Bulk Modulus, Shear Modulus and Poisson's Ratio

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take a metal wire and stretch it. It gets longer, right? And a little thinner too. So, does its volume go up, go down, or stay the same? Most students say: the same, obviously! Longer but thinner, so it balances out. Hold that thought; by the end, you will know the real answer, with numbers. Today: bulk modulus, shear modulus and Poisson's ratio. Questions on these come up often in JEE, so let's begin.

0:35Bulk modulus: squeezing from all sides

So, first, the bulk modulus. Squeeze a body equally from all sides, like a ball deep under water. The pressure rises by delta P; the little triangle, delta, means change in. Capital B, the bulk modulus, equals minus delta P, over delta V over V, the volume strain or fractional change in volume. Stress over strain again, like Young's modulus Y, valid within the elastic limit. Okay? Now, why is there a minus sign here? Yes, exactly! Pressure goes up, volume goes down. So delta V is negative, and the minus sign keeps the bulk modulus positive. Compressibility is just one over B: large bulk modulus, hard to squeeze. Clear? Now gases. At constant temperature, the isothermal case, P times V stays constant. Then the bulk modulus simply equals the gas pressure, capital P, okay? With no heat going in or out, the adiabatic case, P times V to the power gamma stays constant. Gamma, the Greek letter, is the ratio of the gas's two specific heats. Then the bulk modulus is gamma times the pressure. Got it?

1:51Density and a squeezed sphere

Now, density. Rho, the Greek letter, is the density, and the mass, rho times V, does not change. So rho times V is constant, and the double arrow means so. For small changes, delta rho over rho equals minus delta V over V, which equals delta P over B, okay? Volume down by a fraction, density up by the same fraction. Next, a sphere: its volume goes as capital R cubed, where capital R is the radius. So delta V over V equals three times delta R over R; the power three becomes a factor of three. Quick check. Here's a solid sphere under uniform pressure; its volume falls by zero point three percent. So the radius falls by zero point three percent too, right? No, no, careful! The volume strain is three times the radius strain. So divide by three: the radius falls by only zero point one percent. Clear?

2:53Shear modulus: a sideways push

Alright, now a sideways push. Hold a thick book's bottom fixed and push its top cover sideways. What changes: its length, its volume, or its shape? Its shape, correct! The sides tilt over. The force F is tangential, along the face. F over A is the shear stress, with capital A the face's area. And eta, the Greek letter, is the shear modulus, or modulus of rigidity: shear stress over phi, the shear strain. Okay? Phi, another Greek letter, is the small angle through which a side turns. If the top slides by delta x, then phi is approximately delta x over L, the height. The wavy equals sign means approximately. Clear?

3:45Poisson's ratio: longer means thinner

Now, back to our stretched wire. When it gets longer, what happens to its thickness? It gets thinner, yes! Poisson's ratio, sigma, equals minus lateral strain over longitudinal strain. Careful: in the stress lesson, sigma meant stress; here it is Poisson's ratio. Okay? Lateral strain is sideways, delta r over r for the radius; longitudinal strain is delta L over L, along the length. Stretching makes delta r negative, so the minus sign keeps sigma positive. And no unit: it is a ratio of strains. Right? In theory, sigma lies between minus one and one half; those are less than signs. In practice, between zero and one half. Remember that half. Got it?

4:36Relation between Y, B, eta and Poisson's ratio

These constants are linked. First, Young's modulus Y equals two eta, times one plus sigma, in brackets. Okay? Second, Y equals three B, times one minus two sigma. Y and B are both positive, so this bracket must be positive: sigma stays below one half. That is the upper limit! With me? Topper's trick, in one breath: nine over Y equals one over B, plus three over eta. Nine, one, three. Handy when you know B and eta. Clear? And sigma equals three B minus two eta, over six B plus two eta. Know any two constants, and you can find the rest.

5:18Does a stretched wire change its volume?

Now, our puzzle. For a stretched wire, delta V over V equals one minus two sigma, times delta L over L, the length strain. So tell me: for what value of sigma would the volume not change? One half, exactly! The bracket becomes zero, so delta V is zero. But real materials have sigma below one half. Example two will show what that means.

5:45Example 1: shear strain and top face shift of a 10 cm cube

Example one. Here's a ten centimetre cube with shear modulus eta equal to ten to the power ten pascals. There is a tangential force of one thousand newtons on its top face; the bottom is fixed. Find the shear strain and how far the top face moves. Pause the video and try it. Done? Okay, let's check. The area of the top face: ten times ten, one hundred, right? No, no, wait! Pascals are newtons per square metre, so work in metres. The side L is ten centimetres, or zero point one metre, so the area, L squared, is zero point zero one square metres; the small two above the m means squared. Okay? Now phi equals F, over A times eta: one thousand, over zero point zero one times ten to the power ten. The denominator is ten to the power eight, so phi is ten to the power minus five, with no unit. Got it? The top face moves by phi times L: ten to the power minus five, times zero point one, which is ten to the power minus six metres. That is one micrometre; the Greek letter mu means micro. Tiny, isn't it?

7:04Example 2: radius and volume change of a stretched wire

Now Example two, which answers our puzzle. Here's a wire with Poisson's ratio zero point three, stretched to a longitudinal strain of ten to the power minus three. Find the fractional changes in its radius and volume. Pause the video and try it. Done? Okay, let's check. Lateral strain is minus sigma times longitudinal strain: minus zero point three times ten to the power minus three. That is minus three times ten to the power minus four; the radius shrinks. Okay? Now the volume. Length up by ten to the power minus three, radius down by three times ten to the power minus four. So just subtract: seven times ten to the power minus four, right? Oh no, careful! The cross-section goes as r squared, so the radius change counts twice. That is the two in one minus two sigma. So, one minus two times zero point three, times ten to the power minus three. With me? That is zero point four times ten to the power minus three: plus four times ten to the power minus four. Positive! The volume increases. So, our puzzle: the stretched wire gets longer and thinner, yet its volume grows, because sigma is below one half. That's it!

8:27Quick revision: B, eta, Poisson's ratio and Y relations

So, quick revision. Bulk modulus: minus delta P over volume strain; for a gas, P if isothermal, gamma P if adiabatic. Shear modulus: tangential stress over phi. Poisson's ratio: minus lateral strain over longitudinal strain. Y equals two eta, times one plus sigma, equals three B, times one minus two sigma; and a stretched wire's volume strain is one minus two sigma, times its length strain. Now try the practice problems on this page. All the best!