05 Solutions 46060 5/28/10 1:01 PM Page 214
© 2010 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.
•5–1. A shaft is made of a steel alloy having an allowable
shear stress of tallow = 12 ksi. If the diameter of the shaft is
1.5 in., determine the maximum torque T that can be T
transmitted. What would be the maximum torque T¿ if a T¿
1-in.-diameter hole is bored through the shaft? Sketch the
shear-stress distribution along a radial line in each case.
Allowable Shear Stress: Applying the torsion formula
Tc
tmax = tallow =
J
T (0.75)
12 = p
2 (0.754)
T = 7.95 kip # in. Ans.
Allowable Shear Stress: Applying the torsion formula
T¿c
tmax = tallow =
J
T¿ (0.75)
12 = p
2 (0.754 - 0.54)
T¿ = 6.381 kip # in. = 6.38 kip # in. Ans.
T¿r 6.381(0.5)
tr = 0.5 in = = p = 8.00 ksi
J 2 (0.754 - 0.54)
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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–2. The solid shaft of radius r is subjected to a torque T.
r¿
Determine the radius r¿ of the inner core of the shaft that
resists one-half of the applied torque 1T>22. Solve the
r
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
T
Tc Tr 2T
a) tmax = = p 4 =
J 2 r p r3
(T2 )r¿ T
t = p =
2 (r¿)4 p(r¿)3
a b
r¿ T r¿ 2T
Since t = t ; =
r max p(r¿)3 r pr3
r
r¿ = 1 = 0.841 r Ans.
24
r
2 r¿
b) dT = 2p tr2 dr
L0 L0
r
r¿
2
r
dT = 2p tmax r2 dr
L0 L0 r
r
r¿
a 3 br dr
2
r 2T 2
dT = 2p
L0 L0 r pr
r¿
T 4T
= 4 r3 dr
2 r L0
r
r¿ = 1 = 0.841r Ans.
24
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© 2010 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.
5–3. The solid shaft is fixed to the support at C and
subjected to the torsional loadings shown. Determine the
shear stress at points A and B and sketch the shear stress on 10 kN⭈m
volume elements located at these points. C 75 mm
A B 4 kN⭈m
50 mm 75 mm
The internal torques developed at Cross-sections pass through point B and A are
shown in Fig. a and b, respectively.
p
The polar moment of inertia of the shaft is J = (0.0754) = 49.70(10 - 6) m4. For
2
point B, rB = C = 0.075 Thus,
TB c 4(103)(0.075)
tB = = = 6.036(106) Pa = 6.04 MPa Ans.
J 49.70(10 - 6)
From point A, rA = 0.05 m.
TArA 6(103)(0.05)
tA = = = 6.036(106) Pa = 6.04 MPa. Ans.
J 49.70 (10 - 6)
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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–4. The tube is subjected to a torque of 750 N # m. 75 mm
Determine the amount of this torque that is resisted by the
gray shaded section. Solve the problem two ways: (a) by
using the torsion formula, (b) by finding the resultant of the 100 mm
shear-stress distribution. 750 Nm
25 mm
a) Applying Torsion Formula:
Tc 750(0.1)
tmax = = p = 0.4793 MPa
J 2 (0.14 - 0.0254)
tmax = 0.4793 A 106 B =
T¿(0.1)
p
2 (0.14 - 0.0754)
T¿ = 515 N # m Ans.
b) Integration Method:
t = a b tmax
r
and dA = 2pr dr
c
dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr
0.1m
tmax a br2 dr
r
T¿ = 2ptr2 dr = 2p
L L0.075m c
0.1m
2ptmax
= r3 dr
c L0.075m
2p(0.4793)(106) r4 0.1 m
= c d2
0.1 4 0.075 m
= 515 N # m Ans.
5–5. The copper pipe has an outer diameter of 40 mm and
an inner diameter of 37 mm. If it is tightly secured to the wall A
at A and three torques are applied to it as shown, determine
the absolute maximum shear stress developed in the pipe.
Tmax c 90(0.02) 30 N⭈m
tmax = = p 4 4
J 2 (0.02 - 0.0185 ) 20 N⭈m
= 26.7 MPa Ans..
80 N⭈m
217
© 2010 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.
•5–1. A shaft is made of a steel alloy having an allowable
shear stress of tallow = 12 ksi. If the diameter of the shaft is
1.5 in., determine the maximum torque T that can be T
transmitted. What would be the maximum torque T¿ if a T¿
1-in.-diameter hole is bored through the shaft? Sketch the
shear-stress distribution along a radial line in each case.
Allowable Shear Stress: Applying the torsion formula
Tc
tmax = tallow =
J
T (0.75)
12 = p
2 (0.754)
T = 7.95 kip # in. Ans.
Allowable Shear Stress: Applying the torsion formula
T¿c
tmax = tallow =
J
T¿ (0.75)
12 = p
2 (0.754 - 0.54)
T¿ = 6.381 kip # in. = 6.38 kip # in. Ans.
T¿r 6.381(0.5)
tr = 0.5 in = = p = 8.00 ksi
J 2 (0.754 - 0.54)
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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–2. The solid shaft of radius r is subjected to a torque T.
r¿
Determine the radius r¿ of the inner core of the shaft that
resists one-half of the applied torque 1T>22. Solve the
r
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
T
Tc Tr 2T
a) tmax = = p 4 =
J 2 r p r3
(T2 )r¿ T
t = p =
2 (r¿)4 p(r¿)3
a b
r¿ T r¿ 2T
Since t = t ; =
r max p(r¿)3 r pr3
r
r¿ = 1 = 0.841 r Ans.
24
r
2 r¿
b) dT = 2p tr2 dr
L0 L0
r
r¿
2
r
dT = 2p tmax r2 dr
L0 L0 r
r
r¿
a 3 br dr
2
r 2T 2
dT = 2p
L0 L0 r pr
r¿
T 4T
= 4 r3 dr
2 r L0
r
r¿ = 1 = 0.841r Ans.
24
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© 2010 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.
5–3. The solid shaft is fixed to the support at C and
subjected to the torsional loadings shown. Determine the
shear stress at points A and B and sketch the shear stress on 10 kN⭈m
volume elements located at these points. C 75 mm
A B 4 kN⭈m
50 mm 75 mm
The internal torques developed at Cross-sections pass through point B and A are
shown in Fig. a and b, respectively.
p
The polar moment of inertia of the shaft is J = (0.0754) = 49.70(10 - 6) m4. For
2
point B, rB = C = 0.075 Thus,
TB c 4(103)(0.075)
tB = = = 6.036(106) Pa = 6.04 MPa Ans.
J 49.70(10 - 6)
From point A, rA = 0.05 m.
TArA 6(103)(0.05)
tA = = = 6.036(106) Pa = 6.04 MPa. Ans.
J 49.70 (10 - 6)
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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–4. The tube is subjected to a torque of 750 N # m. 75 mm
Determine the amount of this torque that is resisted by the
gray shaded section. Solve the problem two ways: (a) by
using the torsion formula, (b) by finding the resultant of the 100 mm
shear-stress distribution. 750 Nm
25 mm
a) Applying Torsion Formula:
Tc 750(0.1)
tmax = = p = 0.4793 MPa
J 2 (0.14 - 0.0254)
tmax = 0.4793 A 106 B =
T¿(0.1)
p
2 (0.14 - 0.0754)
T¿ = 515 N # m Ans.
b) Integration Method:
t = a b tmax
r
and dA = 2pr dr
c
dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr
0.1m
tmax a br2 dr
r
T¿ = 2ptr2 dr = 2p
L L0.075m c
0.1m
2ptmax
= r3 dr
c L0.075m
2p(0.4793)(106) r4 0.1 m
= c d2
0.1 4 0.075 m
= 515 N # m Ans.
5–5. The copper pipe has an outer diameter of 40 mm and
an inner diameter of 37 mm. If it is tightly secured to the wall A
at A and three torques are applied to it as shown, determine
the absolute maximum shear stress developed in the pipe.
Tmax c 90(0.02) 30 N⭈m
tmax = = p 4 4
J 2 (0.02 - 0.0185 ) 20 N⭈m
= 26.7 MPa Ans..
80 N⭈m
217
