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Exercises · 8.7

Q.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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

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Distributing the structure's weight among the four identical columns gives the compressive force on each; dividing by the column's cross-sectional area gives the stress; dividing stress by mild steel's Young's modulus gives the compressional strain. The printed answer for this question is 2.8×10−6\boxed{2.8\times10^{-6}}.

This is a direct application of Young's modulus, which describes a material's resistance to elastic deformation under compressive (or tensile) stress:

Y=StressStrain=F/AΔL/L⟹Strain=FAYY = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \quad\Longrightarrow\quad \text{Strain} = \frac{F}{AY}

Here FF is the compressive force on one column, AA is its cross-sectional area, and YY is the Young's modulus of steel (Table 8.1: Y=2×1011 N/m2Y = 2\times10^{11}\ \text{N/m}^2).

Given:

  • Mass of the structure, M=50,000 kgM = 50{,}000\ \text{kg}
  • Number of columns, N=4N = 4 (identical, load shared uniformly)
  • Inner radius, ri=30 cm=0.30 mr_i = 30\ \text{cm} = 0.30\ \text{m}
  • Outer radius, ro=60 cm=0.60 mr_o = 60\ \text{cm} = 0.60\ \text{m}
  • g=9.8 m/s2g = 9.8\ \text{m/s}^2; Young's modulus of steel, Y=2×1011 N/m2Y = 2\times10^{11}\ \text{N/m}^2
  1. Total force (weight) supported.

Ftotal=Mg=50,000×9.8=490,000 NF_{\text{total}} = Mg = 50{,}000 \times 9.8 = 490{,}000\ \text{N}

  1. Force on each column. Since the load is distributed uniformly among the four identical columns:

F=FtotalN=490,0004=122,500 NF = \frac{F_{\text{total}}}{N} = \frac{490{,}000}{4} = 122{,}500\ \text{N}

  1. Cross-sectional area of each hollow column. A hollow cylinder's cross-sectional area is the outer circle's area minus the inner circle's:

A=π(ro2−ri2)=π((0.60)2−(0.30)2)=π(0.36−0.09)=0.27π≈0.848 m2A = \pi(r_o^2 - r_i^2) = \pi\big((0.60)^2 - (0.30)^2\big) = \pi(0.36 - 0.09) = 0.27\pi \approx 0.848\ \text{m}^2

  1. Compressional stress on each column. …

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