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

Plan the Solution
A
A free-body diagram cut through the assembly will expose the internal
torques in the tube and the core. Since there are two internal torques and T
only one equilibrium equation, the assembly is statically indeterminate.
The tube and the core are attached to rigid end plates; therefore, as the
assembly twists, both the tube and the core will twist by the same amount.
This relationship will be used to derive a compatibility equation in terms
of the unknown internal torques. Information about the allowable shear stresses will
then be used to determine which of the two components controls the torque capacity of
the composite shaft assembly.
(1)
(2)
T 1
T 2
SOLUTION
Step 1 — Equilibrium Equation: Cut a free-body diagram through the assembly
around rigid end plate A. From this free-body diagram, the following equilibrium
equation can be obtained:
Σ Mx = − T + T1 + T2
= 0
(a)
Since there are three unknowns—T 1 , T 2 , and the external torque T—this assembly is
statically indeterminate.
Step 2 — Geometry of Deformation: The tube and the core are both attached to
rigid end plates. Therefore, when the assembly is twisted, both components must twist
the same amount:
φ
= φ
(b)
1 2
Step 3 — Torque–Twist Relationships: The angles of twists in tube (1) and core (2)
can be expressed as
φ
1
TL 1 1
= and φ2
=
JG
1 1
TL
JG
2 2
2 2
(c)
Step 4 — compatibility Equation: Substitute the torque–twist relationships [Equation
(c)] into the geometry-of-deformation equation [Equation (b)] to obtain the
compatibility equation:
TL
JG
1 1
1 1
TL 2 2
= (d)
JG
2 2
Step 5 — Solve the Equations: Two equations have been derived in terms of the
three unknown torques (T 1 , T 2 , and the external torque T). Additional information is
needed to solve for the unknown torques.
Allowable Shear Stresses
The maximum shear stress in the tube and in the core will be determined by the elastic
torsion formula. Since allowable shear stresses are specified for both components, the
elastic torsion formula can be written for each component and rearranged to solve for the
torque. For brass tube (1),
τ
1
Tc 1 1
τ J
= ∴ T1
=
J
c
1
1 1
1
(e)
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