MECHANICAL PROPERTIES OF SOLIDS-Notes
Physics - Notes
Elastic and Plastic Behaviour of Solids
Definition of Elasticity
Elasticity is the property of a material by virtue of which it completely regains its original shape and size after the removal of the deforming force, provided the deformation is within certain limits.
In simpler terms, if a body returns to its initial configuration once the applied force is removed, it is said to exhibit elastic behaviour.
Definition of Plasticity
Plasticity is the property of a material by virtue of which it does not regain its original shape and size after the removal of the deforming force.
If a body undergoes permanent deformation even after the external force is removed, it is said to exhibit plastic behaviour.
| Elastic Properties | Plastic Properties |
|---|---|
| Elasticity is the property of a material by which it completely regains its original shape and size after the deforming force is removed, provided the elastic limit is not exceeded. | Plasticity is the property of a material by which it does not regain its original shape and size after the deforming force is removed. |
| The deformation produced in an elastic material is temporary and reversible. | The deformation produced in a plastic material is permanent and irreversible. |
| Elastic behaviour is observed only up to a certain limit called the elastic limit. | Plastic behaviour begins once the elastic limit is crossed. |
| Elastic materials obey Hooke’s law, where stress is directly proportional to strain within the elastic limit. | Plastic materials do not obey Hooke’s law once yielding begins. |
| Elastic deformation involves only slight changes in interatomic distances without permanent rearrangement of atoms. | Plastic deformation involves slipping and permanent rearrangement of atomic layers. |
| Energy supplied during elastic deformation is stored as elastic potential energy and is recoverable. | Energy supplied during plastic deformation is not recoverable and is mostly dissipated. |
| Elastic properties are important for structures and devices where shape recovery is essential, such as springs and bridges. | Plastic properties are important in processes like forging, rolling, and moulding, where permanent shaping is required. |
STRESS AND STRAIN
Stress
When a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force is equal in magnitude but opposite in direction to the applied force. The restoring force per unit area is known as stress.
Mathematically, If F is the force applied normal to the cross–section and A is the area of cross section of the body,
\[ \begin{aligned} \text{Stress} &= \frac{F}{\text{Area of cross-section}} \end{aligned} \]
The SI unit of stress is the pascal (Pa), where \(1\,\text{Pa} = 1\,\text{N m}^{-2}\). Stress is a scalar quantity in magnitude but is associated with direction depending on the nature of the applied force.
Depending on how the force acts, stress is classified into longitudinal stress, volume stress, and shear stress. These classifications arise naturally from the geometry of deformation and help in analysing different mechanical responses of solids.
Strain
Strain is defined as the measure of deformation produced in a body due to applied stress. It is the ratio of change in dimension to the original dimension of the body. Strain provides a dimensionless measure of how much deformation has occurred relative to the size of the object.
In general form, strain is written as
\[ \begin{aligned} \text{Strain} &= \frac{\text{Change in dimension}}{\text{Original dimension}} \end{aligned} \]
Since both numerator and denominator have the same dimensions, strain has no unit. Like stress, strain is classified into longitudinal strain, volumetric strain, and shear strain depending on the nature of deformation.
To understand the physical connection between stress and strain, consider a uniform rod of original length \(L\) subjected to a tensile force \(F\) applied along its length. The rod elongates by a small amount \(\Delta L\). The longitudinal strain produced is
\[ \begin{aligned} \text{Longitudinal strain} &= \frac{\Delta L}{L} \end{aligned} \]
The corresponding longitudinal stress developed in the rod is
\[ \begin{aligned} \text{Longitudinal stress} &= \frac{F}{A} \end{aligned} \]
where \(A\) is the cross-sectional area of the rod. Experimental observations show that, for small deformations, the ratio of stress to strain remains constant for a given material. This leads to the concept of elastic constants and forms the experimental basis of Hooke’s law.
Proof of Dimensional Consistency of Stress and Strain can be established by examining their units. Stress is force per unit area, so its dimensions are
\[ \begin{aligned} [\text{Stress}] &= \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] \end{aligned} \]
Strain is a ratio of two lengths and therefore has no dimensions:
\[ \begin{aligned} [\text{Strain}] &= 1 \end{aligned} \]
This distinction highlights an important physical insight: stress represents the cause of deformation, while strain represents the effect produced in the material.
Shearing Strain
Shearing strain is the strain produced in a body when it is subjected to forces that act tangentially to its surface and tend to change its shape without significantly changing its volume.
Consider a rectangular block fixed at its base. When a tangential force \(F\) is applied to the top surface parallel to the base, the upper face gets displaced sideways while the lower face remains fixed. As a result, the block changes its shape and the originally right angles between its faces are distorted.
If the height of the block is \(L\) and the top surface is displaced by a small distance \(\Delta x\), the angle of deformation produced is very small. This angular distortion represents the shearing strain.
Shearing strain is defined as the ratio of the lateral displacement to the original height of the body. It is mathematically expressed as
\[ \begin{aligned} \text{Shearing strain} &= \frac{\Delta x}{L} \end{aligned} \]
For very small deformations, this ratio is equal to the tangent of the angle of shear \(\theta\). Since \(\theta\) is small, \(\tan\theta \approx \theta\) (in radians). Hence, shearing strain may also be written as
\[ \begin{aligned} \text{Shearing strain} &= \tan\theta \approx \theta \end{aligned} \]
Shearing strain is a dimensionless quantity because it is the ratio of two lengths or an angular measure expressed in radians. It indicates the extent to which the shape of a body is distorted under shear stress.
Within the elastic limit, shearing strain is directly proportional to the applied shear stress. The constant of proportionality between shear stress and shearing strain is known as the shear modulus or modulus of rigidity, which measures the material’s resistance to shape deformation.
Hydraulic Stress
Hydraulic stress is the stress developed in a solid body when it is subjected to a uniform pressure acting normally on all its surfaces. In NCERT Class XI Physics, hydraulic stress is discussed in connection with volume deformation and is also referred to as volume stress.
When a solid is immersed in a fluid or placed under the action of an external fluid pressure, every point on its surface experiences a force perpendicular to the surface. Since the pressure acts equally in all directions, the shape of the body does not change, but its volume decreases uniformly.
Hydraulic stress is defined as the normal force per unit area acting uniformly over the entire surface of a body due to fluid pressure. If a pressure \(P\) is applied, the hydraulic stress developed in the body is numerically equal to this pressure.
Mathematically, hydraulic stress is expressed as
\[ \begin{aligned} \text{Hydraulic stress} &= \frac{\text{Normal force}}{\text{Area}} = P \end{aligned} \]
The SI unit of hydraulic stress is the pascal (Pa), the same as that of pressure. Since the stress acts equally in all directions, it does not produce any shear or change in shape, but only alters the volume of the body.
As a result of hydraulic stress, the body undergoes a volume strain, defined as the fractional change in volume. If the original volume of the body is \(V\) and the decrease in volume is \(\Delta V\), the volume strain produced is
\[ \begin{aligned} \text{Volume strain} &= \frac{\Delta V}{V} \end{aligned} \]
Within the elastic limit, hydraulic stress is directly proportional to the volume strain produced. The constant of proportionality between hydraulic stress and volume strain is known as the bulk modulus of elasticity, which measures the resistance of a material to uniform compression.
HOOKE’S LAW
Hooke’s law states that within the elastic limit of a material, the strain produced in the body is directly proportional to the applied stress. This means that if the applied stress is doubled, the resulting strain also doubles, provided the material is not stretched beyond its elastic range.
Mathematically, Hooke’s law is expressed as
\[ \begin{aligned} \text{Stress} &\propto \text{Strain} \end{aligned} \]
Introducing a constant of proportionality, the relation becomes
\[ \begin{aligned} \text{Stress} &= k \times \text{Strain} \end{aligned} \]
Here, \(k\) is known as the modulus of elasticity. Its value depends on the nature of the material and the type of deformation involved. Different forms of deformation give rise to different elastic moduli such as Young’s modulus, bulk modulus, and shear modulus.
Hooke’s law is valid only up to a certain limit called the elastic limit. Beyond this limit, the proportionality between stress and strain no longer holds, and the material may undergo plastic deformation. Once this happens, the body does not return completely to its original shape after removal of the force.
Thus, Hooke’s law provides a simple yet powerful principle that connects stress and strain, forming the foundation for understanding elasticity, defining elastic constants, and analysing the mechanical behaviour of solids.
STRESS-STRAIN CURVE
Axes and initial linear (elastic) region
The vertical axis represents stress \(\sigma\) (force per unit area) and the horizontal axis represents strain \(\epsilon\) (fractional change in length). For small deformations, stress is proportional to strain, obeying Hooke’s law \[\sigma=E\epsilon\] where \(E\) is Young’s modulus.
The initial straight part of the red curve from the origin to point A corresponds to the elastic region; in this range, if the load is removed, the wire or rod regains its original length completely (no permanent set), as emphasized while defining elastic behavior and Young’s modulus.
Yield point and permanent set
At point B, labeled yield point with stress \(\sigma_y\), the material just begins to yield; beyond this stress the strain increases without much increase in stress and the specimen does not return to its original length when unloaded, leaving a permanent set (shown on the strain axis near \(\lt 1%)\).
The segment from B to C shows the end of the yield region; which is treated as the limit of proportional and elastic behavior, after which the deformation becomes plastic and is used to explain the difference between elastic and plastic deformation in metals.
Plastic region and ultimate stress
From C to D the curve rises again but is now curved: this is the plastic region where strain increases considerably with increasing stress, and the material undergoes permanent elongation even if the load is removed; This regime is used to introduce concepts of ductility and plastic flow.
Point D is marked “Ultimate stress” corresponding to the maximum stress \(\sigma_u\) ; in the NCERT context this is called ultimate tensile strength and is important for choosing materials that can safely withstand large loads in structures and machines.
Beyond D, the stress shown by the engineering curve decreases up to point E (Fracture), even though the specimen is still being stretched; this is because necking reduces the cross‑sectional area, so the same or even lower load gives a smaller stress value when calculated with the original area.
Point E represents fracture of the specimen; the total strain at E (up to about 30% in the figure) is a measure of ductility, and this region is used to contrast ductile materials (large strain at fracture) with brittle materials (very small strain before breaking).
Young’s Modulus
The ratio of tensile (or compressive) stress \((\sigma)\) to the longitudinal strain \((\epsilon)\) is defined as Young’s modulus and is denoted by the symbol Y.
\[Y=\dfrac{\sigma}{\epsilon}\] \[ \begin{aligned} Y&=\dfrac{F/A}{\Delta L/L}\\\\ &=\dfrac{F\times L}{A\times \Delta L} \end{aligned} \]Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress i.e., \(\mathrm{N\ m^{–2}}\) or Pascal (Pa).
Shear Modulus
The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by G. It is also called the modulus of rigidity.
\[ \begin{aligned} G&=\dfrac{\text{shearing stress }(\sigma_s)}{\text{shearing strain}}\\\\ &=\dfrac{F/A}{\Delta x/L}\\\\ &=\dfrac{F\times L}{A\times\Delta x} \end{aligned} \] But, \[\dfrac{\Delta x}{L}=\theta\] \[G=\dfrac{F}{A\times\theta}\]The shearing stress \(\sigma_s\) can also be expressed as \[\sigma_s=G\times \theta\]
SI unit of shear modulus is \(\mathrm{N\ m^{–2}}\) or Pa.
Bulk Modulus
when a body is submerged in a fluid, it undergoes a hydraulic stress (equal in magnitude to the hydraulic
pressure). This leads to the decrease in the volume of the body thus producing a strain called volume
strain
The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus. It is denoted by
symbol B.
The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs. That
is, if \(p\) is positive, \(\Delta V\) is negative.
Thus for a system in equilibrium, the value of bulk modulus B is always positive. SI unit of bulk modulus is
the same as that of pressure i.e., \(\mathrm{N\ m^{–2}}\) or Pa.
Compressibility
The reciprocal of the bulk modulus is called compressibility and is denoted by \(k\). It is defined as the fractional change in volume per unit increase in pressure.
\[ \begin{aligned} k&=\dfrac{1}{B}\\\\ &=-\dfrac{1}{\Delta p}\times\dfrac{\Delta V}{V} \end{aligned} \]POISSON’S RATIO
The strain perpendicular to the applied force is called lateral strain. Simon Poisson pointed out that within the elastic limit, lateral strain is directly proportional to the longitudinal strain. The ratio of the lateral strain to the longitudinal strain in a stretched wire is called Poisson’s ratio.
If the original diameter of the wire is \(d\) and the contraction of the diameter under stress is \(\Delta d\), the lateral strain is \[\dfrac{\Delta d}{d}\] If the original length of the wire is \(L\) and the elongation under stress is \(\Delta L\), the longitudinal strain is \[\Delta L/L\]
Poisson’s ratio is then
\[\begin{aligned} \dfrac{\Delta d/d}{\Delta L/L}\\\\ \dfrac{\Delta d}{\Delta d}\times\dfrac{L}{d} \end{aligned} \]Elastic Potential Energy in a Stretched Wire
Elastic Potential Energy in a Stretched Wire refers to the energy stored within a wire when it is elongated under the action of an external stretching force, provided the deformation remains within the elastic limit. This energy arises because work is done against the internal restoring forces of the material, which attempt to bring the wire back to its original length once the applied force is removed.
When a wire obeys Hooke’s law, the extension produced is directly proportional to the applied force. This proportionality ensures that the energy stored is recoverable and purely elastic in nature, distinguishing it from permanent (plastic) deformation.
Consider a wire of original length \(L\), cross-sectional area \(A\), and
Young’s modulus \(Y\).
Let the wire be stretched by a small extension \(l\)
within its elastic limit. If \(F\) is the stretching force, then from the definition of
Young’s modulus,
\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{l/L} \]
Rearranging this expression, the force required to produce an extension \(l\) is
\[ F = \frac{YA}{L}\,l \]
This equation shows that the force increases linearly with extension, a hallmark of elastic behavior. During stretching, the force does not jump instantly to its maximum value; instead, it gradually increases from zero to \(F\). Therefore, the work done in stretching the wire must be calculated using the average force.
The work done \(dW\) in producing a small additional extension \(dl\) is
\[ dW = F\,dl \]
Substituting \(F = \frac{YA}{L}l\) and integrating from zero extension to final extension \(l\),
\[ \begin{aligned} W &= \int\limits_0^{l} \frac{YA}{L} \, l \, dl \\\\ &= \frac{YA}{L} \int_0^{l} l \, dl \\\\ &= \frac{YA}{L} \left[ \frac{l^2}{2} \right]_0^{l} \\\\ &= \frac{1}{2}\frac{YA}{L} l^2 \end{aligned} \]
This work done is stored in the wire as elastic potential energy. Hence, the elastic potential energy \(U\) of a stretched wire is
\[ U = \frac{1}{2}\frac{YA}{L} l^2 \]
The above result can also be expressed in alternative forms using stress and strain. Since stress \(\sigma = \frac{F}{A}\) and strain \(\epsilon = \frac{l}{L}\), the energy stored per unit volume, known as elastic energy density, becomes
\[ u = \frac{U}{AL} = \frac{1}{2} \sigma \epsilon \]
Important Aspects and Physical Significance: The elastic potential energy exists only as long as the wire remains within its elastic limit; once this limit is crossed, part of the work done is lost in permanent deformation and cannot be recovered. The quadratic dependence on extension indicates that doubling the extension results in four times the stored energy. This concept is crucial in understanding the behavior of suspension cables, piano wires, and structural elements, where energy storage and release play a key role in mechanical stability.
Thus, elastic potential energy in a stretched wire represents the reversible mechanical energy stored due to elastic deformation, fully consistent with the principles and formulations presented in the NCERT Class XI chapter on Mechanical Properties of Solids.
APPLICATIONS OF ELASTICBEHAVIOUR OF MATERIALS
Applications of Elastic Behaviour of Materials arise from the fundamental property of solids to regain their original shape and size after the removal of deforming forces, provided the deformation remains within the elastic limit. In the NCERT framework, elastic behaviour is not treated as an abstract idea but as a practical principle that governs the design, safety, and functioning of real-world structures and devices.
Elasticity allows materials to store mechanical energy, withstand external loads, and respond predictably to applied stresses. The quantitative measures such as Young’s modulus, bulk modulus, and shear modulus help engineers and physicists choose suitable materials for specific applications.
- 1. Design of Bridges, Buildings, and Structural Elements: Beams, pillars, and girders used in buildings and bridges are continuously subjected to tensile, compressive, and bending stresses. Materials with a high Young’s modulus are preferred because they undergo very small strain for a given stress, ensuring dimensional stability.
- For a wire or rod of length \(L\), area \(A\), and Young’s modulus \(Y\), subjected to a force \(F\), the extension \(l\) is
- \[ \begin{aligned} Y &= \frac{F/A}{l/L} \\\\ \Rightarrow l &= \frac{FL}{YA} \end{aligned} \]
- A smaller extension \(l\) for the same load ensures that the structure does not deform dangerously, thereby providing safety and durability.
- 2. Suspension Cables and Lifting Machines: Cranes, elevators, and suspension bridges rely on steel cables that must support large loads without excessive elongation. Elastic behaviour ensures that when a load is applied, the cable stretches only slightly and returns to its original length when the load is removed.
- The elastic potential energy stored in a stretched cable also acts as a buffer against sudden jerks, reducing the risk of mechanical failure.
- 3. Springs and Shock Absorbers: Springs are classic examples of elastic materials obeying Hooke’s law within limits. They are widely used in vehicle suspensions, weighing machines, and mechanical clocks. \[ \begin{aligned} F &\propto x \ F &= kx \end{aligned} \]
- Here, \(k\) is the spring constant. The predictable linear relation between force and extension allows springs to absorb shocks and vibrations smoothly, converting kinetic energy into elastic potential energy and releasing it gradually.
- 4. Use of Bulk Modulus in Hydraulic Systems: Liquids are often assumed incompressible, but in reality, they exhibit slight compressibility described by bulk modulus \(K\). In hydraulic presses and brakes, a large bulk modulus ensures that pressure applied at one point is transmitted efficiently throughout the fluid. \[ \begin{aligned} K &= -\frac{\Delta P}{\Delta V / V} \end{aligned} \]
- A high value of \(K\) means a very small change in volume for a large pressure, which is essential for the effective functioning of hydraulic devices.
- 5. Selection of Materials for Musical Instruments: Strings of musical instruments such as guitars and violins depend on elastic behaviour. The tension in the string, along with its elastic properties, determines the frequency of vibration and hence the pitch of sound produced. Stable elasticity ensures consistent sound quality.
- 6. Safety in Mechanical Design: Knowledge of elastic limits helps prevent permanent deformation and fracture. Engineers always design machines so that operational stresses remain well below the elastic limit, ensuring that components return to their original shape after use.
Important Aspects and Physical Significance: Elastic behaviour enables energy storage, shock absorption, and structural stability. Materials with appropriate elastic constants are chosen based on whether rigidity, flexibility, or compressibility is required. The linear stress–strain relationship within the elastic limit provides a reliable mathematical foundation for predicting material response.
In conclusion, applications of elastic behaviour form the backbone of modern engineering and technology. From towering buildings to delicate instruments, the principles of elasticity discussed in NCERT Class XI Physics ensure safety, efficiency, and reliability in real-life systems.
Example-1
A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is \(2.0 × 10^{11}\ N\ m^{-2}\)
Solution
Given data:
Radius of steel rod, \(r = 10\,\text{mm} = 10 \times 10^{-3}\,\text{m}\)
Length of rod, \(L = 1.0\,\text{m}\)
Force applied, \(F = 100\,\text{kN} = 100 \times 10^{3}\,\text{N}\)
Young’s modulus of structural steel, \(Y = 2.0 \times 10^{11}\,\text{N m}^{-2}\)
(a) Stress on the rod:
Cross-sectional area of the rod,
\[
A = \pi r^{2} = \pi (10 \times 10^{-3})^{2}
= \pi \times 10^{-4}\,\text{m}^{2}
\]
Stress,
\[
S = \dfrac{F}{A}
= \dfrac{100 \times 10^{3}}{\pi \times 10^{-4}}
= \dfrac{100 \times 10^{7}}{\pi}
\approx 3.2 \times 10^{8}\,\text{N m}^{-2}
\]
(b) Elongation of the rod:
Using \(Y = \dfrac{S}{\varepsilon} = \dfrac{FL}{A \Delta L}\),
\[
\begin{aligned}
\Delta L &= \dfrac{F L}{A Y}\\\\
&= \dfrac{100 \times 10^{3} \times 1}{\pi \times 10^{-4} \times 2.0 \times 10^{11}}\\\\
&= \dfrac{100}{2\pi} \times 10^{-4}\,\text{m}\\\\
&\approx 1.6 \times 10^{-4}\,\text{m}\\\\
&= 0.16\,\text{mm}
\end{aligned}
\]
(c) Strain produced in the rod:
\[
\begin{aligned}
\text{Strain},\ \varepsilon &= \dfrac{\Delta L}{L}\\\\
&= \dfrac{1.6 \times 10^{-4}}{1}\\\\
&= 1.6 \times 10^{-4}\\\\
\end{aligned}
\]
In percentage,
\[
\varepsilon (\%) = 1.6 \times 10^{-4} \times 100
= 0.016\%
\]
Example-2
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Solution
Given:
Length of copper wire, \(L_{c} = 2.2\,\text{m}\)
Length of steel wire, \(L_{s} = 1.6\,\text{m}\)
Diameter of each wire \(= 3.0\,\text{mm}\), therefore radius,
\(r = 1.5\,\text{mm} = 1.5 \times 10^{-3}\,\text{m}\)
Net elongation of the combination,
\(\Delta L = 0.70\,\text{mm} = 0.70 \times 10^{-3}\,\text{m}\)
Young’s modulus of steel, \(Y_{s} = 200 \times 10^{9}\,\text{N m}^{-2}\)
Young’s modulus of copper, \(Y_{c} = 110 \times 10^{9}\,\text{N m}^{-2}\)
Net elongation is the sum of elongations in the two wires, \[ \Delta L = \Delta L_{c} + \Delta L_{s} \] Under the same load and hence same tension, the stress in each wire is the same, so \[ S_{s} = S_{c} = \dfrac{F}{A} \] and therefore, using \(S = Y \dfrac{\Delta L}{L}\), \[ \begin{aligned} Y_{s} \dfrac{\Delta L_{s}}{L_{s}} = Y_{c} \dfrac{\Delta L_{c}}{L_{c}} \end{aligned} \] This gives \[ \begin{aligned} \dfrac{\Delta L_{s}}{\Delta L_{c}} &= \dfrac{Y_{c}}{Y_{s}} \dfrac{L_{s}}{L_{c}}\\\\ &= \dfrac{110}{200} \cdot \dfrac{1.6}{2.2}\\\\ &= \dfrac{11}{20} \cdot \dfrac{16}{22}\\\\ &= \dfrac{4}{10}\\\\ &= \dfrac{2}{5}\\\\ \end{aligned} \] Hence \[ \Delta L_{s} = \dfrac{2}{5} \Delta L_{c}. \]
Using the condition on total elongation, \[ \begin{aligned} \Delta L_{c} + \Delta L_{s} &= \Delta L\\ &= 0.70 \times 10^{-3}\,\text{m}, \end{aligned} \] so \[ \begin{aligned} \Delta L_{c} + \dfrac{2}{5} \Delta L_{c} &= 0.70 \times 10^{-3}, \\\\ \dfrac{7}{5} \Delta L_{c} &= 0.70 \times 10^{-3}\\\\ \Delta L_{c} &= \dfrac{5}{7} \times 0.70 \times 10^{-3}\\\\ &= 0.50 \times 10^{-3}\,\text{m} \end{aligned} \] Therefore \[ \begin{aligned} \Delta L_{s} &= \dfrac{2}{5} \Delta L_{c}\\\\ &= \dfrac{2}{5} \times 0.50 \times 10^{-3}\\\\ &= 0.20 \times 10^{-3}\,\text{m}\\\\ \end{aligned} \]
To find the load, use the steel wire (same load passes through both). For steel, \[ \begin{aligned} \dfrac{F}{A} &= Y_{s} \dfrac{\Delta L_{s}}{L_{s}}\\\\ \Rightarrow F &= Y_{s} A \dfrac{\Delta L_{s}}{L_{s}} \end{aligned} \] where the common cross-sectional area is \[ \begin{aligned} A &= \pi r^{2}\\\\ &= \pi (1.5 \times 10^{-3})^{2}\\\\ &= \pi \times 2.25 \times 10^{-6}\,\text{m}^{2} \end{aligned} \] Thus \[ \begin{aligned} F &= 200 \times 10^{9} \times \pi \times 2.25 \times 10^{-6} \times \dfrac{0.20 \times 10^{-3}}{1.6}\\\\ &= 200 \times 10^{9} \times \pi \times 2.25 \times 10^{-6} \times 1.25 \times 10^{-4}\\\\ &\approx 180\,\text{N} \end{aligned} \] Hence, the load applied on the combined wire system is \[ F \approx 1.8 \times 10^{2}\,\text{N}. \]
Example-3
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig. 8.4). The combined mass of all the persons performing the act, and the tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.
Solution
Total mass of the performers, tables and plaques
\(= 280\,\text{kg}\)
Mass of the performer lying on the ground
\(= 60\,\text{kg}\)
Effective mass acting on the legs of the performer at the bottom
\[
m = 280 - 60 = 220\,\text{kg}
\]
Therefore, total weight supported by his legs
\[
W = mg = 220 \times 9.8\,\text{N} = 2156\,\text{N}
\]
Since the weight is shared equally by the two legs,
\[
F = \dfrac{W}{2} = \dfrac{2156}{2} = 1078\,\text{N}
\]
is the force on each thighbone.
Young’s modulus of bone
\(Y = 9.4 \times 10^{9}\,\text{N m}^{-2}\)
Length of each femur
\(L = 0.50\,\text{m}\)
Radius of femur
\(r = 2.0\,\text{cm} = 2.0 \times 10^{-2}\,\text{m}\)
Cross-sectional area of each femur
\[
\begin{aligned}
A &= \pi r^{2}\\\\
&= \pi (2.0 \times 10^{-2})^{2}\\\\
&= \pi \times 4.0 \times 10^{-4}\,\text{m}^{2}
\end{aligned}
\]
Using \[ \begin{aligned} Y &= \dfrac{F}{A} \dfrac{L}{\Delta L}\\\\ \Rightarrow \Delta L &= \dfrac{F L}{Y A} \end{aligned} \] the compression of each thighbone is \[ \begin{aligned} \Delta L &= \dfrac{1078 \times 0.50} {9.4 \times 10^{9} \times \pi \times 4.0 \times 10^{-4}}\\\\ \Delta L &= \dfrac{539} {9.4 \times 10^{9} \times 4\pi \times 10^{-4}}\\\\ &\approx 4.6 \times 10^{-5}\,\text{m} \end{aligned} \] Hence, each thighbone gets compressed by \[ \begin{aligned} \Delta L &\approx 4.6 \times 10^{-5}\,\text{m}\\ &(\approx 0.046\,\text{mm}) \end{aligned} \]
Example-4
A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of (9.0 ×10^4\ N\). The lower edge is riveted to the floor. How much will the upper edge be displaced?
Solution
Side of square lead slab \(= 50\,\text{cm} = 0.50\,\text{m}\)
Thickness (distance between the two faces) \(L = 10\,\text{cm} = 0.10\,\text{m}\)
Shearing force applied on the narrow face
\(F = 9.0 \times 10^{4}\,\text{N}\)
Shear modulus of lead
\(G = 5.6 \times 10^{9}\,\text{N m}^{-2}\)
Area of the face on which the shearing force acts is \[ \begin{aligned} A &= 0.50 \times 0.50\\\\ &= 0.25\,\text{m}^{2} \end{aligned} \] Using the relation for shear modulus, \[ \begin{aligned} G &= \dfrac{F}{A} \dfrac{L}{\Delta x}\\\\ \Rightarrow \Delta x &= \dfrac{F L}{G A} \end{aligned} \] the displacement of the upper edge is \[ \begin{aligned} \Delta x &= \dfrac{9.0 \times 10^{4} \times 0.10} {5.6 \times 10^{9} \times 0.25}\\\\ &= \dfrac{9.0 \times 10^{3}} {1.4 \times 10^{9}}\\\\ &\approx 6.4 \times 10^{-6}\,\text{m} \end{aligned} \] Thus, the upper edge of the slab is displaced by \[ \begin{aligned} \Delta x &\approx 6.4 \times 10^{-6}\,\text{m}\\\\ &(\approx 6.4\,\mu\text{m}) \end{aligned} \]
Example-5
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, ∆V/V, of water at the bottom of the ocean, given that the bulk modulus of water is \(2.2 \times 10^9\ N\ m^{–2}\). (Take g = \(10\ m\ s^{–2}\))
Solution
Average depth of Indian Ocean \(h = 3000\,\text{m}\)
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 \(\rho \approx 1000\,\text{kg m}^{-3}\)
The pressure at a depth \(h\) is \[ \begin{aligned} P &= \rho g h\\ &= 1000 \times 10 \times 3000\\ &= 3.0 \times 10^{7}\,\text{N m}^{-2} \end{aligned} \] From the definition of bulk modulus, \[ \begin{aligned} B &= \dfrac{P}{\Delta V / V}\\\\ \Rightarrow \dfrac{\Delta V}{V} &= \dfrac{P}{B} \end{aligned} \] Hence, the fractional compression is \[ \begin{aligned} \dfrac{\Delta V}{V} &= \dfrac{3.0 \times 10^{7}} {2.2 \times 10^{9}}\\\\ &\approx 1.36 \times 10^{-2} \end{aligned} \] Therefore, the water at the bottom is compressed by a fraction \(\Delta V / V \approx 1.4 \times 10^{-2}\).
| Basis of Comparison | Tensile / Compressive | Shearing | Hydraulic (Volumetric) |
|---|---|---|---|
| Nature of Deforming Force | Force acts along the length of the body, either pulling or pushing | Force acts parallel to the surface | Force acts normally and equally in all directions |
| Type of Stress | Tensile Stress / Compressive Stress | Shearing Stress | Hydraulic Stress |
| Definition of Stress | Ratio of longitudinal force to cross-sectional area | Ratio of tangential force to surface area | Ratio of normal force to area acting uniformly |
| Expression of Stress | \(\sigma = \dfrac{F}{A}\) | \(\tau = \dfrac{F}{A}\) | \(p = \dfrac{F}{A}\) |
| Type of Strain Produced | Longitudinal Strain | Shearing Strain | Volumetric Strain |
| Definition of Strain | Ratio of change in length to original length | Ratio of lateral displacement to original dimension | Ratio of change in volume to original volume |
| Expression of Strain | \(\varepsilon = \dfrac{\Delta L}{L}\) | \(\gamma = \tan\theta \approx \theta\) | \(\varepsilon_v = \dfrac{\Delta V}{V}\) |
| Change Produced in Body | Change in length | Change in shape (angular distortion) | Change in volume |
| Associated Elastic Modulus | Young’s Modulus (Y) | Modulus of Rigidity (G or η) | Bulk Modulus (K) |
| Definition of Elastic Modulus | Ratio of longitudinal stress to longitudinal strain | Ratio of shearing stress to shearing strain | Ratio of hydraulic stress to volumetric strain |
| Expression of Elastic Modulus | \(Y = \dfrac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}\) | \(G = \dfrac{\text{Shearing Stress}}{\text{Shearing Strain}}\) | \(K = \dfrac{\text{Hydraulic Stress}}{\text{Volumetric Strain}}\) |
| SI Unit of Modulus | Pascal (Pa) | Pascal (Pa) | Pascal (Pa) |
| Typical Example | Stretching a metal wire | Deforming a cube by tangential force | Compression of a fluid in a closed container |
| Applicability | Solids | Solids only | Solids and fluids |
