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

For stainless steel core (2),
τ
1
Tc 2 2
τ J
= ∴ T2
=
J
c
2
2 2
2
(f)
Substitute Equations (e) and (f) into the compatibility equation [Equation (d)], and simplify:
T
1
L 1
JG
1 1
= T
2
L 2
JG
2 2
τ J
c
1 1
1
L1
JG
1 1
τ J
=
c
2 2
2
L2
JG
2 2
τ L
cG
1 1
1 1
τ2L2
= (g)
cG
2 2
Note: Equation (g) is simply Equation (6.13) written for tube (1) and core (2). Since the
tube and the core are both the same length, Equation (g) can be simplified to
τ1
cG
1 1
τ2
= (h)
cG
2 2
We cannot know beforehand which component will control the capacity of the torsional
assembly. Let us assume that the maximum shear stress in the stainless steel core (2) will
control; that is, τ 2 = 18 ksi. In that case, the corresponding shear stress in brass tube (1)
can be calculated from Equation (h):
⎛ c1
⎞
τ = τ
⎝
⎜
⎠
⎟ ⎛ ⎝ ⎜
G1
⎞
⎠
⎟ = ⎛ 2.75 in./2⎞
⎝
⎜
⎠
⎟ ⎛ ⎝ ⎜
5,600 ksi ⎞
1 2
(18ksi)
c G
1.50 in./2 12,500 ksi ⎠
⎟ = 14.784 ksi > 12ksi N.G.
2
2
This shear stress exceeds the 12 ksi allowable shear stress for the brass tube. Therefore,
our initial assumption is proved incorrect—the maximum shear stress in the brass tube
actually controls the torque capacity of the assembly.
Equation (h) is then solved for τ 2 , given that the allowable shear stress of the brass
tube is τ 1 = 12 ksi:
⎛ c2
⎞
τ = τ
⎝
⎜
⎠
⎟ ⎛ ⎝ ⎜
G2
⎞
⎠
⎟ = ⎛ 1.50 in./2⎞
(12ksi)
⎝
⎜
⎠
⎟ ⎛ 2 1
c G
2.75 in./2 ⎝ ⎜
1
1
12,500 ksi
5,600 ksi
⎞
⎠
⎟ = 14.610 ksi < 18ksi O.K.
Allowable Torques
On the basis of the compatibility equation, we now know the maximum shear stresses that
will be developed in each of the components. From these shear stresses, we can determine
the torques in each component by using Equations (e) and (f ).
The polar moments of inertia for each component are required. For the brass tube (1),
J
1
π
= [(2.75 in.) − (2.50 in.) ] = 1.779801 in.
32
4 4 4
and for the stainless steel core (2),
J
2
π
= (1.50 in.) = 0.497010 in.
32
4 4
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