HW 7 - Homework set 7 Solutions PDF

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PROBLEM 3.32

A

5000 ft

The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft. Knowing that the top of the 8-in.-diameter steel drill pipe (G  11.2  10 6psi) rotates through two complete revolutions before the drill bit at B starts to operate, determine the maximum shearing stress caused in the pipe by torsion.

B

SOLUTION TL GJ  T  GJ L Tc GJ c G c     J JL L

 

  2 rev  (2)(2 )  12.566 rad,

c 

1 d  4.0 in. 2

L  5000 ft  60, 000 in.

 

(11.2  106 )(12.566)(4.0)  9.3826  103 psi 60,000

  9.38 ksi 

C

PROBLEM 3.39 B

The solid spindle AB has a diameter ds  1.75 in. and is made of a steel with G  11.2  10 6 psi and  all  12 ksi, while sleeve CD is made of a brass with G  5.6  106 psi and  all  7 ksi. Determine (a) the largest torque T that can be applied at A if the given allowable stresses are not to be exceeded and if the angle of twist of sleeve CD is not to exceed 0.375, (b) the corresponding angle through which end A rotates.

3 in. 8 in.

t5

1 4

in.

D ds

4 in. A

T

SOLUTION Spindle AB:

c  J 

Sleeve CD:

 2

c4 

 2

L  12 in., all  12 ksi, G  11.2 106 psi

(0.875) 4  0.92077 in 4

c1  1.25 in., c2  1.5 in., L  8 in., all  7 ksi J 

(a)

1 (1.75 in.)  0.875 in. 2

 2

c

4 2



 c14  4.1172 in4 ,

G  5.6  106 psi

Largest allowable torque T. Ciriterion: Stress in spindle AB.

Critrion: Stress in sleeve CD.

J c

 

Tc J

T

(0.92077)(12)  12.63 kip  in. 0.875

T

T 

J  4.1172 in 4 (7 ksi)  c2 1.5 in.

T  19.21 kip  in.

Criterion: Angle of twist of sleeve CD.  0.375  6.545 10 3 rad

 

TL JG

T 

JG (4.1172)(5.6  106 )   (6.545 10 3 ) L 8

T  18.86 kip  in. T  12.63 kip  in. 

The largest allowable torque is (b)

Angle of rotation of end A.

 A   A / D   A /B   C / D  

Ti Li L T i J i Gi Ji Gi

 12 8 3   (12.63  10 )  6  6   (0.92077)(11.2  10 ) (4.1172)(5.6  10 ) 

 0.01908 radians

 A  1.093 

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PROBLEM 3.56

250 mm

Solve Prob. 3.55, assuming that the shaft AB is replaced by a hollow shaft of the same outer diameter and 25-mm inner diameter.

C 200 mm B

PROBLEM 3.55 Two solid steel shafts (G  77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. For the loading shown, determine (a) the reaction at each support, (b) the maximum shearing stress in shaft AB, (c) the maximum shearing stress in shaft BC.

38 mm A 1.4 kN · m 50 mm

SOLUTION Shaft AB:

T  TAB , LAB  0.200 m, c2  25 mm  0.025 m c1  12.5 mm  0.0125 m

TAB

Shaft BC:









c 42  c14  (0.0254  0.01254 )  575.24  10 9 m4 2 2 GJ AB (77.2  109 )(575.24  109 )  B   B  222.04  103  B LAB 0.200

J AB 

T  TBC , LBC  0.250 m, c 



c 4  (0.019)4  204.71  10 9 m 2 2  77.2  109 )(204.71  109 ) GJ BC  B  B  63.214  103B LBC 0.250

J BC  TBC



1 d  19 mm  0.019 m 2

Equilibrium of coupling disk:

T  T AB  T BC

1.4  103  222.04  103 B  63.214  103 B

 B  4.9079  103 rad T AB  (222.04  103 )(4.9079  103 )  1.08975  103 N  m TBC  (63.214  103 )(4.9079  103 )  310.25 N  m

(a)

T A  TAB  1090 N  m 

Reactions at supports:

TC  TBC  310 N  m  (b)

Maximum shearing stress in AB:

AB  (c)

TABc 2 (1.08975  103 )(0.025)   47.4  106 Pa J AB 575.243  10 9

 AB  47.4 MPa 

Maximum shearing stress in BC:

BC 

(310.25)(0.019) TBCc   28.8 106 Pa J BC 204.71  10 9

 BC  28.8 MPa ...