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Torsion Torsional Deformations
Problem 3.21 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?
Solution 3.21 d
T T L Copper rod in torsion
d
T T L L 18.0 in. f 3.0 ( 3.0) ¢ 0.0006 rad gmax
dmax rad Find dmax rf
L 2Lgallow
f dmax 180 0.05236 rad
allow ≤ From Eq. (33):
df
2L 0.413 in. — ( 2)(18.0 in.)(0.0006 rad)
0.05236 rad Problem 3.22 A plastic bar of diameter d 50 mm is to be twisted by
torques T (see figure) until the angle of rotation between the ends of the
bar is 5.0°.
If the allowable shear strain in the plastic is 0.012 rad, what is the
minimum permissible length of the bar?
Solution 3.22
d
f Plastic bar in torsion 50 mm
5.0 allow ( 5.0) ¢ 180 0.012 rad ≤ d
T rad L Lmin Find Lmin
From Eq. (33): gmax T 0.08727 rad rf
L df
2L df
2gallow Lmin 182 mm — ( 50 mm)(0.08727 rad)
(2)(0.012 rad) 181 182 CHAPTER 3 Torsion Problem 3.23 A circular aluminum tube subjected to pure torsion
by torques T (see figure) has an outer radius r2 equal to twice the inner
radius r1. T (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 0.15 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?
Solution 3.23
r2
max uallow r2
6 10 rad r1 0.15 ft 10 6 g1 200 Problems 3.23, 3.24, and 3.25 1
(400
2
10 6 10 6 rad) rad — Problem 3.24 A circular steel tube of length L
torsion by torques T (see figure). (b) MINIMUM OUTER RADIUS
From Eq. (35a):
r2 f
L r2u
400 10 6 rad
218.2 10 6 rad in. 0.90 m is loaded in (a) If the inner radius of the tube is r1 40 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 0.0005 rad and the angle
of twist is to be kept at 0.5° by adjusting the torque T, what is the
maximum permissible outer radius (r 2)max? gmax
uallow ( r2 ) min rad in. From Eq. (35b):
1
g
22 r1 ( r2 ) min (a) SHEAR STRAIN AT INNER SURFACE g1 r2 gmax rad
1 ft
( 0.15 ft) ¢
≤¢
≤
180 degree 12 in. 218.2 L Circular aluminum tube 2r1
400 T 1.83 in. — Solution 3.24 Circular steel tube
r2 L 0.90 m r1 40 mm f 0.5 ( 0.5 ) ¢ 0.008727 rad
gmax (b) MAXIMUM OUTER RADIUS r1 rad
≤
180 degree 0.0005 rad (a) SHEAR STRAIN AT INNER SURFACE From Eq. (35a):
gmax g2 r2 f
; r2
L gmax L
f ( r2 ) max ( 0.0005 rad)(900 mm)
0.008727 rad ( r2 ) max 51.6 mm — From Eq. (35b):
gmin g1 g1 f
L r1 388 ( 40 mm)(0.008727 rad)
900 mm
6 10 rad — Problem 3.25 Solve the preceding problem if the length L 50 in.,
the inner radius r1 1.5 in., the angle of twist is 0.6°, and the allowable
shear strain is 0.0004 rad. Solution 3.25 Circular steel tube
r2 L 50 in. r1 1.5 in. f 0.6 r1 ( 0.6 ) ¢ 0.010472 rad
gmax rad
≤
180 degree 0.0004 rad (a) SHEAR STRAIN AT INNER SURFACE
From Eq. (35b):
gmin
g1 g1 r1 314 10 f
L
6 ( 1.5 in.)(0.010472 rad)
50 in.
rad — (b) MAXIMUM OUTER RADIUS
From Eq. (35a):
gmax g2 r2 f
;r
L2 gmaxL
f ( r2 ) max ( 0.0004 rad)(50 in.)
0.010472 rad ( r2 ) max 1.91 in. — 184 CHAPTER 3 Torsion Circular Bars and Tubes
Problem 3.31 A prospector uses a handpowered 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? P W d
b
W Solution 3.31 Handpowered winch Problem 3.32 When drilling a hole in a table leg, a furniture maker
uses a handoperated 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)? SECTION 3.3 Circular Bars and Tubes Problem 3.316 A hollow aluminum tube used in a roof structure has
an outside diameter d2 100 mm and an inside diameter d1 80 mm
(see figure). The tube is 2.5 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 50 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? Solution 3.316 Hollow aluminum tube
d2 80 mm L 2.5 m G d1
d2 100 mm d1 28 GPa max FOR THE SOLID SHAFT: tmax Td2
,
2Ip
¢ d2 d3 ( 100 mm)
( 80 mm) 4
100 mm d L
≤¢
≤
GIp f 2tmaxL
Gd2 f 2 (50 MPa)(2.5 m )
(28 GPa)(100 mm) f 5.12 83.9 mm 590,400 mm3 — (c) RATIO OF WEIGHTS 0.08929 rad Torque is the same.
2IPtmax
d2 2tmax
4
¢ ≤(d
d2 32 2 ( 100 mm) 2 ( 80 mm) 2
(83.9 mm) 2 d2
0.51 — The weight of the tube is 51% of the weight of the
solid shaft, but they resist the same torque. is the same as for tube. T 2
d1 Atube
Asolid Wtube
Wsolid — For the tube: T 2
d2 Wtube
Wsolid (b) DIAMETER OF A SOLID SHAFT d 4
d1 d2 f max 4
d2 4
d1 ) 4 2Iptmax TL
GIp d2 16 2tmax
4
¢
≤¢
≤(d
32 2
d 3 d2 Solve for d 3: d 3 50 MPa T 2Iptmax 16T
d3 tmax (a) ANGLE OF TWIST FOR THE TUBE
Tr
Ip d d1
d2 4
d1 ) 195 196 CHAPTER 3 Torsion Problem 3.317 A circular tube of inner radius r1 and outer radius r2
is subjected to a torque produced by forces P 900 lb (see figure). The
forces have their lines of action at a distance b 5.5 in. from the outside
of the tube.
If the allowable shear stress in the tube is 6300 psi and the inner
radius r1 1.2 in., what is the minimum permissible outer radius r2? P P
P
r2
r1
P
b Solution 3.317 Circular tube in torsion
P SOLUTION OF EQUATION
r2
r1 UNITS: Pounds, Inches
Substitute numerical values: P
b b 2 r2 2 r2 b 6300 psi 4 (900 lb)(5.5 in. r2 )( r2 )
[ ( r4 ) ( 1.2 in.) 4 ]
2 P 900 lb or b 5.5 in. r4 2.07360
2
r2 ( r2 5.5)
or allow r1 6300 psi
1.2 in. Find minimum permissible radius r2 r4
2 0.181891 2
0.181891 r2 0 1.000402 r2 Solve numerically:
TORSION FORMULA
T r2 2P(b r2) 1.3988 in. MINIMUM PERMISSIBLE RADIUS
IP
tmax 2 ( r4
2
Tr2
IP r4 )
1
2P ( b
4
2 ( r2 r2
r2 ) r2
r4 )
1 4P ( b r2 ) r2
( r4 r4 )
2
1 All terms in this equation are known except r2. 1.40 in. — 2.07360 0 SECTION 3.4 Nonuniform Torsion 197 Nonuniform Torsion
Problem 3.41 A stepped shaft ABC consisting of two solid
circular segments is subjected to torques T1 and T2 acting in
opposite directions, as shown in the figure. The larger segment
of the shaft has diameter d1 2.25 in. and length L1 30 in.;
the smaller segment has diameter d2 1.75 in. and length
L2 20 in. The material is steel with shear modulus
G 11 106 psi, and the torques are T1 20,000 lbin.
and T2 8,000 lbin.
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 3.41 T1
d1 d2
B A C L1 L2 Stepped shaft
T1
d1 L1 A 2.25 in. L1 1.75 in. L2 11 T1 20,000 lbin. T2 8,000 lbin. tBC 16 TBC
3
d2
TBC L2
G ( Ip ) BC 20 in. G C T2 fBC 30 in. d2 TBC L2 B d1 SEGMENT BC T2 d2 8,000 lbin.
16(8,000 lbin.)
( 1.75 in.) 3 0.015797 rad 106 psi 7602 psi ( 8,000 lbin.)(20 in.)
(11 106 psi) ¢ 32 ≤ (1.75 in.) 4 (a) MAXIMUM SHEAR STRESS
Segment BC has the maximum stress
tmax SEGMENT AB
TAB T2 T1 12,000 lbin. tAB 16 TAB
`
`
d3
1 16(12,000 lbin.)
( 2.25 in.) 3 fAB TABL1
G ( Ip ) AB ( 12,000 lbin.)(30 in.)
(11 0.013007 rad 106 psi) ¢ 7600 psi (b) ANGLE OF TWIST AT END C
5365 psi 32 ≤ (2.25 in.) 4 C fC AB BC 0.002790 ( 0.013007
rad 0.16 — T2 0.015797) rad 198 CHAPTER 3 Torsion Problem 3.42 A circular tube of outer diameter d3 70 mm and
inner diameter d2 60 mm is welded at the righthand end to a fixed
plate and at the lefthand 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. Tube
Fixed
plate
End
plate Bar
T
A Tube (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. Bar d1
d2
d3 Solution 3.42 Bar and tube
TORQUE Tube T 1000 N m (a) MAXIMUM SHEAR STRESSES Bar Bar: tbar
T
A Tube: ttube 70 mm Ltube d2 32 4
( d3 60 mm G 0.5 m ( Ip ) tube 27 GPa Bar: fbar
Tube: ftube 4
d2 ) A 106 mm4 BAR
40 mm ( Ip ) bar 4
d1 32 Lbar
251.3 1.0 m
103 mm4 G 27 GPa bar fA 1.0848 d1 79.6 MPa T ( d3 2 )
( Ip ) tube — 32.3 MPa — (b) ANGLE OF TWIST AT END A TUBE
d3 16T
d3
1 9.43 TL bar
G ( Ip ) bar 0.1474 rad TL tube
G ( Ip ) tube
tube — 0.1474 0.0171 rad
0.0171 0.1645 rad Problem 3.43 A stepped shaft ABCD consisting of solid circular
segments is subjected to three torques, as shown in the figure. The
torques have magnitudes 12.0 kin., 9.0 kin., and 9.0 kin. The
length of each segment is 24 in. and the diameters of the segments
are 3.0 in., 2.5 in., and 2.0 in. The material is steel with shear
modulus of elasticity G 11.6 103 ksi. 12.0 kin.
3.0 in. Solution 3.43 24 in. 24 in. Stepped shaft
12.0 kin. 9.0 kin.
2.5 in. 9.0 kin.
2.0 in. (a) SHEAR STRESSES
tAB TAB rAB
( Ip ) AB 5660 psi tBC TBC rBC
( Ip ) BC 5870 psi tCD TCD rCD
( Ip ) CD 5730 psi tmax 5870 psi — 3.0 in.
A G C B 103 11.6 rAB 1.25 in. LAB LBC ksi 1.5 in. rBC D rCD LCD 1.0 in. 24 in.
(b) ANGLE OF TWIST AT END D TORQUES
TAB 12.0 9.0 TBC 9.0 TCD fAB 18 kin. POLAR MOMENTS OF INERTIA
( Ip ) AB
( Ip ) BC
( Ip ) CD 32
32
32 (3.0 in.) 4 7.952 in.4 4 4 0.007805 rad TBC LBC
G ( Ip ) BC 0.009711 rad fCD 30 kin. TAB LAB
G ( Ip ) AB fBC 9.0 kin. 9.0 9.0 TCD LCD
G ( Ip ) CD 0.011853 rad (2.0 in.) 4 3.835 in. 1.571 in.4 AB fD
(2.5 in.) D 1.68 BC — CD 9.0 kin.
2.0 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. 9.0 kin.
2.5 in. 0.02937 rad 24 in. 200 CHAPTER 3 Torsion Problem 3.44 A solid circular bar ABC consists of two segments,
as shown in the figure. One segment has diameter d1 50 mm and
length L1 1.25 m; the other segment has diameter d2 40 mm
and length L2 1.0 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 1.5°? (Assume G 80 GPa.) Solutions 3.44 d1 = 50 mm
A allow G L 1 = 1.25 m 30 MPa
1.5 C B
L1 d2 = 40 mm
B 0.02618 rad L 2 = 1.0 m TiLi
a GI f 80 GPa Pi Segment BC has the smaller diameter and hence the
larger stress.
16T
d3 A T L2 Tallow T
C ALLOWABLE TORQUE BASED UPON ANGLE OF TWIST ALLOWABLE TORQUE BASED UPON SHEAR STRESS tmax d2 Bar consisting of two segments
T allow d1
T 3
d2tallow
16 f 32T L1
¢4
G d1 Tallow 3.77 N # m TL1
GIP1
L2
4≤
d2 T L1
¢
G IP1 TL2
GIP2 fallowG
L4 L4
32 ¢ 1
2≤
d1 d2 L2
≤
IP2 348 N m ANGLE OF TWIST GOVERNS
Tallow Problem 3.45 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 lbin., T2 T4 500 lbin., and T3 T5 800 lbin.
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. 348 N m — T2 =
T1 =
1000 lbin. 500 lbin. A B T3 =
T4 =
800 lbin. 500 lbin. C D
d2 = 1.0 in. T5 =
800 lbin. E SECTION 3.4 Solution 3.45 Nonuniform Torsion Hollow tube of monel metal
REQUIRED POLAR MOMENT OF INERTIA BASED UPON d2 1.0 in. Tmaxr
IP tmax 12,000 psi allow 2 /ft allow ALLOWABLE SHEAR STRESS d2 d1 0.16667 /in. IP Tmax ( d2 2 )
tallow 0.05417 in.4 REQUIRED POLAR MOMENT OF INERTIA BASED UPON 0.002909 rad/in. ALLOWABLE ANGLE OF TWIST From Table H2, Appendix H: G 9500 ksi u TORQUES Tmax
GIP Tmax
Guallow IP 0.04704 in.4 SHEAR STRESS GOVERNS
T2 T1
A T3 B T4
D C T5
E IP
T1 1000 lbin. T2 500 lbin. T4 500 lbin. 800 lbin. T5 T3 0.05417 in.4 Required IP 800 lbin.
4
d1 INTERNAL TORQUES 32
4
d2 4
( d2 4
d1 ) 32IP ( 1.0 in.) 4 32(0.05417 in.4 ) 0.4482 in.4 TAB T1 1000 lbin. TBC T1 T2 TCD T1 T2 T3 TDE T1 T2 T3 d1 500 lbin. 0.818 in. — (Maximum permissible inside diameter) 1300 lbin.
T4 800 lbin. Largest torque (absolute value only):
Tmax 1300 lbin. Problem 3.46 A shaft of solid circular cross section consisting of two
segments is shown in the first part of the figure. The lefthand segment
has diameter 80 mm and length 1.2 m; the righthand segment has
diameter 60 mm and length 0.9 m.
Shown in the second part of the figure is a hollow shaft made of the
same material and having the same length. The thickness t of the hollow
shaft is d/10, where d is the outer diameter. Both shafts are subjected to
the same torque.
If the hollow shaft is to have the same torsional stiffness as the solid
shaft, what should be its outer diameter d? 80 mm 1.2 m 60 mm 0.9 m
d 2.1 m d
t=—
10 201 202 CHAPTER 3 Solution 3.46 Torsion Solid and hollow shafts SOLID SHAFT CONSISTING OF TWO SEGMENTS
80 mm TORSIONAL STIFFNESS 60 mm T
f kT Torque T is the same for both shafts. ‹ For equal stiffnesses,
1.2 m f1 © TLi
GIPi G¢ 0.9 m T (1.2 m )
32 32T
( 29,297 m
G 3 ≤ (80 mm) 4 G¢ 98,741 m
T (0.9 m )
32 69,444 m 3 ) ≤ (60 3 3.5569
98,741 d mm) d4 0.0775 m 4 1 2 3.5569 m
d4
36.023 10 6 m4 77.5 mm — 32T
(98,741 m 3 )
G
HOLLOW SHAFT
d = outer diameter
d
t=—
10
2.1 m d0 inner diameter f2 TL
GIp T (2.1 m ) ≤ [ d 4 ( 0.8d ) 4 ]
32
32T
2.1 m
32T 3.5569 m
¢
≤
¢
≤
G 0.5904 d 4
G
d4 UNITS: d G¢ 0.8d meters Problem 3.47 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.
(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.? 8,000 lbin.
19,000 lbin.
4,000 lbin.
A 7,000 lbin.
B
C
D ...
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This note was uploaded on 10/01/2009 for the course MEGR 2144 taught by Professor Sharpe during the Fall '08 term at UNC Charlotte.
 Fall '08
 Sharpe

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