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Unformatted text preview: Sec_3.53.7.qxd 9/27/08 1:10 PM Page 287 287 SECTION 3.5 Pure Shear Pure Shear
Problem 3.51 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. 3.51, 3.52, and 3.53 Solution 3.51 Hollow aluminum shaft d2 4.0 in. G 4.0 d1 2.0 in. (a) MAXIMUM TENSILE STRESS 0.54°/ft 106 psi max
max MAXIMUM SHEAR STRESS
max Gr (from Eq. 37a) r d2 /2 u max (4.0 max 6280 psi ; 2.0 in. 10 6 6283.2 psi Use the torsion formula tmax
T rad/in. 106 psi)(2.0 in.)(785.40 max. (b) APPLIED TORQUE 1 ft
prad
(0.54°/ft) a
ba
b
12 in. 180 degree
785.40 occurs on a 45° plane and is equal to 10 6 tmaxIP
r IP p
[(4.0 in.)4
32 23.562 in.4 rad/in.)
T (6283.2 psi) (23.562 in.4)
2.0 in.
74,000 lbin. Tr
IP ; (2.0 in.)4] Sec_3.53.7.qxd 290 9/27/08 1:10 PM Page 290 CHAPTER 3 Torsion Solution 3.54 Bar in a testing machine Strain gage at 45°:
max 339 10 d tmax 50 mm T SHEAR STRESS (FROM EQ. 312) 6 500 N m 2 678 max pd 3 p(0.050 m)3 20.372 MPa SHEAR MODULUS SHEAR STRAIN (FROM EQ. 333)
max 16(500 N # m) 16T 10 G 6 tmax
gmax 20.372 MPa
678 * 10 6 30.0 GPa 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1
When twisted by a torque T, the tube develops a maximum normal strain of 170 10 6.
What is the magnitude of the applied torque T ? Problem 3.55 A steel tube (G Solution 3.55 G
max IP 11.5 106 psi d2 170 10 p2
1d
32 2 2.0 in. d1 1.5 in. 4
d1 2 p
[(2.0 in.)4
32 Equate expressions:
Td2
2IP 6 1.07379 in. (1.5 in.)4] 340 max 10 6 SHEAR STRESS (FROM TORSION FORMULA)
tmax
Also, Tr
IP
max Td2
2IP
G max Ggmax SOLVE FOR TORQUE
T SHEAR STRAIN (FROM EQ. 333)
2 1.5 in. Steel tube 4 max ; 2GIPgmax
d2
2(11.5 * 106 psi)(1.07379 in.4)(340 * 10
2.0 in.
4200 lbin. ; 6 ) Sec_3.8.qxd 9/27/08 310 1:13 PM Page 310 CHAPTER 3 Torsion Solution 3.810 Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (344A AND B)
Bar: T1 Ta IP1
b
IP1 + IP2 Ta Tube: T2 100.2783 N # m IP2
b
IP1 + IP2 299.7217 N # m (a) MAXIMUM SHEAR STRESSES
Bar: t1
Tube: t2
d1 25.0 mm G d2 30.0 mm d3 37.5 mm POLAR MOMENTS OF INERTIA
Bar: IP1
Tube: IP2 p
1d 4
32 3 38.3495 * 10
4
d22 T2(d3/2)
IP2 4 m 114.6229 * 10 T1L
GIP1 T2L
GIP2 1.03° f 9 32.7 MPa
49.0 MPa ;
; (b) ANGLE OF ROTATION OF END PLATE 80 GPa p4
d
32 1 T1(d1/2)
IP1 ; 0.017977 rad (c) TORSIONAL STIFFNESS
9 m4 kT T
f Problem 3.811 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 lbin.
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. 344a and b to find the torques in the bar and tube.) 22.3 kN # m ; Tube
A B
T Bar End
plate L d1
d2
d3 Sec_3.8.qxd 9/27/08 1:13 PM Page 311 SECTION 3.8 Statically Indeterminate Torsional Members Solution 3.811 311 Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (344A AND B)
Bar: T1 Ta IPI
b
IPI + IPI Ta Tube: T2 1187.68 lbin. IP2
b
IP1 + IP2 3812.32 lbin. (a) MAXIMUM SHEAR STRESSES
Bar: t1 T1(d1/2)
IP1
T2(d3/2)
IP2 Tube: t2 1790 psi
2690 psi ;
; (b) ANGLE OF ROTATION OF END PLATE
d1 1.50 in. d2 1.75 in. d3 2.25 in. T1L
GIP1 T2L
GIP2 POLAR MOMENTS OF INERTIA 0.354° ; p4
d
32 1 (c) TORSIONAL STIFFNESS G 11.6 Bar: IP1
Tube: IP2 f 6 10 psi p4
1d
32 3 0.497010 in.4
4
d2 2 kT 4 1.595340 in. T
f 809 k in. T Problem 3.812 The composite shaft shown in the figure is
manufactured by shrinkfitting 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. 0.006180015 rad ; Steel sleeve
Brass core
T Assuming that the allowable shear stresses in the brass and steel are
48 MPa and s 80 MPa, respectively, determine the maximum
permissible torque Tmax that may be applied to the shaft. (Hint: Use
Eqs. 344a and b to find the torques.)
b d1 d2
Probs. 3.812 and 3.813 Solution 3.812 Composite shaft shrink fit
d1
d2
GB 40 mm
50 mm
36 GPa GS 80 GPa Allowable stresses:
48 MPa S 80 MPa
B Sec_3.8.qxd 9/27/08 312 1:13 PM CHAPTER 3 Page 312 Torsion BRASS CORE (ONLY) Ta Eq. (344b): TS GSIPS
b
GBIPB + GSIPS
0.762082 T T TB TS (CHECK) ALLOWABLE TORQUE T BASED UPON BRASS CORE
p4
d
32 1 IPB 251.327 * 10 9 TB(d1/2)
2tBIPB
TB
IPB
d1
Substitute numerical values: m4 tB 9047.79 N m2 GBIPB TB STEEL SLEEVE (ONLY) 0.237918 T
2(48 MPa)(251.327 * 10
40 mm T 9 m4) 2535 N m ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE IPS
GSIPS p4
(d
32 2 4
d1 ) 362.265 * 10 9 m4 2 28,981.2 N m tS Eq. (344a): TB TB
Ta TS 2tSIPS
d2 0.762082 T
2(80 MPa)(362.265 * 10
50 mm TS
GBIPB
b
GBIPB + GS IPS TS SUBSTITUTE NUMERICAL VALUES: TORQUES
Total torque: T TS(d2/2)
IPS T 9 m4) 1521 N m STEEL SLEEVE GOVERNS Tmax 1520 N m ; 0.237918 T Problem 3.813 The composite shaft shown in the figure is manufactured by shrinkfitting 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. 344a and b to find
the torques.) Sec_3.8.qxd 314 9/27/08 1:13 PM Page 314 CHAPTER 3 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 3.814
(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 statindet; use Ts
as the internal redundant; see equ. 344a in text
example
Ts 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
T1 a
b
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
+
+
Gb IPb
Gs IPs
Gs IPs
T1 fa L
L
L
T1
T1
4
4
2
+
+
Gb IPb
Gs IPs + Gb IPb
Gs IPs
T1 fa
fa 1
L
1
2
+
T1 a
+
b
4 Gb IPb
GsIPs + Gb IPb
Gs IPs T1, allow 4a
Gb IPb(GsIPs + Gb IPb)Gs IPs
c22
d
2
L Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb Sec_3.8.qxd 9/27/08 1:13 PM Page 315 SECTION 3.8 8a p
b rad
180 NUMERICAL VALUES fa Gs 80 GPa Gb 40 GPa L d1 70 mm d2 p4
d1
32 IPs 2.357 315 13.69 kN m T2, allow 90 mm IPs Statically Indeterminate Torsional Members so T2 for hollow segment controls
(c) ALLOWABLE p
A d 4 d14 B
32 2
T1, allow 9.51 kN m
TORQUE STRESS IN BRASS, T2 IPb 10 6 t d2
2 IPb 4.084 10 6 m4 ; T2, allow t Tb d2
2 IPb
T2 a T2, allow t 2tbIPb
d2 T3 a BASED ON ALLOWABLE SHEAR where from statindet analysis above
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
IPs T3
t T3, allow
where from statindet analysis above d1
2 IPS Ts controls over T2 below also check segment 2 with
brass sleeve over steel shaft
Tb Ts BASED ON ALLOWABLE SHEAR b 6.35 kN m T2, allow T3 s 110 MPa
First check segment 2 with brass sleeve over steel
shaft m4 70 MPa
b
First check hollow segment 1 (brass sleeve only)
T2 TORQUE STRESS IN STEEL,
s IPb (b) ALLOWABLE 3.0 m T3, allow
7.41 kN m 2ts IPs
d1
; controls over T3 above (d) Tmax IF ALL PRECEDING CONDITIONS MUST BE
CONSIDERED GbIPb
b
Gs IPS + Gb IPb
2tb(Gs IPS + Gb IPb)
d2 Gb from (b) above
Tmax 6.35 kN m ; max. shear stress in
hollow brass sleeve in
segment 1 controls overall ...
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This note was uploaded on 11/26/2011 for the course EM 319 taught by Professor Kennethm.liechti during the Spring '08 term at University of Texas at Austin.
 Spring '08
 KennethM.Liechti
 Shear

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