## PEC EPE Exam Solid Mechanics Numerical Questions

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### 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (10 mm/2)ยฒ โ 78.54 mmยฒ

Stress = Force / Area of the rod

Stress = 50 kN / (78.54 mmยฒ / 1000) โ 636.62 MPa

Answer: Stress developed in the rod โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (5 mm/2)ยฒ โ 19.63 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for brass)

Strain = (500 N ร 1.5 m) / (ฯ ร (5 mm/2)ยฒ ร 105 GPa) โ 0.00476

Answer: Strain developed in the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (20 mm/2)ยฒ โ 314.16 mmยฒ

Elongation = (Force ร Length) / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for steel)

Elongation = (100 kN ร 2 m) / (ฯ ร (20 mm/2)ยฒ ร 200 GPa) โ 0.1592 mm

Answer: Elongation of the bar โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (4 mm/2)ยฒ โ 12.57 mmยฒ

Stress = Force / Area of the wire

Stress = 300 N / (12.57 mmยฒ / 1000) โ 23.87 MPa

Answer: Stress developed in the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (8 mm/2)ยฒ โ 50.27 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for steel)

Strain = (80 kN ร 1 m) / (ฯ ร (8 mm/2)ยฒ ร 200 GPa) โ 0.1273

Answer: Strain developed in the rod โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (6 mm/2)ยฒ โ 28.27 mmยฒ

Elongation = (Force ร Length) / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for brass)

Elongation = (200 N ร 2 m) / (ฯ ร (6 mm/2)ยฒ ร 100 GPa) โ 0.2127 mm

Answer: Elongation of the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (12 mm/2)ยฒ โ 113.10 mmยฒ

Stress = Force / Area of the bar

Stress = 150 kN / (113.10 mmยฒ / 1000) โ 1326.68 MPa

Answer: Stress developed in the bar โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (5 mm/2)ยฒ โ 19.63 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for copper)

Strain = (400 N ร 4 m) / (ฯ ร (5 mm/2)ยฒ ร 120 GPa) โ 0.1064

Answer: Strain developed in the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (16 mm/2)ยฒ โ 201.06 mmยฒ

Elongation = (Force ร Length) / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for steel)

Elongation = (120 kN ร 2 m) / (ฯ ร (16 mm/2)ยฒ ร 210 GPa) โ 0.1808 mm

Answer: Elongation of the rod โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (10 mm/2)ยฒ โ 78.54 mmยฒ

Stress = Force / Area of the wire

Stress = 250 N / (78.54 mmยฒ / 1000) โ 3.18 MPa

Answer: Stress developed in the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (6 mm/2)ยฒ โ 28.27 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for steel)

Strain = (40 kN ร 1 m) / (ฯ ร (6 mm/2)ยฒ ร 200 GPa) โ 0.0449

Answer: Strain developed in the bar โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (4 mm/2)ยฒ โ 12.57 mmยฒ

Elongation = (Force ร Length) / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for copper)

Elongation = (150 N ร 2 m) / (ฯ ร (4 mm/2)ยฒ ร 120 GPa) โ 0.7943 mm

Answer: Elongation of the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (8 mm/2)ยฒ โ 50.27 mmยฒ

Stress = Force / Area of the rod

Stress = 80 kN / (50.27 mmยฒ / 1000) โ 1591.74 MPa

Answer: Stress developed in the rod โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (5 mm/2)ยฒ โ 19.63 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for copper)

Strain = (200 N ร 1 m) / (ฯ ร (5 mm/2)ยฒ ร 120 GPa) โ 0.1692

Answer: Strain developed in the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (12 mm/2)ยฒ โ 113.10 mmยฒ

Elongation = (Force ร Length) / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for steel)

Elongation = (100 kN ร 2 m) / (ฯ ร (12 mm/2)ยฒ ร 210 GPa) โ 0.2508 mm

Answer: Elongation of the bar โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (8 mm/2)ยฒ โ 50.27 mmยฒ

Stress = Force / Area of the wire

Stress = 300 N / (50.27 mmยฒ / 1000) โ 5.96 MPa

Answer: Stress developed in the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (10 mm/2)ยฒ โ 78.54 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for steel)

Strain = (60 kN ร 1 m) / (ฯ ร (10 mm/2)ยฒ ร 200 GPa) โ 0.0956

Answer: Strain developed in the rod โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (6 mm/2)ยฒ โ 28.27 mmยฒ

Elongation = (Force ร Length) / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for copper)

Elongation = (250 N ร 2 m) / (ฯ ร (6 mm/2)ยฒ ร 120 GPa) โ 0.6635 mm

Answer: Elongation of the wire โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (16 mm/2)ยฒ โ 201.06 mmยฒ

Stress = Force / Area of the bar

Stress = 120 kN / (201.06 mmยฒ / 1000) โ 597.34 MPa

Answer: Stress developed in the bar โ 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 = ฯ ร (Diameter/2)ยฒ = ฯ ร (4 mm/2)ยฒ โ 12.57 mmยฒ

Strain = Change in length / Original length

Change in length = Force ร Length / (ฯ ร (Diameter/2)ยฒ ร Modulus of elasticity for copper)

Strain = (100 N ร 1 m) / (ฯ ร (4 mm/2)ยฒ ร 120 GPa) โ 0.3971

Answer: Strain developed in the wire โ 0.3971