Mechanical Properties of Solids - Comprehensive Notes

Chapter 8: Mechanical Properties of Solids

8.1 Introduction

Rigid bodies are hard solid objects with definite shape and size, but in reality, bodies can be deformed by stretching, compressing, or bending.
Even steel bars can be deformed by sufficient external force, meaning solid bodies are not perfectly rigid.
Solids have definite shape and size, and force is required to change their shape or size.
Elasticity is the property of a body to regain its original size and shape when the applied force is removed, resulting in elastic deformation.
Plasticity is the property of substances like putty or mud to be permanently deformed without regaining their original shape.
Elastic behavior is important in engineering design for materials like steel and concrete in buildings, bridges, and automobiles.
Questions addressed include designing light yet strong airplanes or artificial limbs, understanding railway track shapes, and why glass is brittle while brass is not.

8.2 Stress and Strain

When forces are applied to a body in static equilibrium, it deforms depending on the material and force magnitude.
Deformation may not always be visually noticeable.
A restoring force develops in the body, equal in magnitude but opposite in direction to the applied force.
Stress is the restoring force per unit area, defined as: $Stress = F/A$ , where $F$ is the force and $A$ is the area.
The SI unit of stress is N/m² or pascal (Pa), with a dimensional formula of $[ML^{-1}T^{-2}]$ .
There are three ways a solid changes dimensions under external force:
- Tensile Stress: Cylinder stretched by forces normal to the cross-sectional area, restoring force per unit area is tensile stress.
- Compressive Stress: Cylinder compressed under applied forces, restoring force per unit area is compressive stress. Tensile and compressive stresses are also termed longitudinal stress.
- Longitudinal Strain: The change in length $\Delta L$ to the original length $L$ , defined as: $Longitudinal \, strain = \frac{\Delta L}{L}$ .
- Shearing Stress: Equal and opposite forces applied parallel to the cross-sectional area, causing relative displacement between opposite faces.
- Tangential or Shearing Stress: Restoring force per unit area due to tangential force.
- Shearing Strain: Ratio of relative displacement of faces $\Delta x$ to the length $L$ , defined as: $Shearing \, strain = \frac{\Delta x}{L} = \tan \theta \approx \theta$ , where $\theta$ is the angular displacement; usually $\theta$ is very small.
- Hydraulic Stress: Solid sphere in fluid under high pressure compressed uniformly on all sides.
- Hydraulic Stress: Internal restoring force per unit area, equal to hydraulic pressure.
- Volume Strain: Ratio of change in volume $\Delta V$ to the original volume $V$ , defined as: $Volume \, strain = \frac{\Delta V}{V}$ .
Strain is dimensionless, having no units or dimensional formula.

8.3 Hooke’s Law

For small deformations, stress and strain are proportional.
Hooke’s Law: $stress \propto strain$
Expressed as: $stress = k \times strain$ , where $k$ is the modulus of elasticity.
Hooke’s law is an empirical law valid for most materials, but some materials do not exhibit this linear relationship.

8.4 Stress-Strain Curve

The stress-strain relationship for a material under tensile stress is found experimentally by stretching a test cylinder or wire and recording fractional change in length (strain) and applied force.
The applied force is gradually increased, and the change in length is noted, plotting a graph of stress versus strain.
Stress-strain curves vary by material, aiding in understanding deformation under increasing loads.
Key points in the curve:
- O to A: Linear region where Hooke’s law is obeyed, and the body regains original dimensions when force is removed (elastic behavior).
- A to B: Stress and strain are not proportional, but the body still returns to its original dimension when the load is removed.
- B: Yield point (elastic limit), with corresponding yield strength $\sigma_y$ .
- B to D: Stress exceeds yield strength, and strain increases rapidly even with small stress changes. Removing load at point C results in a permanent set (plastic deformation).
- D: Ultimate tensile strength $\sigma_u$ , beyond which additional strain occurs even with reduced force, leading to fracture at point E.
Brittle materials have close ultimate strength and fracture points, while ductile materials have them far apart.
Elastomers (e.g., rubber) can be stretched to several times their original length and still return to their original shape, with a large elastic region but without obeying Hooke’s law over most of the region and no well-defined plastic region.

8.5 Elastic Moduli

The proportional region of the stress-strain curve (OA in Fig. 8.2) is crucial for structural and manufacturing engineering designs.
Modulus of elasticity is the ratio of stress to strain, characteristic of the material.

8.5.1 Young’s Modulus

For a given material, the magnitude of strain is the same whether the stress is tensile or compressive.
Young’s modulus (Y) is the ratio of tensile (or compressive) stress $\sigma$ to the longitudinal strain $\epsilon$ .
$Y = \frac{\sigma}{\epsilon}$
$Y = \frac{(F/A)}{(\Delta L/L)} = \frac{F \times L}{A \times \Delta L}$
The unit of Young’s modulus is N/m² or Pascal (Pa), the same as stress, since strain is dimensionless.
Metals have large Young’s moduli, requiring large force for small length changes.
Steel is more elastic than copper, brass, and aluminum; preferred in heavy-duty machines and structural designs.
Wood, bone, concrete, and glass have smaller Young’s moduli.
Example 8.1: Calculation of stress, elongation, and strain on a structural steel rod with radius 10 mm and length 1.0 m under a 100 kN force, given Young’s modulus of structural steel is $2.0 \times 10^{11} \, N/m^2$ .
- Stress: $\frac{F}{A} = \frac{100 \times 10^3 \, N}{\pi (10 \times 10^{-3} \, m)^2} = 3.18 \times 10^8 \, N/m^2$
- Elongation: $\Delta L = \frac{(F/A)L}{Y} = \frac{3.18 \times 10^8 \, N/m^2 \times 1 \, m}{2 \times 10^{11} \, N/m^2} = 1.59 \times 10^{-3} \, m = 1.59 \, mm$
- Strain: $\frac{\Delta L}{L} = \frac{1.59 \times 10^{-3} \, m}{1 \, m} = 1.59 \times 10^{-3} = 0.16 \%$ .
Example 8.2: Copper wire of length 2.2 m and steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched, the net elongation is 0.70 mm. Obtain the load applied.
- The copper and steel wires are under the same tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A.
- $\frac{W}{A} = Yc \times \frac{\Delta Lc}{Lc} = Ys \times \frac{\Delta Ls}{Ls}$
- $\frac{\Delta Lc}{\Delta Ls} = \frac{Ys}{Yc} \times \frac{Lc}{Ls}$
- Given $Lc = 2.2$ m, $Ls = 1.6$ m, $Yc = 1.1 \times 10^{11} N.m^{–2}$ , and $Ys = 2.0 \times 10^{11} N.m^{–2}$ .
- $\frac{\Delta Lc}{\Delta Ls} = (\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}) \times (\frac{2.2}{1.6}) = 2.5$
- The total elongation is given to be $\Delta Lc + \Delta Ls = 7.0 \times 10^{-4} m$
- Solving the above equations, $\Delta Lc = 5.0 \times 10^{-4} m$ , and $\Delta Ls = 2.0 \times 10^{-4} m$
- Therefore $W = (A \times Yc \times \Delta Lc)/L_c$
- $W = \pi (1.5 \times 10^{-3})^2 \times [(5.0 \times 10^{-4} \times 1.1 \times 10^{11})/2.2] = 1.8 \times 10^2 N$
Example 8.3: Human pyramid supported by the legs of a performer. The combined mass is 280 kg, and the performer's mass is 60 kg. Each thighbone is 50 cm long with a 2.0 cm radius. Determine the compression in each thighbone.
- The mass supported by the legs is 280 kg – 60 kg = 220 kg.
- The weight supported is $220 \times 9.8 N = 2156 N$ .
- Weight supported by each thighbone is $\frac{1}{2} (2156) N = 1078 N$ .
- From Table 9.1, the Young’s modulus for bone is given by $Y = 9.4 \times 10^9 N m^{–2}$ .
- Length of each thighbone L = 0.5 m and radius = 2.0 cm.
- The cross-sectional area of the thighbone is $A = \pi \times (2 \times 10^{-2})^2 m^2 = 1.26 \times 10^{-3} m^2$ .
- Using $\Delta L = \frac{F \times L}{Y \times A}$ , the compression in each thighbone can be computed as:
- $\Delta L = \frac{1078 \times 0.5}{9.4 \times 10^9 \times 1.26 \times 10^{-3}} = 4.55 \times 10^{-5} m$
- This is a very small change! The fractional decrease in the thighbone is $\frac{\Delta L}{L} = 0.000091$ or 0.0091%.

8.5.2 Shear Modulus

Shear modulus (G) is the ratio of shearing stress to shearing strain, also called the modulus of rigidity.
$G = \frac{shearing \, stress}{shearing \, strain}$
$G = \frac{(F/A)}{(\Delta x/L)} = \frac{F \times L}{A \times \Delta x}$
$G = \frac{(F/A)}{\theta} = \frac{F}{A \times \theta}$
Shearing stress can be expressed as $\sigma_s = G \times \theta$ .
The SI unit is N/m² or Pa.
Shear modulus is generally less than Young’s modulus, with $G \approx Y/3$ for most materials.
Example 8.4: A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force of $9.0 \times 10^4 N$ . The lower edge is riveted to the floor. Find the displacement of the upper edge.
- The area of the face parallel to which this force is applied is $A = 50 \, cm \times 10 \, cm = 0.5 \, m \times 0.1 \, m = 0.05 \, m^2$
- Therefore, the stress applied is $= (9.4 \times 10^4 N/0.05 m^2) = 1.80 \times 10^6 N.m^{–2}$
- Shearing strain = $\frac{\Delta x}{L} = \frac{Stress}{G}$ .
- Therefore the displacement $\Delta x = \frac{Stress \times L}{G}$
- $\Delta x = (\frac{1.8 \times 10^6 N m^{–2} \times 0.5m}{5.6 \times 10^9 N m^{–2}}) = 1.6 \times 10^{–4} m = 0.16 mm$

8.5.3 Bulk Modulus

When a body is submerged in a fluid, it undergoes hydraulic stress, leading to a decrease in volume and producing volume strain.
Bulk modulus (B) is the ratio of hydraulic stress to the corresponding hydraulic strain.
$B = -\frac{p}{(\Delta V/V)}$
The negative sign indicates that with an increase in pressure, a decrease in volume occurs.
The SI unit is N/m² or Pa.
Compressibility (k) is the reciprocal of the bulk modulus.
$k = \frac{1}{B} = -\frac{1}{\Delta p} \times \frac{\Delta V}{V}$
Bulk moduli for solids are much larger than for liquids, which are again much larger than for gases.
Solids are the least compressible, and gases are the most compressible.
Example 8.5: The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, $\frac{\Delta V}{V}$ , 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} $<ul><li>The pressure exerted by a 3000 m column of water on the bottom layer$ p = h \rho g = 3000 m \times 1000 kg m^{–3} \times 10 m s^{–2} = 3 \times 10^7 kg m^{–1} s^{-2} = 3 \times 10^7 N m^{–2} $</li><li>Fractional compression$ \frac{\Delta V}{V} $, is$ \frac{\Delta V}{V} = \frac{stress}{B} = \frac{3 \times 10^7 N m^{-2}}{2.2 \times 10^9 N m^{–2}} = 1.36 \times 10^{-2} $or 1.36 %</li></ul></li></ul><h5 id="34e0c560-fff0-47b0-8fca-b98527181eb3" data-toc-id="34e0c560-fff0-47b0-8fca-b98527181eb3" collapsed="false" seolevelmigrated="true">8.5.4 Poisson’s Ratio</h5><ul><li>Lateral strain is the strain perpendicular to the applied force.</li><li>Poisson’s ratio is the ratio of lateral strain to longitudinal strain within the elastic limit.</li><li>If the original diameter of the wire is d and the contraction of the diameter under stress is$ \Delta d $, the lateral strain is$ \frac{\Delta d}{d} $.</li><li>If the original length of the wire is L and the elongation under stress is$ \Delta L $, the longitudinal strain is$ \frac{\Delta L}{L} $.</li><li>Poisson’s ratio is then$ (\frac{\Delta d}{d})/(\frac{\Delta L}{L}) $or$ (\frac{\Delta d}{\Delta L}) \times (\frac{L}{d}) $.</li><li>Poisson’s ratio is a pure number with no dimensions or units, depending only on the material. It is between 0.28 and 0.30 for steels, and about 0.33 for aluminum alloys.</li></ul><h5 id="280f8fd8-4550-4c6f-b882-faed44ebe44a" data-toc-id="280f8fd8-4550-4c6f-b882-faed44ebe44a" collapsed="false" seolevelmigrated="true">8.5.5 Elastic Potential Energy in a Stretched Wire</h5><ul><li>When a wire is put under tensile stress, work is done against inter-atomic forces, stored as elastic potential energy.</li><li>For a wire of original length L and area of cross-section A subjected to force F along the length, let the length be elongated by l. Then from Eq. (8.8), we have$ F = YA \times (l/L) $.</li><li>For a further elongation of infinitesimal small length dl, work done dW is F dl or$ \frac{YAldl}{L} $.</li><li>The amount of work done (W) in increasing the length of the wire from L to L + l, that is from l = 0 to l = l is<ul><li>$ W = \int{0}^{l} \frac{YAl}{L} dl = \frac{YA}{L} \int{0}^{l} l dl = \frac{YA}{L} [\frac{l^2}{2}]_{0}^{l} = \frac{YA l^2}{2L} $</li></ul></li><li>$ W = \frac{1}{2} \times \frac{Y}{L} AL l^2 = \frac{1}{2} \times Young’s \, modulus \times strain^2 \times volume \, of \, the \, wire $</li><li>$ W = \frac{1}{2} \times stress \times strain \times volume \, of \, the \, wire $</li><li>This work is stored in the wire in the form of elastic potential energy (U).</li><li>Therefore the elastic potential energy per unit volume of the wire (u) is<ul><li>$ u = \frac{1}{2} \sigma \epsilon $</li></ul></li></ul><h4 id="822c1c41-c63d-46db-ae18-b8e671f0f52d" data-toc-id="822c1c41-c63d-46db-ae18-b8e671f0f52d" collapsed="false" seolevelmigrated="true">8.6 Applications of Elastic Behavior of Materials</h4><ul><li>The elastic behavior of materials plays an important role in everyday life.</li><li>All engineering designs require precise knowledge of the elastic behavior of materials.</li><li>For example while designing a building, the structural design of the columns, beams and supports require knowledge of strength of materials used.</li><li>Examples include bridge and building design, crane construction, etc.</li><li>Cranes use thick metal ropes to lift heavy loads.</li><li>To make a crane with a lifting capacity of 10 metric tons (1 metric ton = 1000 kg), the steel rope thickness is calculated to avoid permanent deformation, ensuring extension does not exceed the elastic limit.</li><li>Mild steel has a yield strength$ \sigma_y $of about$ 300 \times 10^6 N m^{–2} $.</li><li>Area of cross-section (A) of the rope:<ul><li>$ A ≥ \frac{W}{\sigmay} = \frac{Mg}{\sigmay} $</li><li>$ A = \frac{10^4 kg \times 9.8 m s^{-2}}{300 \times 10^6 N m^{-2}} = 3.3 \times 10^{-4} m^2 $</li><li>This corresponds to a radius of about 1 cm for a rope of circular cross-section.</li></ul></li><li>A large margin of safety (of about a factor of ten in the load) is generally provided; thus a thicker rope of radius about 3 cm is recommended.</li><li>Ropes are made of braided thin wires for ease of manufacture, flexibility, and strength.</li><li>Bridges and buildings use beams and columns, addressing the problem of bending under load.</li><li>A bar of length l, breadth b, and depth d when loaded at the centre by a load W sags by an amount given by<ul><li>$ \delta = \frac{W l^3}{4bd^3Y} $</li></ul></li><li>To reduce bending for a given load, use a material with a large Young’s modulus Y.</li><li>Increasing the depth d rather than the breadth b is more effective in reducing the bending, since$ \delta $is proportional to$ d^{-3} $and only to$ b^{-1} $(of course the length l of the span should be as small as possible).</li><li>A common compromise is the cross-sectional shape which provides a large load-bearing surface and enough depth to prevent bending.</li><li>The use of pillars or columns is also very common in buildings and bridges.</li><li>A pillar with rounded ends supports less load than that with a distributed shape at the ends.</li><li>The precise design of a bridge or a building has to take into account the conditions under which it will function, the cost and long period, reliability of usable material, etc.</li></ul><h3 id="1e6eecc7-8b81-4b9f-8d20-3bd1c37985dd" data-toc-id="1e6eecc7-8b81-4b9f-8d20-3bd1c37985dd" collapsed="false" seolevelmigrated="true">Summary</h3><ol><li>Stress is the restoring force per unit area and strain is the fractional change in dimension.In general there are three types of stresses (a) tensile stress — longitudinal stress (associated with stretching) or compressive stress (associated with compression), (b) shearing stress, and (c) hydraulic stress.</li><li>For small deformations, stress is directly proportional to the strain for many materials. This is known as Hooke’s law. The constant of proportionality is called modulus of elasticity. Three elastic moduli viz., Young’s modulus, shear modulus and bulk modulus are used to describe the elastic behaviour of objects as they respond to deforming forces that act on them. A class of solids called elastomers does not obey Hooke’s law.</li><li>When an object is under tension or compression, the Hooke’s law takes the form$ F/A = Y\Delta L/L $where$ \Delta L/L $is the tensile or compressive strain of the object, F is the magnitude of the applied force causing the strain, A is the cross-sectional area over which F is applied (perpendicular to A) and Y is the Young’s modulus for the object. The stress is F/A.</li><li>A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement$ \Delta L $of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. This type of stress is possible only in solids. In this kind of deformation the Hooke’s law takes the form$ F/A = G \times \Delta L/L $where$ \Delta L $is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus.</li><li>When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke’s law takes the form$ p = B (\Delta V/V) $, where p is the pressure (hydraulic stress) on the object due to the fluid,$ \Delta V/V $(the volume strain) is the absolute fractional change in the object’s volume due to that pressure and B is the bulk modulus of the object.</li></ol><ul><li>The answer to the question why the maximum height of a mountain on earth is ~10 km can also be provided by considering the elastic properties of rocks. A mountain base is not under uniform compression and this provides some shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow. At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg where ρ is the density of the material of the mountain and g is the acceleration due to gravity. The material at the bottom experiences this force in the vertical direction, and the sides of the mountain are free. Therefore, this is not a case of pressure or bulk compression. There is a shear component, approximately hρg itself. Now the elastic limit for a typical rock is$ 30 \times 10^7 N m^{-2} $. Equating this to hρg, with ρ =$ 3 \times 10^3 kg m^{-3} $gives$ hρg = 30 \times 10^7 N m^{-2} $.</li><li>$ h = 30 \times 10^7 N m^{-2}/(3 \times 10^3 kg m^{-3} \times 10 m s^{-2}) $</li><li>$ = 10 km $which is more than the height of Mt. Everest!</li></ul><h3 id="6d1df733-b86c-4814-baa8-c3efcfafe2b4" data-toc-id="6d1df733-b86c-4814-baa8-c3efcfafe2b4" collapsed="false" seolevelmigrated="true">Points to Ponder</h3><ol><li>In the case of a wire, suspended from celing and stretched under the action of a weight (F) suspended from its other end, the force exerted by the ceiling on it is equal and opposite to the weight. However, the tension at any cross-section A of the wire is just F and not 2F. Hence, tensile stress which is equal to the tension per unit area is equal to F/A.</li><li>Hooke’s law is valid only in the linear part of stress-strain curve.</li><li>The Young’s modulus and shear modulus are relevant only for solids since only solids have lengths and shapes.</li><li>Bulk modulus is relevant for solids, liquid and gases. It refers to the change in volume when every part of the body is under the uniform stress so that the shape of the body remains unchanged.</li><li>Metals have larger values of Young’s modulus than alloys and elastomers. A material with large value of Young’s modulus requires a large force to produce small changes in its length.</li><li>In daily life, we feel that a material which stretches more is more elastic, but it a is misnomer. In fact material which stretches to a lesser extent for a given load is considered to be more elastic.</li><li>In general, a deforming force in one direction can produce strains in other directions also. The proportionality between stress and strain in such situations cannot be described by just one elastic constant. For example, for a wire under longitudinal strain, the lateral dimensions (radius of cross section) will undergo a small change, which is described by another elastic constant of the material (called Poisson ratio).</li></ol><h3 id="2aa57eab-7382-4bf0-80cf-c1be3c402516" data-toc-id="2aa57eab-7382-4bf0-80cf-c1be3c402516" collapsed="false" seolevelmigrated="true">Exercises</h3>8.1 A steel wire of length 4.7 m and cross-sectional area$ 3.0 \times 10^{-5} m^2 $stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of$ 4.0 \times 10^{–5} m^2 $under a given load. What is the ratio of the Young’s modulus of steel to that of copper?8.2 Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?8.3 The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale.(a) Which of the materials has the greater Young’s modulus?(b) Which of the two is the stronger material?8.4 Read the following two statements below carefully and state, with reasons, if it is true or false.(a) The Young’s modulus of rubber is greater than that of steel;(b) The stretching of a coil is determined by its shear modulus.8.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.8.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?8.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.8.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?8.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed$ 10^8 N m^{–2} $, what is the maximum load the cable can support ?8.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.8.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is$ 0.065 cm^2 $. Calculate the elongation of the wire when the mass is at the lowest point of its path.8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm =$ 1.013 \times 10^5 Pa $), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.8.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is$ 1.03 \times 10^3 kg m^{–3} $?8.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.8.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of$ 7.0 \times 10^6 Pa$$.
8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?