A steel bar 2m in length elongates by 1.5 mm when subject to an axial pUll of 1100kn the bar is 40mm in diameter for a length of 1m 30 mm in diameter for the length of.6m 20mm in diameter for the remaining length.calculate the value of modulus of elasticity?

Question

Answer

Given:

- Length of the steel bar (L) = 2 m
- Elongation of the bar (ΔL) = 1.5 mm
- Axial pull on the bar (P) = 1100 kN
- Diameter of the bar (d1) = 40 mm for the length of 1 m
- Diameter of the bar (d2) = 30 mm for the length of 0.6 m
- Diameter of the bar (d3) = 20 mm for the remaining length

Assumption:

- The steel bar is homogeneous and isotropic.
- The deformation is elastic, i.e., the steel bar regains its original shape after the load is removed.
- The stress is uniformly distributed across the cross-section of the bar.

Calculating the Cross-sectional Areas:

- Cross-sectional area (A1) for the length of 1 m = π * (d1/2)^2 = π * (40/2)^2 = 1256.64 mm^2
- Cross-sectional area (A2) for the length of 0.6 m = π * (d2/2)^2 = π * (30/2)^2 = 706.86 mm^2
- Cross-sectional area (A3) for the remaining length = π * (d3/2)^2 = π * (20/2)^2 = 314.16 mm^2

Calculating the Strains:

- Strain (ε) = ΔL / L

Calculating the Stresses:

- Stress (σ1) for the length of 1 m = P / A1
- Stress (σ2) for the length of 0.6 m = P / A2
- Stress (σ3) for the remaining length = P / A3

Calculating the Elongations:

- Elongation (ΔL1) for the length of 1 m = σ1 / E
- Elongation (ΔL2) for the length of 0.6 m = σ2 / E
- Elongation (ΔL3) for the remaining length = σ3 / E

Calculating the Total Elongation:

- ΔL = ΔL1 + ΔL2 + ΔL3

Calculating the Modulus of Elasticity:

- E = P * L / ΔL

Substituting the Given Values:

- ΔL = 1.5 mm = 0.0015 m
- P = 1100 kN = 1100000 N
- L = 2 m

Substituting the Calculated Values:

- A1 = 1256.64 mm^2 = 0.00125664 m^2
- A2 = 706.86 mm^2 = 0.00070686 m^2
- A3

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