Exercises
1. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10⁻⁵ m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Answer
Given:
Steel wire: Ls = 4.7 m, As = 3.0 × 10⁻⁵ m²
Copper wire: Lc = 3.5 m, Ac = 4.0 × 10⁻⁵ m²
Same load F and same elongation ΔL.
For a wire: ΔL = FL / (AY)
Since elongations are equal and force is the same:
F·Ls / (As·Ys) = F·Lc / (Ac·Yc)
⇒ Ys/Yc = (Ls/As) / (Lc/Ac) = (Ls × Ac) / (Lc × As)
⇒ Ys/Yc = (4.7 × 4.0 × 10⁻⁵) / (3.5 × 3.0 × 10⁻⁵)
→ Ys/Yc = 18.8 / 10.5
→ Ys/Yc ≈ 1.79
2. Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
Answer
From the stress-strain graph (Fig. 8.9), reading the linear (elastic) region:
At strain = 0.002, stress ≈ 3 × 10⁸ N/m²
(a) Young's modulus Y = Stress/Strain = (3 × 10⁸) / 0.002 = 1.5 × 10¹¹ N/m² = 150 GPa
(b) The yield point corresponds to a stress of approximately:
Yield strength ≈ 3 × 10⁸ N/m² = 300 MPa
3. The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young's modulus?
(b) Which of the two is the stronger material?
Answer
(a) Young's modulus is the slope of the stress-strain curve in the linear elastic region. Material A has a steeper slope, so material A has the greater Young's modulus.
(b) The strength is measured by the maximum stress before fracture. From the graph, material B can withstand a higher maximum stress. Therefore, material B is the stronger material.
4. Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young's modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Answer
(a) False. For the same applied stress, rubber undergoes a much larger strain than steel. Therefore Y = stress/strain is much smaller for rubber. Steel has Y ≈ 2 × 10¹¹ Pa while rubber has Y ≈ 10⁷ Pa. Young's modulus of rubber is far less than that of steel.
(b) True. When a coil spring is stretched, each element of the wire undergoes torsion (twisting), which is a form of shearing. Therefore, the extension of a coil spring is determined by the shear modulus (modulus of rigidity) of the material, not Young's modulus.
5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Answer
Given:
Diameter d = 0.25 cm
⇒ radius r = 1.25 × 10⁻³ m
Cross-sectional area A = πr² = π × (1.25 × 10⁻³)² = 4.91 × 10⁻⁶ m²
Lsteel = 1.5 m, Lbrass = 1.0 m
From Fig. 8.11: steel wire (upper) supports 10 kg total; brass wire (lower) supports 6 kg.
Tension in steel wire Ts = 10 × 9.8 = 98 N
Tension in brass wire Tb = 6 × 9.8 = 58.8 N
Ysteel = 2.0 × 10¹¹ Pa, Ybrass = 0.91 × 10¹¹ Pa
Elongation: ΔL = TL / (AY)
Steel wire elongation
ΔLs = (98 × 1.5) / (4.91 × 10⁻⁶ × 2.0 × 10¹¹)
= 147 / (9.82 × 10⁵)
= 1.50 × 10⁻⁴ m = 0.15 mm
Brass wire elongation
ΔLb = (58.8 × 1.0) / (4.91 × 10⁻⁶ × 0.91 × 10¹¹)
= 58.8 / (4.47 × 10⁵)
= 1.32 × 10⁻⁴ m = 0.13 mm
6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Answer
Given:
Edge L = 0.1 m, mass m = 100 kg
⇒ F = 100 × 9.8 = 980 N
G = 25 × 10⁹ Pa, Area A = (0.1)² = 0.01 m²
Shearing stress = F/A = 980 / 0.01 = 9.8 × 10⁴ Pa
Shearing strain = Stress/G = (9.8 × 10⁴) / (25 × 10⁹) = 3.92 × 10⁻⁶
Vertical deflection Δx = Shearing strain × L = 3.92 × 10⁻⁶ × 0.1
= 3.92 × 10⁻⁷ m ≈ 4 × 10⁻⁷ m
7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Answer
Given:
Total weight = 50000 × 9.8 = 4.9 × 10⁵ N
⇒ Load per column = 1.225 × 10⁵ N
r₁ = 0.30 m, r₂ = 0.60 m, Ymild steel = 2.0 × 10¹¹ Pa
Cross-sectional area of each column:
A = π(r₂² - r₁²) = π × (0.36 - 0.09) = π × 0.27 = 0.848 m²
Stress = F/A = 1.225 × 10⁵ / 0.848 = 1.445 × 10⁵ Pa
Compressional strain = Stress/Y = (1.445 × 10⁵) / (2.0 × 10¹¹) = 7.22 × 10⁻⁷
8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Answer
Given:
A = 15.2 × 10⁻³ × 19.1 × 10⁻³ = 2.903 × 10⁻⁴ m²
F = 44500 N, Ycopper = 1.2 × 10¹¹ Pa
Tensile stress = F/A = 44500 / (2.903 × 10⁻⁴) = 1.533 × 10⁸ Pa
Longitudinal strain = Stress/Y = (1.533 × 10⁸) / (1.2 × 10¹¹) = 1.28 × 10⁻³
9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N m⁻², what is the maximum load the cable can support?
Answer
Given:
r = 1.5 × 10⁻² m, Maximum stress = 10⁸ Pa
A = πr² = π × (1.5 × 10⁻²)² = 7.069 × 10⁻⁴ m²
Maximum load = Stress × A = 10⁸ × 7.069 × 10⁻⁴ = 7.07 × 10⁴ N
10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Answer
For a rigid bar supported symmetrically, all wires have the same elongation ΔL (same length L = 2.0 m).
Strain = ΔL/L is the same for all.
For equal tension T: T = Y × A × (ΔL/L)
→ YCu × ACu = YFe × AFe
→ YCu × (π dCu²/4) = YFe × (π dFe²/4)
⇒ (dCu/dFe)² = YFe/YCu = (1.9 × 10¹¹) / (1.2 × 10¹¹) = 1.583
→ dCu/dFe = √1.583 = 1.26
11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Answer
Given:
m = 14.5 kg, L = 1.0 m, ω = 2 × 2π = 4π rad/s
A = 0.065 cm² = 6.5 × 10⁻⁶ m², Ysteel = 2.0 × 10¹¹ Pa
At the lowest point, tension provides centripetal force + supports weight:
T = m(g + ω²L) = 14.5 × (9.8 + (4π)² × 1.0)
= 14.5 × (9.8 + 157.91) = 14.5 × 167.71 = 2431.8 N
ΔL = TL / (AY) = (2431.8 × 1.0) / (6.5 × 10⁻⁶ × 2.0 × 10¹¹)
= 2431.8 / (1.3 × 10⁶) = 1.87 × 10⁻³ m ≈ 1.87 mm
12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Answer
Given:
V₀ = 100.0 L = 0.1 m³, ΔP = 100 × 1.013 × 10⁵ = 1.013 × 10⁷ Pa
|ΔV| = 0.5 L = 5 × 10⁻⁴ m³
B = ΔP / (|ΔV|/V₀) = (1.013 × 10⁷) / (5 × 10⁻⁴ / 0.1)
= (1.013 × 10⁷) / (5 × 10⁻³) = 2.026 × 10⁹ Pa ≈ 2.0 × 10⁹ Pa
Bair at STP (isothermal) = Patm = 1.013 × 10⁵ Pa
Ratio Bwater/Bair = 2.0 × 10⁹ / 1.013 × 10⁵ ≈ 2 × 10⁴
Reason for large ratio: Air molecules are very far apart and easily compressed. Water molecules are closely packed with strong cohesive forces that strongly resist further compression. Thus water requires ~20,000 times more pressure than air for the same fractional volume change.
13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m⁻³?
Answer
Given: P = 80 × 1.013 × 10⁵ = 8.104 × 10⁶ Pa, ρ₀ = 1.03 × 10³ kg/m³, B = 2.2 × 10⁹ Pa
ΔV/V = -ΔP/B = -8.104 × 10⁶ / (2.2 × 10⁹) = -3.684 × 10⁻³
Since mass is constant: ρ = ρ₀ / (1 + ΔV/V) ≈ ρ₀ × (1 - ΔV/V)
= 1.03 × 10³ × (1 + 3.684 × 10⁻³)
= 1.034 × 10³ kg/m³
14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Answer
P = 10 × 1.013 × 10⁵ = 1.013 × 10⁶ Pa, Bglass = 37 × 10⁹ Pa
|ΔV/V| = P/B = (1.013 × 10⁶) / (37 × 10⁹) = 2.74 × 10⁻⁵
15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.
Answer
V = (0.1)³ = 10⁻³ m³, P = 7.0 × 10⁶ Pa, Bcopper = 140 × 10⁹ Pa
ΔV/V = -P/B = -(7.0 × 10⁶) / (140 × 10⁹) = -5.0 × 10⁻⁵
|ΔV| = 5.0 × 10⁻⁵ × 10⁻³ = 5.0 × 10⁻⁸ m³ = 0.05 cm³
16. How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer
ΔV/V = 0.10% = 1.0 × 10⁻³, Bwater = 2.2 × 10⁹ Pa
ΔP = B × (ΔV/V) = 2.2 × 10⁹ × 1.0 × 10⁻³ = 2.2 × 10⁶ Pa = 2.2 MPa
Important Extra Questions
Define stress and strain. What are the SI units of each?
Answer
Stress: The restoring force developed per unit cross-sectional area of a body when subjected to a deforming force.
Stress = F/A; SI unit: N/m² = Pascal (Pa); Dimensional formula: [ML⁻¹T⁻²]
Strain: The fractional change in dimension produced by the deforming force.
Strain = Change in dimension / Original dimension; it is dimensionless (no unit).
Types:
• Longitudinal strain = ΔL/L
• Shearing strain = Δx/L ≈ θ
• Volume strain = ΔV/V
State Hooke's law and define the modulus of elasticity.
Answer
Hooke's Law: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it.
Stress ∝ Strain → Stress = k × Strain
The proportionality constant k is the modulus of elasticity = Stress/Strain.
SI unit: N/m² or Pascal (Pa).
Three moduli:
• Young's modulus Y = (F/A) / (ΔL/L) for longitudinal stress
• Shear modulus G = (F/A) / θ for shearing stress
• Bulk modulus B = -P / (ΔV/V) for hydraulic stress
A wire of Young's modulus 2 × 10¹¹ Pa and length 2 m has a cross-sectional area of 4 × 10⁻⁶ m². What force is required to stretch it by 1 mm?
Answer
Given: Y = 2 × 10¹¹ Pa, L = 2 m, A = 4 × 10⁻⁶ m², ΔL = 10⁻³ m
Y = FL / (A·ΔL) → F = Y·A·ΔL / L
= (2 × 10¹¹ × 4 × 10⁻⁶ × 10⁻³) / 2
= 800 / 2 = 400 N
Distinguish between ductile and brittle materials using the stress-strain curve.
Answer
The stress-strain curve plots stress (y-axis) vs strain (x-axis) for increasing load. Key regions:
• O
⇒ A: Linear, elastic, Hooke's law obeyed (proportional limit).
• A
⇒ B: Non-linear but still elastic (returns to original shape).
• B (Yield point): Permanent deformation begins; stress = yield strength σy.
• B
⇒ D: Plastic region, large strain for small stress increase.
• D: Ultimate tensile strength σu.
• E: Fracture point.
Ductile materials (steel, copper, aluminium): D and E are far apart — large plastic region, material gives warning before fracture, can be drawn into wires.
Brittle materials (glass, cast iron, ceramics): D and E are very close — fracture occurs suddenly with almost no plastic deformation, no warning before breaking.
Why is the I-shaped cross-section preferred for load-bearing beams in construction?
Answer
For a beam loaded at the centre and supported at ends, the sag is:
δ = Wl³ / (4bd³Y)
Sag is proportional to d⁻³ (depth) but only b⁻¹ (breadth). So increasing depth d is much more effective at reducing bending.
However, a very deep but thin rectangular bar buckles sideways under off-centre loads.
The I-shaped section solves this by having:
• Wide top and bottom flanges (large d
⇒ resists bending, large A
⇒ load-bearing surface)
• Thin vertical web (saves material
⇒ reduces weight and cost)
• Flanges prevent lateral buckling
Result: maximum strength with minimum material — making I-beams the standard choice for bridges, buildings, cranes, and structural frameworks.