PLAIN AND CIVIL: EXAMPLE 3.1.1. SOLID SHAFT

Thursday 19 March 2020

1. Determine the torque, T, that causes a maximum shearing stress of 100 MPa in the steel cylindrical shaft shown.

2. For the cylindrical shaft shown, determine the maximum shearing stress caused by a torque of magnitude T=1kN-m.

1. Required: Torque, T:

The general formula to be used is as follows, where all are given except the torque, T, and the polar moment of inertia, J so compute for J first.
   $\tau _{max}=\frac{Tc}{J}$

For polar moment of inertia, use the solid shaft:

$J=\frac{\pi }{2}c^{4}$ where c will be taken as the radius to have the maximum shearing stress.

So,

   $J=\frac{\pi }{2}c^{4} = \frac{\pi }{2}\left (18mm \right )^{4}=52488\pi$

Thus,

$T=\frac{\tau _{max}\left ( J \right )}{r}=\frac{100\frac{N}{mm^{2}}\left ( 52488\pi mm^{4} \right )}{18mm}$

${\color{Red} T=916088N-mm \simeq 916.09N-m}$

2. Required: Shearing stress:

   $\tau _{max}=\frac{Tc}{J}$

$\tau _{max}=\frac{1000N-m\left ( 1000mm \right )\left ( 18mm \right )}{52488\pi mm^{4}}$

${\color{Red} \tau _{max}=1077.36\frac{N}{mm^{2}}}$

PLAIN AND CIVIL

Pages

Thursday 19 March 2020

EXAMPLE 3.1.1. SOLID SHAFT

No comments:

Post a Comment