Answer:
(a) 2.5 ksi
(b) 0.1075 in
Explanation:
(a)
[tex]E=\frac {\sigma}{\epsilon}[/tex]
Making [tex]\sigma[/tex] the subject then
[tex]\sigma=E\epsilon[/tex]
where [tex]\sigma[/tex] is the stress and [tex]\epsilon[/tex] is the strain
Since strain is given as 0.025% of the length then strain is [tex]\frac {0.025}{100}=0.00025[/tex]
Now substituting E for [tex]10\times 10^{6} psi[/tex] then
[tex]\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi[/tex]
(b)
Stress, [tex]\sigma= \frac {F}{A}[/tex] making A the subject then
[tex]A=\frac {F}{\sigma}[/tex]
[tex]A=\frac {\pi(d_o^{2}-d_i^{2})}{4}[/tex]
where d is the diameter and subscripts o and i denote outer and inner respectively.
We know that [tex]2t=d_o - d_i[/tex] where t is thickness
Now substituting
[tex]\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}[/tex]
[tex]\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4[/tex]
[tex](d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4[/tex]
But the outer diameter is given as 2 in hence
[tex](2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4[/tex]
[tex]2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}[/tex]
[tex]d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in[/tex]
As already mentioned, [tex]2t=d_o - d_i hence t=0.5(d_o - d_i)[/tex]
[tex]t=0.5(2-1.785)=0.1075 in[/tex]
