Based on maximum allowable shearing stress:
$\tau_{max} = \dfrac{16TD}{\pi(D^4 - d^4)}$
$60 = \dfrac{16T(100)}{\pi (100^4 - 80^4)}$
$T = 6\,955 486.14 \, \text{N}\cdot\text{mm}$
$T = 6\,955.5 \, \text{N}\cdot\text{m}$
Based on maximum allowable angle of twist:
$\theta = \dfrac{TL}{JG}$
$0.5^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{T(1000)}{\frac{1}{32}\pi (100^4 - 80^4)(83\,000)}$
$T = 4\,198\,282.97 \, \text{N}\cdot\text{mm}$
$T = 4\,198.28 \, \text{N}\cdot\text{m}$
Use the smaller torque, T = 4 198.28 N·m. answer
Based on maximum allowable
The maximum allowable twist
In reply to Based on maximum allowable by fatigue01
The maximum allowable twist is 0.5 deg/meter, that is, 0.5 degree for every L = 1 meter length.
if u see here 0.5 degree per
In reply to Based on maximum allowable by fatigue01
if u see here 0.5 degree per meter given,but we have to calculate mm unit thats why here 1000 multiplied
how is MPa got canceled by mm
Note: 1 MPa = 1 MN/m2 = 1 N
In reply to how is MPa got canceled by mm by fatigue01
Note: 1 MPa = 1 MN/m2 = 1 N/mm2.