TST1501 Assignment 01 2026 | Theory of structures

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄


TST1501: Engineering Science

Assignment 01 — Semester 1, 2026

⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄




TST1501
Module Code:
Engineering Science
Module Name:
Assignment 01
Assignment:
2026
Due Date:
70
Total Marks:




Submitted in partial fulfilment of the requirements for TST1501 — UNISA 2026

,UNISA | TST1501 Assignment 01 – 2026



Question 1: Stress, Strain and Deformation of Structural Members [25 Marks]


1.1 – Strut under Compressive Load [10 Marks]


Question: 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.


Step 1: Identify 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 2: Calculate Cross-Sectional Area


The strut has a circular cross-section, implying:


πd2 π × (30)2 π × 900
A= = = = 706.86 mm2
4 4 4


Step 3: Calculate Compressive Stress

F 60 000
σ= = = 84.88 N/mm2 (compressive)
A 706.86

Therefore the compressive stress is σ = 84.88 MPa .


Step 4: Calculate Strain


Using Hooke’s Law, stress and strain are related by the modulus of elasticity:


σ 84.88
ε= = = 4.042 × 10−4
E 210 × 103

Therefore ε = 4.042 × 10−4 (dimensionless).




Page 2 of 20

, UNISA | TST1501 Assignment 01 – 2026



Step 5: Calculate Change in Length


δL = ε × L = 4.042 × 10−4 × 4000 = 1.617 mm


Therefore the strut shortens by δL = 1.617 mm .

Implementation Insight
Summary of Results:
Cross-sectional area: A = 706.86 mm2
Compressive stress: σ = 84.88 MPa
Strain: ε = 4.042 × 10−4
Change in length: δL = 1.617 mm (shortening)




Page 3 of 20