Answer:
the normal stress induced by the concrete post 
the normal stress induced by the steel 
Explanation:
Given that:
Modulus for elasticity for concrete post
=
psi
Thermal coefficient for concrete post
= 
Modulus for elasticity of steel bar
= 
Thermal coefficient of steel bar
= 
Change in temperature ΔT = 80°F
Diameter of the steel rood = 7/8-in
Area of the steel rod
= 
= 
= 3.61 in²
Area of concrete parts
= (10)(10) - 
= (100 - 3.61) in²
= 96.39 in²
The total strain developed in the concrete post can be expressed as:
= ![[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7BE_cA_c%7D%2B%5Cfrac%7B1%7D%7BE_sA_s%7D%5DP%3D%28%5Calpha_s-%5Calpha_c%29%28%5Cdelta%20T%29)
= ![[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B%283.6%2A10%5E6%29%2896.39%29%7D%2B%5Cfrac%7B1%7D%7B%2829%2A10%5E6%29%283.61%29%7D%5DP%3D%286.5%2A10%5E%7B-6%7D-5.5%2A10%5E%7B-6%7D%29%2880%29)
= ![[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}](https://tex.z-dn.net/?f=%5B%282.88%2A10%5E%7B-9%7D%29%20%2B%289.55%2A10%5E%7B-9%7D%5DP%20%3D%208.0%2A10%5E%7B-5%7D)
= 

P = 6482.98 lb
Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :



Thus, the normal stress induced by the concrete post 
Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:



Thus, the normal stress induced by the steel 