Chapter Notes

Introduction

While we often think of solid objects as being perfectly rigid, with a fixed shape and size, this is not entirely true. In reality, all solid bodies can be stretched, compressed, or bent when a force is applied to them. Even a strong steel bar will deform if the force is large enough. The study of how solid bodies deform under external forces is crucial for understanding their mechanical properties.

Elasticity and Plasticity

When you apply a force to an object, you can change its shape or size. How the object responds depends on its material properties.

• Elasticity is the property of a body that allows it to return to its original size and shape after the deforming force is removed. The deformation that occurs in this case is called elastic deformation. A helical spring is a good example; when you stretch it and then let go, it springs back to its original length.

• Plasticity is the property of a substance to undergo permanent deformation. When the applied force is removed, the body does not regain its original shape. Such substances are called plastic. A lump of putty or mud are examples of materials that are nearly ideal plastics.

The elastic behavior of materials is extremely important in engineering and design. For instance, when designing buildings, bridges, or automobiles, engineers must know the elastic properties of materials like steel and concrete to ensure the structures are safe and durable. This knowledge helps answer questions like: Why are railway tracks I-shaped? Why is glass brittle while brass is not?

Stress and Strain

When an external force is applied to a solid body, it deforms. In response to this deforming force, the body develops an internal restoring force that is equal in magnitude and opposite in direction to the applied force.

Stress is defined as this internal restoring force per unit area. If a force $F$ is applied normal (perpendicular) to a cross-sectional area $A$, the magnitude of the stress is: $\text{Stress} = \frac{F}{A}$

• The SI unit of stress is $\text{N m}^{-2}$ or pascal (Pa).
• The dimensional formula for stress is $[\text{ML}^{-1}\text{T}^{-2}]$.

Strain is the measure of the deformation of the body. It is defined as the fractional change in the dimension of the body. $\text{Strain} = \frac{\text{Change in dimension}}{\text{Original dimension}}$

• Since strain is a ratio of two similar quantities, it has no units or dimensional formula.

Types of Stress and Strain

There are three main ways a solid can be deformed, leading to three types of stress and strain.

1. Longitudinal Stress and Strain

This occurs when a force is applied perpendicular to the cross-sectional area, causing a change in the object's length.

• Tensile Stress: If the force stretches the object (e.g., pulling a cylinder from both ends), the stress is called tensile stress.
• Compressive Stress: If the force compresses the object, it is called compressive stress.
• Longitudinal Strain: The change in length ($\Delta L$) relative to the original length ($L$) is called longitudinal strain. $\text{Longitudinal strain} = \frac{\Delta L}{L}$

2. Shearing Stress and Strain

This occurs when deforming forces are applied parallel (tangential) to the cross-sectional area, causing a relative displacement between opposite faces.

• Shearing Stress (or Tangential Stress): The restoring force per unit area developed due to the tangential applied force.
• Shearing Strain: The ratio of the relative displacement of the faces ($\Delta x$) to the length of the cylinder ($L$). $\text{Shearing strain} = \frac{\Delta x}{L} = \tan \theta$ Here, $\theta$ is the angular displacement of the cylinder from its original vertical position. For very small angles, $\tan \theta \approx \theta$, so the shearing strain is approximately equal to $\theta$.

3. Hydraulic Stress and Volume Strain

This occurs when a body is subjected to a uniform force from all sides, such as when it is placed in a fluid under high pressure. The force acts perpendicular to the entire surface of the body.

• Hydraulic Stress: The internal restoring force per unit area. In this case, its magnitude is equal to the hydraulic pressure ($p$).
• Volume Strain: The ratio of the change in volume ($\Delta V$) to the original volume ($V$). This type of stress changes the volume of the body without changing its geometrical shape. $\text{Volume strain} = \frac{\Delta V}{V}$

Hooke's Law

For most materials, when the deformation is small, there is a simple relationship between stress and strain. Hooke's Law states that for small deformations, stress is directly proportional to strain. $\text{stress} \propto \text{strain}$ $\text{stress} = k \times \text{strain}$ The constant of proportionality, $k$, is called the modulus of elasticity. This constant is a characteristic property of the material and indicates its stiffness.

Stress-Strain Curve

To understand how a material behaves under increasing load, we can perform a tensile test on a sample (like a wire or cylinder) and plot a graph of stress versus strain. This graph is called a stress-strain curve. A typical curve for a metal is shown below.

<img src="https://cdn.mathpix.com/cropped/ad64b4ec-8b0b-4bd9-b392-083a1a362583-03.jpg?height=606&width=623&top_left_y=714&top_left_x=1106" alt="Stress-Strain Curve for a Metal" width="450">

The curve shows several distinct regions:

• Region O to A (Proportional Limit): The curve is a straight line, which means stress is directly proportional to strain. In this region, the material obeys Hooke's law. If the load is removed, the material returns to its original dimensions. This is the region of elastic behaviour.
• Region A to B (Elastic Limit): In this region, stress is no longer proportional to strain. However, the material is still elastic—it will return to its original shape if the load is removed. Point B is called the yield point or elastic limit, and the corresponding stress is the yield strength ($\sigma_y$).
• Region B to D (Plastic Behaviour): If the load is increased beyond the yield point, the strain increases rapidly for even a small change in stress. If the load is removed (e.g., at point C), the material does not regain its original shape. It is said to have a permanent set. This is called plastic deformation.
• Point D (Ultimate Tensile Strength): This is the maximum stress ($\sigma_u$) that the material can withstand before it starts to fail.
• Point E (Fracture Point): Beyond point D, the material begins to thin and eventually breaks at point E.

Materials can be classified based on their stress-strain curves:

• Ductile Materials: Materials where the fracture point E is far from the ultimate strength point D. These materials can be stretched into thin wires (e.g., copper, steel).
• Brittle Materials: Materials where the fracture point E is very close to the ultimate strength point D. These materials break with very little plastic deformation (e.g., glass, concrete).

Elastomers

Some substances, like rubber or the tissue of the aorta, can be stretched to cause very large strains and still return to their original shape. These materials are called elastomers. Their stress-strain curve is not a straight line, meaning they do not obey Hooke's law over most of their range.

Elastic Moduli

The modulus of elasticity ($k$ from Hooke's law) is a measure of a material's resistance to being deformed elastically. There are three types of elastic moduli, corresponding to the three types of stress and strain.

Young's Modulus

Young's Modulus (Y) is the ratio of longitudinal stress ($\sigma$) to longitudinal strain ($\varepsilon$). It measures a material's stiffness or resistance to a change in length. $Y = \frac{\sigma}{\varepsilon}$ Substituting the formulas for stress and strain: $Y = \frac{(F/A)}{(\Delta L/L)} = \frac{F \times L}{A \times \Delta L}$

• The unit of Young's modulus is the same as stress: $\text{N m}^{-2}$ or Pa.
• A material with a large Young's modulus (like steel) requires a large force to produce a small change in length, making it very elastic and strong. Materials like wood or bone have smaller Young's moduli.

Given

• Radius, $r = 10 \text{ mm} = 10 \times 10^{-3} \text{ m} = 10^{-2} \text{ m}$
• Length, $L = 1.0 \text{ m}$
• Force, $F = 100 \text{ kN} = 100 \times 10^3 \text{ N}$
• Young's modulus, $Y = 2.0 \times 10^{11} \text{ N m}^{-2}$

To Find

(a) Stress (b) Elongation, $\Delta L$ (c) Strain

Formula

$\text{Stress} = \frac{F}{A} = \frac{F}{\pi r^2}$ $\Delta L = \frac{(\text{Stress}) \times L}{Y}$ $\text{Strain} = \frac{\Delta L}{L}$

Solution

(a) Calculate the stress The cross-sectional area is $A = \pi r^2$. $\text{Stress} = \frac{100 \times 10^3 \text{ N}}{3.14 \times (10^{-2} \text{ m})^2} = \frac{10^5 \text{ N}}{3.14 \times 10^{-4} \text{ m}^2}$ $\text{Stress} = 3.18 \times 10^8 \text{ N m}^{-2}$ Answer for part (a) = $3.18 \times 10^8 \text{ N m}^{-2}$

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(b) Calculate the elongation $\Delta L = \frac{(3.18 \times 10^8 \text{ N m}^{-2}) \times (1 \text{ m})}{2 \times 10^{11} \text{ N m}^{-2}}$ $\Delta L = 1.59 \times 10^{-3} \text{ m} = 1.59 \text{ mm}$ Answer for part (b) = $1.59 \text{ mm}$

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(c) Calculate the strain $\text{Strain} = \frac{1.59 \times 10^{-3} \text{ m}}{1 \text{ m}} = 1.59 \times 10^{-3}$ This can also be expressed as a percentage: $1.59 \times 10^{-3} \times 100\% = 0.159\% \approx 0.16\%$. Answer for part (c) = $1.59 \times 10^{-3}$ or $0.16\%$

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Given

• Length of copper wire, $L_c = 2.2 \text{ m}$
• Length of steel wire, $L_s = 1.6 \text{ m}$
• Diameter, $d = 3.0 \text{ mm}$, so radius $r = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}$
• Total elongation, $\Delta L_{total} = \Delta L_c + \Delta L_s = 0.70 \text{ mm} = 7.0 \times 10^{-4} \text{ m}$
• Young's modulus for copper, $Y_c = 1.1 \times 10^{11} \text{ N m}^{-2}$
• Young's modulus for steel, $Y_s = 2.0 \times 10^{11} \text{ N m}^{-2}$

To Find

The applied load, $W$

Formula

$\text{Stress} = \frac{W}{A} = Y \times \text{Strain} = Y \times \frac{\Delta L}{L}$

Solution

Since the wires are connected end to end, they experience the same tension (load $W$) and have the same cross-sectional area $A$. $\frac{W}{A} = Y_c \frac{\Delta L_c}{L_c} \quad \text{and} \quad \frac{W}{A} = Y_s \frac{\Delta L_s}{L_s}$ Therefore, $Y_c \frac{\Delta L_c}{L_c} = Y_s \frac{\Delta L_s}{L_s} \implies \frac{\Delta L_c}{\Delta L_s} = \frac{Y_s}{Y_c} \times \frac{L_c}{L_s}$ Substituting the values: $\frac{\Delta L_c}{\Delta L_s} = \left(\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}\right) \times \left(\frac{2.2}{1.6}\right) = 1.818 \times 1.375 = 2.5$ So, $\Delta L_c = 2.5 \Delta L_s$.

We also know that $\Delta L_c + \Delta L_s = 7.0 \times 10^{-4} \text{ m}$. Substituting $\Delta L_c$: $2.5 \Delta L_s + \Delta L_s = 7.0 \times 10^{-4} \implies 3.5 \Delta L_s = 7.0 \times 10^{-4}$ $\Delta L_s = \frac{7.0 \times 10^{-4}}{3.5} = 2.0 \times 10^{-4} \text{ m}$ And, $\Delta L_c = 2.5 \times (2.0 \times 10^{-4}) = 5.0 \times 10^{-4} \text{ m}$ Now we can find the load $W$ using the data for the copper wire: $W = \frac{A \times Y_c \times \Delta L_c}{L_c}$ The area $A = \pi r^2 = \pi (1.5 \times 10^{-3})^2 = 7.07 \times 10^{-6} \text{ m}^2$. $W = \frac{(7.07 \times 10^{-6} \text{ m}^2) \times (1.1 \times 10^{11} \text{ N m}^{-2}) \times (5.0 \times 10^{-4} \text{ m})}{2.2 \text{ m}}$ $W \approx 1.8 \times 10^2 \text{ N}$

Final Answer The load applied is approximately $1.8 \times 10^2 \text{ N}$.

Shear Modulus

Shear Modulus (G), also known as the modulus of rigidity, is the ratio of shearing stress to shearing strain. It measures a material's resistance to a change in shape. $G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{(F/A)}{(\Delta x/L)} = \frac{F \times L}{A \times \Delta x}$ Since shearing strain can also be expressed as $\theta$: $G = \frac{F}{A \times \theta}$

• The SI unit is $\text{N m}^{-2}$ or Pa.
• Shear modulus is only relevant for solids.
• For most materials, the shear modulus is about one-third of the Young's modulus ($G \approx Y/3$).

Given

• Side length = 50 cm = 0.5 m
• Thickness = 10 cm = 0.1 m
• Shearing force, $F = 9.0 \times 10^4 \text{ N}$ (The source text has a typo of $9.4 \times 10^4$ in the calculation, we use the value from the problem statement)
• Length of the side perpendicular to the force, $L = 50 \text{ cm} = 0.5 \text{ m}$
• Shear modulus, $G = 5.6 \times 10^9 \text{ N m}^{-2}$

To Find

The displacement of the upper edge, $\Delta x$

Formula

$G = \frac{F \times L}{A \times \Delta x} \implies \Delta x = \frac{F \times L}{A \times G}$

Solution

The force is applied to the narrow face, so the area $A$ is: $A = \text{side} \times \text{thickness} = (0.5 \text{ m}) \times (0.1 \text{ m}) = 0.05 \text{ m}^2$ Now we can calculate the displacement $\Delta x$: $\Delta x = \frac{(9.0 \times 10^4 \text{ N}) \times (0.5 \text{ m})}{(0.05 \text{ m}^2) \times (5.6 \times 10^9 \text{ N m}^{-2})}$ $\Delta x = \frac{4.5 \times 10^4}{0.28 \times 10^9} = 1.6 \times 10^{-4} \text{ m} = 0.16 \text{ mm}$ (Note: The source text calculates stress as $1.80 \times 10^6 \text{ N m}^{-2}$ which implies a force of $9.0 \times 10^4 \text{ N}$, not $9.4 \times 10^4 \text{ N}$. We follow the calculation logic.)

Final Answer The upper edge will be displaced by $0.16 \text{ mm}$.

Bulk Modulus

Bulk Modulus (B) is the ratio of hydraulic stress (pressure) to the corresponding volume strain. It measures a material's resistance to a change in volume (compression). $B = \frac{\text{Hydraulic Stress}}{\text{Volume Strain}} = \frac{-p}{(\Delta V/V)}$

• The negative sign indicates that as pressure ($p$) increases, the volume decreases ($\Delta V$ is negative), ensuring that the value of $B$ is always positive.
• The SI unit is $\text{N m}^{-2}$ or Pa.
• Bulk modulus is relevant for solids, liquids, and gases.
• The reciprocal of the bulk modulus is called compressibility ($k = 1/B$). A material with a high bulk modulus has low compressibility.

Solids are the least compressible (highest B), while gases are the most compressible (lowest B). Gases are about a million times more compressible than solids.

Given

• Depth, $h = 3000 \text{ m}$
• Bulk modulus of water, $B = 2.2 \times 10^9 \text{ N m}^{-2}$
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$
• Density of water, $\rho = 1000 \text{ kg m}^{-3}$

To Find

Fractional compression, $\frac{\Delta V}{V}$

Formula

$p = h \rho g$ $B = \frac{p}{(\Delta V/V)} \implies \frac{\Delta V}{V} = \frac{p}{B}$

Solution

First, calculate the pressure at the bottom of the ocean: $p = (3000 \text{ m}) \times (1000 \text{ kg m}^{-3}) \times (10 \text{ m s}^{-2})$ $p = 3 \times 10^7 \text{ N m}^{-2}$ Now, calculate the fractional compression: $\frac{\Delta V}{V} = \frac{3 \times 10^7 \text{ N m}^{-2}}{2.2 \times 10^9 \text{ N m}^{-2}}$ $\frac{\Delta V}{V} = 1.36 \times 10^{-2}$ As a percentage, this is $1.36\%$.

Final Answer The fractional compression of water at the bottom of the ocean is $1.36 \times 10^{-2}$ or $1.36\%$.

Poisson's Ratio

When a wire is stretched along its length (longitudinal strain), it gets slightly thinner. The strain perpendicular to the applied force is called lateral strain. Poisson's Ratio is the ratio of the lateral strain to the longitudinal strain, within the elastic limit. If a wire has an original diameter $d$ which changes by $\Delta d$, and an original length $L$ which changes by $\Delta L$:

• Lateral Strain = $\Delta d / d$
• Longitudinal Strain = $\Delta L / L$ $\text{Poisson's Ratio} = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = \frac{(\Delta d/d)}{(\Delta L/L)}$ Poisson's ratio is a pure number with no dimensions or units. For steel, its value is typically between 0.28 and 0.30.

Elastic Potential Energy in a Stretched Wire

When a wire is stretched, work is done against the inter-atomic forces. This work is stored in the wire as elastic potential energy. The elastic potential energy ($U$) stored in a wire is given by: $W = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume of the wire}$ The elastic potential energy per unit volume ($u$) of the wire is: $u = \frac{1}{2} \times \sigma \times \varepsilon$

Applications of Elastic Behaviour of Materials

Understanding elasticity is vital for many real-world engineering applications.

Cranes

Cranes use thick metal ropes to lift heavy loads. The rope must be thick enough so that the stress produced by the load does not exceed the material's elastic limit, preventing permanent stretching. For a crane lifting 10 tonnes ($10^4$ kg), the minimum cross-sectional area ($A$) of a steel rope with a yield strength ($\sigma_y$) of $300 \times 10^6 \text{ N m}^{-2}$ should be: $A \geq \frac{W}{\sigma_y} = \frac{Mg}{\sigma_y} = \frac{(10^4 \text{ kg} \times 9.8 \text{ m s}^{-2})}{300 \times 10^6 \text{ N m}^{-2}} = 3.3 \times 10^{-4} \text{ m}^2$ This corresponds to a radius of about 1 cm. In practice, a large safety margin is used, and a thicker rope (e.g., radius 3 cm) is recommended. These ropes are made of many thin wires braided together for flexibility and strength.

Beams in Bridges and Buildings

Beams are designed to support loads without bending too much or breaking. The amount a beam sags ($\delta$) when loaded at the center is given by: $\delta = \frac{W l^3}{4 b d^3 Y}$ where $l$ is the length, $b$ is the breadth, $d$ is the depth, and $Y$ is the Young's modulus. To reduce bending:

• Use a material with a large Young's modulus ($Y$).
• Make the beam shorter ($l$).
• Increase the depth ($d$) of the beam. Bending is inversely proportional to $d^3$, so increasing depth is very effective.

However, a very deep, thin beam can bend sideways, a phenomenon called buckling. To prevent this while keeping the beam strong and lightweight, a compromise is the I-shaped cross-section. This shape has a large depth to prevent sagging and wide surfaces to prevent buckling, all while using less material than a solid rectangular beam.

Maximum Height of a Mountain

The elastic properties of rocks limit the maximum height of mountains on Earth. The pressure at the base of a mountain of height $h$ is $h \rho g$. This pressure creates a shearing stress. If this stress exceeds the elastic limit of the rock at the base, the rock will flow and the mountain will sink. The elastic limit for a typical rock is about $30 \times 10^7 \text{ N m}^{-2}$. Setting this equal to the stress: $h \rho g = 30 \times 10^7 \text{ N m}^{-2}$ Using the density of rock ($\rho \approx 3 \times 10^3 \text{ kg m}^{-3}$) and $g \approx 10 \text{ m s}^{-2}$: $h = \frac{30 \times 10^7 \text{ N m}^{-2}}{(3 \times 10^3 \text{ kg m}^{-3}) \times (10 \text{ m s}^{-2})} = 10^4 \text{ m} = 10 \text{ km}$ This calculation shows that the maximum possible height for a mountain on Earth is about 10 km, which is in the range of the height of Mt. Everest.