Question

A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in length is 0.025%, determine

(a) the maximum normal stress in the tube,
(b) the minimum wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.

Answer 1

(a) 2.5 ksi

(b) 0.1075 in

Step-by-step explanation:

(a)

`E=\frac {\sigma}{\epsilon}`

Making
`\sigma` the subject then

`\sigma=E\epsilon`

where
`\sigma` is the stress and
`\epsilon` is the strain

Since strain is given as 0.025% of the length then strain is
`\frac {0.025}{100}=0.00025`

Now substituting E for
`10* 10^(6) psi` then

`\sigma=(10* 10^(6) psi)* 0.00025=2500 si= 2.5 ksi`

(b)

Stress,
`\sigma= \frac {F}{A}` making A the subject then

`A=\frac {F}{\sigma}`

`A=\frac {\pi(d_o^(2)-d_i^(2))}{4}`

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that
`2t=d_o - d_i` where t is thickness

Now substituting

`\frac {\pi(d_o^(2)-d_i^(2))}{4}=\frac {1600}{2500}`

`\pi(d_o^(2)-d_i^(2))=\frac {1600}{2500}* 4`

`(d_o^(2)-d_i^(2))=\frac {1600}{2500* \pi}* 4`

But the outer diameter is given as 2 in hence

`(2^(2)-d_i^(2))=\frac {1600}{2500* \pi}* 4`

`2^(2)-(\frac {1600}{2500* \pi}* 4)=d_i^(2)`

`d_i=\sqrt {2^(2)-(\frac {1600}{2500* \pi}* 4)}=1.784692324 in\approx 1.785 in`

As already mentioned,
`2t=d_o - d_i hence t=0.5(d_o - d_i)`

`t=0.5(2-1.785)=0.1075 in`