Question

A short reinforced concrete column is subjected to a 750 kips axial compressive load. The Moduli of elasticity of plain concrete and steel are 5.5 and 30 million psi, respectively. The cross-sectional area of the steel is 1.8% of the reinforced concrete. Considering the column is made of composite material and subjected to a load parallel to the steel rebars, calculate the following:

a. the modulus of elasticity of the reinforced concrete.
b. the load carried by each of the steel and plain concrete.
c. the minimum required cross-sectional area of the column, given that the allowable compressive stress of the plain concrete is 5,500 psi and that the allowable compressive stress of plain concrete will be reached before that of steel.

Answer 1

a) 5.94* 10^6 psi

b) Pc = 683.06 kips ( load carried by concrete )

Ps = 0.098 * 683 .06 = 66.94 kips ( load carried by steel )

c) 126.47 In^2

Step-by-step explanation:

A)

Er = Ec * Ac + Es *As ------- ( 1 )

where : As ( cross sectional area of steel )= 1.8% = 0.018

Ac = 1 - 0.018 = 0.982

back to equation 1

modulus of elasticity = ( 5.5 * 10^6 * 0.982 ) + ( 30 * 10^6 * 0.018 )

= 5.94* 10^6 psi

B)

Ps = 0.098 Pc

Given that : Ps + Pc = 750 kips

0.098 Pc + pc = 750 kips

therefore ; Pc = 750 kips / 1.098

Pc = 683.06 kips ( load carried by concrete )

Ps = 0.098 * 683 .06 = 66.94 kips ( load carried by steel )

c) Determine the minimum required cross-sectional area of the column (Ag)

given that allowable compressive force ( Pac ) = 5500 psi

Ac = 0.982 * Ag

Pc = Pac * Ac

683.06 = 5500 * 0.982 * Ag

∴ Ag = 126.47 In^2