Numericals on Tapered Rectangular Bar



1- Extension of a Tapered Rectangular Steel Bar

A rectangular steel bar is 2.8 m long and 15 mm thick. It carries an axial tensile load of 40 kN. The bar tapers in width from 75 mm at one end to 30 mm at the other end. If the modulus of elasticity of steel is E = 2 × 105 N/mm², find the extension of the bar.

Given:

  • Length, L = 2.8 m = 2800 mm
  • Thickness, t = 15 mm
  • Load, P = 40 kN = 40,000 N
  • Width at bigger end, a = 75 mm
  • Width at smaller end, b = 30 mm
  • Modulus of Elasticity, E = 2 × 105 N/mm²

Formula Used:

For a tapered rectangular bar,

dL = (P × L) / (E × t × (a − b)) × ln(a/b)

Substituting Values:

dL = (40,000 × 2800) / (2 × 105 × 15 × (75 − 30)) × ln(75/30)

= 0.8296 × 0.9163

Extension, dL = 0.76 mm

✔ Final Answer: 0.76 mm

2- Find Axial Load from Extension in a Tapered Bar

A rectangular steel bar 400 mm long and 10 mm thick extends by 0.21 mm when loaded. The width tapers uniformly from 100 mm to 50 mm. Given E = 2 × 105 N/mm², find the axial load acting on the bar.

Given:

  • Extension, dL = 0.21 mm
  • Length, L = 400 mm
  • Thickness, t = 10 mm
  • Width at bigger end, a = 100 mm
  • Width at smaller end, b = 50 mm
  • Modulus of Elasticity, E = 2 × 105 N/mm²

Formula Used:

dL = (P × L) / (E × t × (a − b)) × ln(a/b)

Substituting Values:

0.21 = (P × 400) / (2 × 105 × 10 × (100 − 50)) × ln(100/50)

= (P × 400) / 1,000,000 × 0.6931

0.21 = 0.000004 × P × 0.6931

P = 75,746 N

✔ Final Answer: 75.746 kN


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