Mechanics of Materials by Timothy A. Philpot, P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

SOLUTION:

The cross-sectional area of the stainless tube is 

A=\frac{\pi }{4}\left(D^2-d^2\right)

A=\frac{\pi }{4}\left[\left(60\:mm\right)^2-\left(50\:mm\right)^2\right]

A=863.938\:mm^2

The normal stress in the tube can be expressed as 

\sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa.  Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

P_{max}= \sigma _{allow}A

P_{max}= \left(200\:N/mm^2\right)\left(863.938\:mm^2\right)

P_{max}= 172\:788\:N

P_{max}=172.8\:kN

You can now buy the complete solution manual for Mechanics of Materials: An Integrated Learning System 3rd Edition by Timothy A. Philpot for a very cheap price.

Buy Now button for Mechanics of Materials: An Integrated Learning System 3rd Edition

Advertisements