© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The solid 30-mm-diameter shaft is used to transmit
the torques applied to the gears. Determine the absolute
maximum shear stress oil the shaft.
300 N-m
500 N-m
200 N-m
Internal Torque: As shown on torque diagram.
Maximum Shear Siresi : From the torque diagram
7^ = 400 N • m, Then, applying torsion Formula.
1
400(0.015)
= 75.5MPa
f(0.015-i)
Ans
5-6. The solid 1.25-in.-diameter shaft is used to transmit the
torques applied to the gears. If it is supported by smooth
bearings at A and B, which do not resist torque, determine the
shear stress developed in the shaft at points C and D. Indicate
the shear stress on volume elements located at these points.
6001b'in.
1500 lb-in.
Tc = 1500(0.625)
' J'~ f(0.6254)
Tc _ 600(0.625)
= 1.56 ksi
' Y ~ f(0.6254)
Ans
163
© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The assembly consists of two sections of galvanized
steel pipe connected together using a reducing coupling
at B, The smaller pipe has an outer diameter of 0.75 in. and
an inner diameter of 0.68 in., whereas the larger pipe has an
outer diameter of 1 in. and an inner diameter of 0.86 in. If
the pipe is tightly secured to the wall at C, determine the
maximum shear stress developed in each section of the pipe
when the couple shown is applied to the handles of the
wrench.
15 Ib
15 Ib
7c_
210(0.375)
= 7.82 ksi
J ~ f (0.3754 - 0.344)
Tc
/
210(0.5)
= 2.36 ksi
f (0.54 - 0.434)
Ans
210 Ik-in
=e)T Aa «2/o' iirr)
#0*'"**.
Ans
5-10. The link acts as part of the elevator control for a
small airplane. If the attached aluminum tube has an inner
diameter of 25 mm and a wall thickness of 5 mm, determine
the maximum shear stress in the tube when the cable force
of 600 N is applied to the cables. Also, sketch the shearstress distribution over the cross section.
600 N
600 N
r= 600(0.15) = 90 N - m
t°"
_ Tc _
90(0.0175)
= 14.5 MPa
~ T ~ f[(0.0175)4-(0.0125)4]
Ans
Tp _
90(0.0125)
= 10.3 MPa
J
f[(0.0175)4-(0.0125)4]
165
© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5-34. The gear motor can develop 1/10 tip when it turns at
300 rev/min. If the shaft has a diameter of |in., determine
the maximum shear stress that will be developed in the
start
Internal Torque :
rev fin rad^
rad\n1V.min
rev )
= 10.07T rad/s
60s
w =300-
10
v
r -co ,
- i.75i
Maximum Shear Stress : Applying the torsion formula
Tc
1.751(12)(0.25)
= 856 psi
*(0.25<)
Ans
5-^5. The gear motor can develop 1/10 hp when it turns at
80 rev/min. If the allowable shear stress for the shaft is
ranow = 4 ksij determine the smallest diameter of the shaft
to the nearest |in. that can be used.
Allowable Shear Stress : Applying the torsion formula
Internal Torque:
a = 80
10
rev f"lxttti\ man
= 2.667W rad/s
min V rev ) 60 s
. 6.565(121 (f)
4(103) =
j ,
Ihp
d = 0.4646 in.
r = - = r-^- = 6_5651b.ft
o>
Use d= - in. diameter of shaft
177
Anj
Am
© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle Paver, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5-58. The two shafts are made of A-36 steel. Each has a
diameter of 1 in., and they are supported by bearings at A,
S, and C, which allow free rotation. If the support at D is
fixed, determine the angle of twist of end B when the
torques are applied to the assembly as shown.
Internal Torque : As shown on FED.
Angle of Tw if I :
[-60.0(12X30)+20.0(12)(10)]
= - 0.01778 rad = 0.01778 rad
6
6
-7f£ = -(0.01778) = 0.02667 nd
Since there is no torque applied between Fand-B then
j>t = $f = 0.02667 rad = 1.53°
Arts
The two shafts are made of A-36 steel. Each has a
diameter of 1 hi., and they are supported by bearings at A,
B, and C, which allow free rotation. If the support at D is
fixed, determine the angle of twist of end A when the
torques are applied to the assembly as shown.
Internal Torque : As shown on FBD.
Angle ofTvist:
, TL
JJG
[-60.0(12X30)+20.0(12)(10)]
f(OJ<)CU.O}(10»)
= -0.01778 rad = 0.01778 rad
6
6
<*£ = ^(0.01778) = 0.02667 rad
-40(12X10)
f(0^)(11.0)(10«)
= -0.004445 rad = 0.004445 rad
= 0.02667+0.004445
= 0.03111 rad= 1.78"
Ans
189
© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5~7<>. The steel shaft is made from two segments: AC has
a diameter of 0.5 in, and CB has a diameter of 1 in. If it is
fixed at its ends A and B and subjected to a torque of
500 Ib • ft, determine the maximum shear stress in the shaft.
Grt = 10.S(103) ksi.
;oo ib-ft
lin.
Equilibrium :
£ + T, - 500 = 0
0)
CompaabfliEy condition:
$DIA = ^oia
T^S)
j
f(0.25<)G
Tr
7X8)
3 (12)
f(0.54)G
f(O.S<)G
1408 Tt = 1925
(2)
Solving Eqs. (1) and (2) yields
73 = 601b-ft
T, = 440ft.ft
Tc= 6002X005) =
J
.
-<
7c = 44002X05)
J
5-77. The shaft is made of L2 tool steel, has a diameter of
40 mm, and is fixed at its ends A and B. If it is subjected to
the couple, deterrnine the maximum shear stress in regions
AC and CB.
2kN
2kN
Equilibrium :
-2(0.1) =0
[1]
Compatibility :
JG
JG
[2]
TA = 1.507-,
Solving Eqs. [1] and [2] yields :
•^=0.080kN-m
TJ =0,120 kN-m
Maximum Shear stress .*
T^c
0.12(103)(0.02)
Kc
0.08(10J)(0.02)
Ans
An,
198