Four identical hollow cylindrical columns of mild steel support a big structure of a mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Asked by Abhisek | 1 year ago |  94

##### Solution :-

Mass of the big structure, M = 50,000 kg

Total force exerted on the four columns= total weight of the structure=50000×9.8N

The compressional force on each column

=$$\dfrac{Mg}{4}$$ =$$\dfrac{(50000×9.8)}{4N}$$= 122500 N

Therefore, Stress = 122500 N

Young’s modulus of steel, Y=2×1011 Pa

Young’s modulus, Y= $$\dfrac{Stress}{Strain}$$

Strain = $$\dfrac{Young’s\; modulus}{Stress}$$

Strain = (F/A)/Y

Inner radius of the column, r = 30 cm = 0.3 m

Outer radius of the column, R = 60 cm = 0.6 m

Where,

Area, A=π(R2−r2)=π((0.6)2−(0.3)2) = 0.27 π m2

Strain =$$\dfrac{122500}{0.27 \times 3.14×2×10^{11}}$$

=7.22×10−7

Hence, the compressional strain of each column is 7.22×10−7.

Answered by Pragya Singh | 1 year ago

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