Introduction
Solids, though seemingly rigid, can be deformed under the application of external forces. This chapter explores the mechanical properties of solids, focusing on elasticity and plasticity.
Elastic Behavior
Stress and Strain: When a force is applied to a solid, it undergoes deformation. Stress is the restoring force per unit area, while strain is the relative change in size or shape.
Hooke’s Law: Hooke’s law states that within the elastic limit, stress is directly proportional to strain.
Stress-Strain Curve: This curve graphically represents the relationship between stress and strain. It can be divided into three regions: elastic region, plastic region, and fracture region.
Elastic Moduli
Young’s Modulus: Measures the resistance of a material to longitudinal stress.
Bulk Modulus: Measures the resistance of a material to volumetric stress.
Shear Modulus: Measures the resistance of a material to shear stress.
Applications of Elasticity
Elasticity in Structures: Understanding elastic properties is crucial in designing structures like bridges, buildings, and machines.
Materials Science: The study of elastic behavior helps in developing new materials with desired properties.
Plastic Behavior
Beyond the elastic limit, a solid undergoes plastic deformation, where it doesn’t return to its original shape after the removal of the deforming force. This property is essential in shaping metals and other materials.
Factors Affecting Elastic Behavior
Nature of the material: Different materials have different elastic properties.
Temperature: Elastic properties can change with temperature.
Impurities: Impurities can affect the elastic behavior of a material.
By understanding the mechanical properties of solids, we can design and engineer structures and devices that can withstand various stresses and strains.
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?
Ans : Let’s denote:
* The Young’s modulus of steel as Y_steel
* The Young’s modulus of copper as Y_copper
* The length of the steel wire as L_steel = 4.7 m
* The length of the copper wire as L_copper = 3.5 m
* The cross-sectional area of the steel wire as A_steel = 3.0 × 10⁻⁵ m²
* The cross-sectional area of the copper wire as A_copper = 4.0 × 10⁻⁵ m²
We know that Young’s modulus is defined as:
Y = (Stress) / (Strain) = (F/A) / (ΔL/L)
Where:
* F is the applied force
* A is the cross-sectional area
* ΔL is the change in length
* L is the original length
Since both wires are subjected to the same load (force) and experience the same elongation (ΔL), we can equate the stress-strain ratios for both wires:
(F/A_steel) / (ΔL/L_steel) = (F/A_copper) / (ΔL/L_copper)
Simplifying and canceling common terms:
(Y_steel * A_steel) / L_steel = (Y_copper * A_copper) / L_copper
Now, we can solve for the ratio of Young’s moduli:
Y_steel / Y_copper = (A_copper * L_steel) / (A_steel * L_copper)
Substituting the given values:
Y_steel / Y_copper = (4.0 × 10⁻⁵ m² * 4.7 m) / (3.0 × 10⁻⁵ m² * 3.5 m)
≈ 1.8
Therefore, the ratio of the Young’s modulus of steel to that of copper is approximately 1.8.
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?
Ans : (a) Young’s modulus (Y) is defined as the ratio of stress to strain within the elastic limit. From the given stress-strain curve, we can calculate Y using the slope of the linear region
Y = (Stress) / (Strain)
= (200 – 100) × 10⁶ N/m² / (0.002 – 0.001)
= 100 × 10⁶ N/m²
= 1 × 10¹¹ N/m²
(b) The yield strength of a material is the stress at which it begins to deform plastically. From the graph, we can approximate the yield strength as the stress value corresponding to the end of the linear region, which is around 300 × 10⁶ N/m²
or 3 × 10⁸ N/m².
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?
Ans :
(a) Young’s modulus is the ratio of stress to strain within the elastic limit. It represents the stiffness of a material. A steeper slope on a stress-strain curve indicates a higher Young’s modulus.
Comparing the two graphs, we can see that material B has a steeper slope in the linear elastic region. Therefore, **material B has a greater Young’s modulus**.
(b) Strength refers to the ability of a material to withstand stress without breaking or deforming plastically. While both materials exhibit similar behavior in the elastic region, material A shows a larger plastic deformation region before failing. This indicates that **material A is stronger**.
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.
Ans :(a) False. Young’s modulus measures a material’s resistance to elastic deformation under tension or compression. Despite stretching more than steel under the same force, rubber is less elastic due to its lower Young’s modulus.
(b) True. The elasticity of a coil is primarily governed by its shear modulus. Applying forces to opposite ends of a coil alters both its length and the shape of its helical turns, which is a shearing deformation.
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. Fig. 8.11
Ans : **Understanding the Problem:**
We have a steel wire and a brass wire connected in series. A 4 kg mass is suspended from the steel wire, and a 6 kg mass is suspended from the brass wire.
find the elongation of each wire.
Forces on the Wires:
Steel Wire:
– Force: F_steel = (4 + 6) kg * 9.8 m/s² = 98 N
– Original length: L_steel = 1.5 m
– Radius: r_steel = 0.125 cm = 0.00125 m
– Young’s modulus: Y_steel = 2.0 × 10¹¹ Pa
Brass Wire
– Force: F_brass = 6 kg * 9.8 m/s² = 58.8 N
– Original length: L_brass = 1.0 m
– Radius: r_brass = 0.125 cm = 0.00125 m
– Young’s modulus: Y_brass = 0.91 × 10¹¹ Pa
**Calculating Elongation:**
We can use the formula for Young’s modulus to calculate the elongation of each wire:
Y = (F * L)
———
(A * ΔL)
Rearranging to solve for ΔL:
ΔL = (F * L)
———
(Y * A)
Where:
* ΔL is the elongation
* F is the force
* L is the original length
* Y is Young’s modulus
* A is the cross-sectional area (πr²)
Calculating Elongation of Steel Wire:
ΔL_steel = (98 N * 1.5 m) / (2.0 × 10¹¹ Pa * π * (0.00125 m)²) ≈ 1.49 × 10⁻⁴ m
**Calculating Elongation of Brass Wire:**
ΔL_brass = (58.8 N * 1.0 m) / (0.91 × 10¹¹ Pa * π * (0.00125 m)²)
≈ 1.3 × 10⁻⁴ m
Therefore, the elongation of the steel wire is approximately 1.49 × 10⁻⁴ m, and the elongation of the brass wire is approximately 1.3 × 10⁻⁴ m.
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?
Ans : *Understanding the Problem:**
We have a cube of aluminum with a side length of 10 cm. A force is applied to one face due to the weight of the 100 kg mass. This force causes shear stress on the cube, leading to a vertical deflection.
Solution:
*1. Calculate the Force:
The force applied to the cube is the weight of the 100 kg mass:
F = m * g = 100 kg * 9.8 m/s² = 980 N
2. Calculate the Shear Stress:
Shear stress (σ) is the force applied per unit area:
σ = F / A
Where:
* F is the force (980 N)
* A is the area of the face (10 cm × 10 cm = 0.01 m²)
σ = 980 N / 0.01 m² = 98,000 Pa
*3. Calculate the Shear Strain:
Shear strain (γ) is the ratio of the lateral displacement (x) to the original length (L):
γ = x / L
*4. Use Shear Modulus
Shear modulus (G) is defined as the ratio of shear stress to shear strain:
G = σ / γ
Substituting the values:
25 × 10⁹ Pa = 98,000 Pa / (x / 0.1 m)
Solving for x:
x = (98,000 Pa * 0.1 m) / (25 × 10⁹ Pa)
= 3.92 × 10⁻⁶ m
Therefore, the vertical deflection of the face is approximately 3.92 micrometers.
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.
Ans : Understanding the Problem:**
We have four identical cylindrical columns supporting a 50,000 kg mass. We need to find the compressional strain of each column.
Solution:
1. Load on Each Column:
Since the load is distributed uniformly among the four columns, each column supports a load of:
F = 50,000 kg / 4 = 12,500 kg
= 122,500 N
2. Cross-sectional Area of Each Column:
The cross-sectional area of each column is the difference between the areas of the outer and inner circles:
A = π(R² – r²) = π(0.6² – 0.3²) m²
≈ 0.2827 m²
3. Stress on Each Column:
Stress (σ) is the force per unit area:
σ = F / A = 122,500 N / 0.2827 m²
≈ 433,723 Pa
4. Strain on Each Column:
Young’s modulus (Y) = ratio of stress to strain:
Y = σ / ε
Where:
* Y is Young’s modulus of steel (approximately 2 × 10¹¹ Pa)
* σ is the stress
* ε is the strain
Rearranging the formula to solve for strain:
ε = σ / Y
Substituting the values:
ε = 433,723 Pa / (2 × 10¹¹ Pa)
≈ 2.17 × 10⁻⁶
Therefore, the compressional strain of each column is approximately 2.17 × 10⁻⁶.
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?
Ans : Understanding the Problem:**
We have a copper piece with a given cross-sectional area subjected to a specific force. We need to calculate the resulting strain.
Solution:
1. Calculate Stress
Stress (σ) is the force per unit area:
σ = F / A
Where:
* F is the force applied (44,500 N)
* A is the cross-sectional area (15.2 mm × 19.1 mm = 2.9032 × 10⁻⁴ m²)
2. Calculate Strain:
Strain (ε) is the ratio of stress to Young’s modulus:
ε = σ / Y
Where:
* Y is Young’s modulus of copper (42 × 10⁹ N/m²)
Combining the two equations:
ε = F / (AY)
Substituting the given values:
ε = 44,500 N / (2.9032 × 10⁻⁴ m² × 42 × 10⁹ N/m²)
≈ 3.65 × 10⁻³
Therefore, the resulting strain in the copper piece is approximately 3.65 × 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 10^8 N m–2, what is the maximum load the cable can support ?
Ans : Understanding the Problem:**
We have a steel cable with a given radius and a maximum allowable stress. We need to find the maximum load it can support.
Solution
1. Calculate the Cross-sectional Area:
The cross-sectional area (A) of the cable is given by the formula for the area of a circle:
A = πr²
Where:
* r = radius of the cable (0.015 m)
A = π × (0.015 m)²
≈ 7.07 × 10⁻⁴ m²
2. Use the Maximum Stress:
The maximum stress (σ) is given by the force (F) divided by the area (A):
σ = F / A
Rearranging to solve for the maximum force (F):
F = σ × A
Substituting the values:
F = (10⁸ N/m²) × (7.07 × 10⁻⁴ m²)
≈ 7.07 × 10⁴ N
Therefore, the maximum load the cable can support is approximately 70,700 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.
Ans : **Understanding the Problem:**
We have a rigid bar supported by three wires: two copper wires and one iron wire. All wires have the same length and are subjected to the same tension. We need to find the ratio of their diameters to ensure equal tension.
1. Tension in Each Wire:
Since the bar is supported symmetrically, each wire carries one-third of the total weight:
Tension in each wire, T = (15 kg * 9.8 m/s²) / 3
= 49 N
2. Elongation of the Wires:
Given,
elongation (ΔL) of a wire under tension
ΔL = (FL) / (AY)
Where:
* F = tension force
* L = original length
* A = cross-sectional area (πr²)
* Y stands for Young’s modulus of the material
For all three wires, the elongation (ΔL) and the length (L) are the same. Therefore, to maintain equal tension, the ratio of their cross-sectional areas must be inversely proportional to the ratio of their Young’s moduli:
A_copper / A_iron = Y_iron / Y_copper
3. Ratio of Diameters:
As the cross-sectional area of a wire is proportional to the square of its diameter,
(d_copper)² / (d_iron)² = Y_iron / Y_copper
Taking the square root of both sides:
d_copper / d_iron = √(Y_iron / Y_copper)
Now, we need to know the Young’s modulus of copper and iron. Let’s assume typical values:
* Young’s modulus of copper (Y_copper) ≈ 110 GPa
* Young’s modulus of iron (Y_iron) ≈ 200 GPa
Substituting these values:
d_copper / d_iron = √(200/110)
≈ 1.35
Therefore, the ratio of the diameters of the copper and iron wires should be approximately 1.35 to ensure equal tension in all three wires.
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.
Ans : Here,
* m = 14.5 kg (mass of the object)
* r = 1 m (radius of the circular path)
* ω = 2 rev/s = 4π rad/s (angular velocity)
* A = 0.065 × 10⁻⁴ m² (cross-sectional area of the wire)
The total force (F) acting on the mass at the lowest point of the vertical circle is the sum of its weight (mg) and the centripetal force (mrω²):
F = mg + mrω² = m(g + rω²)
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.
Ans : Understanding the Problem:
We’re given the initial and final volumes of water, along with the pressure increase. We need to calculate the bulk modulus of water and compare it to that of air.
Solution:
1. Calculate the Volume Change
ΔV = Final volume – Initial volume
= 100.5 L – 100.0 L
= 0.5 L
= 0.0005 m³
2. Calculate the Bulk Modulus:
Bulk modulus (B) is defined as the ratio of volumetric stress to volumetric strain:
B = – (ΔP / (ΔV/V))
Where:
* ΔP is the change in pressure (100 atm = 100 × 1.013 × 10⁵ Pa = 1.013 × 10⁷ Pa)
* ΔV/V is the volumetric strain
Substituting the values:
B = – (1.013 × 10⁷ Pa) / (0.0005 / 100)
≈ 2.026 × 10¹⁰ Pa
3. Comparison with Air:
The bulk modulus of air is significantly smaller than that of water. This is because:
Intermolecular Forces:** Water molecules are strongly attracted to each other due to hydrogen bonding. This strong intermolecular force makes it difficult to compress water. In contrast, air molecules are much farther apart and have weaker intermolecular forces, making it easier to compress.
*Density:* Water is much denser than air. This means that there are more molecules packed into a given volume of water compared to air. The closer packing of molecules in water makes it more resistant to compression.
Therefore, water has a much higher bulk modulus than air.
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 × 10^3 kg m–3?
Ans : Understanding the Problem:
We are given the initial density of water at the surface and the pressure increase at a certain depth. We need to find the density of water at that depth.
Solution:
1. Compressibility of Water:
The compressibility (k) of a substance is the reciprocal of its bulk modulus (B). It is a measure of how much the volume of a substance changes under pressure:
k = 1/B = 45.8 × 10⁻¹¹ Pa⁻¹
2. Change in Pressure:
The change in pressure (ΔP) is given as 80 atm – 1 atm = 79 atm = 7.987 × 10⁶ Pa.
3. Volume Change:
The fractional change in volume (ΔV/V) can be calculated using the compressibility:
ΔV/V = ΔP × k
= 79 × 1.013 × 10⁵ Pa × 45.8 × 10⁻¹¹
= 3.665 × 10⁻⁵
4. Final Density:
The final density (ρ’) can be calculated using the formula:
ρ’ = ρ / (1 – ΔV/V)
Substituting the values:
ρ’ = 1.03 × 10³ kg/m³ / (1 – 3.665 × 10⁻⁵)
≈ 1.034 × 10³ kg/m³
Therefore, the density of water at the given depth is approximately 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.
Ans :Given:
Pressure (P) = 10 atm = 10 x 1.013 x 10^5 Pa
Bulk modulus (K) = 37 x 10^9 Nm^-2
Calculation:
Volumetric strain:
Volumetric strain is the ratio of the change in volume of a material to its original volume. It is also equal to the ratio of pressure (P) to the bulk modulus (K).
Volumetric strain = (ΔV/V) = (P/K)
= (10 x 1.013 x 10^5)
—————————-
(37 x 10^9)
= 2.74 x 10^-5
Fractional change in volume:
This is equivalent to the volumetric strain expressed as a percentage.
Fractional change in volume
= (ΔV/V)
= 2.74 x 10^-5
Conclusion:
The volume of the material decreases by 2.74 x 10^-5 of its original volume under the given pressure.
15 .Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10^6 Pa.
Ans : Understanding the Problem:
We are given a copper cube with a specific edge length and a hydraulic pressure applied to it. We need to calculate the resulting volume contraction.
1. Calculate the Initial Volume:
The initial volume (V₀) of the cube is:
V₀ = (edge length)³
= (10 cm)³ = 1000 cm³ = 0.001 m³
2. Use the Bulk Modulus:
(B)bulk modulus measures a material’s resistance to compression. It is related to the fractional change in volume (ΔV/V) and the pressure change (ΔP) as follows:
B = – (ΔP / (ΔV/V))
Rearranging the formula to find the volume change (ΔV):
ΔV = – (V₀ * ΔP) / B
Where:
* ΔP is the pressure change (7.0 × 10⁶ Pa)
* B is the bulk modulus of copper (approximately 1.4 × 10¹¹ Pa)
Substituting the values:
ΔV = – (0.001 m³ * 7.0 × 10⁶ Pa) / (1.4 × 10¹¹ Pa)
≈ -5.0 × 10⁻⁸ m³
Therefore, the volume contraction of the copper cube is approximately 5.0 × 10⁻⁸ m³.
16 .How much should the pressure on a litre of water be changed to compress it by 0.10%? ( carry one quarter of the load.)
Ans : Understanding the Problem:
We are given the initial volume of water and the fractional change in volume. We need to calculate the required pressure change.
Solution:
1. Calculate the Volume Change (ΔV):
ΔV = 0.1% of 1 L = 0.001 L = 0.000001 m³
2. Use the Bulk Modulus
(B)bulk modulus measures a material’s resistance to compression.. It is related to the fractional change in volume (ΔV/V) and the pressure change (ΔP) as follows:
B = – (ΔP / (ΔV/V))
Determine pressure change (ΔP):
ΔP = -B × (ΔV/V)
The bulk modulus of water is approximately 2.2 × 10¹⁰ Pa.
Substituting the values:
ΔP = – (2.2 × 10¹⁰ Pa) × (-0.000001)
= 2.2 × 10⁴ Pa
Therefore, the pressure on the water should be increased by approximately 2.2 × 10⁴ Pa to compress it by 0.1%.
