MecMovies 4.0 : M6.20: Shear Stresses in End-to-end Shafts

scene 12 of 14

Diagram shows an aluminum shaft A B marked 1 and a brass shaft B C marked 2 with a flange in between at B. Both the shafts are centered on the central axis x. The shaft is supported at A and C. 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 at B of angle phi subscript B, angle of twist. A caption to the bottom left of the shaft 1 reads: phi subscript 1 = phi subscript B. A caption to the bottom left of the shaft 2 reads: phi subscript 2 = negative phi subscript B.

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 $(1)$ :

$ϕ_{1} = \frac{T_{1} L_{1}}{J_{1} G_{1}} = \frac{(424.63 N-m) (1,500 mm) (1,000 mm/m)}{(362,265 {mm}^{4}) (75,000 N/mm^{2})} = + 0.02344 rad = + 1.34 \deg$