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*

## Comments

## 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

The maximum allowable twist is 0.5 deg/meter, that is, 0.5 degree for every

L= 1 meter length.## how is MPa got canceled by mm

how is MPa got canceled by mm?

## Note: 1 MPa = 1 MN/m2 = 1 N

Note: 1 MPa = 1 MN/m

^{2}= 1 N/mm^{2}.