd3
TL
32TL
⫽
GIP
Gpd4
Gpd 4f
, substitute Tinto Eq. (1):
32L
tmax ⫽ a
d ⫽ 0.5 in.
f ⫽ 30° ⫽ (30°)a
From Eq. (3-15): f ⫽
ba
3
16
pd
Gpd4f
Gdf
b ⫽
32L
2L
Gdf
2t allow
(11,600 ksi)(0.5 in.)(0.52360 rad)
2(40 ksi)
Lmin ⫽ 38.0 in.
;
(1)
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.)
255
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 ⫽
d ⫽ 8.0 mm
L ⫽ 200 mm
␶allow ⫽ 60 MPa
G ⫽ 78 GPa
MAXIMUM PERMISSIBLE TORQUE
16T
From Eq. (3-12): tmax ⫽
pd3
3
pd tmax
Tmax ⫽
16
3
Tmax ⫽
p(8.0 mm) (60 MPa)
16
Tmax ⫽ 6.03 N # m
;
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).
TmaxL
GIP
From Eq. (3-12): Tmax ⫽
f⫽ a
f⫽
f⫽
pd3t max
L
ba
b
16
GIP
pd3tmaxL(32)
16G(pd4)
⫽
pd3tmax
16
IP ⫽
pd4
32
2tmaxL
Gd
2(60 MPa)(200 mm)
⫽ 0.03846 rad
(78 GPa)(8.0 mm)
f ⫽ 10.03846 rad2a
180
deg/radb ⫽ 2.20°
p
T
;
T
24 in.
1.25 in.
1.75 in.
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SECTION 3.3
Circular Bars and Tubes
257
Solution 3.3-7
NUMERICAL DATA
L ⫽ 24 in.
f ⫽ 4a
r2 ⫽
1.75
in.
2
p
b radians
180
MAX. SHEAR STRESS
p
Ip ⫽ 1 r24 ⫺ r142
2
tmax ⫽
Tr2
Ip
gmax ⫽
MAX. SHEAR STRAIN
1.25
in.
2
r1 ⫽
T ⫽ 6200 lb-in.
␥max ⫽ 0.00255 radians
r2
f
L
;
SHEAR MODULUS OF ELASTICITY
G
G⫽
G ⫽ 3.129 ⫻ 106 psi
tmax ⫽
Tr2
Ip
or
G⫽
Ip ⫽ 0.681 in.
4
␶max ⫽ 7965 psi
TL
fIp
G ⫽ 3.13 ⫻ 106 psi
tmax
gmax
;
;
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
FIND MAX. TORQUE BASED ON ALLOWABLE RATE OF TWIST
␶a ⫽ 48 MPa
p
2a
b
180 rad
u⫽
3.5
m
Ip ⫽
p 4
d
32
f
u⫽
L
G ⫽ 80 GPa
Ip ⫽ 1.149 ⫻ 107 mm4
Tmax ⫽
GIpf
L
Tmax ⫽ GIp␪
Tmax ⫽ 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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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
135∞
P1
P3
d1
A
D
135∞
P1
90∞
d2
P2
P2
Solution 3.3-9
B
Three circular bars
THE THREE TORQUES MUST BE IN EQUILIBRIUM
T3 is the largest torque
T3 ⫽ T1 12 ⫽ P1d2 12
MAXIMUM SHEAR STRESS (Eq. 3-12)
16T3
16P1d2 12
16T
tmax ⫽
⫽
⫽
3
3
pd
pd1
pd31
d1 ⫽ diameter of bars
⫽ 0.5 in.
tmax ⫽
d2 ⫽ diameter of disks
⫽ 3.0 in.
P1 ⫽ 28 lb
T1 ⫽ P1d2
T2 ⫽ P2d2
T3 ⫽ P3d2
C
16(28 lb)(3.0 in.) 12
p(0.5 in.)3
⫽ 4840 psi
;
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SECTION 3.3
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.)
Circular Bars and Tubes
T
259
d
T
Solution 3.3-10
NUMERICAL DATA
␶a ⫽ 48 MPa
T ⫽ 1.65 kN # m
u a ⫽ 0.75 a
G ⫽ 80 GPa
p
rad
b
180 m
32T
d ⫽
pGua
4
Ip ⫽
T
Gu
dmin
;
MIN. REQUIRED DIAMETER OF SHAFT BASED ON ALLOWABLE
ALLOWABLE RATE OF TWIST
T
GIp
dmin ⫽ 63.3 mm
^ governs
SHEAR STRESS
MIN. REQUIRED DIAMETER OF SHAFT BASED ON
u⫽
dmin ⫽ 0.063 m
t⫽
Td
2Ip
dmin
16T 3
⫽c
d
pta
p 4
T
d ⫽
32
Gu
32T
⫽a
b
pGua
1
4
t⫽
Td
p 4
2a
d b
32
1
dmin ⫽ 0.056 m
dmin ⫽ 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 ␶2 at the outer surface of the shaft,
(b) shear stress ␶1 at the inner surface, and
(c) rate of twist ␪ (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 ⫽ 6.0 in.
r2 ⫽ 3.0 in.
d1 ⫽ 4.5 in.
r1 ⫽ 2.25 in.
(c) RATE OF TWIST
u⫽
G ⫽ 11 ⫻ 106 psi
u ⫽ 157 * 10⫺6 rad/ in. ⫽ 0.00898°/ in.
T ⫽ 150 k-in.
IP ⫽
(150 k-in.)
T
⫽
GIP
(11 * 106 psi)(86.98 in.)4
p 4
(d ⫺ d14) ⫽ 86.98 in.4
32 2
;
(d) SHEAR STRESS DIAGRAM
(a) SHEAR STRESS AT OUTER SURFACE
t2 ⫽
(150 k-in.)(3.0 in.)
Tr2
⫽
IP
86.98 in.4
⫽ 5170 psi
;
(b) SHEAR STRESS AT INNER SURFACE
t1 ⫽
Tr1
r1
⫽
t ⫽ 3880 psi
IP
r2 2
;
Problem 3.3-12 Solve the preceding problem if the shaft has outer diameter d2 ⫽ 150 mm and inner diameter d1 ⫽ 100 mm.
Also, the steel has shear modulus of elasticity G ⫽ 75 GPa and the applied torque is 16 kN ⭈ m.
Solution 3.3-12
Construction auger
d2 ⫽ 150 mm
r2 ⫽ 75 mm
d1 ⫽ 100 mm
r1 ⫽ 50 mm
G ⫽ 75 GPa
T ⫽ 16 kN # m
p 4
IP ⫽
(d2 ⫺ d14) ⫽ 39.88 * 106 mm4
32
(a) SHEAR STRESS AT OUTER SURFACE
(16 kN # m)(75 mm)
Tr2
⫽
IP
39.88 * 106 mm4
⫽ 30.1 MPa
;
t2 ⫽
(b) SHEAR STRESS AT INNER SURFACE
t1 ⫽
Tr1
r1
⫽ t 2 ⫽ 20.1 MPa
IP
r2
;
(c) RATE OF TWIST
T
16 kN # m
u⫽
⫽
GIP
(75 GPa)(39.88 * 106 mm4)
␪ ⫽ 0.005349 rad/m ⫽ 0.306°/m
(d) SHEAR STRESS DIAGRAM
;
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SECTION 3.3
Circular Bars and Tubes
Problem 3.3-13 A vertical pole of solid circular cross section is twisted by
c
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?
Solution 3.3-13
261
P
c
B
A
P
d
Vertical pole
tmax ⫽
P(2c + d)d
4
pd /16
⫽
16P(2c + d)
pd3
(␲␶max)d3 ⫺ (16P)d ⫺ 32Pc ⫽ 0
P ⫽ 1100 lb
SUBSTITUTE NUMERICAL VALUES:
c ⫽ 5.0 in.
UNITS: Pounds, Inches
␶allow ⫽ 4500 psi
(␲)(4500)d3 ⫺ (16)(1100)d ⫺ 32(1100)(5.0) ⫽ 0
Find dmin
or
d3 ⫺ 1.24495d ⫺ 12.4495 ⫽ 0
Solve numerically:
TORSION FORMULA
Tr
Td
tmax ⫽
⫽
IP
2IP
T ⫽ P12c + d2
d ⫽ 2.496 in.
dmin ⫽ 2.50 in.
IP ⫽
;
pd 4
32
Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P ⫽ 5.0 kN, the distance c ⫽ 125 mm,
and the allowable shear stress is 30 MPa.
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Torsion
Solution 3.3-14
Vertical pole
TORSION FORMULA
tmax ⫽
Tr
Td
⫽
IP
2IP
T ⫽ P(2c + d)
tmax ⫽
P(2c + d)d
4
pd /16
IP ⫽
⫽
pd 4
32
16P (2c + d)
pd 3
(␲␶max)d 3 ⫺ (16P)d ⫺ 32Pc ⫽ 0
SUBSTITUTE NUMERICAL VALUES:
P ⫽ 5.0 kN
UNITS: Newtons, Meters
c ⫽ 125 mm
(␲)(30 ⫻ 106)d3 ⫺ (16)(5000)d ⫺ 32(5000)(0.125) ⫽ 0
␶allow ⫽ 30 MPa
or
Find dmin
d3 ⫺ 848.826 ⫻ 10⫺6d ⫺ 212.207 ⫻ 10⫺6 ⫽ 0
d ⫽ 0.06438 m
Solve numerically:
dmin ⫽ 64.4 mm
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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SECTION 3.3
263
Circular Bars and Tubes
Solution 3.3-15
␶a ⫽ 12 ksi
d2 ⫽ 1.25 in. d1 ⫽ 0.625 in.
p
1 d24 ⫺ d142
32
⫽
d2
2
d24 ⫺d12
1
⫽ tap
16
d2
ta
(a)
MAX. PERMISSIBILE VALUE OF TORQUE
T1max ⫽
ta
taIp
T1max ⫽
d
2
1
T1max ⫽
tapd3
16
1
T1max ⫽
(12)p (1.25)3
16
;
T1max ⫽ 4.60 in.-k
T1 – SOLID BAR
p 4
d
32
d
2
(b) MAX. PERMISSIBILE VALUE OF TORQUE T2 –
HOLLOW BAR
T2max
T2max
T2max ⫽ 4.31 in.-k
(c) PERCENT
;
DECREASE IN TORQUE
IN WEIGHT DUE TO HOLE IN
&
PERCENT DECREASE
(b)
percent decrease in torque
T1max ⫺ T2max
(100) ⫽ 6.25%
T1max
;
percent decrease in weight (weight is proportional to
x-sec area)
A1 ⫽
d1
d2
p 2
d
4 2
A2 ⫽
p 2
1d ⫺ d122
4 2
A1 ⫺ A2
(100) ⫽ 25 %
A1
;
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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CHAPTER 3
Page 264
Torsion
Solution 3.3-16
Td22
NUMERICAL DATA
set tmax expression equal to
d2 ⫽ 104 mm
d1 ⫽ 82 mm
⫽
L ⫽ 2.75 ⫻ 103 mm
G ⫽ 28 GPa
I p ⫽ (p/32)(d 2 4 ⫺ d 14 )
32Td2
p ad24 ⫺ d14 b
d3 ⫽
Ip ⫽ 7.046 ⫻ 106 mm4
p
a d42 ⫺ d41 b
32
then solve for d
d24 ⫺ d14
d2
dreqd ⫽ a
1
d24 ⫺ d14 3
b dreqd⫽88.4 mm
d2
;
(c) RATIO OF WEIGHTS OF HOLLOW & SOLID SHAFTS
(a) FIND ANGLE OF TWIST
f⫽
TL
GIp
f ⫽ (tmax)
f⫽ a
␶max ⫽ 48 MPa
WEIGHT IS PROPORTIONAL TO CROSS SECTIONAL AREA
p 2
A d ⫺ d12 B
4 2
Ah
p
A s ⫽ d reqd 2
⫽ 0.524
4
As
Td2 2L
b
2Ip Gd2
Ah ⫽
2L
Gd2
So the weight of the tube is 52% of the solid shaft,
but they resist the same torque.
␾ ⫽ 0.091 radians
␾ ⫽ 5.19°
;
;
(b) REPLACE HOLLOW SHAFT WITH SOLID SHAFT - FIND
DIAMETER
d
2
p
T
tmax ⫽
32d 4
tmax ⫽
16T
d3p
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SECTION 3.3
265
Circular Bars and Tubes
Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is
P
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
r2
r1
P
b
Solution 3.3-17
2r2
b
Circular tube in torsion
SOLUTION OF EQUATION
UNITS: Pounds, Inches
Substitute numerical values:
6300 psi ⫽
P ⫽ 900 lb
b ⫽ 5.5 in.
␶allow ⫽ 6300 psi
r1 ⫽ 1.2 in.
4(900 lb)(5.5 in. + r2)(r2)
p[(r42) ⫺ (1.2 in.)4]
or
r42 ⫺ 2.07360
⫺ 0.181891 ⫽ 0
r2(r2 + 5.5)
or
r42 ⫺ 0.181891 r22 ⫺ 1.000402 r2 ⫺ 2.07360 ⫽ 0
Find minimum permissible radius r2
Solve numerically:
TORSION FORMULA
r2 ⫽ 1.3988 in.
T ⫽ 2P(b ⫹ r2)
IP ⫽
p 4
(r ⫺ r41)
2 2
2P(b + r2)r2
4P(b + r2)r2
Tr2
⫽
⫽
p 4
IP
p (r42 ⫺ r41)
(r2 ⫺ r41)
2
All terms in this equation are known except r2.
tmax ⫽
MINIMUM PERMISSIBLE RADIUS
r2 ⫽ 1.40 in.
;
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Torsion
Nonuniform Torsion
T1
Problem 3.4-1 A stepped shaft ABC consisting of two solid
d1
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.
Solution 3.4-1
d2
T2
B
A
C
L1
L2
Stepped shaft
SEGMENT BC
TBC T2 8,000 lb-in.
tBC fBC d1 2.25 in.
L1 30 in.
d2 1.75 in.
L2 20 in.
T2 8,000 lb-in.
` TABL1
G(Ip)AB
p(1.75 in.)3
7602 psi
(8,000 lb-in.)(20 in.)
(11 * 106 psi)a
p
b(1.75 in.)4
32
16(12,000 lb-in.)
p(2.25 in.)3
C AB BC (0.013007 0.015797) rad
5365 psi
(12,000 lb-in.)(30 in.)
(11 * 106 psi)a
0.013007 rad
;
(b) ANGLE OF TWIST AT END C
TAB T2 T1 12,000 lb-in.
fAB TBCL2
G(Ip)BC
16(8,000 lb-in.)
max 7600 psi
SEGMENT AB
pd31
(a) MAXIMUM SHEAR STRESS
Segment BC has the maximum stress
T1 20,000 lb-in.
tAB `
pd32
0.015797 rad
G 11 10 psi
6
16 TAB
16 TBC
p
b(2.25 in.)4
32
C 0.002790 rad 0.16°
;
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Page 267
SECTION 3.4
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.
267
Nonuniform Torsion
Tube
Fixed
plate
End
plate
Bar
T
A
(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
Bar
d1
d2
d3
Solution 3.4-2
Bar and tube
TORQUE
T 1000 N m
(a) MAXIMUM SHEAR STRESSES
Bar: t bar 16T
79.6 MPa
;
pd31
T(d3/2)
Tube: t tube 32.3 MPa
(Ip) tube
;
TUBE
d3 70 mm
Ltube 0.5 m
(Ip) tube (b) ANGLE OF TWIST AT END A
d2 60 mm
G 27 GPa
Bar: fbar p 4
(d d24)
32 3
Tube: ftube 1.0848 * 106 mm4
A 9.43°
(Ip) bar pd14
32
Lbar 1.0 m
TL tube
0.0171 rad
G(Ip) tube
A bar tube 0.1474 0.0171 0.1645 rad
BAR
d1 40 mm
TL bar
0.1474 rad
G(Ip) bar
G 27 GPa
251.3 * 103 mm4
;
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CHAPTER 3
Page 268
Torsion
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.
12.5 k-in.
3.5 in.
25 in.
D
C
B
A
(a) Calculate the maximum shear stress max in the shaft.
(b) Calculate the angle of twist D (in degrees) at end D.
9.8 k-in.
9.2 k-in.
2.75 in.
2.5 in.
25 in.
25 in.
Solution 3.4-3
NUMERICAL DATA (INCHES, KIPS)
TB 12.5 k-in.
TC 9.8k-in.
TD 9.2 k-in.
tBC Check BC:
1TC + TD2
p
d 4
32 BC
L 25 in.
dAB 3.5 in.
dBC 2.75 in.
dCD 2.5 in.
G 11.6 (103) ksi
dBC
2
BC 4.65 ksi
; controls
(b) ANGLE OF TWIST AT END D
(a) MAX. SHEAR STRESS IN SHAFT
T1 |RA|
torque reaction at A: RA (TB TC TD)
RA 31.5 in.-kip
|RA|
tAB dAB
2
p
d 4
32 AB
max 3.742 ksi
TD
Check CD:
tCD dCD
2
p
d 4
32 CD
CD 2.999 ksi
T2 TC TD
p
d 4
32 AB
p
IP3 d 4
32 CD
T3 TD
IP1 IP2 p
d 4
32 BC
TiLi
fD a
GIpi
fD T2
T3
L T1
a
+
+
b
G IP1
IP2
IP3
D 0.017 radians
fD 0.978 degrees
;
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Page 269
SECTION 3.4
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.)
269
Nonuniform Torsion
d1
d2
T
A
C
B
L1
T
L2
Solution 3.4-4
Tallow based on angle of twist
NUMERICAL DATA
d1 56 mm
d2 48 mm
L1 1450 mm
L2 1200 mm
f a 1.25 a
a 30 MPa
G 80 GPa
fmax p
b radians
180
Allowable torque
Tallow 16T
d32p
tapd23
Tallow 16
L1
J 32
a
p
d14 b
L2
+
a
Gf a
L1
a
Tallow based on shear stress
tmax T
G
p 4
d1 b
32
+
p 4
d b
32 2 K
L2
a
p 4
d2 b
32
Tallow 459 N # m
;
T1 =
T2 =
1000 lb-in. 500 lb-in.
T3 =
T4 =
800 lb-in. 500 lb-in.
governs
Tallow 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.
A
B
C
D
d2 = 1.0 in.
T5 =
800 lb-in.
E
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CHAPTER 3
Page 270
Torsion
Solution 3.4-5
Hollow tube of monel metal
REQUIRED POLAR MOMENT OF INERTIA BASED UPON
ALLOWABLE SHEAR STRESS
tmax d2 1.0 in.
allow 12,000 psi
allow 2°/ft 0.16667°/in.
Tmaxr
Ip
REQUIRED
IP T max(d2/2)
0.05417 in.4
tallow
POLAR MOMENT OF INERTIA BASED UPON
ALLOWABLE ANGLE OF TWIST
0.002909 rad/in.
From Table H-2, Appendix H: G 9500 ksi
TORQUES
u
Tmax
GIP
IP Tmax
0.04704 in.4
Gu allow
SHEAR STRESS GOVERNS
Required IP 0.05417 in.4
IP T1 1000 lb-in.
T2 500 lb-in.
T4 500 lb-in.
T5 800 lb-in.
T3 800 lb-in.
INTERNAL TORQUES
p 4
(d d41)
32 2
d41 d43 32(0.05417 in.4)
32IP
(1.0 in.)4 p
p
0.4482 in.4
TAB T1 1000 lb-in.
d1 0.818 in.
TBC T1 T2 500 lb-in.
(Maximum permissible inside diameter)
;
TCD T1 T2 T3 1300 lb-in.
TDE T1 T2 T3 T4 800 lb-in.
Largest torque (absolute value only):
Tmax 1300 lb-in.
80 mm
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?
1.2 m
60 mm
0.9 m
d
2.1 m
d
t=—
10
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SECTION 3.4
Solution 3.4-6
Nonuniform Torsion
271
Solid and hollow shafts
SOLID SHAFT CONSISTING OF TWO SEGMENTS
TORSIONAL STIFFNESS
kT T
f
Torque T is the same for both shafts.
⬖ For equal stiffnesses, 1 2
98,741 m3 TLi
f1 g
GIPi
T(1.2 m)
p
Ga b(80 mm)4
32
T(0.9 m)
+
p
Ga b (60 mm)4
32
32T
(29,297 m3 + 69,444 m3)
pG
d4 3.5569 m
d4
3.5569
36.023 * 106 m4
98,741
d 0.0775 m 77.5 mm
;
32T
(98,741 m3)
pG
HOLLOW SHAFT
d0 inner diameter 0.8d
f2 TL
GIp
T(2.1 m)
Ga
p
b[d4 (0.8d)4]
32
2.1 m
32T 3.5569 m
32T
a
b
a
b
4
pG 0.5904 d
pG
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.?
19,000 lb-in.
4,000 lb-in.
A
7,000 lb-in.
B
C
D
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CHAPTER 3
Page 272
Torsion
Solution 3.4-7
Shaft with four gears
(b) HOLLOW SHAFT
Inside diameter d0 1.0 in.
allow 10,000 psi
TBC 11,000 lb-in.
TAB 8000 lb-in.
TCD 7000 lb-in.
Tr
tmax Ip
10,000 psi (a) SOLID SHAFT
tmax d3 t allow 16T
pd3
d
Tmax a b
2
Ip
d
(11,000 lb-in.) a b
2
a
p
b[d4 (1.0 in.)4]
32
UNITS: d inches
16(11,000 lb-in.)
16Tmax
5.602 in.3
pt allow
p(10,000 psi)
Required d 1.78 in.
10,000 ;
56,023 d
d4 1
or
d 4 5.6023 d 1 0
Solving, d 1.832
Required d 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
;
B
A
T
L
dA
dB
Problems 3.4-8, 3.4-9 and 3.4-10
Solution 3.4-8
Tapered bar AB
ANGLE OF TWIST
TAPERED BAR (From Eq. 3-27)
b2 + b + 1
TL
f1 a
b
G(IP)A
3b 3
PRISMATIC BAR
TL
f2 G(IP)A
dB
b
dA
b2 + b + 1
f1 1
f
2 2
or
3
3 2
2 2
2 0
3b 3
SOLVE NUMERICALLY:
b
dB
1.45
dA
;
1
2
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SECTION 3.4
Nonuniform Torsion
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 3–5).
Solution 3.4-9
Tapered bar
MINIMUM DIAMETER BASED
(From Eq. 3-27)
dB 1.5 dA
T 36,000 lb-in.
b dB/dA 1.5
L 4.0 ft 48 in.
b2 + b + 1
TL
TL
a
b
(0.469136)
G(IP)A
G(IP)A
3b 3
(36,000 lb-in.)(48 in.)
(0.469136)
p
(3.9 * 106 psi)a bd4A
32
f
G 3.9 106 psi
allow 15,000 psi
allow 3.0°
0.0523599 rad
MINIMUM
DIAMETER BASED UPON ALLOWABLE SHEAR
STRESS
tmax 16T
pd3A
d3A UPON ALLOWABLE ANGLE OF
TWIST
16(36,000 lb-in.)
16 T
ptallow
p(15,000 psi)
12.2231 in.3
dA 2.30 in.
d4A 2.11728 in.4
d4A
2.11728 in.4
2.11728 in.4
0.0523599 rad
fallow
40.4370 in.4
dA 2.52 in.
ANGLE OF TWIST GOVERNS
Min. dA 2.52 in.
;
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CHAPTER 3
Page 274
Torsion
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 25 mm
L 300 mm
G 82 GPa
T 180 N m
allow 0.3°
p
(0.3°)a 180
rad
b
degrees
(180 N # m)(0.3 m)
p
(82 GPa)a b (25 mm)4
32
b2 + b + 1
Find dB
0.304915 DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
0.914745
1 0
(From Eq. 3-27)
dB
b
dA
f
b2 + b + 1
TL
p 4
a
b(IP)A d
G(IP)A
32 A
3b 3
3
3b 3
2
SOLVE NUMERICALLY:
1.94452
Min. dB dA 48.6 mm
;
a
b2 + b + 1
3b 3
b
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SECTION 3.4
Nonuniform Torsion
Segment 1
Segment 2
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.
x
7
—Ip
8
R1
Ip
T
1
2
T
—
2
3
x
L–x
T1
T2
TMD 0
φ2
TDD 0
0
φ3
0
Solution 3.4-11
(a) REACTION TORQUE R1
T
3
a Mx 0 R1 a T + 2 b R1 2 T ;
(b) INTERNAL MOMENTS IN SEGMENTS 1 & 2
T1 R1
T1 1.5 T
T2 T
2
(c) FIND X REQUIRED TO OBTAIN TRWIST AT JT 3
L
17
1
x + L
14
2
x
14 L
a b
17 2
TL
GIP
L
T1x
7
Ga IP b
8
+
3
a Tbx
2
7
Ga IP b
8
3
a bx
2
7
a b
8
+
T2(L x)
GIP
T
a b(L x)
2
+
1
(L x)
2
GIP
;
(d) ROTATION AT JOINT 2 FOR X VALUE IN (C)
f2 TiLi
f3 a
GIPi
TL
GIP
7
L
17
x
f2 T1x
7
Ga Ip b
8
12TL
17GIP
f2 3
7
a Tb a Lb
2
17
7
Ga Ip b
8
;
(e) TMD & TDD – SEE PLOTS ABOVE
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
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Page 276
Torsion
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
length L. The average diameters at the ends are dA and dB 2dA. The polar
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.
B
A
T
T
L
t
t
dA
dB = 2dA
Solution 3.4-12
Tapered tube
t thickness (constant)
dA, dB average diameters at the ends
dB 2dA
Ip pd3t
(approximate formula)
4
ANGLE OF TWIST
Take the origin of coordinates at point O.
d(x) Lp(x) x
x
(d ) dA
2L B
L
p[d(x)]3t
ptd3A 3
x
4
4L3
For element of length dx:
df Tdx
GIP(x)
Tdx
ptd3A
G a 3 bx3
4TL3dx
pGtdA3 x3
4L
2L
f
LL
df 4TL3
2L
pGtd3A
LL x3
dx
3TL
2pGtd3A
;
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Page 277
SECTION 3.4
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.)
L
dB
(a)
L
—
2
dA
A
from B to centerline
outer and inner diameters as function of x
0 … x …
L
2
d0(x) 2d A d0(x) dB a
xdA
L
di (x) (dB 2t)
di (x) dB dA
bx
L
[(2dA 2t)(dA 2t)]
x
L
1 9L + 5x
d
5 A
L
solid from centerline to A
L
… x … L
2
d0(x) 2dA L
x dA
L
L
T 32
1
1
2
f a b
dx + L 4 dx
4
4
G p P L0 d0 d i
L2 d0
Q
dA
B
T
L
(b)
PART (A) - CONSTANT THICKNESS
use x as integration variable measured from B toward A
B
T
dA
T
Solution 3.4-13
t constant
dB – 2t
L
—
2
A
T
277
Nonuniform Torsion
dB
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1:09 PM
CHAPTER 3
Page 278
Torsion
L
T 32
2
f a b≥
G p
L0
f 32
L
1
xdA 4
1
9L + 5x 4
a 2dA b a
dA
b
L
5
L
dx +
L2
L
1
a2dA xdA 4
b
L
T 125 3ln(2) + 2ln(7) ln(197)
125 2ln(19) + ln(181)
19
a
L
L
+
Lb
4
Gp
2
2
dA
dA 4
81dA 4
Simplifying: f 16TL
81GpdA4
a38 + 10125 lna
71117
bb
70952
Use numerical properties as follows L 48 in.
a 0.049 radians
fa 2.79°
or fa 3.868
G 3.9 106 psi
TL
GdA4
dA 2.5 in.
;
PART (B) - CONSTANT HOLE DIAMETER
0 … x …
d B dA
bx
L
d0( x) dB a
L
2
L
… x … L
2
d0(x) 2dA L
2
f
T 32
a b
G p
f
2
T 32
a b
G p
L0
1
4
P L0 d0 di
J
d0(x) 2 dA xdA
L
xdA
L
L
dx +
4
1
L d0 4
L
2
L
dx
Q
L
1
1
dx +
dx
1
xdA 4
xdA 4
L
2
4
K
a2dA b dA
a2dA b
L
L
3
ln(5) + 2atan a b
2
T 1
1 ln(3) + 2atan(2)
19
± L
fb 32
L
+
L≤
Gp 4
4
dA 4
dA 4
81dA 4
Simplifying, fb 3.057
TL
GdA4
Use numerical properties given above
b 0.039 radians
fa
fb
dx ¥
1.265
fb 2.21°
;
so tube (a) is more flexible than tube (b)
di (x) dA
t
dA
10
T 40000 in.-lb
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Page 279
SECTION 3.4
Problem 3.4-14 For the thin nonprismatic steel pipe of constant
2d
t
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).
d
Nonuniform Torsion
t
d
T, f3
T/2
R1
2
1
279
L
—
2
x
3
L
—
2
T
T
—
2
0
TMD
Solution 3.4-14
(a) REACTION TORQUE R1
T 0
statics:
R1 L
2
2T
f3 Gpt L0
T
+ T0
2
R1 T
2
;
(b) ROTATION AT JOINT 3
p
3
Ga d12(x) tb
4
L
+
L
2
LL2
T
x 3
bd
L
dx
L
Gpd3t LL2
dx
L
2
2T
f3 Gpt L0
1
c2d a 1 x 3
bd
L
dx
2TL
+
T
2
L
2
L
0
0 … x …
L
… x … L
2
d23(x) d
f3 x
b
L
c2da1 4T
+
d12( x) 2d a1 1
dx
dx
p
Ga d23(x)3tb
4
use IP expression for thin walled tubes
f3 f3 Gpd3t
3TL
2TL
3
8Gp d t
19TL
8Gpd3t
+
Gpd3t
;
(c) TMD
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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CHAPTER 3
Page 280
Torsion
Problem 3.4-15 A mountain-bike rider going uphill applies
Handlebar extension
d01, t01
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.
B
A
E
d03, t03
C
L1 L2
T = Fd
D
L3
d
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.
45∞
Handlebar
extension
F
D
Handlebar
Solution 3.4-15
ASSUME THIN WALLED TUBES
Segments AB & CD
p
p
IP1 d01 3t01
IP3 d03 3t03
4
4
Segment BC
0 x L2
d02(x) d01 a1 d02(x) d01L2 d01x + d03x
L2
t02(x) t01 a1 t02(x) fD x
x
b + d03 a b
L2
L2
x
x
b + t03 a b
L2
L2
t01L2 t01x + t03x
L2
L2
L3
1
Fd L1
+
dx +
p
G IP1
I
L0
P3
P
Q
d (x)3t02(x)
4 02
fD L2
L2 4
L1
4Fd
c 3 dx
Gp d01 t01 L0 (d01L2 d01x + d03x)3
* (t01L2 t01x + t03x)
+
L3
d03 3t03
d
;
NUMERICAL DATA
L1 2 in.
L2 1.2 in.
L3 8.5 in.
t03 0.115 in.
d01 1.25 in.
t01 0.125 in.
F 15 lb
d 4 in.
d03 0.87 in.
G 4 (106) psi
f D 0.142°
;
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Page 281
SECTION 3.4
Nonuniform Torsion
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).
t
A
(a) Determine the maximum shear stress max in the bar.
(b) Determine the angle of twist between the ends of the bar.
B
L
Solution 3.4-16
Bar with distributed torque
(a) MAXIMUM SHEAR STRESS
Tmax tL
tmax 16Tmax
pd3
16tL
pd3
;
(b) ANGLE OF TWIST
T(x) tx
df t intensity of distributed torque
d diameter
IP T(x)dx
32 tx dx
GIP
pGd4
L
f
pd4
32
L0
df 32t
pGd4 L0
L
x dx 16tL2
pGd4
;
G shear modulus of elasticity
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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CHAPTER 3
Solution 3.4-17
Page 282
Torsion
Bar with linearly varying torque
(a) Maximum shear stress
tmax 16Tmax
3
pd
16TA
3
pd
8tAL
pd3
;
(b) ANGLE OF TWIST
T(x) torque at distance x from end B
T(x) t(x)x
tAx2
2
2L
IP pd4
32
T(x) dx
16tAx2 dx
GIP
pGLd4
L
L
16tA
16tA L2
2
f
df x
dx
;
pGLd4 L0
3pGd4
L0
df t(x) intensity of distributed torque
tA maximum intensity of torque
d diameter
G shear modulus
TA maximum torque
12 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).
T
—0
L
A
T0
—
6
Fc
IP
2Ip
RA
C
B
L
—
2
T0
—
3L
L
—
2
2°
2°
0
TMD
–T0
—
12
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Page 283
SECTION 3.4
Nonuniform Torsion
283
Solution 3.4-18
(a) TORQUE REACTION RA
(d) MAXIMUM SHEAR STRESS ALONG BAR
T 0
STATICS:
RA +
1 T0 L
1 T0
L
a ba b a ba b 0
2 L
2
2 3L
2
RA +
1
T 0
6 0
RA T0
6
p
d 4
32 AB
p
For BC IP d 4
32 BC
For AB 2IP 1
;
dBC
1 4
a b dAB
2
(b) INTERNAL TORSIONAL MOMENTS IN AB & BC
T0
TAB (x) 6
TAB (x) a
TBC (x) T0 x
x
a b
L
L 2
P Q
2
T0
x2
2 T0 b
6
L
0 … x …
(L x) T0 (L x)
a b
L
3L
2
2
TBC (x) c a
x L 2 T0
b
d
L
3
L
… x … L
2
L
2
L
TBC(x)
TAB(x)
dx +
dx
L
GIP
L0 G(2IP)
L2
2
T0
x T0
L
2
2 6
3L
dx
fC G(2IP)
L0
L
+
c a
LL2
x L 2 T0
b
d
L
3
GIP
fC T0L
T0L
48GIP
72GIP
fC T0L
144GIP
;
L
2
;
tmax 8T0
3pdAB3
tmax
; controls
Just to right of B, T T0/12
T0 dBC
a
b
12 2
tmax p
d 4
32 BC
T0 0.841dAB
a
b
12
2
tmax p
(0.841dAB)4
32
;
(c) ROTATION AT C
fC At A, T T0/6
T0 dAB
6 2
p
d 4
32 AB
dx
tmax 2.243T0
pdAB 3
(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
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CHAPTER 3
Page 284
Torsion
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.
T0 = torque
Flexible tube
B
d
A
t
(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?
Solution 3.4-19 Wire inside a flexible tube
(b) ANGLE OF TWIST d 4 mm
T0 0.2 N m
t 0.04 N m/m
(a) MAXIMUM LENGTH Lmax
allow 30 MPa
Equilibrium: T tL T0
16T
From Eq. (3-12): tmax pd3
tL + T0 L
Lmax
L 4 m G 15 GPa
1 angle of twist due to distributed torque t
T
pGd4
(from problem 3.4-16)
2 angle of twist due to torque T0
pd3tmax
16
pd 3tmax
16
T0 L
32 T0 L
(from Eq. 3 -15)
GIP
pGd4
total angle of twist
1 2
1
(pd3tmax 16T0)
16t
1
(pd3tallow 16T0)
16t
16tL2
f
;
Substitute numerical values: Lmax 4.42 m
;
16L
pGd 4
(tL + 2T0) ;
Substitute numerical values:
2.971 rad 170°
;
T
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Page 285
SECTION 3.4
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.
B
d3
TA
d2
d1
d2
A
LA
(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?
Solution 3.4-20
Pin at B is in shear at interface between the two
tubes; force couple V # d2 T0
V
T0
d2
tpin tpin T0
d2
a
pdp 2
4
b
V
AS
tpin T0,max tp,allow a
ttubeAB pd2dp 2
b
IPAC
p
ad3 4 d2 4 b
32
p
ad 4 d1 4 b
32 2
ttubeAB d3
T0 a b
2
IPAB
p
(d 4 d2 4)
32 3
16T0d3
p(d3 4 d2 4)
T0,max tt,allow c
p(d3 4 d2 4)
d
16d3
;
and based on tube BC:
;
(b) T0,max BASED ON ALLOWABLE SHEAR IN TUBES
AB & BC
IPAB d3
b
2
so based on tube AB:
4T0
p d2 d2p
4
ttubeAB T0 a
ttubeBC ttubeBC T0 a
d2
b
2
p
(d 4 d1 4)
32 2
16T0d2
p(d2 4 d1 4)
T0,max tt,allow
J
p(d2 4 d1 4)
16d2
K
T0, Fc
C
Pin dp LB
Cross-section at B
(a) T0,max BASED ON ALLOWABLE SHEAR IN PIN AT B
285
Nonuniform Torsion
;
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1:09 PM
CHAPTER 3
(c)
Page 286
Torsion
MAX. ROTATION AT
SHEAR IN PIN AT
C
B
BASED ON EITHER ALLOWABLE
OR ALLOWABLE SHEAR STRESS IN
TUBES
MAX. ROTATION BASED ON ALLOWABLE SHEAR IN
B
MAX.
ROTATION BASED ON ALLOWABLE SHEAR STRESS
IN TUBE
AB
fCmax tt, allow c
2(d3 4 d2 4)
d
Gd3
PIN AT
fC T0,max
fCmax G
a
LA
IPAB
t p,allow a
+
LB
IPBC
p d2d2p
4
b
b
MAX.
LA
4
4
J 32 (d3 d2 )
p
fCmax tp, allow a
J
8d2dp 2
G
LB
p
(d 4 d1 4) K
32 2
b
LA
(d3 d2 )
4
+
4
+
LB
(d2 d1 4)
4
K
;
(d3 d2 )
4
+
LB
K
;
K
;
(d2 d1 4)
4
ROTATION BASED ON ALLOWABLE SHEAR STRESS
IN TUBE
G
J
LA
4
BC
fCmax tt, allow c
J
2(d2 4 d1 4)
d
Gd2
LA
(d3 d2 )
4
4
+
LB
(d2 d1 4)
4
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Page 287
SECTION 3.5
Pure Shear
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.
d2
T
T
L
(a) Determine the maximum tensile stress max in the shaft.
(b) Determine the magnitude of the applied torques T.
d1 d2
Probs. 3.5-1, 3.5-2, and 3.5-3
Solution 3.5-1
d2 4.0 in.
Hollow aluminum shaft
d1 2.0 in.
0.54°/ft
(a) MAXIMUM TENSILE STRESS
G 4.0 106 psi
max occurs on a 45° plane and is equal to max.
MAXIMUM SHEAR STRESS
max max 6280 psi
max Gr (from Eq. 3-7a)
r d2 /2 2.0 in.
1 ft
prad
u (0.54°/ft)a
ba
b
12 in. 180 degree
785.40 106 rad/in.
max (4.0 106 psi)(2.0 in.)(785.40 106 rad/in.)
6283.2 psi
;
(b) APPLIED TORQUE
Use the torsion formula tmax T
tmaxIP
r
IP p
[(4.0 in.)4 (2.0 in.)4]
32
23.562 in.4
T
(6283.2 psi) (23.562 in.4)
2.0 in.
74,000 lb-in.
Tr
IP
;
Sec_3.5-3.7.qxd
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CHAPTER 3
1:10 PM
Page 288
Torsion
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 640 106 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 80 GPa max 640 106 rad
d2 150 mm
IP max Gmax (80 GPa)(640 106)
d1 120 mm
51.2 MPa
p 4
(d d 41)
32 2
max max 51.2 MPa
p
[(150 mm)4 (120 mm)4]
32
Torsion formula: tmax (a) MAXIMUM TENSILE STRAIN
gmax
320 * 106
2
;
(c) APPLIED TORQUES
29.343 * 106 mm4
âmax (b) MAXIMUM TENSILE STRESS
T
;
Td2
Tr
IP
2IP
2(29.343 * 106 mm4)(51.2 MPa)
2IPtmax
d2
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 70.0 k-in. (see figure). Under the
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 4.0 106 psi, what is the angle of
twist (in degrees) between the ends of the bar?
(c) Determine the maximum shear strain max (in radians)?
Sec_3.5-3.7.qxd
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Page 289
SECTION 3.5
Solution 3.5-3
d2 4.0 in.
T 70.0 k-in. 70,000 lb-in.
f
Torsion formula: tmax TL
GIp
From torsion formula, T (a) INSIDE DIAMETER d1
Td2
Tr
IP
2IP
‹ f
(70.0 k-in.)(4.0 in.)
Td2
2tmax
2(6400 psi)
21.875 in.4
Also, Ip 289
Tubular bar
max 6400 psi max max 6400 psi
IP Pure Shear
2(48 in.)(6400 psi)
(4.0 * 106 psi)(4.0 in.)
0.03840 rad
;
(c) MAXIMUM SHEAR STRAIN
gmax Equate formulas:
p
[256 in.4 d14] 21.875 in.4
32
Solve for d1: d1 2.40 in.
2Ltmax
2IPtmax
L
a
b
d2
GIP
Gd2
f 2.20°
p 4
p
(d d 14) [(4.0 in.)4 d14]
32 2
32
2IP tmax
d2
6400 psi
tmax
G
4.0 * 106 psi
1600 * 106 rad
;
;
(b) ANGLE OF TWIST L 48 in.
G 4.0 106 psi
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 339 106.
What is the shear modulus G of the material?
d = 50 mm
Strain gage
T
45°
T = 500 N·m
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Solution 3.5-4
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Torsion
Bar in a testing machine
Strain gage at 45°:
SHEAR STRESS (FROM EQ. 3-12)
6
max 339 10
tmax d 50 mm
16T
pd
3
16(500 N # m)
p(0.050 m)3
20.372 MPa
SHEAR MODULUS
T 500 N m
G
SHEAR STRAIN (FROM EQ. 3-33)
6
max 2max 678 10
tmax
20.372 MPa
30.0 GPa
gmax
678 * 106
;
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 1.5 in.
When twisted by a torque T, the tube develops a maximum normal strain of 170 106.
What is the magnitude of the applied torque T?
Solution 3.5-5
Steel tube
G 11.5 106 psi d2 2.0 in.
d1 1.5 in.
max 170 106
IP p 2
p
1d d142 [(2.0 in.)4 (1.5 in.)4]
32 2
32
1.07379 in.
Equate expressions:
Td2
Ggmax
2IP
SOLVE FOR TORQUE
4
T
SHEAR STRAIN (FROM EQ. 3-33)
2GIPgmax
d2
2(11.5 * 106 psi)(1.07379 in.4)(340 * 106)
2.0 in.
max 2max 340 106
SHEAR STRESS (FROM TORSION FORMULA)
4200 lb-in.
Td2
Tr
tmax IP
2IP
Also, max Gmax
;
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SECTION 3.5
Pure Shear
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
220 106.
Determine the minimum required diameter d of the bar.
Solution 3.5-6
Solid circular bar of steel
DIAMETER BASED UPON ALLOWABLE TENSILE STRAIN
T 360 N m G 78 GPa
ALLOWABLE STRESSES
Tension: 90 MPa Compression: 70 MPa
Shear: 40 MPa
Allowable tensile strain: max 220 106
max 2max; max Gmax 2Gmax
tmax 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 40 MPa
tmax 16T
pd
3
d3 16T
3
pd
d3 16T
16T
ptmax
2pGâmax
16(360 N # m)
2p(78 GPa)(220 * 106)
53.423 * 106 m3
d 0.0377 m 37.7 mm
TENSILE STRAIN GOVERNS
d3 16(360 N # m)
16T
pt allow
p(40 MPa)
dmin 37.7 mm
;
d 45.837 106 m3
3
d 0.0358 m 35.8 mm
Problem 3.5-7 The normal strain in the 45° direction on the
surface of a circular tube (see figure) is 880 106 when the
torque T 750 lb-in. The tube is made of copper alloy with
G 6.2 106 psi.
Strain gage
T
45°
If the outside diameter d2 of the tube is 0.8 in., what is the
inside diameter d1?
Solution 3.5-7
Circular tube with strain gage
d2 0.80 in. T 750 lb-in.
G 6.2 106 psi
Strain gage at 45°: max 880 106
T = 750 lb-in.
d 2 = 0.8 in.
MAXIMUM SHEAR STRAIN
max 2max
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Torsion
MAXIMUM SHEAR STRESS
INSIDE DIAMETER
tmax Ggmax 2Gâmax
Substitute numerical values:
tmax IP T(d2/2)
IP
IP Td2
Td2
2tmax
4Gâmax
8Td2
pGâmax
d14 d24 8(750 lb-in.) (0.80 in.)
p (6.2 * 106 psi) (880 * 10 6)
0.4096 in.4 0.2800 in.4 0.12956 in.4
Td2
p 4
(d d14) 32 2
4Gâmax
d24 d14 d42 (0.8 in.)4 d1 0.60 in.
;
8Td2
pGâmax
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 900 106.
Determine the required outside diameter d2.
Solution 3.5-8
d1 50 mm
Aluminum tube
G 27 GPa
T 4.0 kN m allow 50 MPa allow 900 106
NORMAL STRAIN GOVERNS
allow 48.60 MPa
Determine the required diameter d2.
REQUIRED DIAMETER
ALLOWABLE SHEAR STRESS
t
(allow)1 50 MPa
Tr
IP
48.6 MPa ALLOWABLE SHEAR STRESS BASED ON NORMAL STRAIN
âmax g
t
2
2G
Rearrange and simplify:
t 2Gâmax
(allow)2 2Gallow 2(27 GPa)(900 106)
48.6 MPa
(4000 N # m)(d2/2)
p 4
[d2 (0.050 m)4]
32
d42 (419.174 * 106)d2 6.25 * 106 0
Solve numerically:
d2 0.07927 m
d2 79.3 mm
;
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SECTION 3.5
293
Pure Shear
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.
Solution 3.5-9
Solid steel bar
T 8.0 k-in.
(b) MAXIMUM STRAINS
G 11.8 10 psi
6
gmax (a) MAXIMUM STRESSES
tmax T = 8.0 k-in.
d = 2.0 in.
T
16T
pd
3
432 * 106 rad
16(8000 lb-in.)
5093 psi
3
p(2.0 in.)
âmax ;
t 5090 psi c 5090 psi
gmax
5093 psi
G
11.8 * 106 psi
Problem 3.5-10 A solid aluminum bar (G 27 GPa) of
(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.
gmax
216 * 106
2
t 216 106 c 216 106
;
diameter d 40 mm is subjected to torques T 300 N m
acting in the directions shown in the figure.
;
d = 40 mm
T
;
T = 300 N·m
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Torsion
Solution 3.5-10
Solid aluminum bar
(b) MAXIMUM STRAINS
(a) MAXIMUM STRESSES
tmax 16T
pd
3
16(300 N #
m)
gmax 3
p(0.040 m)
23.87 MPa
884 * 106 rad
;
t 23.9 MPa c 23.9 MPa
tmax
23.87 MPa
G
27 GPa
;
âmax ;
gmax
442 * 106
2
t 442 106
c 442 106
;
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).
120 rpm
d
(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?
Solution 3.7-1
Generator shaft
n 120 rpm H 50 hp d diameter
TORQUE
H
2pnT
H hp n rpm T 1b-ft
33,000
(33,000)(50 hp)
33,000 H
T
2pn
2p(120 rpm)
2188 1b-ft 26,260 1b-in.
(a) MAXIMUM SHEAR STRESS max
d 3.0 in.
50 hp
tmax 16T
pd
3
16(26,260 1b-in.)
tmax 4950 psi
p (3.0 in.)3
;
(b) MINIMUM DIAMETER dmin
allow 4000 psi
d3 16(26,260 1b-in.)
16T
33.44 in.3
ptallow
p (4000 psi)
dmin 3.22 in.
;
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SECTION 3.7
Transmission of Power
Problem 3.7-2 A motor drives a shaft at 12 Hz and delivers 20 kW
of power (see figure).
12 Hz
d
(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?
Solution 3.7-2
f 12 Hz
20 kW
Motor-driven shaft
P 20 kW 20,000 N m/s
16T
tmax pd3
TORQUE
T Newton meters
20,000 W
P
265.3 N # m
2pf
2p(12 Hz)
(a) MAXIMUM SHEAR STRESS max
16(265.3 N # m)
p(0.030 m)3
50.0 MPa
P 2fT P watts f Hz s1
T
;
(b) MINIMUM DIAMETER dmin
allow 40 MPa
d3 16(265.3 N # m)
16T
pt allow
p(40 MPa)
33.78 106 m3
d 30 mm
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.
12 in.
18 in.
(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?
Hollow propeller shaft
d2 18 in. d1 12 in. allow 4500 psi
p 4
IP (d d42) 8270.2 in.4
32 2
TORQUE
tmax
T(d2/2)
IP
T
2t allowIP
T
d2
2(4500 psi)(8270.2 in.4)
18 in.
4.1351 * 106 1b-in.
344,590 1b-ft.
;
100 rpm
18 in.
(a) If the shaft is turning at 100 rpm, what is the maximum
horsepower that can be transmitted without exceeding the
allowable stress?
Solution 3.7-3
295
(a) HORSEPOWER
n 100 rpm
H
2pnT
33,000
n rpm T lb-ft H hp
H
2p(100 rpm)(344,590 lb-ft)
33,000
6560 hp
;
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Torsion
(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).
2500 rpm
60 mm
(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?
40 mm
60 mm
Solution 3.7-4
d2 60 mm
IP Drive shaft for a truck
d1 40 mm
n 2500 rpm
p 4
(d d14) 1.0210 * 106 m4
32 2
(a) MAXIMUM SHEAR STRESS max
P power (watts)
P 150 kW 150,000 W
T torque (newton meters) n rpm
P
2pnT
60
T
60(150,000 W)
572.96 N # m
2p(2500 rpm)
T
60P
2pn
tmax (572.96 N # m)(0.060 m)
Td2
2 IP
2(1.0210 * 106 m4)
16.835 MPa
tmax 16.8 MPa
;
(b) MAXIMUM POWER Pmax
allow 30 MPa
Pmax P
tallow
30 MPa
(150 kW) a
b
tmax
16.835 MPa
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 3.7
Solution 3.7-5
d0 inside diameter
0.75 d
n 400 rpm
allow 6000 psi
IP H hp
T
n rpm T lb-ft
(33,000)(400 hp)
33,000 H
2pn
2p(400 rpm)
5252.1 lb-ft 63,025 lb-in.
MINIMUM OUTSIDE DIAMETER
p 4
[d (0.75 d)4] 0.067112 d 4
32
TORQUE
H
297
Hollow shaft
d outside diameter
H 400 hp
Transmission of Power
tmax Td
2IP
IP 0.067112 d 4 2pnT
33,000
Td
Td
2tmax
2t allow
(63,025 lb-in.)(d)
2(6000 psi)
d3 78.259 in.3
dmin 4.28 in.
;
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 inside diameter
0.5 d
P 120 kW 120,000 W f 1.75 Hz
allow 45 MPa
IP p 4
[d (0.5 d)4] 0.092039 d 4
32
T
120,000 W
P
10,913.5 N # m
2pf
2p(1.75 Hz)
MINIMUM OUTSIDE DIAMETER
tmax Td
2IP
IP 0.092039 d4 Td
Td
2tmax
2t allow
(10,913.5 N # m)(d)
2(45 MPa)
TORQUE
d3 0.0013175 m3
P 2 fT P watts f Hz
dmin 110 mm
d 0.1096 m
;
T 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
EQUATE TORQUES
SOLID SHAFT
tmax 16 T1
pd3
T1 pd3tmax
16
For the same power, the torques must be the same.
For the same material, both parts can be stressed to the
same maximum stress.
HOLLOW COLLAR
IP T2(d1/2)
T2r
p 4
(d1 d 4) tmax 32
IP
IP
2tmaxIP
2tmax p
a
b(d14 d 4)
T2 d1
d1 32
ptmax 4
(d1 d 4)
16 d1
‹ T1 T2
or a
pd3tmax
ptmax 4
(d d 4)
16
16d1 1
d1
d1 4
b 10
d
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 50 mm
Hollow propeller shaft
d1 40 mm
G 80 GPa n 600 rpm
allow 100 MPa allow 3.0°/m
IP p 4
(d d41) 362.3 * 109 m4
32 2
BASED UPON ALLOWABLE SHEAR STRESS
tmax T1(d2/2)
IP
T1 2t allowIP
d2
2(100 MPa)(362.3 * 109 m4)
T1 0.050 m
1449 N # m
BASED UPON ALLOWABLE RATE OF TWIST
T2
u
T2 GIPallow
GIP
T (80 GPa) (362.3 * 10
2
* a
9 4
m )(3.0°/m)
p
rad /degreeb
180
T2 1517 N m
SHEAR STRESS GOVERNS
Tallow T1 1449 N m
MAXIMUM POWER
2p(600 rpm)(1449 N # m)
2pnT
P
60
60
P 91,047 W
Pmax 91.0 kW
;
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SECTION 3.7
Transmission of Power
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
C
A
d
B
L1
L2
PROBS. 3.7-9 and 3.7-10
Solution 3.7-9
Motor-driven shaft
FREE-BODY DIAGRAM
L1 6 ft
L2 4 ft
TA 17,332 lb-in.
d diameter
TC 9454 lb-in.
n 1000 rpm
d diameter
allow 7500 psi
(AC)allow 1.5° 0.02618 rad
G 11.5 106 psi
TORQUES ACTING ON THE SHAFT
2pnT
H
33,000
T
TB 7878 lb-in.
INTERNAL TORQUES
TAB 17,332 lb-in.
TBC 9454 lb-in.
H hp
n rpm T lb-ft
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS
The larger torque occurs in segment AB
33,000 H
2pn
At point A: TA 33,000(275 hp)
2p(1000 rpm)
1444 lb-ft
17,332 lb-in.
At point B: TB At point C: TC tmax 16TAB
d3 3
pd
16TAB
pt allow
16(17,332 lb-in.)
11.77 in.3
p(7500 psi)
d 2.27 in.
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
125
275
TA 7878 lb-in.
150
275
TA 9454 lb-in.
IP pd4
32
f
TL
32TL
GIP
pGd4
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Torsion
Segment AB:
fAB fAB fBC 32TAB LAB
pGd 4
32(17,330 lb in.)(6 ft)(12 in./ft)
p(11.5 * 106 psi)d 4
1.1052
d
32 TBCLBC
pGd
1.5070
d4
(AC)allow 0.02618 rad
0.02618 1.5070
d4
and
d 2.75 in.
Angle of twist governs
Segment BC:
fBC d4
From A to C: fAC fAB + fBC ⬖
4
0.4018
d 2.75 in.
;
4
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
L1 1.5 m
L2 0.9 m
d diameter
f 32 Hz
At point A: TA 300,000 W
1492 N # m
2p(32 Hz)
At point B: TB 120
T 596.8 N # m
300 A
At point C: TC 180
T 895.3 N # m
300 A
FREE-BODY DIAGRAM
allow 50 MPa
G 75 GPa
(AC)allow 4° 0.06981 rad
TORQUES ACTING ON THE SHAFT
P 2fT P watts f Hz
TA 1492 N m
T newton meters
TB 596.8 N m
T
P
2pf
TC 895.3 N m
d diameter
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SECTION 3.7
Segment BC:
INTERNAL TORQUES
TAB 1492 N m
fBC TBC 895.3 N m
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS
fBC 32 TBCLBC
pGd 4
tmax 16(1492 N # m)
16 TAB
d pt allow
p(50 MPa)
3
pd 3
d3 0.0001520 m3
d 0.0534 m 53.4 mm
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
IP pd 4
32
f
TL
32TL
GIP
pGd 4
fAB 32(895.3 N # m)(0.9 m)
From A to C: fAC fAB + fBC (AC)allow 0.06981 rad
⬖ 0.06981 0.1094 * 106
d4
49.3mm
SHEAR STRESS GOVERNS
32 TABLAB
pGd
4
0.3039 * 10
d4
32(1492 N #
6
m)(1.5 m)
p(75 GPa)d
4
p(75 GPa)d 4
d4
and d 0.04933 m
Segment AB:
fAB 0.1094 * 106
The larger torque occurs in segment AB
16 TAB
Transmission of Power
d 53.4mm
;
0.4133 * 106
d4
301
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Torsion
Statically Indeterminate Torsional Members
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
T0
2T0
B
C
3L
—
10
3L
—
10
D
TD
D
TD
4L
—
10
L
Solution 3.8-1
Circular bar with fixed ends
ANGLE OF TWIST AT SECTION B
From Eqs. (3-46a and b):
TA(3L/10)
9T0 L
⫽
GIP
20GIP
TA ⫽
T0 LB
L
fB ⫽ fAB ⫽
TB ⫽
T0 LA
L
ANGLE OF TWIST AT SECTION C
APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:
TA ⫽ T0 a
15T0
7
4
b + 2T0 a b ⫽
10
10
10
15T0
3
6
TD ⫽ T0 a b + 2T0 a b ⫽
10
10
10
fC ⫽ fCD ⫽
TD(4L/10)
3T0 L
⫽
GIP
5GIP
MAXIMUM ANGLE OF TWIST
fmax ⫽ fC ⫽
3T0 L
5GIP
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.)
TA
A
;
T0
T0
B
C
x
x
L
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SECTION 3.8
Solution 3.8-2
303
Statically Indeterminate Torsional Members
Circular bar with fixed ends
(a) ANGLE OF TWIST AT SECTIONS B AND C
φB
φ max
0
From Eqs. (3-46a and b):
fB ⫽ fAB ⫽
L/A
T0 LB
L
dfB
T0
⫽
(L ⫺ 4x)
dx
GIPL
TB ⫽
T0 LA
L
dfB
⫽ 0; L ⫺ 4x ⫽ 0
dx
or x ⫽
L
4
x
TAx
T0
⫽
(L ⫺ 2x)(x)
GIP
GIPL
TA ⫽
APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:
L/2
;
(b) MAXIMUM ANGLE OF TWIST
fmax ⫽ (fB)max ⫽ (fB)x⫽ L4 ⫽
TA ⫽
T0L
8GIP
;
T0(L ⫺ x)
T0x
T0
⫺
⫽ (L ⫺ 2x) TD ⫽ TA
L
L
L
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
Solution 3.8-3
b
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⫹b
a⬎b
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tmax ⫽
T0 ⫽
Page 304
Torsion
ANGLE OF ROTATION OF THE DISK (FROM Eq. 3-49)
TB(d/2)
T0 ad
⫽
IP
2LIP
2LIPtmax
ad
(T0)max ⫽
2L IPt allow
ad
f⫽
T0 ab
GLIP
(T0)maxab
2bt allow
⫽
GLIp
Gd
fmax ⫽
;
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
200 mm
C
B
P
600 mm
400 mm
Solution 3.8-4
Hollow shaft with fixed ends
GENERAL FORMULAS:
T0 ⫽ P(400 mm)
LB ⫽ 400 mm
LA ⫽ 600 mm
L ⫽ LA ⫹ LB ⫽ 1000 mm
d2 ⫽ 50 mm
d1 ⫽ 40 mm
␶allow ⫽ 45 MPa
APPLY THE ABOVE FORMULAS TO THE GIVEN SHAFT
TA ⫽
P(0.4 m)(400 mm)
T0 LB
⫽
⫽ 0.16 P
L
1000 mm
TB ⫽
P(0.4 m)(600 mm)
T0 LA
⫽
⫽ 0.24 P
L
1000 mm
UNITS: P ⫽ Newtons T ⫽ Newton meters
From Eqs. (3-46a and b):
TA ⫽
T0 LB
L
T0 LA
L
The larger torque, and hence the larger shear
stress, occurs in part CB of the shaft.
TB ⫽
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SECTION 3.8
Tmax ⫽ TB ⫽ 0.24 P
Tmax(d/2)
IP
Tmax ⫽
2tmaxIP
d
0.24P ⫽
(Eq. 1)
P⫽
␶max ⫽ 45 ⫻ 10 N/m
Ip ⫽
2(45 * 106 N/m2)(362.26 * 10⫺9 m4)
0.05 m
⫽ 652.07 N # m
UNITS: Newtons and meters
6
305
Substitute numerical values into (Eq. 1):
SHEAR STRESS IN PART CB
tmax ⫽
Statically Indeterminate Torsional Members
2
652.07 N # m
⫽ 2717 N
0.24 m
Pallow ⫽ 2720 N
p 4
(d ⫺d 4) ⫽ 362.26 * 10⫺9m4
32 2 1
;
d ⫽ d2 ⫽ 0.05 mm
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.)
Solution 3.8-5
C
A
B
T0
6.0 in.
15.0 in.
Stepped shaft ACB
Combine Eqs. (1) and (3) and solve for T0:
dA ⫽ 0.75 in. dB ⫽ 1.50 in.
LA ⫽ 6.0 in. LB ⫽ 15.0 in.
␶allow ⫽ 6000 psi
(T0)AC ⫽
LAIPB
1
pd3 t
a1 +
b
16 A allow
LBIPA
⫽
LAdB4
1
pd3At allow a1 +
b
16
LBdA4
Find (T0)max
REACTIVE TORQUES (from Eqs. 3-45a and b)
LBIPA
b
TA ⫽ T0 a
LBIPA + LAIPB
TB ⫽ T0 a
1.50 in.
0.75 in.
LAIPB
b
LBIPA + LAIPB
(1)
(4)
Substitute numerical values:
(T0)AC ⫽ 3678 lb-in.
(2)
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
CB
IN SEGMENT
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
AC
tCB ⫽
IN SEGMENT
tAC ⫽
16TA
pdA3
1
1
TA ⫽
pd3 t ⫽
pd3 t
16 A AC 16 A allow
TB ⫽
(3)
16TB
pdB3
1
1
pd 3 t ⫽
pd 3 t
16 B CB 16 B allow
(5)
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CHAPTER 3
Page 306
Torsion
SEGMENT AC GOVERNS
Combine Eqs. (2) and (5) and solve for T0:
(T0)CB ⫽
(T0)max ⫽ 3680 lb-in.
LBIPA
1
pd3 t
a1 +
b
16 B allow
LAIPB
LBdA4
1
⫽
pd3Bt allow a1 +
b
16
LAdB4
Substitute numerical values:
(T0)CB ⫽ 4597 lb-in.
NOTE: From Eqs. (4) and (6) we find that
(6)
(T0)AC
LA dB
⫽ a ba b
(T0)CB
LB dA
which can be used as a partial check on the results.
20 mm
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.)
Solution 3.8-6
;
25 mm
B
C
A
T0
225 mm
450 mm
Stepped shaft ACB
1
1
pd 3 t ⫽
pd 3 t
16 A AC 16 A allow
dA
⫽ 20 mm
dB
⫽ 25 mm
LA
⫽ 225 mm
Combine Eqs. (1) and (3) and solve for T0:
LB
⫽ 450 mm
(T0)AC ⫽
TA ⫽
␶allow ⫽ 43 MPa
⫽
Find (T0)max
REACTIVE TORQUES (from Eqs. 3-45a and b)
LAIPB
1
pdA3 t allow a 1 +
b
16
LBIPA
LAdB4
1
pdA3t allow a 1 +
b
16
LBdA4
(1)
(T0)AC ⫽ 150.0 N ⭈ m
TB ⫽ T0 a
(2)
IN SEGMENT
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
IN SEGMENT AC
tAC ⫽
16TA
pd3A
(4)
Substitute numerical values:
LBIPA
b
TA ⫽ T0 a
LBIPA + LAIPB
LAIPB
b
LBIPA + LAIPB
(3)
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
CB
tCB ⫽
TB ⫽
16TB
pd3B
1
1
pd3 t ⫽
pd3 t
16 B CB 16 B allow
(5)
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Page 307
SECTION 3.8
(T0)max ⫽ 150 N ⭈ m
LBIPA
1
pd 3 t
a1 +
b
16 B allow
LAIPB
LBdA4
1
b
⫽ pdB3t allow a 1 +
16
LAdB4
(6)
(T0)AC
LA dB
⫽ a ba b
(T0)CB
LB dA
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.
dA
dB
IPA
A
C
T0
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.)
L
Stepped shaft
SEGMENT AC: dA, IPA LA ⫽ a
or
SEGMENT CB: dB, IPB LB ⫽ L ⫺ a
REACTIVE TORQUES (from Eqs. 3-45a and b)
Solve for a/L:
TA ⫽ T0 a
;
NOTE: From Eqs. (4) and (6) we find that
Substitute numerical values:
(T0)CB ⫽ 240.0 N ⭈ m
Solution 3.8-7
LBIPA
LAIPB
b ; TB ⫽ T0 a
b
LBIPA + LAIPB
LBIPA + LAIPB
(a) EQUAL SHEAR STRESSES
TA(dA/2)
TB(dB/2)
tCB ⫽
tAC ⫽
IPA
IPB
␶AC ⫽ ␶CB or
TAdA
TBdB
⫽
IPA
IPB
Substitute TA and TB into Eq. (1):
LBIPAdA
LAIPBdB
⫽
IPA
IPB
or
307
SEGMENT AC GOVERNS
Combine Eqs. (2) and (5) and solve for T0:
(T0)CB ⫽
Statically Indeterminate Torsional Members
LBdA ⫽ LAdB
(L ⫺ a)dA ⫽ adB
(b) EQUAL TORQUES
TA ⫽ TB or LBIPA ⫽ LAIPB
or
(L ⫺ a)IPA ⫽ aIPB
Solve for a/L:
(Eq. 1)
dA
a
⫽
L
dA + dB
or
IPA
a
⫽
L
IPA + IPB
dA4
a
⫽ 4
L
dA + dB4
;
;
IPB
B
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CHAPTER 3
Page 308
Torsion
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
TB
A
B
x
L
Solution 3.8-8
Fixed-end bar with triangular load
ELEMENT OF DISTRIBUTED LOAD
dTA ⫽ Elemental reactive torque
t(x) ⫽
dTB ⫽ Elemental reactive torque
t0x
L
From Eqs. (3-46a and b):
T0 ⫽ Resultant of distributed torque
L
L
t0 x
t0 L
t(x)dx ⫽
dx ⫽
T0 ⫽
2
L0
L0 L
EQUILIBRIUM
t0L
TA + TB ⫽ T0 ⫽
2
L⫺x
x
b dTB ⫽ t(x)dxa b
L
L
REACTIVE TORQUES (FIXED-END TORQUES)
dTA ⫽ t(x)dxa
L
t0L
x
L⫺x
TA ⫽ dTA ⫽
a t0 b a
bdx ⫽
L
L
6
L
L0
L
t0 L
x
x
TB ⫽ dTB ⫽ a t0 b a bdx ⫽
;
L L
3
L
L0
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?
t0L
(check)
2
25 in.
A
;
25 in.
T0
3.0 in.
B
x
2.4
in.
3.0
in.
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Page 309
SECTION 3.8
Solution 3.8-9
Statically Indeterminate Torsional Members
309
Bar with a hole
IPA ⫽ Polar moment of inertia at left-hand end
IPB ⫽ Polar moment of inertia at right-hand end
fB ⫽
T0(L/2)
GIPA
⫺
L ⫽ 50 in.
L/2 ⫽ 25 in.
(2)
Substitute Eq. (1) into Eq. (2) and simplify:
d2 ⫽ outer diameter
fB ⫽
⫽ 3.0 in.
d1 ⫽ diameter of hole
T0 L
L
x
L
L
c
+
⫺
+
⫺
d
G 4IPB
4IPA
IPB
2IPB
2IPA
COMPATIBILITY ␾B ⫽ 0
⫽ 2.4 in.
x
T0 ⫽ Torque applied at distance x
‹
Find x so that TA ⫽ TB
IPB
⫽
3L
L
⫺
4IPB 4IPA
soLVE FOR x:
EQUILIBRIUM
TA ⫹ TB ⫽ T0 ‹ TA ⫽ TB ⫽
TB(L/2)
TB(L/2)
T0(x ⫺ L/2)
+
⫺
GIPB
GIPA
GIPB
T0
2
(1)
REMOVE THE SUPPORT AT END B
x⫽
IPB
L
a3 ⫺
b
4
IPA
d24 ⫺ d14
d1 4
IPB
⫽
⫽
1
⫺
a
b
IPA
d2
d24
d1 4
L
x ⫽ c2 + a b d ;
4
d2
SUBSTITUTE NUMERICAL VALUES:
x⫽
50 in.
2.4 in. 4
c2 + a
b d ⫽ 30.12 in.
4
3.0 in.
;
␾B ⫽ 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
B
T
Bar
End
plate
L
d1
d2
d3
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CHAPTER 3
Page 310
Torsion
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 ⫽ T a
IP1
b ⫽ 100.2783 N # m
IP1 + IP2
Tube: T2 ⫽ T a
IP2
b ⫽ 299.7217 N # m
IP1 + IP2
(a) MAXIMUM SHEAR STRESSES
Bar: t1 ⫽
T1(d1/2)
⫽ 32.7 MPa
IP1
Tube: t2 ⫽
d1 ⫽ 25.0 mm
d2 ⫽ 30.0 mm
d3 ⫽ 37.5 mm
T2L
T1L
⫽
⫽ 0.017977 rad
GIP1
GIP2
f⫽
POLAR MOMENTS OF INERTIA
␾ ⫽ 1.03°
Bar: IP1
Tube: IP2 ⫽
;
(b) ANGLE OF ROTATION OF END PLATE
G ⫽ 80 GPa
p 4
⫽
d ⫽ 38.3495 * 10⫺9 m4
32 1
T2(d3/2)
⫽ 49.0 MPa
IP2
;
;
(c) TORSIONAL STIFFNESS
p
1d 4 ⫺ d242 ⫽ 114.6229 * 10⫺9 m4
32 3
kT ⫽
T
⫽ 22.3 kN # m
f
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
B
T
Bar
End
plate
L
d1
d2
d3
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SECTION 3.8
Solution 3.8-11
Statically Indeterminate Torsional Members
311
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (3-44A AND B)
Bar: T1 ⫽ T a
IPI
b ⫽ 1187.68 lb-in.
IPI + IPI
Tube: T2 ⫽ T a
IP2
b ⫽ 3812.32 lb-in.
IP1 + IP2
(a) MAXIMUM SHEAR STRESSES
Bar: t1 ⫽
T1(d1/2)
⫽ 1790 psi
IP1
Tube: t2 ⫽
T2(d3/2)
⫽ 2690 psi
IP2
;
;
(b) ANGLE OF ROTATION OF END PLATE
d1 ⫽ 1.50 in. d2 ⫽ 1.75 in. d3 ⫽ 2.25 in.
f⫽
G ⫽ 11.6 ⫻ 10 psi
6
␾ ⫽ 0.354°
POLAR MOMENTS OF INERTIA
Bar: IP1 ⫽
p 4
d ⫽ 0.497010 in.4
32 1
T2L
T1L
⫽
⫽ 0.006180015 rad
GIP1
GIP2
;
(c) TORSIONAL STIFFNESS
kT ⫽
p
Tube: IP2 ⫽ 1d34 ⫺ d242 ⫽ 1.595340 in.4
32
T
⫽ 809 k- in.
f
T
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.
;
Steel sleeve
Brass core
T
Assuming that the allowable shear stresses in the brass and steel are
␶b ⫽ 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.)
d1 d2
Probs. 3.8-12 and 3.8-13
Solution 3.8-12
Composite shaft shrink fit
d1 ⫽ 40 mm
d2 ⫽ 50 mm
GB ⫽ 36 GPa GS ⫽ 80 GPa
Allowable stresses:
␶B ⫽ 48 MPa ␶S ⫽ 80 MPa
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CHAPTER 3
Page 312
Torsion
BRASS CORE (ONLY)
Eq. (3-44b): TS ⫽ T a
GSIPS
b
GBIPB + GSIPS
⫽ 0.762082 T
T ⫽ TB ⫹ TS (CHECK)
ALLOWABLE TORQUE T BASED UPON BRASS CORE
IPB ⫽
p 4
d ⫽ 251.327 * 10⫺9 m4
32 1
GBIPB ⫽ 9047.79 N ⭈ m2
TB(d1/2)
2tBIPB
TB ⫽
IPB
d1
Substitute numerical values:
tB ⫽
TB ⫽ 0.237918 T
STEEL SLEEVE (ONLY)
⫽
2(48 MPa)(251.327 * 10⫺9 m4)
40 mm
T ⫽ 2535 N ⭈ m
ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE
IPS
p 4
⫽
(d ⫺ d14) ⫽ 362.265 * 10⫺9 m4
32 2
GSIPS ⫽ 28,981.2 N ⭈ m
2
TORQUES
Total torque: T ⫽ TB ⫹ TS
Eq. (3-44a): TB ⫽ T a
GBIPB
b
GBIPB + GS IPS
tS ⫽
TS(d2/2)
IPS
TS ⫽
2tSIPS
d2
SUBSTITUTE NUMERICAL VALUES:
TS ⫽ 0.762082 T
⫽
2(80 MPa)(362.265 * 10⫺9 m4)
50 mm
T ⫽ 1521 N ⭈ m
STEEL SLEEVE GOVERNS
Tmax ⫽ 1520 N ⭈ m
;
⫽ 0.237918 T
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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Page 313
SECTION 3.8
Solution 3.8-13
Statically Indeterminate Torsional Members
Composite shaft shrink fit
TORQUES
Total torque: T ⫽ TB ⫹ TS
Eq. (3-44 a): TB ⫽ T a
GBIPB
b
GBIPB + GSIPS
⫽ 0.237918 T
d1 ⫽ 1.6 in.
d2 ⫽ 2.0 in.
Eq. (3-44 b): TS ⫽ T a
GB ⫽ 5,400 psi GS ⫽ 12,000 psi
GSIPS
b
GBIPB + GSIPS
⫽ 0.762082 T
Allowable stresses:
T ⫽ TB ⫹ TS (CHECK)
␶B ⫽ 4500 psi ␶S ⫽ 7500 psi
ALLOWABLE TORQUE T BASED UPON BRASS CORE
BRASS CORE (ONLY)
TB(d1/2)
2tBIPB
TB ⫽
IPB
d1
Substitute numerical values:
tB ⫽
TB ⫽ 0.237918 T
⫽
IPB ⫽
p 4
d1 ⫽ 0.643398 in.4
32
GBIPB ⫽ 3.47435 ⫻ 106 lb-in.2
STEEL SLEEVE (ONLY)
2(4500 psi)(0.643398 in.4)
1.6 in.
T ⫽ 15.21 k-in.
ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE
TS(d2/2)
2tSIPS
TS ⫽
IPS
d2
Substitute numerical values:
tS ⫽
tS ⫽ 0.762082 T ⫽
2(7500 psi)(0.927398 in.4)
2.0 in.
T ⫽ 9.13 k-in.
STEEL SLEEVE GOVERNS
IPS ⫽
Tmax ⫽ 9.13 k-in.
;
p
(d 4 ⫺ d14) ⫽ 0.927398 in.4
32 2
GSIPS ⫽ 11.1288 ⫻ 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.
313
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Torsion
(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
Steel
shaft
d2 = 90 mm
d1 = 70 mm
T
T
A
B
1.0 m
L
= 2.0 m
2
d1
C
L
= 2.0 m
2
d1
Brass
sleeve
d2
d1
Steel
shaft
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
statics
Gb IPb
b
Gs IPs + GbIPb
now find twist of 3 segments:
Tb ⫽ T1 ⫺ Ts
Tb ⫽ T1 a
L
L
L
Ts
T1
4
4
2
␾⫽
+
+
Gb IPb
Gs IPs
Gs IPs
T1
Gs IPs
b
Ts ⫽ T1 a
Gs IPs + Gb IPb
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
fa ⫽
+
+
Gb IPb
Gs IPs
Gs IPs
T1
L
L
L
T1
T1
4
4
2
fa ⫽
+
+
Gb IPb
Gs IPs + Gb IPb
Gs IPs
T1
1
L
1
2
+
fa ⫽ T1 a
+
b
4 Gb IPb
GsIPs + Gb IPb
Gs IPs
T1, allow ⫽
4␾ a
Gb IPb(GsIPs + Gb IPb)Gs IPs
c 2 2
d
2
L Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb
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SECTION 3.8
f a ⫽ 8a
NUMERICAL VALUES
p
b rad
180
Gs ⫽ 80 GPa Gb ⫽ 40 GPa L ⫽ 3.0 m
d1 ⫽ 70 mm
IPs ⫽
p 4
d1
32
d2 ⫽ 90 mm
IPs ⫽ 2.357 ⫻ 10⫺6 m4
p
A d 4 ⫺ d14 B
32 2
T1, allow ⫽ 9.51 kN⭈ m
TORQUE
STRESS IN BRASS,
␶b
T2
;
BASED ON ALLOWABLE SHEAR
␶b ⫽ 70 MPa
First check hollow segment 1 (brass sleeve only)
T2
t⫽
d2
2
IPb
T2, allow ⫽
T2, allow ⫽ 6.35 kN⭈m
2tbIPb
d2
;
controls over T2 below also check segment 2 with
brass sleeve over steel shaft
Tb
t⫽
d2
2
IPb
Tb ⫽ T2 a
T2, allow
so T2 for hollow segment controls
(c) ALLOWABLE
TORQUE
STRESS IN STEEL,
␶s
T3
GbIPb
b
Gs IPS + Gb IPb
2tb(Gs IPS + Gb IPb)
⫽
d2 Gb
BASED ON ALLOWABLE SHEAR
␶s ⫽ 110 MPa
First check segment 2 with brass sleeve over steel
shaft
Ts
t⫽
d1
2
where from stat-indet analysis above
IPS
Ts ⫽ T3 a
GsIPS
b
Gs IPS + Gb IPb
T3, allow ⫽
2ts (Gs IPs + Gb IPb)
d1Gs
T3, allow ⫽ 13.83 kN⭈m
also check segment 3 with steel shaft alone
d1
2
t⫽
IPs
T3
T3, allow ⫽
T3, allow ⫽ 7.41 kN⭈m
where from stat-indet analysis above
315
T2, allow ⫽ 13.69 kN⭈m
IPb ⫽ 4.084 ⫻ 10⫺6 m4
IPb ⫽
(b) ALLOWABLE
Statically Indeterminate Torsional Members
2ts IPs
d1
; controls over T3 above
(d) Tmax IF ALL PRECEDING CONDITIONS MUST BE
CONSIDERED
from (b) above
Tmax ⫽ 6.35 kN⭈m
; max. shear stress in
hollow brass sleeve in
segment 1 controls overall
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Torsion
Problem 3.8-15 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is fixed against rota-
tion 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 ⫽ 2.5 in., L ⫽ 48 in.,
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
L
—
2
A
d(x)
t constant
TB
T0
dA
Fixed against
rotation
B
dB
L
(a)
TA
A
Fixed against
rotation
B
L
—
2
dA
TB
T0
dA
Fixed against
rotation
dB
L
(b)
Solution 3.8-15
Solution approach-superposition: select TB as the redundant (1° SI )
L
—
2
TA1
A
B
ϕ1(same results for parts a & b)
T0
f1 ⫽
L
+
TA2
L
—
2
A
81GpdA
4
f1 ⫽ 2.389
See Prob. 3.4-13 for results
for w2 for Parts a & b
B
ϕ2
TB
L
608T0L
f2a ⫽ 3.868
f2a ⫽ 3.057
T0 L
Gd A 4
T0L
Gd A 4
T0L
GdA 4
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SECTION 3.8
CONSTANT THICKNESS OF HOLLOW SECTION OF TUBE
CONSTANT DIAMETER OF HOLE
␾1 ⫺ ␾2 ⫽ 0
TB ⫽ a
TB ⫽ 24708 in-lb
TA ⫽ T0 ⫺ TB TA ⫽ 15292 in-1b
ba
4
608T0 L
81GpdA
TB ⫽ 2.45560
GdA4
TB ⫽ a
b
a
b
81G pdA4 3.86804L
608T0 L
T0
TB ⫽ 1.94056
p
317
(b) REACTIVE TORQUES, TA & TB, FOR CASE OF
(a) REACTIVE TORQUES, TA & TB, FOR CASE OF
compatibility equation:
TB ⫽ redundant
T0 ⫽ 40000 in.-lb
Statically Indeterminate Torsional Members
T0
p
GdA4
b
3.05676L
TB ⫽ 31266 in.-lb
TA ⫽ T0 ⫺ TB TA ⫽ 8734 in.-lb
;
;
TA ⫹ TB ⫽ 40,000 in.-1b (check)
;
;
TA ⫹ TB ⫽ 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
IPB
TA
Tube A
A
B
C
L
b Pin at C
TB
Tube B
L
Tube A
Tube B
Cross-section at C
Solution 3.8-16
(a) SUPERPOSITION
TO FIND TORQUE REACTIONS
-
USE
TB
AS THE REDUNDANT
compatibility: ␾B1 ⫹ ␾B2 ⫽ 0
␾B1 ⫽ ⫺␤ ⬍ joint tubes by pin then release end B
fB2 ⫽
TBL 1
1
a
+
b
G IPA
IPB
fB2 ⫽
TBL IPB + IPA
a
b
G
IPAIPB
TB ⫽
Gb IPAIPB
a
b
L IPA ⫹IPB
TA ⫽ ⫺TB
; statics
(b) ALLOWABLE SHEAR IN PIN RESTRICTS MAGNITUDE OF ␤
TB ⫽ FORCE COUPLE VdB WITH V ⫽ SHEAR IN
C
TB
V
V⫽
tp ⫽
dB
As
TORQUE
PIN AT
TB
dB
tp, allow ⫽
p 2
d
4 P
;
b max ⫽ tp, allow
ca
tp, allow
Gb IPAIPB
a
b
L IPA ⫹IPB
⫽
p
dB dp2
4
L
4G
IPB + IPA
b dBpdP 2 d
IPAIPB
;
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Torsion
(c) ALLOWABLE SHEAR IN TUBES RESTRICTS MAGNITUDE
OF ␤
dA
TB
2
tmax ⫽
IPA
tmax ⫽
or
tmax
Bearing stresses from tubes A & B are:
sbA ⫽
dB
TB
2
⫽
IPB
TB
dA ⫺ tA
sbA ⫽
dPtA
Gb IPAIPB dA
a
b
L IPA ⫹IPB 2
Gb
IPAIPB
a
b
L IPA + IPB
Gb IPAIPB dB
a
b
L IPA ⫹IPB 2
sbA ⫽
IPB
simplifying these two equ., then solving for ␤
gives:
tmax ⫽
Gb
IPB
dA
a
b
L IPA + IPB 2
sbA ⫽ b
Gb
dB
IPA
a
b
L IPA + IPB 2
b max ⫽ tt, allow a
IPA + IPB
2L
ba
b
GdA
IPB
sbB ⫽ b
;
or
b max
dA ⫺tA
dPtA
Gb
IPAIPB
a
b
L IPA + IPB
IPA + IPB
2L
⫽ tt, allow a
ba
b
GdB
IPA
dB ⫺ tB
dPtB
sbB ⫽
or
tmax ⫽
TB
dB ⫺ tB
sbB ⫽
dP tB
substitute TB expression from part (a), then simplify
␧ solve for b
IPA
or
tmax ⫽
FA
FB
sbB ⫽
dPtA
dPtB
;
where lesser value of ␤ controls
(d) ALLOWABLE BEARING STRESS IN PIN RESTRICTS
MAGNITUDE OF ␤
Torque TB ⫽ force couple FB (dB ⫺ tB) or
FA(dA ⫺ tA), with F ⫽ ave. bearing force on
pin at C
L(IPB
IPAIPB
G
L (IPB + IPA)(dB ⫺ tB)dPtB
b max ⫽ sb, allow
c
L
G
(IPB + IPA)(dA ⫺ tA)dPtA
d
IPAIPB
b max ⫽ s b, allow
c
GIPAIPB
+ IPA)(dA ⫺ tA) dP tA
;
L
G
(IPB + IPA)(dB ⫺ tB)dPtB
d
IPAIPB
where lesser value controls
;
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SECTION 3.9
Strain Energy in Torsion
319
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).
d
T
T
(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).
Solution 3.9-1
L
Steel bar
⫽
pd2Lt2max
16G
(Eq. 2)
Substitute numerical values:
U ⫽ 32.0 in.-lb
G ⫽ 11.4 ⫻ 106 psi
(b) ANGLE OF TWIST
L ⫽ 30 in.
U⫽
d ⫽ 1.75 in.
␶max ⫽ 4500 psi
tmax ⫽
IP ⫽
16 T
pd3
Tf
2
f⫽
2U
T
Substitute for T and U from Eqs. (1) and (2):
pd3tmax
T⫽
16
pd 4
32
;
f⫽
(Eq. 1)
2Ltmax
Gd
(Eq. 3)
Substitute numerical values:
␾ ⫽ 0.013534 rad ⫽ 0.775°
;
(a) STRAIN ENERGY
U⫽
pd3tmax 2 L
T2L
32
⫽a
b a
b a 4b
2GIP
16
2G pd
Problem 3.9-2 A solid circular bar of copper (G ⫽ 45 GPa) with length L ⫽ 0.75 m and diameter d ⫽ 40 mm 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 32 MPa.
(b) From the strain energy, calculate the angle of twist ␾ (in degrees)
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Torsion
Solution 3.9-2
Copper bar
(a) STRAIN ENERGY
U⫽
pd3tmax 2 L
T2L
32
⫽a
b a
ba
b
2GIP
16
2G pd 4
L ⫽ 0.75 m
pd2Lt2max
16G
Substitute numerical values:
d ⫽ 40 mm
U ⫽ 5.36 J
⫽
G ⫽ 45 GPa
␶max ⫽ 32 MPa
tmax ⫽
IP ⫽
pd
T⫽
3
;
(b) ANGLE OF TWIST
3
16T
pd tmax
16
pd4
32
Tf
2U
f⫽
2
T
Substitute for T and U from Eqs. (1) and (2):
2Ltmax
(Eq. 3)
f⫽
Gd
Substitute numerical values:
U⫽
(Eq. 1)
␾ ⫽ 0.026667 rad ⫽ 1.53°
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
;
d1
T
T
L
—
2
L
—
2
Solution 3.9-3
(Eq. 2)
Stepped shaft
⫽
8T2L 1
1
a 4 + 4b
pG d2
d1
Also, U ⫽
(Eq. 1)
Tf
2
(Eq. 2)
Equate U from Eqs. (1) and (2) and solve for T:
d1 ⫽ 1.0 in.
T⫽
d2 ⫽ 1.2 in.
L ⫽ 45 in.
pGd14 d24f
16L(d41 + d42)
pGf2 d14 d24
Tf
⫽
a
b
2
32L d41 + d42
G ⫽ 5.6 ⫻ 10 psi (brass)
U⫽
␾ ⫽ 3.0° ⫽ 0.0523599 rad
SUBSTITUTE NUMERICAL VALUES:
STRAIN ENERGY
U ⫽ 22.6 in.-lb
6
16 T2(L/2)
16 T2(L/2)
T2L
U⫽ a
⫽
+
4
2GIP
pGd2
pGd41
;
f ⫽ radians
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SECTION 3.9
Strain Energy in Torsion
321
Problem 3.9-4 A stepped shaft of solid circular cross sections (see figure) has length L ⫽ 0.80 m, diameter d2 ⫽ 40 mm,
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°.
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 ⫽ 30 mm
d2 ⫽ 40 mm
L ⫽ 0.80 m
G ⫽ 80 GPa (steel)
U⫽
pG d14 d24f
16L(d14 + d24)
pGf2 d14 d24
Tf
⫽
a
b
2
32L d14 + d24
f ⫽ radians
␾ ⫽ 1.0° ⫽ 0.0174533 rad
SUBSTITUTE NUMERICAL VALUES:
STRAIN ENERGY
2
TL
U⫽ a
⫽
2GIP
⫽
U ⫽ 1.84 J
16T2(L/2)
pGd24
8T2L 1
1
a
+ 4b
pG d24
d1
;
16T2(L/2)
+
pGd14
(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?
Solution 3.9-5
Cantilever bar with distributed torque
G ⫽ shear modulus
IP ⫽ polar moment of inertia
T ⫽ torque acting at free end
t ⫽ torque per unit distance
t
L
T
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Torsion
(c) BOTH LOADS ACT SIMULTANEOUSLY
(a) LOAD T ACTS ALONE (Eq. 3-51a)
U1 ⫽
T2L
2GIP
;
(b) LOAD t ACTS ALONE
From Eq. (3-56) of Example 3-11:
At distance x from the free end:
t2L3
U2 ⫽
6GIP
T(x) ⫽ T + tx
;
L
L
[T(x)]2
1
dx ⫽
(T + tx)2dx
2GI
2GI
L0
P
PL
0
T2L
TtL2
t2L3
⫽
+
+
;
2GIP
2GIP
6GIP
U3 ⫽
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.
Solution 3.9-6
3L
b
4
L
T0 a b
4
+
L
L
⫽
7T0
4
C
L
—
4
D
L
—
2
⫽
L
L
L
1
2
2
cT 2 a b + TCD
a b + TDB
a bd
2GIp AC 4
2
4
⫽
7T0 2 L
1
ca⫺
b a b
2GIP
4
4
+ a
INTERNAL TORQUES
7T0
4
B
n
Ti2Li
U⫽ a
i⫽1 2GiIPi
5T0
TB ⫽ 3T0 ⫺ TA ⫽
4
TAC ⫽ ⫺
A
STRAIN ENERGY (from Eq. 3-53)
From Eq. (3-46a):
TA ⫽
T0
Statically indeterminate bar
REACTIVE TORQUES
(2T0)a
2T0
TCD ⫽
T0
4
TDB ⫽
5T0
4
U⫽
T0 2 L
5T0 2 L
b a b + a
b a bd
4
2
4
4
19T02L
32GIP
;
L
—
4
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SECTION 3.9
Strain Energy in Torsion
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
B
LB
Solution 3.9-7
Statically indeterminate bar
WORK DONE BY THE TORQUE T0
W⫽
T0f
2
EQUATE U AND W AND SOLVE FOR ␾
T0f
Gf2 IPA
IPB
a
+
b⫽
2
LA
LB
2
f⫽
STRAIN ENERGY (FROM EQ. 3-51B)
;
(This result agrees with Eq. (3-48) of Example 3-9,
Section 3.8.)
GIPAf2
GIPBf2
GIPif2i
⫽
+
U⫽ a
2LA
2LB
i⫽1 2Li
n
⫽
T0LALB
G(LBIPA + LAIPB)
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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Torsion
Solution 3.9-8
Cantilever bar with distributed torque
x ⫽ distance from right-hand end of the bar
ELEMENT d␰
STRAIN ENERGY OF ELEMENT dx
Consider a differential element dj at distance j from the
right-hand end.
dU ⫽
[T(x)]2dx
t0 2
1
⫽
a b x4dx
2GIP
2GIP 2L
⫽
t20
8L2GIP
x4 dx
STRAIN ENERGY OF ENTIRE BAR
L
U⫽
L0
dT ⫽ external torque acting on this element
dT ⫽ t(j)dj
j
⫽ t0 a bdj
L
U⫽
ELEMENT dx AT DISTANCE x
T(x) ⫽ internal torque acting on this element
T(x) ⫽ total torque from x ⫽ 0 to x ⫽ x
x
T(x) ⫽
⫽
L0
t0x2
2L
dT ⫽
x
L0
t0 a
j
bdj
L
t20
dU ⫽
t20L3
40GIP
L
x4 dx
8L2GIP L0
t20
L5
⫽ 2
a b
8L GIP 5
;
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SECTION 3.9
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).
B
A
T
T
(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.
L
t
Note: Use the approximate formula IP ⬇ ␲d3t/4 for a thin circular
ring; see Case 22 of Appendix D.
t
dB
dA
Solution 3.9-9
325
Strain Energy in Torsion
Thin-walled, hollow tube
Therefore,
L
dx
3
dB ⫺ dA
L0
cdA + a
bx d
L
⫽⫺
t ⫽ thickness
dA ⫽ average diameter at end A
dB ⫽ average diameter at end B
⫽⫺
d(x) ⫽ average diameter at distance x from end A
d(x) ⫽ dA + a
dB ⫺ dA
bx
L
⫽
pd3t
4
U⫽
3
p[d(x)]3t
dB ⫺ dA
pt
IP(x) ⫽
⫽ cdA + a
bx d
4
4
L
L
T2dx
L0 2GIP(x)
L
dx
2T2
⫽
3
pGt L0
dB ⫺ dA
cdA + a
bx d
L
From Appendix C:
⫽⫺
2(dB ⫺ dA)(dB)
2
+
2(dB ⫺ dA)(dA)2
L(dA + dB)
2dA2 dB2
1
2b(a + bx)2
2T2 L(dA + dB)
T2L dA + dB
⫽
a 2 2 b
pGt 2dA2dB2
pGt
dA dB
Work of the torque T: W ⫽
U⫽
dx
L
L
(b) ANGLE OF TWIST
(a) STRAIN ENERGY (FROM EQ. 3-54)
L (a + bx)3
2 †
2(dB ⫺ dA)
dB ⫺ dA
cdA + a
bx d
L
L
0
Substitute this expression for the integral into the equation
for U (Eq. 1):
POLAR MOMENT OF INERTIA
IP ⫽
L
1
W⫽U
(Eq. 1)
T2L(dA + dB)
Tf
⫽
2
pGt d2Ad2B
Solve for ␾:
f⫽
Tf
2
2TL(dA + dB)
pGt d2A d2B
;
;
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Torsion
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
IPB
Tube A
Bar B
L
L
b
Tube A
Bar B
Solution 3.9-10
Circular tube and bar
TUBE A
COMPATIBILITY
␾A ⫹ ␾B ⫽ ␤
FORCE-DISPLACEMENT RELATIONS
fA ⫽
T ⫽ torque acting on the tube
␾A ⫽ angle of twist
BAR B
TL
GIPA
fB ⫽
TL
GIPB
Substitute into the equation of compatibility and solve
for T:
T⫽
bG
IPAIPB
a
b
L IPA + IPB
STRAIN ENERGY
U⫽ g
⫽
T 2L
T 2L
T 2L
⫽
+
2GIP
2GIPA
2GIPB
T2L 1
1
a
+
b
2G IPA
IPB
Substitute for T and simplify:
U⫽
T ⫽ torque acting on the bar
␾B ⫽ angle of twist
b 2G IPA IPB
a
b
2L IPA + IPB
;
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SECTION 3.9
Strain Energy in Torsion
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.)
A
d
n (rpm)
B
C
Hint: Equate the kinetic energy of the rotating flywheel to the strain
energy of the shaft.
Solution 3.9-11
Rotating flywheel
p 4
d
32
IP ⫽
d ⫽ diameter of shaft
U⫽
pGd 4f2
64L
UNITS:
d ⫽ diameter
n ⫽ rpm
IP ⫽ (length)4
␾ ⫽ radians
KINETIC ENERGY OF FLYWHEEL
K.E. ⫽
G ⫽ (force)/(length)2
1
Imv2
2
2pn
v⫽
60
n ⫽ rpm
2pn 2
1
b
K.E. ⫽ Im a
2
60
L ⫽ length
U ⫽ (length)(force)
EQUATE KINETIC ENERGY AND STRAIN ENERGY
Solve for ␾:
f⫽
2 2
⫽
p n Im
1800
Im ⫽ (force)(length)(second)2
␻ ⫽ radians per second
K.E. ⫽ (length)(force)
STRAIN ENERGY OF SHAFT (FROM EQ. 3-51b)
U⫽
GIPf
2L
2
2n
2A
15d
2pImL
G
;
MAXIMUM SHEAR STRESS
t⫽
UNITS:
pGd 4f2
p2n2Im
⫽
1800
64 L
K.E. ⫽ U
T(d/2)
IP
f⫽
TL
GIP
Eliminate T:
t⫽
Gdf
2L
2pImL
2L15d A G
2pGIm
n
tmax ⫽
15d A L
tmax ⫽
Gd2n
2
;
327
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Torsion
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?
10.0 in.
1.0 in.
Solution 3.10-1
Hollow circular tube
APPROXIMATE THEORY (EQ. 3-63)
t1 ⫽
T
2
2pr t
⫽
1200 k-in.
2p(5.5 in.)2(1.0 in.)
␶approx ⫽ 6310 psi
⫽ 6314 psi
;
EXACT THEORY (EQ. 3-11)
T ⫽ 1200 k-in.
t ⫽ 1.0 in.
t2 ⫽
T(d2/2)
⫽
IP
r ⫽ radius to median line
r ⫽ 5.5 in.
d2 ⫽ outside diameter ⫽ 12.0 in.
d1 ⫽ inside diameter ⫽ 10.0 in.
⫽
Td2
p
2a
b 1d24 ⫺ d142
32
16(1200k-in.)(12.0 in.)
p[(12.0 in.)4 ⫺ (10.0 in.)4]
⫽ 6831 psi
t exact ⫽ 6830 psi
;
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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SECTION 3.10
Solution 3.10-2
Thin-Walled Tubes
329
Bar and tube
SOLID BAR
tmax ⫽
16T
pd3
(Eq. 3-12)
Am ⫽ (d)(2d) ⫽ 2d2
(Eq. 3-64)
T
T
⫽
2tAm
4td2
(Eq. 3-61)
tmax ⫽
EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t
RECTANGULAR TUBE
16T
pd3
⫽
T
4td2
tmin ⫽
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.
t
h
(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.
b
Probs. 3.10-3 and 3.10-4
Solution 3.10-3
Thin-walled tube
Eq. (3-64): Am ⫽ bh ⫽ 24.0 in.2
J⫽
Eq. (3-71) with t1 ⫽ t2 ⫽ t:
J ⫽ 28.8 in.4
(a) SHEAR STRESS (EQ. 3-61)
t⫽
b ⫽ 6.0 in.
h ⫽ 4.0 in.
t ⫽ 0.25 in.
T ⫽ 15 k-in.
L ⫽ 50 in.
G ⫽ 4.0 ⫻ 106 psi
T
⫽ 1250 psi
2tAm
;
(b) ANGLE OF TWIST (EQ. 3-72)
f⫽
TL
⫽ 0.0065104 rad
GJ
⫽ 0.373°
;
2b2h2t
b + h
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Torsion
Problem 3.10-4 A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions b ⫽ 150 mm
and h ⫽ 100 mm. The wall thickness t is constant and equal to 6.0 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 ⫽ 150 mm
(a) SHEAR STRESS (Eq. 3-61)
h ⫽ 100 mm
t⫽
t ⫽ 6.0 mm
L ⫽ 1.2 m
f⫽
G ⫽ 75 GPa
TL
⫽ 0.002444 rad
GJ
⫽ 0.140°
Eq. (3-64): Am ⫽ bh ⫽ 0.015 m2
J⫽
;
(b) ANGLE OF TWIST (Eq. 3-72)
T ⫽ 1650 N ⭈ m
Eq. (3-71) with t1 ⫽ t2 ⫽ t:
T
⫽ 9.17 MPa
2tAm
;
2b2h2t
b + h
J ⫽ 10.8 ⫻ 10⫺6 m4
Tube (1)
Problem 3.10-5 A thin-walled circular tube and a solid circular bar of
Bar (2)
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 ⫽ ␲r2
tmax ⫽
J ⫽ 2␲r3t A ⫽ 2␲rt
T
T
⫽
2tAm
2pr 2t
T ⫽ 2␲r 2t␶max
U1 ⫽
12pr2ttmax22L
T2L
⫽
2GJ
2G(2pr3t)
prtt2maxL
G
A
But rt ⫽
2p
⫽
At2max L
‹ U1 ⫽
2G
SOLID BAR (2)
A ⫽ pr22
IP ⫽
p 4
r
2 2
Tr2
pr23tmax
2T
⫽ 3 T⫽
IP
2
pr2
2
3
2 2
2
(pr2 tmax) L
pr2 tmaxL
TL
⫽
U2 ⫽
⫽
2GIP
4G
p 4
8Ga r2 b
2
tmax ⫽
But pr22 ⫽ A
RATIO
U1
⫽2
U2
;
‹ U2 ⫽
2
L
Atmax
4G
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SECTION 3.10
Thin-Walled Tubes
t = 8 mm
Problem 3.10-6 Calculate the shear stress ␶ and the angle of twist ␾ (in
r = 50 mm
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.
r = 50 mm
b = 100 mm
Solution 3.10-6
Steel tube
SHEAR STRESS
G ⫽ 76 GPa.
t⫽
10 kN # m
T
⫽
2tAm
2(8 mm)(17,850 mm2)
L ⫽ 1.5 m
T ⫽ 10 kN ⭈ m
Am ⫽ ␲r2 ⫹ 2br
Am ⫽ ␲ (50 mm)2 ⫹ 2(100 mm)(50 mm)
⫽ 17,850 mm2
Lm ⫽ 2b ⫹ 2␲r
⫽ 35.0 MPa
;
ANGLE OF TWIST
f⫽
(10 kN # m)(1.5 m)
TL
⫽
GJ
(76 GPa)(19.83 * 106 mm4)
⫽ 0.00995 rad
⫽ 0.570°
;
⫽ 2(100 mm) ⫹ 2␲(50 mm)
⫽ 514.2 mm
J⫽
4(8 mm)(17,850 mm2)2
4tA2m
⫽
Lm
514.2 mm
⫽ 19.83 * 106 mm4
Problem 3.10-7 A thin-walled steel tube having an elliptical cross
331
t
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.)
2b
2a
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Torsion
Solution 3.10-7
Elliptical tube
FROM APPENDIX D, CASE 16:
Am ⫽ pab ⫽ p(3.0 in.)(2.0 in.) ⫽ 18.850 in.2
Lm L p[1.5(a + b) ⫺ 1ab]
⫽ p[1.5(5.0 in.) ⫺ 26.0 in.2] ⫽ 15.867 in.
J⫽
⫽ 17.92 in.4
T ⫽ 18 k-in.
SHEAR STRESS
G ⫽ 12 ⫻ 106 psi
t⫽
t ⫽ constant
t ⫽ 0.2 in
4(0.2 in.)(18.850 in.2)2
4tA2m
⫽
Lm
15.867 in.
a ⫽ 3.0 in.
b ⫽ 2.0 in.
18 k-in.
T
⫽
2tAm
2(0.2 in.)(18.850 in.2)
⫽ 2390 psi
;
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
u⫽
f
T
18 k-in.
⫽
⫽
L
GJ
(12 * 106 psi)(17.92 in.)4
u ⫽ 83.73 * 10⫺6 rad/in. ⫽ 0.0048°/in.
Problem 3.10-8 A torque T is applied to a thin-walled tube having
;
t
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 ␪.
b
Solution 3.10-8
Regular hexagon
b ⫽ Length of side
t ⫽ Thickness
Lm ⫽ 6b
FROM APPENDIX D, CASE 25:
␤ ⫽ 60° n ⫽ 6
Am ⫽
⫽
b
nb2
6b2
cot ⫽
cot 30°
4
2
4
3 13b2
2
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SECTION 3.10
SHEAR STRESS
T
T13
t⫽
⫽
2tAm
9b2t
u⫽
;
Thin-Walled Tubes
T
2T
2T
⫽
⫽
3
GJ
G(9b t)
9Gb3t
333
;
(radians per unit length)
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
4A2mt
J⫽
Lm
L0
ds
t
⫽
4A2mt
9b3t
⫽
Lm
2
Problem 3.10-9 Compare the angle of twist ␾1 for a thin-walled circular tube
t
(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.
r
(a) Express the ratio ␾1/␾2 in terms of the nondimensional ratio ␤ ⫽ r/t.
(b) Calculate the ratio of angles of twist for ␤ ⫽ 5, 10, and 20. What conclusion
about the accuracy of the approximate theory do you draw from these results?
Solution 3.10-9
C
Thin-walled tube
(a) RATIO
f1
f2
⫽
4r 2 + t 2
Let b ⫽
APPROXIMATE THEORY
TL
f1 ⫽
GJ
J ⫽ 2␲r 3t
(b)
f1 ⫽
TL
GIP
2pGr3t
From Eq. (3-17): Ip ⫽
TL
2TL
f2 ⫽
⫽
GIP
pGrt(4r 2 + t 2)
r
t
f1
f2
t2
4r 2
1
⫽1 +
;
4b 2
␤
␾1/␾2
5
10
20
1.0100
1.0025
1.0006
TL
EXACT THEORY
f2 ⫽
4r 2
⫽1 +
prt
(4r 2 + t 2)
2
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).
How does the shear stress in the tube vary with the ratio ␤ ⫽ a/b if
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
tube is square (␤ ⫽ 1).
t
b
a
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Torsion
Solution 3.10-10
Rectangular tube
T, t, and Lm are constants.
Let k ⫽
2T
tL2m
⫽ constant t ⫽ k
(1 + b)2
b
t ⫽ thickness (constant)
a, b ⫽ dimensions of the tube
b⫽
a
b
t
⫽4
a b
k min
Lm ⫽ 2(a ⫹ b) ⫽ constant
T ⫽ constant
T
2tAm
tL2m
ALTERNATE SOLUTION
Am ⫽ ab ⫽ ␤b2
t⫽
Lm ⫽ 2b(1 ⫹ ␤) ⫽ constant
2T (1 + b)2
c
d
b
tL2m
2T b(2)(1 + b) ⫺ (1 + b)2(1)
dt
⫽ 2c
d ⫽0
db
tLm
b2
2
Lm
Am ⫽ b c
d
2(1 + b)
Lm
b⫽
2(1 + b)
Am ⫽
8T
From the graph, we see that ␶ is minimum when ␤ ⫽ 1
and the tube is square.
SHEAR STRESS
t⫽
tmin ⫽
or 2␤ (1 ⫹ ␤) ⫺ (1 ⫹ ␤)2 ⫽ 0
bL2m
⬖␤ ⫽ 1
2
4(1 + b)
T(4)(1 + b)2
2T(1 + b)2
T
⫽
⫽
t⫽
2
2tAm
2tbLm
tL2m b
;
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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SECTION 3.10
Solution 3.10-11
335
Thin-Walled Tubes
Square aluminum tube
THICKNESS t BASED UPON SHEAR STRESS
t⫽
T
2tAm
tAm ⫽
UNITS: t ⫽ in.
T
2t
T ⫽ lb-in. ␶ ⫽ psi
b ⫽ in.
t(2.0 in. ⫺ t)2 ⫽
T
2t
t(b ⫺ t)2 ⫽
3000 lb-in.
1
⫽ in.3
2(4500 psi)
3
3t(2 ⫺ t)2 ⫺ 1 ⫽ 0
Solve for t: t ⫽ 0.0915 in.
Outer dimensions:
2.0 in. ⫻ 2.0 in.
THICKNESS t BASED UPON RATE OF TWIST
G ⫽ 4 ⫻ 106 psi
T ⫽ 3000 lb-in.
u⫽
␶allow ⫽ 4500 psi
u allow ⫽ 0.01 rad/ft ⫽
0.01
rad/in.
12
T
T
⫽
GJ
Gt(b ⫺ t)3
3000 lb-in
t(2.0 in. ⫺ t)3 ⫽
6
(4 * 10 psi)(0.01/12 rad/in.)
9
⫽
10
⫽ 2.0 in.
Centerline dimension ⫽ b ⫺ t
10t(2 ⫺ t)3 ⫺ 9 ⫽ 0
Am ⫽ (b ⫺ t)2
Solve for t:
J⫽
Lm
Lm ⫽ 4(b ⫺ t)
4t(b ⫺ t)
⫽ t(b ⫺ t)3
4(b ⫺ t)
4
⫽
T
Gu
G ⫽ psi ␪ ⫽ rad/in.
UNITS: t ⫽ in.
Let b ⫽ outer dimension
4tA2m
t(b ⫺ t)3 ⫽
t ⫽ 0.140 in.
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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Torsion
Solution 3.10-12
Thin tube
(b) EXACT THEORY
T ⫽ 5,000 N ⭈ m d1 ⫽ inner diameter ⫽ 100 mm
t⫽
Tr2
Ip
Ip ⫽
p 4
p
(r ⫺ r41) ⫽ [(50 + t)4 ⫺(50)4]
2 2
2
42 MPa ⫽
␶allow ⫽ 42 MPa
t is in millimeters.
(5000 N # m)(2)
(50 + t)4 ⫺ (50)4
⫽
50 + t
(p)(42 MPa)
r ⫽ Average radius
⫽ 50 mm +
t
2
⫽
r1 ⫽ Inner radius
t ⫽ 7.02 mm
r2 ⫽ Outer radius
⫽ 50 mm ⫹ t Am ⫽ ␲r2
(a) APPROXIMATE THEORY
T
T
T
⫽
⫽
2tAm
2t(pr 2)
2pr 2 t
5,000 N # m
42 MPa ⫽
t 2
2pa50 + b t
2
t⫽
or
5,000 N # m
t 2
5 * 106
b ⫽
⫽
mm3
2
2p(42 MPa)
84p
Solve for t:
t ⫽ 6.66 mm
5 * 106
mm3
21p
Solve for t:
⫽ 50 mm
ta50 +
(5,000 N # m)(50 + t)
p
[(50 + t)4 ⫺ (50)4]
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.
Sec_3.9-3.11.qxd
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Page 337
SECTION 3.10
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⫽
T
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.
Solution 3.10-13
B
A
2TL dA + dB
a 2 2 b
pGt
dAdB
337
Thin-Walled Tubes
dB
dA
Thin-walled tapered tube
For entire tube:
f⫽
4T
pGT L0
L
dx
3
dB ⫺ dA
cdA + a
bx d
L
From table of integrals (see Appendix C):
1
t ⫽ thickness
dx
(a + bx)3
⫽⫺
1
2b(a + bx)2
dA ⫽ average diameter at end A
dB ⫽ average diameter at end B
T ⫽ torque
d(x) ⫽ average diameter at distance x from end A.
d(x) ⫽ dA + a
J ⫽ 2pr 3t ⫽
dB ⫺ dA
bx
L
3
pd t
4
3
dB ⫺ dA
pt
pt
J(x) ⫽ [d(x)]3 ⫽ cdA + a
bx d
4
4
L
For element of length dx:
df ⫽
Tdx
⫽
GJ(x)
4Tdx
3
dB ⫺ dA
GptcdA + a
bx d
L
L
4T
f⫽
pGt
⫽
f⫽
J
1
⫺
2a
2
dB ⫺ dA
dB ⫺ dA
# xb K 0
b adA +
L
L
4T
L
L
c⫺
+
d
pGt
2(dB ⫺ dA)d2B
2(dB ⫺ dA)d2A
2TL dA + dB
a 2 2 b
pGt
dAdB
;
Sec_3.9-3.11.qxd
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CHAPTER 3
1:14 PM
Page 338
Torsion
Stress Concentrations in Torsion
D2
The problems for Section 3.11 are to be solved by considering the
stress-concentration factors.
Problem 3.11-1 A stepped shaft consisting of solid circular
D1
T
T
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
R
Probs. 3.11-1 through 3.11-5
Stepped shaft in torsion
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
D2
2.4 in.
⫽
⫽ 1.2
D1
2.0 in.
R
0.1 in.
⫽
⫽ 0.05
D1
2.0 in.
K ⬇ 1.52
tmax ⫽ Kt nom ⫽ K a
16 Tmax
pD31
b
pD31tmax
16K
p(2.0 in.)3(6000 psi)
⫽
⫽ 6200 lb-in.
16(1.52)
D1 ⫽ 2.0 in.
Tmax ⫽
D2 ⫽ 2.4 in.
R ⫽ 0.1 in.
␶allow ⫽ 6000 psi
⬖ Tmax ⬇ 6200 lb-in.
;
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 ⫽ Kt nom ⫽ Ka
16T
pD31
b
p(40 mm)3(120 MPa)
pD31tmax
⫽
⫽ 1.37
16 T
16(1100 N # m)
D2
60 mm
⫽
⫽ 1.5
D1
40 mm
K⫽
D1 ⫽ 40 mm
D2 ⫽ 60 mm
T ⫽ 1100 N ⭈ m
From Fig. (3-48) with
␶allow ⫽ 120 MPa
we get
D2
⫽ 1.5 and K ⫽ 1.37,
D1
R
L 0.10
D1
⬖ Rmin ⬇ 0.10(40 mm) ⫽ 4.0 mm
;
Sec_3.9-3.11.qxd
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Page 339
SECTION 3.11
339
Stress Concentrations in Torsion
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.)
D2/D1
R(in.)
R/D1
K
␶max(psi)
0.7
0.8
0.9
1.43
1.25
1.11
0.15
0.10
0.05
0.214
0.125
0.056
1.20
1.29
1.41
8900
6400
4900
D2 ⫽ 1.0 in.
T ⫽ 500 lb-in.
D1 ⫽ 0.7, 0.8, and 0.9 in.
Full quarter-circular fillet (D2 ⫽ D1 ⫹ 2R)
R⫽
D 2 ⫺ D1
D1
⫽ 0.5 in. ⫺
2
2
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
tmax ⫽ Kt nom ⫽ K a
⫽K
16 T
pD31
16(500 lb-in.)
pD31
b
⫽ 2546
K
D31
NOTE that ␶max gets smaller as D1 gets larger, even
though K is increasing.
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?
Sec_3.9-3.11.qxd
340
9/27/08
1:14 PM
CHAPTER 3
Torsion
Solution 3.11-4
P ⫽ 600 kW
Page 340
Stepped shaft in torsion
D1 ⫽ 100 mm
n ⫽ 400 rpm ␶allow ⫽ 100 MPa
Use the dashed line for a full quarter-circular fillet.
R
L 0.075
D1
R ⬇ 0.075 D1 ⫽ 0.075 (100 mm)
Full quarter-circular fillet
⫽ 7.5 mm
2pnT
POWER P ⫽
( Eq. 3-42 of Section 3.7)
60
P ⫽ watts n ⫽ rpm
T ⫽ Newton meters
60(600 * 103 W)
60P
⫽
⫽ 14,320 N # m
T⫽
2pn
2p(400 rpm)
D2 ⫽ D1 ⫹ 2R ⫽ 100 mm ⫹ 2(7.5 mm) ⫽ 115 mm
⬖ D2 ⬇ 115 mm
;
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 ⫽ Kt nom ⫽ Ka
K⫽
⫽
16T
pD31
b
tmax(pD31)
16T
(100 MPa)(p)(100 mm)3
⫽ 1.37
16(14,320 N # m)
Problem 3.11-5 A stepped shaft (see figure) has diameter D2 ⫽ 1.5 in. and a full quarter-circular fillet. The allowable shear
stress is 15,000 psi and the load T ⫽ 4800 lb-in.
What is the smallest permissible diameter D1?
Sec_3.9-3.11.qxd
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Page 341
SECTION 3.11
Solution 3.11-5
Use trial-and-error. Select trial values of D1
␶allow ⫽ 15,000 psi
T ⫽ 4800 lb-in.
Full quarter-circular fillet D2 ⫽ D1 ⫹ 2R
D1
D 2 ⫺ D1
⫽ 0.75 in. ⫺
2
2
D1 (in.)
R (in.)
R/D1
K
␶max(psi)
1.30
1.35
1.40
0.100
0.075
0.050
0.077
0.056
0.036
1.38
1.41
1.46
15,400
14,000
13,000
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
tmax ⫽ Kt nom ⫽ Ka
16T
pD31
K 16(4800 lb-in.)
⫽ 3c
d
p
D1
⫽ 24,450
341
Stepped shaft in torsion
D2 ⫽ 1.5 in.
R⫽
Stress Concentrations in Torsion
b
K
D31
From the graph, minimum D1 ⬇ 1.31 in.
;
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Page 342