A shaft is made of a steel alloy having an allowable
shear stress of If the diameter of the shaft is
1.5 in., determine the maximum torque T that can be
transmitted. What would be the maximum torque if a
1-in.-diameter hole is bored through the shaft? Sketch the
shear-stress distribution along...
•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)
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
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)
*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
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
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