Problem 1.1 – Stress, Strain and Elongation of a Rod
A rod of 150 cm length and 2.0 cm diameter is subjected to an axial tensile force of 20 kN. The modulus of elasticity of the rod material is:
E = 2 × 105 N/mm²
Determine:
- (i) The stress
- (ii) The strain
- (iii) The elongation of the rod
Given Data
Length, L = 150 cm
Diameter, D = 2.0 cm = 20 mm
Load, P = 20 kN = 20,000 N
Modulus of Elasticity, E = 2 × 105 N/mm²
1. Calculate Cross-Sectional Area
For a circular rod:
A = (Ï€/4) × D²
A = (Ï€/4) × (20)² = 100Ï€ mm²
2. Calculate Stress (σ)
Using the formula:
σ = P / A
σ = 20,000 / 100Ï€ = 63.662 N/mm²
Answer: Stress = 63.662 N/mm²
3. Calculate Strain (e)
Using Hooke’s Law ratio:
e = σ / E
e = 63.662 / (2 × 105) = 0.000318
Answer: Strain = 0.000318
4. Calculate Elongation (ΔL)
Using the strain formula:
e = ΔL / L
Therefore:
ΔL = e × L
ΔL = 0.000318 × 150 = 0.0477 cm
Answer: Elongation = 0.0477 cm
Final Answers Summary
| Quantity | Value | Unit |
|---|---|---|
| Stress | 63.662 | N/mm² |
| Strain | 0.000318 | – |
| Elongation | 0.0477 | cm |
Problem 1.2 – Minimum Diameter of a Steel Wire
We need to determine the minimum diameter of a steel wire required to lift a load of 4000 N without exceeding the allowable stress of 95 N/mm².
Given:
Load, P = 4000 N
Permissible stress, σ = 95 N/mm²
Let the diameter of the wire be:
D (in mm)
For a circular cross-section:
A = (Ï€ / 4) × D²
Using the stress formula:
σ = P / A
Substituting the values:
95 = 4000 / [(Ï€/4) × D²]
Rearranging:
95 = (4000 × 4) / (Ï€ × D²)
D² = (4000 × 4) / (Ï€ × 95) = 53.61
Taking square root:
D = 7.32 mm
Final Answer:
The minimum diameter of the steel wire required is 7.32 mm.
Problem 1.3 – Young’s Modulus of a Brass Rod
Question:
Find the Young’s Modulus of a brass rod of diameter 25 mm and length 250 mm when it is subjected to a tensile load of 50 kN, and the extension produced in the rod is 0.3 mm.
Given:
Diameter, D = 25 mm
Length, L = 250 mm
Load, P = 50 kN = 50,000 N
Extension, ΔL = 0.3 mm
1. Cross-Sectional Area
A = (Ï€/4) × (25)² = 490.87 mm²
2. Stress
σ = P / A = 101.86 N/mm²
3. Strain
e = ΔL / L = 0.0012
4. Young’s Modulus
E = σ / e = 84883.33 N/mm² = 84.883 GN/m²
Final Answer: E = 84.883 GN/m²
Problem 1.4 – Tensile Test on Mild Steel Bar
Question:
A tensile test is performed on a mild steel bar with the following data:
- Diameter of steel bar = 3 cm
- Gauge length = 20 cm
- Load at elastic limit = 250 kN
- Extension at 150 kN load = 0.21 mm
- Maximum load = 380 kN
- Total extension = 60 mm
- Final diameter after fracture = 2.25 cm
Determine:
- (a) Young’s Modulus
- (b) Stress at elastic limit
- (c) Percentage elongation
- (d) Percentage reduction in area
(a) Young’s Modulus
Area = (Ï€/4)(3)² = 7.0685 cm² = 7.0685 × 10⁻⁴ m²
Stress at 150 kN = 212209.9 × 10⁴ N/m²
Strain = 0.21 / (20 × 10) = 0.00105
E = 202.095 GN/m²
(b) Stress at Elastic Limit
σ = (250 × 1000) / (7.0685 × 10⁻⁴) = 353.68 MN/m²
(c) Percentage Elongation
(60 mm / 200 mm) × 100 = 30%
(d) Percentage Reduction in Area
Original area = (Ï€/4)(3)² = 7.0685 cm²
Final area = (Ï€/4)(2.25)² = 3.9765 cm²
Reduction = ((7.0685 − 3.9765) / 7.0685) × 100 = 43.75%
Final Answers:
- Young’s Modulus = 202.095 GN/m²
- Stress at elastic limit = 353.68 MN/m²
- Percentage elongation = 30%
- Percentage decrease in area = 43.75%
Problem 1.5 – Internal Diameter of a Hollow Steel Column
Question
The safe working stress for a hollow steel column carrying an axial load of 2.1 × 10³ kN is 125 MN/m². If the external diameter of the column is 30 cm, determine the internal diameter of the column.
Given
Safe stress, σ = 125 MN/m² = 125 × 10⁶ N/m² = 125 N/mm²
Axial load, P = 2.1 × 10³ kN = 2.1 × 10⁶ N
External diameter, D = 30 cm = 0.30 m
Let internal diameter = d (m)
1. Area of Cross-Section
For a hollow circular section:
A = (Ï€/4) (D² − d²) = (Ï€/4) (0.30² − d²) m²
2. Use Stress Formula
σ = P / A
125 × 10⁶ = 2.1 × 10⁶ / [ (Ï€/4) (0.30² − d²) ]
Rearranging:
125 × 10⁶ = 2.1 × 10⁶ × 4 / [ Ï€ (0.30² − d²) ]
0.09 − d² = 213.9 × 10⁻³ (after simplification)
0.09 − d² = 0.02139
d² = 0.09 − 0.02139 = 0.06861
Therefore:
d = √0.06861 = 0.2619 m = 26.19 cm
Final Answer: Internal diameter of the hollow column is 26.19 cm.
Question
A stepped bar, shown in Fig. 1.6, is subjected to an axially applied compressive load of
35 kN. The upper part of the bar has a diameter of 2 cm and the lower part
has a diameter of 3 cm. Determine the maximum and minimum stresses
produced in the bar.
Given
Axial load, P = 35 kN = 35 × 10³ N
Diameter of upper part, D₁ = 2 cm = 20 mm
Diameter of lower part, D₂ = 3 cm = 30 mm
1. Areas of the Two Sections
Area of upper part:
A₁ = (Ï€/4) D₁² = (Ï€/4)(20)² = 100Ï€ mm²
Area of lower part:
A₂ = (Ï€/4) D₂² = (Ï€/4)(30)² = 225Ï€ mm²
2. Stress in Each Section
Stress = Load / Area. Since the same load passes through both sections, stress will be highest where area is smallest.
Maximum stress (upper part):
σmax = P / A₁ = (35 × 10³) / (100Ï€) = 111.408 N/mm²
Minimum stress (lower part):
σmin = P / A₂ = (35 × 10³) / (225Ï€) = 49.5146 N/mm²
Final Answers:
Maximum stress = 111.408 N/mm² (in the upper, smaller-diameter portion).
Minimum stress = 49.5146 N/mm² (in the lower, larger-diameter portion).