Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
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Answer:
Mass of the big structure, M = 50,000 kg
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 m
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 m
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 Pa
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 N
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 N
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 NYoung’s modulus, Y = Stress / Strain
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 NYoung’s modulus, Y = Stress / StrainStrain = (F/A) / Y
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 NYoung’s modulus, Y = Stress / StrainStrain = (F/A) / YWhere,
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 NYoung’s modulus, Y = Stress / StrainStrain = (F/A) / YWhere,Area, A = π (R2 – r2) = π ((0.6)2 – (0.3)2)
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 NYoung’s modulus, Y = Stress / StrainStrain = (F/A) / YWhere,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
Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y = 2 × 1011 PaTotal force exerted, F = Mg = 50000 × 9.8 NStress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 NYoung’s modulus, Y = Stress / StrainStrain = (F/A) / YWhere,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 × 10-7Hence, the compressional strain of each column is 7.22 × 10–7.
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