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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.

User Mkorszun
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1 Answer

7 votes

Answer:

(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

User Harley Lang
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