Question

A) A cross-section of a solid circular rod is subject to a torque of T = 3.5 kNâ‹…m. If the diameter of the rod is D = 5 cm, what is the maximum shear stress? include units.B) The maximum stress in a section of a circular tube subject to a torque is Tmax = 37 MPa . If the inner diameter is Di = 4.5 cm and the outer diameter is Do = 6.5 cm , what is the torque on the section? include units.

Answer 1

Answer:

$\tau_{max} = 142.6$ MPa

T = 1536.8 N m

Explanation:

Given data:

Torque = 3.5 k N m = 3.5*10^3 N.m

Diameter D = 5 cm = 0.05 m

a) from torsional equation we have

$\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}$

$\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}$

solving for $\tau_{max}$

$\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}$

$\tau_{max} = 142.6$ MPa

B)

$\tau = 37 MPa = 37 \times 10^6$ Pa

$D_i = 4.5 cm = 0.045$ m

$D_o = 6.5 cm = 0.065$ m

$\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}$

$\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}$

T = 1536.8 N m