Answer:
d= 2.80inch
Explanation:
Given:
Axial force= 30kip
d= 1inch
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
A) The average normal stress in the concrete and in each bar are; σ_st = 15.52 kpi ; σ_con = 2.25 kpi
B) The required diameter of each bar so that 60% of the axial force is carried by concrete is; 0.94 inches
We are told that;
Column has eight A992 steel reinforcing bars.
Column is subjected to an axial force of 200 kip.
A) Diameter of each bar is 1 inch.
Using equations of equilibrium, we have;
∑fy = 0;
8P_st + P_con = 200 ------(eq 1)
Using compatibility concept, we know from the image attached that;
δ_st = δ_con
where δ_st is change in length of steel and δ_con is change in length of concrete.
Thus;
δ_st = (P_st * L)/(A_st * E_st)
where;
P_st is tensile force of steel
L is length of steel = 3 ft = 36 inches
A_st is area of steel = π/4 * 1² = 0.7854 in²
E_st is young's modulus of steel = 29000 ksi
Similarly;
δ_con = (P_con * L)/(A_con * E_con)
where;
P_con is tensile force of concrete
L is length of concrete = 3 ft = 36 inches
E_con is young's modulus of concrete = 4200 ksi
A_con is area of concrete with diameter of 8 inches = (π/4 * 8²) - 6(π/4 * 1²) = 45.5531 in²
Thus;
From δ_st = δ_con;
(P_st * 36)/(0.7854 * 29000) = (P_con * 36)/(45.5531 * 4200)
Solving this gives;
P_st = 0.119P_con -----(eq 2)
Put 0.119P_con for P_st in eq 1 to get;
8(0.119P_con) + P_con = 200
1.952P_con = 200
P_con = 102.459 kip
Thus; P_st = 12.193 kip
Thus, average normal stress is;
Steel; σ_st = P_st/A_st
σ_st = 12.193/0.7854
σ_st = 15.52 kpi
Concrete; σ_con = P_con/A_con
σ_con = 102.459/45.5531
σ_con = 2.25 kpi
B) Since 60% of the axial force is carried by the concrete. Then it means that 40% will be carried by the steel.
Thus;
P_con = 60% * 200 = 120 kip
P_st = 40% * 200 = 80 kip
Using compatibility again;
δ_st = δ_con
Thus;
(P_st * L)/(A_st * E_st) = (P_con * L)/(A_con * E_con)
6(π/4 * d²)) = (80 * ((π/4 * 8²) - 6(π/4 * d²)) * 4200)/(120 * 29000)
⇒ 4.712d² = 0.09655(50.2655 - 4.712d²)
⇒ 4.712d²/0.09655 = 50.2655 - 4.712d²
⇒ 48.8037d² = 50.2655 - 4.712d²
Solving this gives;
d = 0.94 inches
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