MecMovies 4.0 : M6.19: Shear Stresses in Coaxial Shafts

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Diagram shows a shaft A B with an aluminum segment on the outside and a brass shaft in the inside. Both the shafts are centered on the central axis x. A torque T is applied at B. The radial and longitudinal lines are shown on the shaft. The deformation of the shaft is shown with a counterclockwise segment of angle phi subscript B, angle of twist.

The rotation angle at $B$ can be found from the geometry of deformation relationship:

$ϕ_{1} = ϕ_{B} ∴ ϕ_{B} = ϕ_{1}$

From the torque-twist relationship for shaft $(1)$ :

$ϕ_{1} = \frac{T_{1} L_{1}}{J_{1} G_{1}} = \frac{(675.7 N - m) (1,000 mm) (1,000 mm/m)}{(308, 102 {mm}^{4}) (26, 000 N/mm^{2})} = + 0.08435 rad = + 4.83 \deg$