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
Based on maximum allowable angle of twist:
where did you get the value of L?
im still confuse about the value of L being 1000
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
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.