MecMovies 4.0 : M6.22: Maximum Torque Applied to Coaxial Shaft

Diagram shows a shaft A B with aluminum segment marked 1 of length 400 millimeters inside which is a brass shaft marked 2. Both the shafts are centered on the central axis x. A torque T is applied at B.

$\frac{τ_{1} L_{1}}{c_{1} G_{1}} = \frac{τ_{2} L_{2}}{c_{2} G_{2}}$

Let's assume that shaft $(1)$ controls. We will assume that the shear stress in shaft $(1)$ is equal to its $85 MPa$ allowable stress.

Based on this assumption, the shear stress in shaft $(2)$ must be:

$τ_{2} = (\frac{c_{2}}{c_{1}}) (\frac{G_{2}}{G_{1}}) (\frac{L_{1}}{L_{2}}) τ_{1} = (\frac{21 mm}{25 mm}) (\frac{38 GPa}{26 GPa}) {(\frac{400 mm}{400 mm})}^{1} (85 MPa) = 104.35 MPa$

Since the $104.35 MPa$ shear stress is greater than the $100 MPa$ allowable shear stress for shaft $(2)$ , our initial assumption is incorrect. Shaft $(2)$ will actually control the capacity of the composite shaft. The shear stress magnitude in shaft $(2)$ will equal its $100 MPa$ allowable stress.