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

`\tau_(max)  = 142.6` MPa

T = 1536.8 N m

Step-by-step 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)} = (\tau_(max))/(D/2)`

`(T)/(\pi/32 D^4) = (\tau_(max))/(D/2)`

solving for
`\tau_(max)`

`\tau_(max) = (16 T)/(\pi D^3) =(16 * 3.5*10^3)/(\pi 0.05^3)`

`\tau_(max)  = 142.6` MPa

B)

`\tau = 37 MPa = 37 *  10^6` Pa

`D_i = 4.5 cm = 0.045` m

`D_o = 6.5 cm = 0.065` m

`(T)/(J_(hollow)) = (\tau_(max))/(D_o /2)`

`(T)/((\pi/32) (0.065^4 - 0.045^4)) =(37*10^6)/(0.065/2)`

T = 1536.8 N m