Question

The step shaft is subjected to a torque of 710 lb·in. If the allowable shear stress for the material is τallow = 12 ksi, determine the smallest radius at the junction between the cross sections that can be used to transmit the torque.

Answer 1

The smallest radius at the junction between the cross section that can be used to transmit the torque is 0.167 inches.

Step-by-step explanation:

Torsional shear stress is determined by the following expression:

`\tau = (T\cdot r)/(J)`

Where:

`T` - Torque, measured in
`lbf\cdot in`.

`r` - Radius of the cross section, measured in inches.

`J` - Torsion module, measured in quartic inches.

`\tau` - Torsional shear stress, measured in pounds per square inch.

The radius of the cross section and torsion module are, respectively:

`r = (D)/(2)`

`J = (\pi)/(32)\cdot D^(4)`

Where
`D` is the diameter of the cross section, measured in inches.

Then, the shear stress formula is now expanded and simplified as a function of the cross section diameter:

`\tau = T \cdot (D)/((\pi)/(16)\cdot D^(4) )`

`\tau = (16\cdot T)/(\pi \cdot D^(3))`

In addition, diameter is cleared:

`D^(3) = (16\cdot T)/(\pi \cdot \tau)`

`D = 2\cdot \sqrt[3] {\frac {2\cdot T}{\pi\cdot \tau}}`

If
`T = 710\,lb\cdot in` and
`\tau = 12000\,psi`, then:

`D = \sqrt[3]{(2\cdot (710\,lbf\cdot in))/(\pi \cdot (12000\,psi)) }`

`D \approx 0.335\,in`

`r \approx 0.167\,in`

The smallest radius at the junction between the cross section that can be used to transmit the torque is 0.167 inches.