A 25 mm diameter steel rod, 6.0 m long, carries 120 kN in tension. With E = 200 GPa, find the stress, the strain and the elongation.

Start with the thinking

• Area first — every axial answer depends on it.
• Keep units consistent: N and mm give MPa directly.

Step-by-step solution

1. Area — A = πd²/4 = π(25)²/4 = 491 mm²

2. Stress

   $$\sigma =P/A=120,000/491=244MPa$$

3. Strain

   $$\epsilon =\sigma /E=244/200,000=0.00122$$

4. Elongation — δ = PL/(AE) = εL = 0.00122(6,000)

5. Result

   $$\delta =7.33mm$$

Why the other options are there

• δ = 0.733 mm (metres and millimetres mixed)
• σ = 61 MPa (diameter used as area)

Reference: FE Reference Handbook — Mechanics of Materials — Uniaxial loading

A steel member is fully restrained between rigid abutments and heated 40 °C. With α = 11.7 × 10⁻⁶ /°C and E = 200 GPa, what stress develops?

Start with the thinking

• Full restraint means the free thermal strain is cancelled by an equal mechanical strain.
• Length cancels out — the answer does not depend on the span.

Step-by-step solution

1. Free thermal strain — ε_T = αΔT = 11.7×10⁻⁶(40) = 4.68×10⁻⁴

2. Restraint condition

   $${\epsilon }_{m}ech=-{\epsilon }_T$$

3. Stress

   $$\sigma =E\epsilon =200,000(4.68\times 10^{-4})$$

4. Result

   $$\sigma =93.6MPacompression$$

Why the other options are there

• 0 MPa (free expansion assumed)
• 93.6 MPa tension (sign of restraint reversed)

Reference: FE Reference Handbook — Mechanics of Materials — Thermal deformations

A metal specimen has E = 10.4×10⁶ psi and Poisson's ratio ν = 0.26. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 14,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=10.4\times 10^6/[3(1-2\times 0.26)]=7.222\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=10.4\times 10^6/[2(1+0.26)]=4.127\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −14000/7,222,222 = -1.9385 × 10⁻³

Why the other options are there

• K = 2.28×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.32. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 27,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=29.0\times 10^6/[3(1-2\times 0.32)]=26.852\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=29.0\times 10^6/[2(1+0.32)]=10.985\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −27000/26,851,852 = -1.0055 × 10⁻³

Why the other options are there

• K = 5.89×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 16.0×10⁶ psi and Poisson's ratio ν = 0.32. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 8,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=16.0\times 10^6/[3(1-2\times 0.32)]=14.815\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=16.0\times 10^6/[2(1+0.32)]=6.061\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −8000/14,814,815 = -0.5400 × 10⁻³

Why the other options are there

• K = 3.25×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 30.0×10⁶ psi and Poisson's ratio ν = 0.31. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 23,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=30.0\times 10^6/[3(1-2\times 0.31)]=26.316\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=30.0\times 10^6/[2(1+0.31)]=11.450\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −23000/26,315,789 = -0.8740 × 10⁻³

Why the other options are there

• K = 6.17×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.30. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 19,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=29.0\times 10^6/[3(1-2\times 0.30)]=24.167\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=29.0\times 10^6/[2(1+0.30)]=11.154\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −19000/24,166,667 = -0.7862 × 10⁻³

Why the other options are there

• K = 6.04×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 10.4×10⁶ psi and Poisson's ratio ν = 0.33. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 34,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=10.4\times 10^6/[3(1-2\times 0.33)]=10.196\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=10.4\times 10^6/[2(1+0.33)]=3.910\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −34000/10,196,078 = -3.3346 × 10⁻³

Why the other options are there

• K = 2.09×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 29.0×10⁶ psi and Poisson's ratio ν = 0.31. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 9,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=29.0\times 10^6/[3(1-2\times 0.31)]=25.439\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=29.0\times 10^6/[2(1+0.31)]=11.069\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −9000/25,438,596 = -0.3538 × 10⁻³

Why the other options are there

• K = 5.97×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity

A metal specimen has E = 30.0×10⁶ psi and Poisson's ratio ν = 0.27. Compute the bulk (volume) modulus of elasticity, the shear modulus, and the volumetric strain produced by a hydrostatic pressure of 40,000 psi.

Start with the thinking

• The bulk modulus links hydrostatic pressure to volume change; it blows up as ν approaches 0.5 (incompressible).
• E, G, K and ν are not independent — any two fix the others.

Step-by-step solution

1. Formula

   $$K=E/[3(1-2\nu )]$$

2. Substituting

   $$K=30.0\times 10^6/[3(1-2\times 0.27)]=21.739\times 10^{6}psi$$

3. Formula

   $$G=E/[2(1+\nu )]$$

4. Substituting

   $$G=30.0\times 10^6/[2(1+0.27)]=11.811\times 10^{6}psi$$

5. Formula — ΔV/V = −p/K

6. Substituting — ΔV/V = −40000/21,739,130 = -1.8400 × 10⁻³

Why the other options are there

• K = 6.49×10⁶ psi (sign in the bracket flipped)
• K = G (moduli confused)

Reference: FE Reference Handbook — Mechanics of Materials → Bulk (Volume) Modulus of Elasticity