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Question
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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Solution 1
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 × 1011 Pa
Total force exerted, F = Mg = 50000 × 9.8 N
Stress = Force exerted on a single column = `(50000 xx 9.8)/4 = 122500 N`
Young’s modulus, Y = Stress/Strain
Strain = `(F/A)/Y`
Where
Area, `A =pi(R^2 - r^2) = pi((0.6)^2 - (0.3)^2)`
`"Strain" = 122500/(pi[(0.6)^2 - (0.3)^2]xx2xx10^11) = 7.22 xx10^(-7)`
Hence, the compressional strain of each column is 7.22 × 10–7.
Solution 2
Here total mass to be supported, M = 50,000 kg
∴ Total weight of the structure to be supported = Mg
= 50000 x 9.8 N
Since the weight is to be supported by 4 columns
∴Compressional force on each column (F) is given by
`F = "Mg"/4 = (50000xx9.8)/4 N`
Inner radius of columns, `r_1 = 30 cm = 0.3 m`
Outer radius of column, r_2 = 60 cm = 0.6 m
∴ Area of cross- section of each column is given by
`A = pi(r_2^2 - r_1^2)`
`=pi[(0.6)^2 - (0.3)^2] = 0.27 pi m^2`
Young's Modulus, `Y = 2 xx 10^11` Pa
Compressional strain of each column = ?
∴`Y = "Compressional force/area"/"Compressional Strain"`
`= "F/A"/"Compressional Strain"`
or Compressional strain of each column
`= F/(AY)= (50000xx9.8xx7)/(4xx0.27xx22xx2xx10^11)`
`= 0.722 xx 10^(6)`
∴ Compressional strain of all columns is given by
`= 0.722 xx 10^(-6) xx 4 = 2.88 xx 10^(-6)`
`= 2.88 xx 10^(-6)`
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