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tresset_1 [31]
3 years ago
11

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 l

ength 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.
Engineering
1 answer:
Papessa [141]3 years ago
3 0

Answer:

(a) 2.5 ksi

(b) 0.1075 in

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\times 10^{6} psi then

\sigma=(10\times 10^{6} psi)\times 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}\times 4

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

But the outer diameter is given as 2 in hence

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

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

d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 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

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