UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
TST1501 ASSIGNMENT 01
Semester 1 Assignment 01 — 2026
⋄
Module Code: TST1501
Module Name: Technical Science
Assignment No.: 01
Due Date: 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for TST1501
at the University of South Africa.
,UNISA | TST1501 Assignment 01 — 2026
Question 1: Stress, Strain and Lateral Deformation
1.1 A strut 4 m long and 30 mm diameter is subjected to a compressive force of 60 kN.
Find the stress, the change in length and the strain if E = 210 GPa.
Given information:
• Length: L = 4 m = 4000 mm
• Diameter: d = 30 mm
• Compressive force: F = 60 kN = 60 000 N
• Young’s modulus: E = 210 GPa = 210 × 103 N/mm2
Step 1: Calculate the cross-sectional area.
πd2 π × (30)2 π × 900
A= = = = 706.86 mm2
4 4 4
Step 2: Calculate the compressive stress.
Stress is defined as force per unit area. Since the load is compressive, the stress is compressive
(negative by convention):
F 60 000
σ= = = 84.88 N/mm2 = 84.88 MPa (compressive)
A 706.86
Step 3: Calculate the strain.
Using the relationship between stress, strain and Young’s modulus:
σ 84.88
ε= = = 4.042 × 10−4
E 210 × 103
Step 4: Calculate the change in length.
δL = ε × L = 4.042 × 10−4 × 4000 = 1.617 mm (decrease)
Page 1 of 15
, UNISA | TST1501 Assignment 01 — 2026
Implementation Insight
Summary of Results:
Stress: σ = 84.88 MPa (compressive)
Strain: ε = 4.042 × 10−4
Change in length: δL = 1.617 mm (shortening)
1.2 A bar of steel having a rectangular cross-section 10 cm by 4 cm carries an axial
tensile load of 150 kN. Estimate the decrease in the length of the sides of the cross-
section if Young’s modulus, E, is 200 GN/m2 and Poisson’s ratio, ν, is 0.2.
Given information:
• Cross-section: width b = 10 cm = 100 mm, height h = 4 cm = 40 mm
• Axial tensile load: F = 150 kN = 150 000 N
• Young’s modulus: E = 200 GN/m2 = 200 × 103 N/mm2
• Poisson’s ratio: ν = 0.2
Step 1: Calculate the cross-sectional area.
A = b × h = 100 × 40 = 4000 mm2
Step 2: Calculate the axial (longitudinal) stress.
F 150 000
σx = = = 37.5 N/mm2 = 37.5 MPa (tensile)
A 4000
Step 3: Calculate the axial (longitudinal) strain.
σx 37.5
εx = = = 1.875 × 10−4
E 200 × 103
Step 4: Calculate the lateral strain using Poisson’s ratio.
Poisson’s ratio relates the lateral strain to the longitudinal strain. Under a tensile axial load,
the bar elongates longitudinally and contracts laterally:
Page 2 of 15
College of Science, Engineering and Technology
⋄
TST1501 ASSIGNMENT 01
Semester 1 Assignment 01 — 2026
⋄
Module Code: TST1501
Module Name: Technical Science
Assignment No.: 01
Due Date: 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for TST1501
at the University of South Africa.
,UNISA | TST1501 Assignment 01 — 2026
Question 1: Stress, Strain and Lateral Deformation
1.1 A strut 4 m long and 30 mm diameter is subjected to a compressive force of 60 kN.
Find the stress, the change in length and the strain if E = 210 GPa.
Given information:
• Length: L = 4 m = 4000 mm
• Diameter: d = 30 mm
• Compressive force: F = 60 kN = 60 000 N
• Young’s modulus: E = 210 GPa = 210 × 103 N/mm2
Step 1: Calculate the cross-sectional area.
πd2 π × (30)2 π × 900
A= = = = 706.86 mm2
4 4 4
Step 2: Calculate the compressive stress.
Stress is defined as force per unit area. Since the load is compressive, the stress is compressive
(negative by convention):
F 60 000
σ= = = 84.88 N/mm2 = 84.88 MPa (compressive)
A 706.86
Step 3: Calculate the strain.
Using the relationship between stress, strain and Young’s modulus:
σ 84.88
ε= = = 4.042 × 10−4
E 210 × 103
Step 4: Calculate the change in length.
δL = ε × L = 4.042 × 10−4 × 4000 = 1.617 mm (decrease)
Page 1 of 15
, UNISA | TST1501 Assignment 01 — 2026
Implementation Insight
Summary of Results:
Stress: σ = 84.88 MPa (compressive)
Strain: ε = 4.042 × 10−4
Change in length: δL = 1.617 mm (shortening)
1.2 A bar of steel having a rectangular cross-section 10 cm by 4 cm carries an axial
tensile load of 150 kN. Estimate the decrease in the length of the sides of the cross-
section if Young’s modulus, E, is 200 GN/m2 and Poisson’s ratio, ν, is 0.2.
Given information:
• Cross-section: width b = 10 cm = 100 mm, height h = 4 cm = 40 mm
• Axial tensile load: F = 150 kN = 150 000 N
• Young’s modulus: E = 200 GN/m2 = 200 × 103 N/mm2
• Poisson’s ratio: ν = 0.2
Step 1: Calculate the cross-sectional area.
A = b × h = 100 × 40 = 4000 mm2
Step 2: Calculate the axial (longitudinal) stress.
F 150 000
σx = = = 37.5 N/mm2 = 37.5 MPa (tensile)
A 4000
Step 3: Calculate the axial (longitudinal) strain.
σx 37.5
εx = = = 1.875 × 10−4
E 200 × 103
Step 4: Calculate the lateral strain using Poisson’s ratio.
Poisson’s ratio relates the lateral strain to the longitudinal strain. Under a tensile axial load,
the bar elongates longitudinally and contracts laterally:
Page 2 of 15
