Question

A bar 500 mm long & 22 mm in diameter is elongated by 1.2 mm under the effect of
axial pull of 105 kN. Calculate the intensities of stress, strain & modulus of elasticity
of thebar.
Ans--- б = 276.22 N/mm2
e = 0.0024
E = 115.09 x 103 N/mm2
please explain it
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Answer 1

Answer:

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Explanation:

1. A steel rod 500mm long and 20mm10mm in cross-section is subjected to axial pull of 300 KN.If modulus of elasticity is 2105 N/mm2 .Calculate the elongation of the rod. Also calculate strain induced in the bar.

(Ans: 3.75mm,e=7.5E-3)

Answer 2

(i.) Therefore the stress intensity is 276.22 N/mm².

(ii.) Therefore the strain in the bar is 0.24% or 0.0024.

(iii.) Therefore the Modulus of Elasticity of the bar is 115.09 × 10³ N/mm².

Given:

Length of the bar = 500 mm

Diameter of the bar = 22 mm

Change in length or Elongation = ΔL = 1.2 mm

Axial Pull = 105 KN

To Find:

(i.) Intensity of stress

(ii.) strain

(iii.) Modulus of Elasticity

Solution:

The given question can be solved very easily as shown below.

Given that,

Length of the bar ( L )= 500 mm

Diameter of the bar ( D ) = 22 mm

Axial Pull ( P ) = 105 KN

Cross-sectional Area ( A ) = ( π/4 ) × D²

⇒ A = ( π/4 ) × 22² = 121π mm²

(i.) The intensity of Stress:

→ Stress intensity ( σ ) = ( Axial Pull ) / ( Cross-sectional Area )

⇒ σ  = P / A = ( 105 × 10³ ) / 121π = 276.22 N/mm²

Therefore the stress intensity is 276.22 N/mm².

(ii.) Strain:

→ Strain ( ∈ ) = ( Change in length ) / ( Original length ) =  ( ΔL ) / L

⇒ ∈ = 1.2 / 22 = 0.0024 = 0.24%

Therefore the strain in the bar is 0.24% or 0.0024.

(iii.) Modulus of Elasticity:

→ Stress ( σ ) = Strain ( ∈ ) × Modulus of Elasticity ( E )

⇒ E = σ / ∈ = 276.22 / 0.0024 = 1,15,091.67 = 115.09 × 10³ N/mm²

Therefore the Modulus of Elasticity of the bar is 115.09 × 10³ N/mm².

(i.) Therefore the stress intensity is 276.22 N/mm².

(ii.) Therefore the strain in the bar is 0.24% or 0.0024.

(iii.) Therefore the Modulus of Elasticity of the bar is 115.09 × 10³ N/mm².

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