Answer:
the normal stress induced by the concrete post [tex]\sigma_c = 67.26 psi[/tex]
the normal stress induced by the steel [tex]\sigma_s = - 1795.84 psi[/tex]
Explanation:
Given that:
Modulus for elasticity for concrete post [tex]E_c[/tex] = [tex]3.6 *10^6[/tex] psi
Thermal coefficient for concrete post [tex]\alpha _c[/tex] = [tex]5.5 *10^{-6}/^0F[/tex]
Modulus for elasticity of steel bar [tex]E_s[/tex] = [tex]29*10^6psi[/tex]
Thermal coefficient of steel bar [tex]\alpha _2[/tex] = [tex]6.5*10^{-6}/^0F[/tex]
Change in temperature ΔT = 80°F
Diameter of the steel rood = 7/8-in
Area of the steel rod [tex]A_s[/tex] = [tex]6(\frac{ \pi}{4} )(d_s)^2[/tex]
= [tex]6(\frac{ \pi}{4} )(\frac{7}{8} )^2[/tex]
= 3.61 in²
Area of concrete parts [tex]A_c[/tex] = (10)(10) - [tex]A_s[/tex]
= (100 - 3.61) in²
= 96.39 in²
The total strain developed in the concrete post can be expressed as:
= [tex][\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)[/tex]
= [tex][\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)[/tex]
= [tex][(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}[/tex]
= [tex]1.234*10^{-8}P = 8.0*10^{-5}[/tex]
[tex]P = \frac {8.0*10^{-5}}{1.234*10^{-8}}[/tex]
P = 6482.98 lb
Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :
[tex]\sigma_c =\frac{P}{A_c}[/tex]
[tex]\sigma_c = \frac{6482.98}{96.39}[/tex]
[tex]\sigma_c = 67.26 psi[/tex]
Thus, the normal stress induced by the concrete post [tex]\sigma_c = 67.26 psi[/tex]
Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:
[tex]\sigma_s = \frac{-P}{A_s}[/tex]
[tex]\sigma_s =\frac{-6482.98}{3.61}[/tex]
[tex]\sigma_s = - 1795.84 psi[/tex]
Thus, the normal stress induced by the steel [tex]\sigma_s = - 1795.84 psi[/tex]
The normal stresses in concrete and steel respectively are;
σ_c = 67.26 psi and σ_s = -1795.84 psi
We are given;
Elastic modulus of Concrete; E_c = 3.6 × 10⁶ psi
Thermal coefficient for concrete post; α_c = 5.5 × 10⁻⁶ /°F
Elastic Modulus of steel; E_s = 29 × 10⁶ psi
Thermal coefficient of steel; α_s = 6.5 × 10⁻⁶ /°F
Change in temperature; ΔT = 80°F
Diameter of the steel bar; d = 7/8-in
Area of steel is; A_s = πd²/4 = π(7/8)²/4
A_s = 3.61 in²
We see the length and width of the post as 10 inches each. Thus;
Area of Concrete is;
A_c = (10 * 10) - A_s
A_c = 100 - 3.61
A_c = 96.39 in²
To calculate the tensile force, we will use the formula;
P[(1/(E_c * A_c)) + (1/(E_s * A_s)) = (α_s - α_s)ΔT
Plugging in the relevant values gives;
P[(1/(3.6 × 10⁶ * 96.39)) + (1/(29 × 10⁶ * 3.61)) = ((6.5 × 10⁻⁶) - (5.5 × 10⁻⁶))80
Solving for P gives us;
P = 6482.98 lb
Thus;
Normal stress in concrete is;
σ_c = P/A_c
σ_c = 6482.98/96.39
σ_c = 67.26 psi
Normal stress in the steel is;
σ_s = -P/A_s
σ_s = -6482.98/3.61
σ_s = -1795.84 psi
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