Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

From Equation (e), the allowable internal torque in brass tube (1) can be calculated as
T
1
4
τ1J1
(12ksi)(1.779801 in. )
= = = 15.533 kip⋅in.
c 2.75 in./2
1
and from Equation (f), the corresponding internal torque in the stainless steel core (2) is
T
2
4
τ 2J2
(14.610 ksi)(0.497010 in. )
= = = 9.682 kip⋅in.
c
1.50 in./2
2
Finally, substitute these results into the equilibrium equation [Equation (a)] to determine
the magnitude of the external torque T that may be applied to the composite shaft
assembly:
T = T1 + T2
= 15.533 kip⋅ in. + 9.682 kip⋅ in. = 25.2 kip⋅ in. Ans.
mecmovies
ExAmpLES
m6.21 A composite shaft consists of a hollow steel [G =
75 GPa] shaft (1) connected to a solid bronze [G = 38 GPa]
shaft (2) at flange B. The outside diameter of shaft (1) is
80 mm, and the inside diameter is 65 mm. The outside diameter
of shaft (2) is 80 mm. The allowable shear stresses for the steel
and bronze materials are 90 MPa and 50 MPa, respectively.
Determine
(a) the maximum torque T that can be applied to
flange B.
(b) the stresses τ 1 and τ 2 developed in the steel and bronze
shafts.
(c) the angle of rotation of flange B.
m6.22 A composite shaft consists of a hollow aluminum
[G = 26 GPa] shaft (1) bonded to a hollow bronze [G =
38 GPa] shaft (2) at flange B. The outside diameter of shaft
(1) is 50 mm, and the inside diameter is 42 mm. The outside
diameter of shaft (2) is 42 mm, and the inside diameter is
30 mm. The allowable shear stresses for the aluminum and
bronze materials are 85 MPa and 100 MPa, respectively.
Determine
(a) the maximum torque T that can be applied to the free
end B of the shaft.
(b) the stresses τ 1 and τ 2 developed in the shafts.
(c) the angle of rotation of end B.
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