Question

1. A load of 300kN is applied on a short concrete column 250 mm x 250 mm. The column is reinforced by steel bars of total area 5600mm2. If the modulus of elasticity for steel is 15 times that of concrete, find the stress in concrete and steel. If the stress in concrete shall not exceed 4N/mm2, find the area of steel required so that the column may support a load of 600kN.​

Answer 1

Answer:

The stress in concrete $=2.74 \;\text{N}/\;\text{mm}^2$.

The stress in steel $=61.65 \;\text{N}/\;\text{mm}^2$.

The area of steel required $= 8291.9 \;\text{mm}^2$.

Explanation:

Given: Column size $=250\times250$, Modular ratio $m=\frac{Es}{Ec}=15$, Load $P=300Kn$, $Asc=5600mm^2$.

To Find: The stress in concrete and steel, and the area of steel required.​

Solution: Assuming compressive stress in concrete $=fc$

Therefore, the stress in steel under compression, $fsc=1.5\times m \times fc$

Now, $P=0.4fc \times Ac + 0.67fsc \times Asc$

$300 \times 1000=0.4 \times fc \times (250 \times 250) + 0.67 \times (1.5 \times m \times fc) \times 5600$

(Ac is the core area of concrete; here assume full area of section as clear cover and bar sizes not specified)

From this, $fc=2.74 N/mm^2$

Stress in concrete $=2.74 N/mm^2$

Stress in steel $=1.5 \times 15 \times 2.74$

                        $=61.65 N/mm^2$

Next,

$P=600KN$

Here, $fc=4 N/mm^2$ (given)

Therefore, $fsc=1.5 \times m \times fc$

                       $= 1.5 \times 15 \times 4$

                       $=90 N/mm^2$

Now, $P=0.4fc \times Ac + 0.67fsc \times Asc$

$600 \times 1000=0.4 \times 4 \times (250 \times 250) + 0.67 \times 90 \times Asc$

Therefore, $Asc=8291.9 mm^2$

Hence, The stress in concrete is $=2.74 \;\text{N}/\;\text{mm}^2$, the stress in steel is $=61.65 \;\text{N}/\;\text{mm}^2$ and the area of steel required is $= 8291.9 \;\text{mm}^2$.