Longitudinal strain _l is the ratio of change in length Delta L = 0.12 mm to original gauge length L = 200 mm. Lateral strain _d is the ratio of contraction in diameter Delta d = 0.0045 mm to original diameter d = 25 mm.

Longitudinal strain _l is the ratio of change in length Delta L = 0.12 mm to original gauge length L = 200 mm. Lateral strain _d is the ratio of contraction in diameter Delta d = 0.0045 mm to original diameter d = 25 mm.

The circular bar cross-sectional area is A = (π)/(4)d² ≈ 490.87 mm². Tensile stress is axial load P = 60 kN = 60000 N divided by area A, yielding 122.23 N/mm². Young's modulus E is then stress divided by longitudinal strain, giving 2.037 × 10^5 N/mm² (or 203.7 GPa).

The circular bar cross-sectional area is A = (π)/(4)d² ≈ 490.87 mm². Tensile stress is axial load P = 60 kN = 60000 N divided by area A, yielding 122.23 N/mm². Young's modulus E is then stress divided by longitudinal strain, giving 2.037 × 10^5 N/mm² (or 203.7 GPa).

Using standard elastic relationships, modulus of rigidity is G = (E)/(2(1 +)) = (203719)/(2.6) ≈ 78353 N/mm² = 78.35 GPa. Bulk modulus is K = (E)/(3(1 - 2)) = (203719)/(1.2) ≈ 169766 N/mm² = 169.77 GPa.

Using standard elastic relationships, modulus of rigidity is G = (E)/(2(1 +)) = (203719)/(2.6) ≈ 78353 N/mm² = 78.35 GPa. Bulk modulus is K = (E)/(3(1 - 2)) = (203719)/(1.2) ≈ 169766 N/mm² = 169.77 GPa.

A bar of 25mm diameter is tested in tension it is observed that when a load of 60 KN applied the extension measured over a gauges length of 200 mm is 0.12 mm and contraction in diameter is 0.0045 mm find poisson s ratio and elastic constant E,G,K

Answer: Poisson's ratio $\mu = 0.3$ , Young's modulus $E = 2.037 \times 10^5\text{ N/mm}^2$ ( $203.7\text{ GPa}$ ), Shear modulus $G = 7.835 \times 10^4\text{ N/mm}^2$ ( $78.35\text{ GPa}$ ), and Bulk modulus $K = 1.698 \times 10^5\text{ N/mm}^2$ ( $169.77\text{ GPa}$ ).

Step-by-step solution

Step 1: Calculate longitudinal and lateral strains

Longitudinal strain $\varepsilon_l$ is the ratio of change in length $\Delta L = 0.12\text{ mm}$ to original gauge length $L = 200\text{ mm}$ . Lateral strain $\varepsilon_d$ is the ratio of contraction in diameter $\Delta d = 0.0045\text{ mm}$ to original diameter $d = 25\text{ mm}$ .

Step 2: Determine Poisson's ratio

Poisson's ratio $\mu$ is defined as the ratio of lateral strain to longitudinal strain under uniaxial tension. Dividing $1.8 \times 10^{-4}$ by $6 \times 10^{-4}$ gives $0.3$ .

Step 3: Calculate cross-sectional area, tensile stress, and Young's modulus

The circular bar cross-sectional area is $A = \frac{\pi}{4}d^2 \approx 490.87\text{ mm}^2$ . Tensile stress $\sigma$ is axial load $P = 60\text{ kN} = 60000\text{ N}$ divided by area $A$ , yielding $122.23\text{ N/mm}^2$ . Young's modulus $E$ is then stress divided by longitudinal strain, giving $2.037 \times 10^5\text{ N/mm}^2$ (or $203.7\text{ GPa}$ ).

Step 4: Compute Modulus of Rigidity and Bulk Modulus

Using standard elastic relationships, modulus of rigidity is $G = \frac{E}{2(1 + \mu)} = \frac{203719}{2.6} \approx 78353\text{ N/mm}^2 = 78.35\text{ GPa}$ . Bulk modulus is $K = \frac{E}{3(1 - 2\mu)} = \frac{203719}{1.2} \approx 169766\text{ N/mm}^2 = 169.77\text{ GPa}$ .

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