MecMovies 4.0 : M15.1: Combined Axial Force and Torsion

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The cross-sectional area of the shaft can be computed as:

$A = \frac{π}{4} [D^{2} - d^{2}]$

$= \frac{π}{4} [{(114 m m)}^{2} - {(102 m m)}^{2}]$

$= 2, 036 m m^{2}$

The axial normal stress is therefore:

$σ_{a x i a l} = \frac{P}{A}$

$= \frac{(40 k N) (1000 N / k N)}{2, 036 m m^{2}}$

$= 19.65 N / m m^{2} = 19.65 M P a$