MecMovies 4.0 : M6.23: Shafts With Specified Rotation Angle

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For bronze shaft (2):

L_{2} = 800 mm

G_{2} = 38 GPa = 38,000 {N/mm}^{2}

J_{2} = 3, 106, 311 {mm}^{4}

Rearrange the torque-twist relationship for shaft (2) to solve for T₂:

T_{2} = ϕ_{2} \frac{J_{2} G_{2}}{L_{2}}

= (-3 ˚) (\frac{π rad}{180 ˚}) \times \frac{(3,106,311 {mm}^{4}) (38,000 {N/mm}^{2})}{800 mm}

= -7,725,688 N-m = -7.73 kN-m