Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass 70,000 kg. The inner and outer radii of each column are 60 and 70 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column ​

Answer 1

ANSWER

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

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

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

Young’s modulus of steel, Y=2×10¹¹Pa

Total force exerted, F=Mg=50000×9.8N

Stress = Force exerted on a single column

=5000x9.8/4=122500N

Young’s modulus, Y= Stress/Strain

Strain =(F/A)/Y

Where,

Area, A=π(R²−r²)=π((0.6)² −(0.3)²)

Strain =122500/[π((0.6)²−(0.3)² )×2×10¹¹ ] =7.22×10^-7

Hence, the compressional strain of each column is 7.22×10^-7

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