Question

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are designated by size, with the size being the diameter in eights of an inch. For example, a #5 bar is 5/8 inch in diameter. A 12 inch square concrete column is constructed with eight #5 bars. A 90000 pound load is applied to this column. The modulus of elasticity of the reinforcing steel is 29x10^6 psi, while the modulus of elasticity of the concrete is 4.1x10^6 psi.

a. What is the area of the concrete?
b. What is the stress in the steel?
c. What is the stress in the concrete?

Answer 1

Answer:

A) The area of concrete is 141.546 in².

B) The stress in the steel bars is 4Kpsi.

C) The stress in the concrete is 566.4psi.

Explanation:

We can consider the concrete column with the 8 iron bars as a composite material with long fiber parallel to the load. The concrete act as the matrix of the composite. The iron bars act as reinforcement.

The total area of the column is:

$A_c=l\cdot l=12in\cdot12in=144in^2$

The total area of the reinforcement is:

$A_f=n_f\cdot\pi\frac{D^2}{4}=2.454in^2$

Therefore the total area of the matrix (the concrete) is:

$A_c=A_m+A_f \rightarrow A_m=A_c-A_f=141.546in^2$

If we consider this material as a long fiber composite, the matrix volume and fiber volume can be obtained as:

$v_f=\frac{A_f}{A_c}$

$v_m=\frac{A_m}{A_c}$

The stress in the material applied longitudinally to the fibers is:

$\sigma_c=v_f\sigma_f+v_m\sigma_m$

But in this case $\epsilon_c=\epsilon_m=\epsilon_f=\epsilon$, Therefore:

$\sigma_c/\epsilon=v_f\sigma_f/\epsilon+v_m\sigma_m/\epsilon$

$E_c=v_fE_f+v_mE_m=4.5233\cdot10^{6}psi$

We can now calculate the strain of the composite:

$\sigma_c=\frac{F}{A_c}=\epsilon E_c \rightarrow \epsilon=\frac{E_cA_c}{F}=1.3817\cdot10^{-4}$

With the strain we can calculate the stress of the iron bars and the concrete with young's equation:

$\sigma_f=E_f\epsilon=4Kpsi$

$\sigma_m=E_m\epsilon=566.467psi$