 Physics ELASTIC MODULI, YOUNG’S MODULUS, SHEAR MODULUS AND BULK MODULUS

ELASTIC MODULI
YOUNG’S MODULUS
SHEAR MODULUS
BULK MODULUS

### ELASTIC MODULI

The proportional region within the elastic limit of the stress-strain curve (region OA in Fig. 9.3) is of great importance for structural and manufacturing engineering designs. The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material.

### Young’s Modulus

Experimental observation show that for a given material, the magnitude of the strain produced is same whether the stress is tensile or compressive.

The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Young’s modulus and is denoted by the symbol Y.

color{navy}(Y= sigma/in)

From Eqs. (9.1) and (9.2), we have

color{orange}(Y = (F//A)/(ΔL//L))

color{navy}(= (F × L) /(A × ΔL))....(9.8)

Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress i.e., N m^(–2) or Pascal (Pa). Table 9.1 gives the values of Young’s moduli and yield strengths of some materials. From the data given in Table 9.1, it is noticed that for metals Young’s moduli are large.

Therefore, these materials require a large force to produce small change in length. To increase the length of a thin steel wire of 0.1 cm2 crosssectional area by 0.1%, a force of 2000 N is required.

The force required to produce the same strain in aluminium, brass and copper wires having the same cross-sectional area are 690 N, 900 N and 1100 N respectively. It means that steel is more elastic than copper, brass and aluminium.

It is for this reason that steel is preferred in heavy-duty machines and in structural designs. Wood, bone, concrete and glass have rather small Young’s moduli.
Q 3139491312 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 x× 1011 N m^(-2).
Class 11 Chapter 9 Example 1 Solution:

We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. Then the stress on the rod is given by

color{green} {"Stress" = F/A = F/(pir^2)}

(100\ \ 10^3\ \ N)/(3.14 10^2m^2)
The elongation,
color{orange} {DeltaL=(F//A)L/Y}

= (38. 10^8Nm^(-2)1m)/(2\ \ 10^(11)Nm^(-2)
= 1.59xx10^(-3)m
= 1.59mm
The strain is given by

color{purple} {"Strain" = ΔL//L}
= (1.59 × 10–3 m)//(1m)
= 1.59 × 10^(–3)
= 0.16 %
Q 3119591410 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.
Class 11 Chapter 11 Example 2 Solution:

The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A. From Eq. (9.7) we have stress = strain × Young’s modulus. Therefore

color{green} {W//A = Y_c × (ΔL_c//L_c) = Y_s × (ΔL_s//L_s)}

where the subscripts c and s refer to copper and stainless steel respectively. Or,

color{purple} {ΔL_c//ΔL_s = (Y_s//Y_c) × (L_c//L)}
Given L_c = 2.2 m, L_s = 1.6 m,
From Table 9.1 Y_c = 1.1 × 1011 N.m–2, and Y_s = 2.0 × 10^(11) N.m^(–2).

ΔL_c//ΔL_s = (2.0 × 10^(11)//1.1 × 10^(11)) × (2.2//1.6) = 2.5.
ΔL_c + ΔL_s = 7.0 × 10^(-4) m
Solving the above equations,

ΔL_c = 5.0 × 10^(-4) m, and ΔL_s = 2.0 × 10^(-4) m.
Therefore

color{orange} {W = (A × Y_c × ΔL_c)//L_c}

= π (1.5 × 10-3)^2 × [(5.0 × 10^(-4) × 1.1 × 10^(11))//2.2]
= 1.8 × 10^2 N
Q 3169691515 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. 9.5). 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.
Class 11 Chapter 11 Example 3 Solution:

Total mass of all the performers,
tables, plaques etc. = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid
= 280 – 60 = 220 kg
Weight of this supported mass
= 220 kg wt. = 220 × 9.8 N = 2156 N.
Weight supported by each thighbone of the performer = 1/2 (2156) N = 1078 N.
From Table 9.1, the Young’s modulus for bone is given by
Y = 9.4 × 10^9 N m^*–2.)
Length of each thighbone L = 0.5 m
the radius of thighbone = 2.0 cm
Thus the cross-sectional area of the thighbone
A = π × (2 × 10^(-2))2 m^2 = 1.26 × 10^(-3) m^2.
Using Eq. (9.8), the compression in each thighbone (ΔL) can be computed as
color{orange} {ΔL = [(F × L)//(Y × A)]}
= [(1078 × 0.5)//(9.4 × 10^9 × 1.26 × 10^(-3))]
4.55 × 10^(-5) m or 4.55 × 10^(-3) cm.
This is a very small change The fractional decrease in the thighbone is color{purple} {ΔL//L = 0.000091} or color{purple} {0.0091%.}

### Determination of Young’s Modulus of the Material of a Wire

A typical experimental arrangement to determine the Young’s modulus of a material of wire under tension is shown in Fig. 9.6. It consists of two long straight wires of same length and equal radius suspended side by side from a fixed rigid support.

The wire A (called the reference wire) carries a millimetre main scale M and a pan to place a weight. The wire B (called the experimental wire) of uniform area of crosssection also carries a pan in which known weights can be placed.

A vernier scale V is attached to a pointer at the bottom of the experimental wire B, and the main scale M is fixed to the reference wire A.

The weights placed in the pan exert a downward force and stretch the experimental wire under a tensile stress. The elongation of the wire (increase in length) is measured by the vernier arrangement.

The reference wire is used to compensate for any change in length that may occur due to change in room temperature, since any change in length of the reference wire due to temperature change will be accompanied by an equal change in
experimental wire. Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted. Now the experimental wire is gradually loaded with more weights to bring it under a tensile stress and the vernier reading is noted again.

The difference between two vernier readings gives the elongation produced in the wire. Let r and L be the initial radius and length of the experimental wire, respectively.

Then the area of cross-section of the wire would be πr^2. Let M be the mass that produced an elongation ΔL in the wire.

Thus the applied force is equal to Mg, where g is the acceleration due to gravity. From Eq. (9.8), the Young’s modulus of the material of the experimental wire is given by

color{blue}(Y = (sigma)/ε = (Mg)/(pir^2).L/(DeltaL))

color{green}(= (Mg xxL)/(pir^2xxDeltaL)).....(9.9)

### 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.

G = shearing stress (σ_s)/shearing strain

color{navy}(G = (F//A)/(Δx//L))

color{orange}(= (F × L)//(A × Δx)).(9.10)

Similarly, from Eq. (9.4)

color{navy}(G = (F//A)//θ)

color{red}(= F//(A × θ))...(9.11)

The shearing stress σ_s can also be expressed as

color{navy}(σ_s = G × θ).(9.12)

SI unit of shear modulus is N m^(–2) or Pa.

The shear moduli of a few common materials are given in Table 9.2. It can be seen that shear modulus (or modulus of rigidity) is generally less than Young’s modulus (from Table 9.1). For most materials G ≈ Y//3. Q 3169791615 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 × 104 N. The lower edge is riveted to the floor. How much will the upper edge be displaced?
Class 11 Chapter 11 Example 4 Solution:

A = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.05 m^2
Therefore, the stress applied is
= (9.4 × 10^4 N//0.05 m^2)
= 1.80 × 10^6 N.m^(–2)
We know that shearing strain color{orange} { = (Δx//L)= "Stress "//G.}

Therefore the displacement color{pink} {Δx = ("Stress" × L)//G}

= (1.8 × 106 N m^(–2) × 0.5m)//(5.6 × 109 N m^(–2))

= 1.6 × 10^(–4) m = 0.16 mm

### Bulk Modulus

As, we've have seen that 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 [Eq. (9.5)].

The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus. It is denoted by symbol B.

color{green}(B = – p/(ΔV//V)) .......(9.13)

The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs.

That is, if p is positive, ΔV 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., N m^(–2) or Pa. The bulk moduli of a few common materials are given in Table 9.3.

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.

color{green}(k = (1//B) = – (1//Δp) × (ΔV//V))........(9.14)

It can be seen from the data given in Table 9.3 that the bulk moduli for solids are much larger than for liquids, which are again much larger than the bulk modulus for gases (air). Thus solids are least compressible whereas gases are most compressible. Gases are about a million times more compressible than solids! Gases large compressibilities, which vary with pressure and temperature.

The incompressibility of the solids is primarily due to the tight coupling between the neighbouring atoms. The molecules in liquids are also bound with their neighbours but not as strong as in solids. Molecules in gases are very poorly coupled to their neighbours.

Table 9.4 shows the various types of stress, strain, elastic moduli, and the applicable state of matter at a glance. Q 3129891711 The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, ΔV//V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m^(–2). (Take g = 10 m s^(–2))
Class 11 Chapter 11 Example 5 Solution:

The pressure exerted by a 3000 m column of water on the bottom layer

color{green} {p = hρ g}

 = 3000 m × 1000 kg m^(–3) × 10 m s^(–2)

= 3 × 10^7 kg m^(–1) s^(-2)

= 3 × 10^7 N m^(–2)

Fractional compression ΔV//V, is

color{orange} {ΔV//V = "stress"//B = (3 × 10^7 N m^(-2))//(2.2 × 10^9 N m^(–2))}

= 1.36 × 10^(-2) or 1.36 %

### APPLICATIONS OF ELASTIC BEHAVIOUR OF MATERIALS

The elastic behaviour of materials plays an important role in everyday life. All engineering designs require precise knowledge of the elastic behaviour of materials.

For example while designing a building, the structural design of the columns, beams and supports require knowledge of strength of materials used.

Cranes used for lifting and moving heavy loads from one place to another have a thick metal rope to which the load is attached. The rope is pulled up using pulleys and motors.

Suppose we want to make a crane, which has a lifting capacity of 10 tonnes or metric tons (1 metric ton = 1000 kg). How thick should the steel rope be? We obviously want that the load does not deform the rope permanently.

Therefore, the extension should not exceed the elastic limit. From Table 9.1, we find that mild steel has a yield strength (Sy) of about 300 × 10^6 N m^(–2).

color{navy}(A ≥ W/S_y = Mg//S_y)....(9.15)

color{blue}(= (10^4 kg × 10 m s^(-2))//(300 × 10^6 N m^(-2)))
color{navy}(= 3.3 × 10^(-4) m^2)

Thus, the area of cross-section (A) of the rope should at least be corresponding to a radius of about 1 cm for a rope of circular cross-section.

Generally a large margin of safety (of about a factor of ten in the load) is provided. Thus a thicker rope of radius about 3 cm is recommended. A single wire of this radius would practically be a rigid rod.

So the ropes are always made of a number of thin wires braided together, like in pigtails, for ease in manufacture, flexibility and strength.

A bridge has to be designed such that it can withstand the load of the flowing traffic, the force of winds and its own weight.

Similarly, in the design of buildings use of beams and columns is very common. In both the cases, the overcoming of the problem of bending of beam under a load is of prime importance.

The beam should not bend too much or break. Let us consider the case of a beam loaded at the centre and supported near its ends as shown in Fig. 9.8.

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

color{orange}(δ = (W l^3)/(4bd^3Y))....(9.16) This relation can be derived using what you have already learnt and a little calculus. From Eq. (9.16), we see that to reduce the bending for a given load, one should use a material with a large Young’s modulus Y.

For a given material, increasing the depth d rather than the breadth b is more effective in reducing the bending, since δ is proportional to d^(-3) and only to b^(-1) (of course the length l of the span should be as small as possible). But on increasing the depth, unless the load is exactly at the right place (difficult to arrange in a bridge with moving traffic), the deep bar may bend as shown in Fig. 9.9(b).

This is called buckling. To avoid this, a common compromise is the cross-sectional shape shown in Fig. 9.9(c). This section provides a large loadbearing surface and enough depth to prevent bending. This shape reduces the weight of the beam without sacrificing the strength and hence reduces the cost. Different cross-sectional shapes of a beam. (a) Rectangular section of a bar; (b) A thin bar and how it can buckle; (c) Commonly used section for a load bearing bar.

Use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends as shown in Fig. 9.10(a) supports less load than that with a distributed shape at the ends [Fig. 9.10(b)].

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 materials etc. 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 × 10^7 N m^(-2).

Equating this to hρg, with ρ = 3 × 10^3 kg m^(-3) gives

color{green}(hρg = 30 × 10^7 N m^(-2) . Or

h = 30 × 10^7 N m^(-2)//(3 × 103 kg m^(-3) × 10 m s^(-2))
= 10 km) which is more than the height of Mt. Everest! 