Question

A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of 300 KNm, as shown. The modulus of rigidity G for the steel is 80 GPa. Determine: (a) The maximum shearing stress in the shaft. (b) The shearing stress on a traverse cross section at the inside surface of the shaft (c) The magnitude of the angle of twist for a (L) -2.5 m length.

Answer 1

a.
`\tau=51.55 MPa`

b.
`\tau=42.95MPa`

c.
`\theta=7.67* 10^(-3)` rad.

Step-by-step explanation:

Given:
`D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa`

We know that

`(\tau)/(J)=(T)/(r)=(G\theta)/(L)`

J for hollow shaft
`J=(\pi (D_o^4-D_i^4))/(64)`

(a)

Maximum shear stress
`\tau =(16T)/(\pi Do^3(1-K^4))`

`K=(D_i)/(D_o)`⇒K=0.83

`\tau =(16* 300* 1000)/(\pi* 0.42^3(1-.88^4))`

`\tau=51.55 MPa`

(b)

We know that
`\tau \alpha r`

So
`(\tau_(max))/(\tau)=(R_o)/(r)`

`(51.55)/(\tau)=(210)/(175)`

`\tau=42.95MPa`

(c)

`(\tau_(max))/(R_(max))=(G\theta )/(L)`

`(51.55)/(210)=(80* 10^3\theta )/(2500)`

`\theta=7.67* 10^(-3)` rad.