# Mechanical Properties of Solids

Notes for Mechanical properties of Solids chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

Matter is usually classified into one of the three states or phases: solid, liquid, or gas. Because they can flow easily, both liquids and gases are called fluids.

A solid has a fixed shape which it tends to retain, whereas fluids have no fixed shape.

A liquid sinks to the bottom of its container, and a gas expands to fill the available volume.

The atoms in a solid vibrate about fixed equilibrium positions, whereas the atoms or molecules in a liquid move about relatively freely  and collide frequently with each other.

The atoms in a solid or liquid are quite closely packed, which makes it difficult to reduce their volume, they are almost incompressible.

On the average, the atoms or molecules in a gas are far apart, typically about ten atomic diameters at room temperature and pressure. They collide much less frequently than those in a liquid. Gases in general are compressible.

STRESS, STRAIN AND ELASTIC MODULII

A force applied to an object can change its dimensions and shape. In general, the response of a material to a given type of deforming force is characterized by an elastic modulus, which is defined as

Elastic modulus = ${{Stress} \over {Strain}}$

The precise definition of stress depends on the particular situation being considered, but in general it is a force per unit area.

The strain indicates some fractional change in a dimension or volume. The unit of stress is N/m2, whereas strain is a dimensionless quantity.

### HOOKE'S LAW

If the deformation is small, the stress in a body is proportional to the corresponding strain.

Tensile Stress $\alpha$ Tensile Strain

Tensile Stress = Y x Tensile Strain

This proportionality constant Y is called Young's modulus for the material. It is constant for a given material.

### Young’s Modulus

Young’s modulus is a measure of the resistance of a solid to a change in its length when a force is applied perpendicular to a face. Consider a rod with an unstressed length Lo and cross-sectional area A, as shown in the figure. When it is subjected to equal and opposite forces ­ along it axis and perpendicular to the end faces its length changes by DL. These forces tend to stretch the rod. The tensile stress on the rod is defined as

$\sigma = {{{F_n}} \over A}$

Forces acting in the opposite direction, as shown in figure, would produce a compressive stress. The resulting strain is defined as the dimensionless ratio.

$\varepsilon = {{\Delta L} \over {{L_o}}}$ Young’s modulus Y for the material of the rod is defined as the ratio

Young’s Modulus =${{Tensile\,\,stress} \over {Tensile\,\,strain}}$

Y =${\sigma \over \varepsilon } = {{{F_n}/A} \over {\Delta L/{L_o}}} = {{{F_n}{L_o}} \over {A\Delta L}}$

Application 1

A metal wire 75 cm long and 0.130 cm in diameter stretches 0.035 cm when a load of 8.0 kg is hung on its end. Find the stress, the strain, and the Young’s modulus of the material of the wire.

Solution

Stress = ${F \over A} = {{\left( {8.0kg} \right)\left( {9.8m/{s^2}} \right)} \over {\pi {{\left( {6.5 \times {{10}^{ - 4}}m} \right)}^2}}}$ = 5.91 x 107 N/m2

Strain = ${{\Delta L} \over L} = {{0.035cm} \over {75cm}} = 4.67 \times {10^{ - 4}}$

Y = ${{Stress} \over {Strain}} = {{5.91 \times {{10}^7}N{m^{ - 2}}} \over {4.67 \times {{10}^{ - 4}}}}$ = 1.27 x 1011 Nm-2

Application 2

A solid cylindrical steel column is 4.0 m long and 9.0 cm in diameter. What will be its decrease in length when carrying a load of 80000 kg? Y = 1.9X1011 Nm-2.

Solution

Let us first calculate the cross-sectional area of column = ${{Stress} \over {Strain}} = {{5.91 \times {{10}^7}N{m^{ - 2}}} \over {4.67 \times {{10}^{ - 4}}}}$r2 ${{Stress} \over {Strain}} = {{5.91 \times {{10}^7}N{m^{ - 2}}} \over {4.67 \times {{10}^{ - 4}}}}$(0.045m)2

= 6.36 x 10-3 m2

Then, from Y = (F/A)/( ${{Stress} \over {Strain}} = {{5.91 \times {{10}^7}N{m^{ - 2}}} \over {4.67 \times {{10}^{ - 4}}}}$L/L) we have ${{Stress} \over {Strain}} = {{5.91 \times {{10}^7}N{m^{ - 2}}} \over {4.67 \times {{10}^{ - 4}}}}$L = ${{FL} \over {AY}} = {{\left[ {\left( {8 \times {{10}^4}} \right)\left( {9.8\,N} \right)} \right]\left( {4.0m} \right)} \over {\left( {6.36 \times {{10}^{ - 3}}{m^2}} \right)\left( {1.9 \times {{10}^{11}}\,\,N{m^{ - 2}}} \right)}}$ = 2.6 x 10-3 m = 2.6 mm

### Shear Modulus (Modulus of Rigidity)

The shear modulus of a solid measures its resistance to a shearing force, which is a force applied tangentially to a surface, as shown in the figure. (Since the bottom of the solid is assumed to be at rest, there is an equal and opposite force on the lower surface.) The top surface is displaced by ${{FL} \over {AY}} = {{\left[ {\left( {8 \times {{10}^4}} \right)\left( {9.8\,N} \right)} \right]\left( {4.0m} \right)} \over {\left( {6.36 \times {{10}^{ - 3}}{m^2}} \right)\left( {1.9 \times {{10}^{11}}\,\,N{m^{ - 2}}} \right)}}$x relative to the bottom surface. The shear stress is defined as

Shear Stress = ${{Tangential\,\,force} \over {Area}}$

t = ${{{F_t}} \over A}$

where A is the area of the surface.

The shear strain is defined as

Shear strain = ${x \over y}$

where y is the separation between the top and the bottom surfaces.

The shear modulus G is defined as

Shear modulus = ${{Shear\,\,Stress} \over {Shear\,Strain}}$

G = ${{{F_t}/A} \over {x/y}} = {F \over A}{y \over x}$ Application 3

A box shaped piece of gelatin dessert has a top area of 15 cm2 and a height of 3 cm. When a shearing force of 0.50 N is applied to the upper surface, the upper surface displaces 4 mm relative to the bottom surface. What are the shearing stress, the shearing strain, and the shear modulus for the gelatin?

Solution

Shear stress = ${{tangential\,\,force} \over {area\,\,of\,\,face}} = {{0.50N} \over {15 \times {{10}^{ - 4}}{m^2}}}$ = 333 Pa

Shear strain = ${{displacement} \over {height}} = {{0.4cm} \over {3cm}} = 0.133$

Shear modulus G = ${{stress} \over {strain}} = {{333\,\,Pa} \over {0.133}} = 2.5\,\,kPa$

### Bulk Modulus

The bulk modulus of a solid or a fluid indicates its resistance to a change in volume. Consider a cube of some material, solid or fluid, as shown in the figure. We assume that all faces experience the same force Fn normal to each face. (One way to accomplish this is to immerse the body in a fluid – as long as the change in pressure over the vertical height of the cube is negligible). The pressure on the cube is defined as the normal force per unit area

p  = ${{{F_n}} \over A}$

The SI unit of pressure is N/m2 and is given the name pascal (Pa).

Pressure is a scalar because on any infinitesimal volume, it acts in all directions; it has no unique direction.

When the pressure on a body is increased, its volume decreases. The change in pressure ${{{F_n}} \over A}$P is called the volume stress and the fractional change in volume ${{{F_n}} \over A}$V/V is called the volume strain. The bulk modulus B of the material is defined as

Bulk modulus = ${{Volume\,\,stress} \over {Volume\,\,strain}}$

or B = ${{ - \Delta p} \over {\Delta V/V}}$

The negative sign is included to make B a positive number since an increase in pressure ( ${{ - \Delta p} \over {\Delta V/V}}$p  > 0) leads to decrease in volume ( ${{ - \Delta p} \over {\Delta V/V}}$V < 0).

The inverse of B is called the compressibility,

k = 1/B

Table: Elastic Properties of Matter

#### State

Shear Modulus Bulk Modulus
Solid Large

#### Large

Liquid Zero Large
Gas Zero Small

### Elastic Potential Energy

Obviously, work has to be done by the applied force in deforming a body, to whatever type of strain it might be subjected. This work done or energy spent remains stored up in the body in the form of elastic potential energy or strain energy. Let us calculate the work done in case of longitudinal strain.

Let F be the force applied (within elastic limit) to a wire of length L and area of cross section A, so that the infinitesimal increase in the length is dl. Then

dW = F.dl

But by definition $Y = {{FL} \over {Al}}$

or $F = {{AY} \over L}l$ , where l  is the total change in length.

$dW = {{AY} \over L}ldl$

$W = \int {dW} = \int\limits_0^l {{{AY} \over L}ldl} = {1 \over 2}{{AY} \over L}{l^2} = {1 \over 2}\left( {{{AYl} \over L}} \right).l = {1 \over 2}Fl$

Since volume of wire = LA, we have work done per unit volume or strain energy per unit volume = 1/2 x Stress x Strain.

### Thermal Stresses

If the ends of a rod are rigidly fixed so as to prevent expansion or contraction and the temperature of the rod is changed, tensile or compressive stresses, called thermal stresses, will be set up in the rod. Hence in the design of many structures which is subject to change in temperature, some provision must be made for expansion to avoid failure of such structures.

Suppose that a rod at a temperature T has its ends rigidly fastened and that while they are thus held, the temperature is reduced to a lower value To.

The fractional change in length if the rod were free to contract would be

${{\Delta L} \over L} = \alpha \left( {T - {T_o}} \right) = \alpha \Delta T$

Since the rod is not free to contract, the tension must increase to produce same fractional change in length. Let F be the tension produced, then

$F = AY{{\Delta L} \over L}$

Substituting for ${{\Delta L} \over L}$, we get,

F = AY $\alpha$ DT

Stress in rod =${F \over A} = Y\alpha \Delta T$

Breaking Stress

If one end of rod or wire is rigidly fixed and a force is applied at the other end, it will stretch. If the force is small, the extension will also be small and upon the withdrawal of the force, the wire will regain its original state. But if the applied force is gradually increased, after some intermediate states, a state is reached when the wire or rod breaks. The stress corresponding to this breaking point is termed as breaking stress.

### STRESS-STRAIN GRAPH

The stress-strain graph of a ductile metal is shown in figure. Initially, the stress-strain graph is linear and it obeys the Hooke’s Law upto the point P called the proportional limit. After the proportional limit the $\sigma - \varepsilon$ graph is non-linear but it still remains elastic upto the yield point Y where the slope of the curve is zero. At the yield point the material starts deforming under constant stress – it behaves like a viscous liquid. The yield point is the beginning of the plastic zone. After the yield point, the material starts gaining strength due to excessive deformation and this phenomenon is called strain hardening. The point U shows the ultimate strength of the material. It is the maximum stress that the material can sustain without failure. After the point U the curve goes down toward the breaking point B because the calculation of the stress is based on the original cross-sectional area whereas the cross-sectional areas of the sample actually decreases. [/vc_column_text][/vc_column][/vc_row]

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