PEC EPE Exam Solid Mechanics Numerical Questions

Click here for MCQs for PEC EPE Exam Solid Mechanics

Other PEC EPE Exam Past Paper Questions

A steel rod of length 2 meters and diameter 10 mm is subjected to a tensile force of 50 kN. Calculate the stress developed in the rod.

Solution:

Area of the rod = π × (Diameter/2)² = π × (10 mm/2)² ≈ 78.54 mm²

Stress = Force / Area of the rod

Stress = 50 kN / (78.54 mm² / 1000) ≈ 636.62 MPa

Answer: Stress developed in the rod ≈ 636.62 MPa

A brass wire of length 1.5 meters and diameter 5 mm is stretched with a force of 500 N. Calculate the strain developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (5 mm/2)² ≈ 19.63 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for brass)

Strain = (500 N × 1.5 m) / (π × (5 mm/2)² × 105 GPa) ≈ 0.00476

Answer: Strain developed in the wire ≈ 0.00476

A steel bar of length 2 meters and diameter 20 mm is subjected to a tensile force of 100 kN. Calculate the elongation of the bar if the modulus of elasticity for steel is 200 GPa.

Solution:

Area of the bar = π × (Diameter/2)² = π × (20 mm/2)² ≈ 314.16 mm²

Elongation = (Force × Length) / (π × (Diameter/2)² × Modulus of elasticity for steel)

Elongation = (100 kN × 2 m) / (π × (20 mm/2)² × 200 GPa) ≈ 0.1592 mm

Answer: Elongation of the bar ≈ 0.1592 mm

A copper wire of length 3 meters and diameter 4 mm is stretched with a force of 300 N. Calculate the stress developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (4 mm/2)² ≈ 12.57 mm²

Stress = Force / Area of the wire

Stress = 300 N / (12.57 mm² / 1000) ≈ 23.87 MPa

Answer: Stress developed in the wire ≈ 23.87 MPa

A steel rod of length 1 meter and diameter 8 mm is subjected to a tensile force of 80 kN. Calculate the strain developed in the rod.

Solution:

Area of the rod = π × (Diameter/2)² = π × (8 mm/2)² ≈ 50.27 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for steel)

Strain = (80 kN × 1 m) / (π × (8 mm/2)² × 200 GPa) ≈ 0.1273

Answer: Strain developed in the rod ≈ 0.1273

A brass wire of length 2 meters and diameter 6 mm is stretched with a force of 200 N. Calculate the elongation of the wire if the modulus of elasticity for brass is 100 GPa.

Solution:

Area of the wire = π × (Diameter/2)² = π × (6 mm/2)² ≈ 28.27 mm²

Elongation = (Force × Length) / (π × (Diameter/2)² × Modulus of elasticity for brass)

Elongation = (200 N × 2 m) / (π × (6 mm/2)² × 100 GPa) ≈ 0.2127 mm

Answer: Elongation of the wire ≈ 0.2127 mm

A steel bar of length 3 meters and diameter 12 mm is subjected to a tensile force of 150 kN. Calculate the stress developed in the bar.

Solution:

Area of the bar = π × (Diameter/2)² = π × (12 mm/2)² ≈ 113.10 mm²

Stress = Force / Area of the bar

Stress = 150 kN / (113.10 mm² / 1000) ≈ 1326.68 MPa

Answer: Stress developed in the bar ≈ 1326.68 MPa

A copper wire of length 4 meters and diameter 5 mm is stretched with a force of 400 N. Calculate the strain developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (5 mm/2)² ≈ 19.63 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for copper)

Strain = (400 N × 4 m) / (π × (5 mm/2)² × 120 GPa) ≈ 0.1064

Answer: Strain developed in the wire ≈ 0.1064

A steel rod of length 2 meters and diameter 16 mm is subjected to a tensile force of 120 kN. Calculate the elongation of the rod if the modulus of elasticity for steel is 210 GPa.

Solution:

Area of the rod = π × (Diameter/2)² = π × (16 mm/2)² ≈ 201.06 mm²

Elongation = (Force × Length) / (π × (Diameter/2)² × Modulus of elasticity for steel)

Elongation = (120 kN × 2 m) / (π × (16 mm/2)² × 210 GPa) ≈ 0.1808 mm

Answer: Elongation of the rod ≈ 0.1808 mm

A brass wire of length 3 meters and diameter 10 mm is stretched with a force of 250 N. Calculate the stress developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (10 mm/2)² ≈ 78.54 mm²

Stress = Force / Area of the wire

Stress = 250 N / (78.54 mm² / 1000) ≈ 3.18 MPa

Answer: Stress developed in the wire ≈ 3.18 MPa

A steel bar of length 1 meter and diameter 6 mm is subjected to a tensile force of 40 kN. Calculate the strain developed in the bar.

Solution:

Area of the bar = π × (Diameter/2)² = π × (6 mm/2)² ≈ 28.27 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for steel)

Strain = (40 kN × 1 m) / (π × (6 mm/2)² × 200 GPa) ≈ 0.0449

Answer: Strain developed in the bar ≈ 0.0449

A copper wire of length 2 meters and diameter 4 mm is stretched with a force of 150 N. Calculate the elongation of the wire if the modulus of elasticity for copper is 120 GPa.

Solution:

Area of the wire = π × (Diameter/2)² = π × (4 mm/2)² ≈ 12.57 mm²

Elongation = (Force × Length) / (π × (Diameter/2)² × Modulus of elasticity for copper)

Elongation = (150 N × 2 m) / (π × (4 mm/2)² × 120 GPa) ≈ 0.7943 mm

Answer: Elongation of the wire ≈ 0.7943 mm

A steel rod of length 3 meters and diameter 8 mm is subjected to a tensile force of 80 kN. Calculate the stress developed in the rod.

Solution:

Area of the rod = π × (Diameter/2)² = π × (8 mm/2)² ≈ 50.27 mm²

Stress = Force / Area of the rod

Stress = 80 kN / (50.27 mm² / 1000) ≈ 1591.74 MPa

Answer: Stress developed in the rod ≈ 1591.74 MPa

A copper wire of length 1 meter and diameter 5 mm is stretched with a force of 200 N. Calculate the strain developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (5 mm/2)² ≈ 19.63 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for copper)

Strain = (200 N × 1 m) / (π × (5 mm/2)² × 120 GPa) ≈ 0.1692

Answer: Strain developed in the wire ≈ 0.1692

A steel bar of length 2 meters and diameter 12 mm is subjected to a tensile force of 100 kN. Calculate the elongation of the bar if the modulus of elasticity for steel is 210 GPa.

Solution:

Area of the bar = π × (Diameter/2)² = π × (12 mm/2)² ≈ 113.10 mm²

Elongation = (Force × Length) / (π × (Diameter/2)² × Modulus of elasticity for steel)

Elongation = (100 kN × 2 m) / (π × (12 mm/2)² × 210 GPa) ≈ 0.2508 mm

Answer: Elongation of the bar ≈ 0.2508 mm

A brass wire of length 3 meters and diameter 8 mm is stretched with a force of 300 N. Calculate the stress developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (8 mm/2)² ≈ 50.27 mm²

Stress = Force / Area of the wire

Stress = 300 N / (50.27 mm² / 1000) ≈ 5.96 MPa

Answer: Stress developed in the wire ≈ 5.96 MPa

A steel rod of length 1 meter and diameter 10 mm is subjected to a tensile force of 60 kN. Calculate the strain developed in the rod.

Solution:

Area of the rod = π × (Diameter/2)² = π × (10 mm/2)² ≈ 78.54 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for steel)

Strain = (60 kN × 1 m) / (π × (10 mm/2)² × 200 GPa) ≈ 0.0956

Answer: Strain developed in the rod ≈ 0.0956

A copper wire of length 2 meters and diameter 6 mm is stretched with a force of 250 N. Calculate the elongation of the wire if the modulus of elasticity for copper is 120 GPa.

Solution:

Area of the wire = π × (Diameter/2)² = π × (6 mm/2)² ≈ 28.27 mm²

Elongation = (Force × Length) / (π × (Diameter/2)² × Modulus of elasticity for copper)

Elongation = (250 N × 2 m) / (π × (6 mm/2)² × 120 GPa) ≈ 0.6635 mm

Answer: Elongation of the wire ≈ 0.6635 mm

A steel bar of length 3 meters and diameter 16 mm is subjected to a tensile force of 120 kN. Calculate the stress developed in the bar.

Solution:

Area of the bar = π × (Diameter/2)² = π × (16 mm/2)² ≈ 201.06 mm²

Stress = Force / Area of the bar

Stress = 120 kN / (201.06 mm² / 1000) ≈ 597.34 MPa

Answer: Stress developed in the bar ≈ 597.34 MPa

A copper wire of length 1 meter and diameter 4 mm is stretched with a force of 100 N. Calculate the strain developed in the wire.

Solution:

Area of the wire = π × (Diameter/2)² = π × (4 mm/2)² ≈ 12.57 mm²

Strain = Change in length / Original length

Change in length = Force × Length / (π × (Diameter/2)² × Modulus of elasticity for copper)

Strain = (100 N × 1 m) / (π × (4 mm/2)² × 120 GPa) ≈ 0.3971

Answer: Strain developed in the wire ≈ 0.3971