Answer:
[tex]21678.47223\ lbf-in^2[/tex]
[tex]383.1109\ lbf-in^2[/tex]
Explanation:
d = Diameter of column = 0.5 inch
[tex]A_c[/tex] = Area of concrete = [tex]119.4\ in^2[/tex]
The strain in the system is conserved
[tex]\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s[/tex]
Now
[tex]F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf[/tex]
[tex]F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf[/tex]
Stress is given by
[tex]\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2[/tex]
The stress in the steel is [tex]21678.47223\ lbf-in^2[/tex]
[tex]\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2[/tex]
The stress in the steel is [tex]383.1109\ lbf-in^2[/tex]