On Wednesday we started this topic, so today we hope to conclude this topic, which is elasticity. Just a review of what we did. We said when a force is applied on a solid material, the solid material will change its shape.
and size. Now when the applied force is removed, it is expected that the solid material will return back to its original shape and size. When that happens, it is called elasticity and if by the removal of the applied force the solid material remain permanently deformed.
That is called plasticity or plastic deformation. So we have two types of deformation, elastic deformation, which is a temporary deformation, and then we have plastic deformation, which is permanent deformation. And we went on to... defined some properties of metals which are also generalized to materials.
Some of them are hardness, brittleness, malleability, ductility, elasticity, density, feasibility, and conductivity. Then we stated Hooke's law, which states that provided the elastic limit is not exceeded. The force applied on a material is directly proportional to the change in length or extension. And we have that expression. F is equal to minus F minus KX.
The negative sign there signifies that the force is a restoring force. And I gave an illustration. I said, if you have a rubber band, for instance, in your hand, and you hold it with your two hands and pull it, stretch it, by the time you remove your hand, the rubber band returns to its original. to its original shape.
That returning, that is what is called restoring force. There is a force that is bringing it back to the way it was before. So that force is an opposite direction to the force that stretched the material. So that is why it is negative, restoring force. And then we use this graph to explain the relationship between force and extension.
Force is directly proportional to extension. And also applying Hooke's law, we have the elastic region, the plastic region. We have the elastic limit.
We have the breaking point. The elastic limit is the limit beyond which the material moves into the plastic region. That is permanent deformation. And then within the permanent deformation, we have what is called dislocation and fracture.
So fracture happens at the breaking point. That is where the material experience is greatest strength or ultimate strength. Beyond the ultimate strength, the material breaks.
Then there's a relationship between stress and strain within. Hooke's law, the relationship between the elastic stress and strain brings us to modulus, young modulus. Then we talked about stress, force over area strain, change in length over original length. We talked about types of stress, tensile stress and compressive stress. Tensile stress.
When you apply opposite and equal force, moving in a direction that is away from each other, that gives you tensile stress. That is, stretching an object. Then compressive stress, when you are trying to bring together a stretched object. So equal and opposite force again.
this time moving in towards each other and we solved some examples up to this point up to this point and then we mentioned the types of elasticity which are basically three we have the young modelers the chair modulus or modulus of rigidity and then we have bulk modulus. We also treated what modulus of elasticity is which is the ratio of stress to strain and I mentioned that we can use a symbol E capital E or capital Y to represent the modulus of elasticity. The modulus of elasticity and this also obeys Hooke's law provided the elastic limit is not exceeded. Then the expression for modulus of elasticity was expanded to what you can see there.
FL over A change in length. F for force, L for regional length, then A for area, and delta L for change in length, which sometimes you can see as small letter E or small letter X. So when you see delta L, It means the same thing as small letter e or small letter x, signifying extension.
extension or change in length. So we also solve this on Wednesday as far as this, as well as this. So this was where we stopped and we want to take time to go through the remaining model and then the relationship between them and energy stored in.
a solid material. So the shear modulus which is denoted by symbol S-G-O, that is a null sign or you can use N sign, is a material property that describes the ability of a material to resist deformation under shear stress. And by formula, by implication, It is given as the shear stress over shear strain. All the three moduli, whether bulk modulus, young modulus, or shear modulus, they are all a ratio of stress over strain. But now, the type of stress and strain differs.
And here we have shear stress. and shear strain. Shear stress occurs when a force is applied parallel to one surface of an object or a material.
That is when shear stress occurs. Okay, this is a diagram that you can easily understand what shear stress is. It's not like... elongation it's not like stretching a rubber band or compressing it shear stress occur when you tend to displace an object from its real original position the object as it is does not change its shape and size but it just that it is displaced tilted let me use the word tilted so look at that box When I apply a force at the top and at the bottom, at the same equal force, the object appears tilted.
Now, if I remove my hand, you'll see the object will return back to the way it is. Just imagine you are reading a table, the table you used to read, and you just decide to push it from the top. trying to tilt it or slant it. That is the kind of shear stress we are talking of.
You are applying a force to a particular area. The object is tilted, is slanted. Then by the time you move it, you are on the table or whatever object you are tilted, it returns back to its original position. So that is where you have a shear stress or call. So that is another form of modulus by shear modulus.
So we can derive expression for the shear modulus based on the generalized modulus formula, which is stress over strain. So now if you look at that figure, you will see that. If we assume that this figure is a square, equal area, equal area, that is length and width are the same. So by tilting it, we have displaced it. If it's the second figure there, you see the shaded part.
So we have labeled. L as the length. The other side also should have been L, but the shaded part is what we are considering.
That is the displaced part. So we have labeled the displaced part with a D. Now, if you cut off that shaded part, it will appear to be a triangle.
And we have an angle there. which is five. So D over L will give us tan theta or tan phi.
So tan phi will be equal to D over L. D over L is equal to tan phi. Now, if you consider that that angle there is very small, like a paroxial angle, angle that is very small, then we can approximate that tan. Phi is equal to phi, approximately equal to phi, which is equal to D over L. Now, remember that initially, all the sides are equal.
We are supposed to have L over L. But now, because of the displacement, we have D. So that D stands for change in position, while L stands for... the original position, original length, change in length over original length. So that will give us what we call strain.
Strain is change in length over original length. So now we have a derivation there, a formula for strain, which is phi is equal to D over L, strain, shear strain. which is equal to phi d over l. So that is what you are seeing on this next slide. So if I want to find the chair modulus, remember that modulus is stress over strain.
So I will substitute all these formulas together, which is stress f over a, strain d over l. Or if I want to just use a single symbol, I can use phi. So if I use a single symbol, my shear stress would be F over A phi. F over A phi.
If I use the double symbol for strain, which is D over A phi, L, that will give the formula of shear modulus to be F L over D A. F L over D A. F L over D A. So that is shear modulus. We can also write it as F over... Shear modulus is an important parameter in material science and engineering, particularly in the analysis of material behavior under various types of load.
That is why we are studying it, because whenever you talk of material science, you must talk about shear modulus, because that is what will help you to determine or to know how different material behave under different conditions. So here we have an example, a steel stud given a shear modulus of 8.27 times 10 raised to 10 Pascal. It's one cm in diameter.
One cm in diameter projects four cm from the wall. A 36,000 Newton sharing force is applied to the end. We are asked to find the defection D of the store.
So you... Look at the figure there, the stud, that is like the rod. coming out through the wall, extend four meters from the wall, as a diameter of one cm. So if we apply a force of 36,000 Newton, the essence of applying the force is to bend the rod.
towards whatever direction we want it to go. So that would give us a defection. So what is that defection?
That is what this question is asking. So now we know our shear modulus. We know our force. We know the length of the rod.
We know... the diameter, but we need the area because from the formula we have F L over A D. So we know force, we know length, we know S, which is the shear modulus. We don't know A, but we know D and we can calculate A from the diameter.
knowing that the cross-sectional area of that rod is circular. So, A becomes pi R squared, or pi D squared over 4. So, if we substitute the parameters, and mind you, if you look at the diameter value and the length value, they are in centimeters. So, you have to first convert them to meters, standard unit, because the...
a shear modulus is given in standard unit pascal and the force is given in standard unit so you must convert every other unit that is not in standard unit to standard unit so one cm will be will be one times 10 raised to minus two four cm becomes four times ten raised to minus two so if you put down the formula and substitute all the parameters the final answer there is 0.22. So the first thing to do is to find the area using that formula. So the area is 7.85 times 7 minus 5 meters squared. So having gotten the area, we make D the subject of the formula.
So D becomes FL over AS. F is 36,000. L is 4 times 10 raised to minus 2, or 0.04, divided by A, which is 7.85 times 10 raised to minus 5, and times the share modulus, which is 8.27 times 10 raised to 10 Pascal. So when you substitute all the power...
values and simplify we have our D to be 0.222 millimeters. Now we move on to the third modular which is bulk modulus. Bulk modulus is a material property that describes the material response to change in volume.
Share modulus change in shape. Young modulus change in length. So just like if you recall your O-level expansivity, you have three of them.
You have the linear, you have the superficial, and you have the volume expansivity. So the same thing is playing out here. The young modulus...
for linear, the superficial for area, which is shear modulus, and then the volume expansivity as the bulk modulus. So it measures the relative change in volume with respect to change in pressure. So this happens when we apply equal force on a on A pump was applied to a solid material of different faces.
For instance, if we have a cube and we apply equal force on all sides of the cube, you will notice that the shape of the cube will remain the same, but the size will reduce, will decrease. Imagine it. You have a cube before you.
Cube has six surfaces, and you apply equal force on all the six. surface, pressing it together, you will notice that the cube will begin to decrease in size, but it will retain that cubic shape because equal forces are applied on all the surface. So that decrease in volume is now our change in volume.
So if we want to find bulk modulus, because bulk modulus would be... Bulk stress over bulk strain. Stress over strain. The stress is force per unit area.
And this time we are talking of volume. So our force per unit area, which is stress, is also the same as pressure. That is why if you look at the unit of stress, it takes the unit of pressure.
So we would rather talk of pressure in bulk modulus. than just force over area or stress. So the bulk stress is the pressure, bulk pressure, or pressure that is applied to the surface. So we add bulk stress as pressure, and then bulk strain as change in volume over original volume. Now, because This time, the material is decreasing, which implies that we are going to introduce a negative sign to denote that the material is decreasing in size.
So that is why you see the negative sign there. The negative sign is included because an increase in pressure typically leads to a decrease in volume and vice. So, this, look at that picture of FIGO.
That is what I was trying to explain to you. If I apply force on equal force, equal force on all sides, the object's size will shrink, but the shape of the object will remain the same. So, now, we have... Our final expression here as a minus PV over change in volume.
For bulk modulus, we can represent bulk modulus with a B, capital B or a K, capital B or capital K. If we take the reciprocal of bulk modulus, that gives us compressibility C. So the reciprocal of bulk modulus of a material is called compressibility, C. So 1 over K or 1 over B is equal to C.
So here we have an example. A hydraulic press contains 5 liters of oil, 5 liters, which is volume. Finally, decrease in volume of the oil. And if...
it is subjected to a pressure of 3000 kilopascal. So we are given the original volume or yes, the original volume. We are given pressure. For bulk modulus, we need three basic terms. Original pressure, original volume, pressure and change in volume.
So now we have the bulk modulus given to us as 170. 0 megapascal. We have to find the decrease in volume. That's the change in volume.
So we rearrange the expression to make change in volume delta V, the subject of the formula. So when we do that, we have minus PV over B, minus PV over B. And P is the pressure, which is 3,000.
Kilo pascal. And the volume is 5 liters, all divided by the bulk modulus, which is 1.7 zeros omega Pascal. So if we substitute all the values and simplify, we have a decrease in volume in millimeter as minus 8.82 million liters.
So the next we'll consider is energy stored in an elastic or in a solid material. Energy stored in a solid material. So when a solid material is stretched or compressed, energy is stored.
When you pull a rubber band, for instance. you have stored energy in that rubber band immediately you remove your hand the energy is released and that is why um the rubber band returns to its original shape so the energy stored in this case is always known as potential energy energy stored is always potential energy So, this energy is released when the applied force is removed, and provided the elastic limit is not exceeded, the object, the material will return to its original shape and size. So, we try to derive an expression for the energy stored in that solid material, and we start from the relationship between work done, which is given as integral boundary taken from 0 to x, f dx. Work done is false times displacement.
So if we substitute in the values, okay. If you look at the second equation, where f is equated to y a over l, l subscript 0x. Now, this expression is gotten from Young-Modulo's.
Remember Young-Modulo's y is equal to f l over a. change in length and i told you that the change in length could be small letter x as you are seeing it there or small letter e for extension so because here we are not dealing with um f the x we chose to harmonize the parameters by using x so in the young modulus if you make f the subject of the formula, you have that expression, Y-A-X over L-naught, L-naught for original length. So if we substitute that into the work done equation, so we replace F with Y-A-X over L.
So that will give us work done is equal to Y-A-L-X-D-X. dx. So y a over l becomes like a constant value.
We can move that out of the integral. We can move that out of the integral. Then we'll have integral x dx, integral x dx. So if we integrate x dx from 0 to x, we are going to have x squared over 2. If we integrate x dx, we are going to have x squared over 2. So that is what you have there.
Now, the constant term that we move out of the integral, which is y a over l, can be replaced with a value k, which we see there by the right side, where... k is equal to y a over l So if we now rewrite the expression, that gives us W equals half kx squared. Half kx squared. So that is an expression for energy stored in a solid material.
And remember I mentioned before that whenever you hear the word stored, energy stored. The energy that is stored is potential energy. Potential energy is energy a body acquires by virtue of its position. Why kinetic energy? Energy a body acquires by virtue of its motion.
So now energy stored is potential energy because it's at a position. So to determine the energy per unit volume, all we need to do is divide the energy by volume. Energy per unit volume is energy over volume.
So if we take that expression and divide it by volume, you see the expression there, E over V, which is equal to E over AL. AL, that is area times length. will give us volume when you multiply area times length you get volume so we are trying to simplify to show how um our energy by unit volume can be represented in terms of stress and strain energy by in volume can be represented in terms of stress and strain so If we now substitute all those parameters you see there from E over V is equal to E over AL.
Now, E, which is energy, we have derived it. Replacing K with YAL, we have YAX squared over 2. Now you'll be wondering why do we have in the denominator, why do we have 2LAL? Why not 2AL squared?
It's because of our target. We are trying to show that the energy per unit volume is half stress times strain. So we are splitting the terms. so that we can get to where we are going to.
So now, if we split it further, open up the equation, y a over l x times 1 over a times x over l. We have also split the x squared. into x times x.
So instead of writing L square, we made it L times L. Instead of writing x square, we made it x times x, and we splitted them. So that y, a, x over L gives us force, F. If you look up the first equation where we have W is equal to F. dx.
We said that where f is equal to yax over l. So by splitting the equation down, we were able to bring up f, yax over l, which is f. So we represent that term with f.
Then if we move to the right, you have 1 over a. So if I have f and 1 over a, That simply means F over A. And F over A is equal to stress. Then the next term there is X over L. X over L.
X for extension or change in length over original length. That gives us strain. So we've gotten our stress times strain.
So if we substitute that back into the equation, we have our... Energy per unit volume as of Stress times strain. So we can also do the same this now. We use our Young modulus to derive this expression. Energy per unit volume using Young modulus relationship.
We got half stress times strain. We can also use shear modulus. Find the energy stored in a shear strain.
We can also find energy stored in a bulk. string using the expression for bulk modular young bulk modulars and share modules to have that expression you are seeing so the relationship between these three modular is represented with that expression that formula nine over y y for young modulars is equal to 3 over 0, 0 for shear modulus, plus 1 over k, which is k for bulk modulus. Now, when we stretch a wire, it becomes longer but thinner. The increase in each length is always accompanied with decrease in cross-sectional area. So, um...
Bringing them together, the Poisson relationship brings these three moduli together. And according to the Poisson ratio, the ratio of transverse contraction strain to longitudinal extension strain in the direction of extraction force. So we have... They are represented with the formula sigma. Sigma is the contraction strain over the extension strain.
Extension strain or longitudinal strain, that is along a length or linear strain. Contraction could be also said to be lateral. strain, then longitudinal could also be said to be the linear strain.
So this is another expression relating all the three formulas together. So the sigma, which is the Poisson ratio, is equal to 3b minus 3b, b for bulk, minus 2s, as for shear modulus, over 6b. 6B, B for bulk, plus 2S, S for shear modulus. And then Y is equal to 2S, Y for young modulus, S for shear modulus, and in bracket 1 plus sigma, the sigma for Poisson ratio.
And the last expression there, which is Y is equal to 3B, in bracket open and close, 1 minus 2. So this is how the three moduli are related. So here we have an example. A load of 50 newton attaches spring angle vertically stretches the spring 5.0 cm.
The spring is now placed horizontally on a table and stretched. 11.0 cm. What force is required to stretch the spring this amount?
So the first thing to consider is your units. Your units are not in standard units, all of them. The force is of course standard unit in Newton, but the stretch spring is not in standard unit, it's 5 cm. So you convert 5 cm to meter, that is 0.05 meters and then the horizontal stretch spring is 11 cm which you also convert to meter 0.11 meters now we are asked to find the force that will be applied to stretch the spring when it is lying horizontally so the the expression to do that That is false is equal to KX. So we know force, we know extension in the first case, vertical case, but we don't know the spring constant.
Being that it is the same spring that was used, we need to find the spring constant. So when the load was hanging vertically, we have a load of 50 newton and the load at the spring stretch 0.05 cm. So we can find the...
spring constant by dividing the force over 0.05. So when we do that, we get our spring constant to be 1,000 newton meter. Now we know our spring constant.
We can find our force, F is equal to KS. This time the X, which is the extension, is 0.11. We know our spring constant to be 1,000. So 1,000 times.
0.11 gives us 110 newton. So that is the force required to stretch the spring horizontally. The next example here, we have a sling shot consists of a light leather cup containing a stone.
that is pulled back against two rubber bands. It takes a force of 30 Newton to stretch the bands one cm. A, what is the potential energy stored in the bands when a 50 gram stone is placed in the cup and pulled back 0.2... zero meter from the equilibrium position. With what speed does it leave the slingshot?
So this is a slingshot like a catapult that we, if you are familiar with those hunters, catapult, they used to shoot birds, kill birds or any small animal that it can kill. So if you put a stone in that catapult cup, and then you pull the band. The band will stretch, and then when you remove your hand, the stone will leave the band and move towards your target.
So we are asked to find the potential energy stored. That is, when you pull the band, remember I said, when you pull an elastic material, you have stored energy in that material. When you remove your hand from that material or the force applied, remove the force applied, the energy will be released.
So now we have stored energy. We are asked to find the energy stored. Our energy stored is half KX squared.
Half KX squared. We know x, we know f, but we don't know k. So the first thing to do is to solve for k.
Using the relationship f is equal to kx. So to solve for k will be k is equal to f over x. k will be equal to f over x.
So if we substitute that value, F being 30 and X 0.01, because from the question, one cm is given, so you have to convert that to standard unit. So that gives us K to be 3,000 newton meters. Now we know our K. we can find our potential energy, which is half kx squared.
k is 3,000. x is 2. Because the catapult or the slingshot is pulled or stretched 0.2 meters. So if we substitute those values into the expression, our...
energy, potential energy, will be 60 joules. 60 joules, that is for A. For B, we are asked to find the speed at which the stone will leave the slingshot. So we consider that potential energy is equal to kinetic energy. Total energy is conserved, so potential energy at a particular point will be equal to kinetic energy.
So if we equate that, kinetic energy being half... mv squared. Our potential energy is 60. So 60 will be equal to half mv squared.
So we make v, the subject of the formula, that will give us 60 times 2 over m all in square root. So if we substitute the parameters, that gives us 49 meters per second squared. So the speed of the stone as it leaves the slingshot is 49 meters per second. So this is another example an aluminium with a bulk modulus of seven point in a bulk modulus of seven point ten.
0.010 raise to 9, aluminum ball with a bulk modulus of 7.0 times 10 raise to 9 Newton per meter with a radius of 0.5 meters falls to the bottom of the sea where the pressure is 150 atmosphere. A, what is the original volume of the ball? What is the change in volume? And C, what is the new volume? D, what is the bulk stress?
And E, what is the bulk strain? Now, since we are told that it is a ball, we consider a spherical shape. we know ball to be spherical. So the volume of a sphere is 4 pi r cubed over 3. 4 pi r cubed over 3. So we are given the radius of the ball as 0.4 or 0.5 meters. So to find the volume of that ball which we serve as the original volume is 4 pi 0.5 cube over 3. So that gives us 0.5236 meter cube.
0.5236 meter cube. So that is the original volume of the aluminum ball. Now, the second question is, what is the change in... volume.
So to find the change in volume, we can use the bulk modulus formula. Bulk modulus is equal to minus pressure, original volume, divided by change in volume. So we make change in volume the subject. And we have change in volume is equal to minus PV over B.
So if you substitute the values minus, we convert our P, atmospheric pressure to Pascal. And that is by multiplying with a standard converter, which is 101 times 325. 101, 325. 150 times 101, 325. 150 times 101, 325. So that gives us 152 times... 10 raised to 7 Pascal. So when we substitute all those parameters, values into the equation, our change in volume will be 1.137 times 10 raised to minus 4 meter cube.
Remember that it's a negative sign, minus 1.13. 7 times 7 is minus 4 meter cubed. So that is our change in volume. Now, the next question is, what is the new volume?
So if I have original volume and then change in volume, I can use the expression of initial volume, a final volume minus initial volume will give me the change in volume. So what is my final volume? That is what I am asked to find. That is a new volume. that is the final volume.
My initial volume is the original volume. My change in volume is what I've already solved in B. So I have the original volume solved in A, change in volume solved in B.
So my final volume is what I'm asked to find. So my final volume will be the expression there you have minus change in volume plus original volume. gives me 0.5235 that should be five not six zero final volume should be 0.5235 meter cube 0.5235 meter cube that's my final volume now knowing that i can find my bulk strip remember that my box stress which is also force over area, is the same thing as the pressure that causes a change in volume. So that pressure is 1.52 times 10 raised to 7. So that is my bulk stress. My bulk strain is change in volume over original volume, which is minus 1.137 times 10 raised to minus 4 over 0. 2.5236.
And that gives me minus 2.17 times 10 raised to minus 4. So that is how to answer a question like this. So the rest of the questions, we have other examples solved there. And then we have exercises.
OBJ exercises because you are both you are CA your test and your exam will be OBJ CBT that is our Multi multiple choice questions. So I put here some multiple choice questions for you to practice I will send this slide to The sub-dean to forward it to you So have a wonderful weekend. So we'll meet again next week, Wednesday, 10 to 11.