Free Physics NEET Video Lectures for Mechanical Properties Of Solids

00:00:02 [Music] [Applause] [Music] [Applause] [Music] Gurukul students we are going to talk about the mechanical properties of solids so by mechanical properties what we mean is that we are going to talk about the deformations of solids or and we are going to talk about the stretching and bending of solids so in the previous chapters you have learnt rigid bodies and rigid bodies are defined as the ones which have their inter particle distances which remain constant throughout the course of motion however that does not stop bending or

00:01:05 stretching or extension or even other kinds of deformation of the body now these deformations can be brought about by the application of forces and you know even the steel rod which is known to be quite strong can be deformed if the applied force is very long large and we are going to talk about two kinds of deformation mainly one in which the body actually regains its a normal configuration once the force is removed and one in which they do not regain the normal configuration mainly we are going

00:01:49 to constrain our constraint our discussion on the forces which after which it if it's removed the body is going to have will regain its normal configuration so these are temporary deformations and mostly we are going to talk about these temporary deformation all right so let's take an example of bungee-jumping have you heard of bungee jumping if you want haven't then you might as well want to look at a video in youtube about bungee jumping in which what happens is that a diver or a jumper

00:02:31 he or she binds himself or herself along with a extensible and then take a dive from a very high distance or large distance and this is typically of the order of maybe even a few hundred feet Stu even more than that this could be a dangerous act so please do not fly without adequate supervision by professionals we are not going to talk about the problems that it could bring if a body suddenly you know jumps from a few thousand feet to the ground because of the pressure difference rather we are going to concentrate on

00:03:18 the amount of or rather the the material which the bungee cord is made off and the bungee cord is made up of of a material which has a great deal of elasticity in this video the boys seem to give a bungee jump so he falls off from a large height he is tied with the rope so this so what happens is that this as a diver or the jumper jumps from a height the cord bungee cord which is elastic it keeps getting elongated more and more till it reaches a point which is this maximum elongation and then the

00:04:06 diver comes to a temporary halt and after that he is going to swing in an oscillatory manner and this is what makes all these bungee jumping very spectacular so so after this initial fall the jumper actually comes to a momentary halt and then the cord as I told that the cord is stretched to its maximum and then there is an oscillatory motion that go to kick over but this mother oscillatory motion does not carry on for ever and viscous our drag due to air and other you know wind etc would take over and then finally the jumper

00:04:52 comes to complete halt so this is a so this bungee cord is made up of a material which has got addthis intrinsic property of getting extended because of this force or which is basically here the weight of the jumper and then the cord is going to regain its original shape and that's why the oscillatory motion occurs so it's from this discussion of this bungee jumping it's clear that it's a property of the body by which it tries to regain its original shape and size after the applied force is removed and

00:05:41 this is called as an elastic material so what could be the examples of elastic material such as we have a rubber band or there could be a spring which is an elastic material however there are examples of materials which are not elastic materials which means that they do not come back to its original to their original shape and size after the force is removed such as a clay or a dough of of wheat and they are called as plastic materials so what actually goes on inside the material which for which

00:06:41 some of these materials actually regain their original shape and some of them do not to understand that let us understand the intrinsic building block of these materials and by building block what we mean are the atoms and the molecules and the way they are bound to each other it will be very a preliminary to say that we could think of these elastic materials to be made of Springs okay where the two balls at the end that you see actually represent maybe atoms or molecules and the spring-like thing that

00:07:25 I have drawn here are the connection between them and when a pressure is exerted on these materials either this way or this way or even these ways are they tend to regain their original configuration so up so having understood the atomic configurations which really give render these elastic properties let us try to understand that the relationship between the forces the elongations and there are terms such as tensile stress and tensile strength that will come to very soon and what are their interrelationships between them

00:08:21 and how do we actually quantify some relations by which assisity can be understood better so let's take about two are an example of two rods okay and consider three different cases in which they have a same initial length but different extension and to make matters even clearer let us call l 0 as the initial length Delta n X n so case number 2 where we have our same initial length which is L 0 but different cross sectional area and in the third case we have different initial length but same extension

00:10:37 so let us draw these three cases one after another let us consider these two rods are which have so these are the unstretched length and and the areas of cross-section are also same same area of cross-section our same extension here the same area of cross section again for the third case so so the unstretched length are same now if I cause an extension and apply a larger force f1 here and a smaller force f2 here I'll have more extension in this rod than a smaller extension in this rock okay so square my f1 is greater

00:12:12 than F 2 so this is case 1 let me draw the same picture to understand the area of cross section so I have a rod like this and I have a rod like this which have same l0 however the cross sectional area is a 1 for this and a cross sectional area is a 2 for this so to create the same extension as it is pointed out in a point number 2 I am going to apply a mod so the force required to create the same extension in the bigger rod is is more in the rod with a bigger cross-sectional area so here we have F 2 is greater than F 1 to

00:13:25 create the same extension going to the third case we have different initial length now to have same extension I need bigger force in this the rod with smaller length than smaller force for the rod bigger length so this is your l0 1 and this is your l0 2 so my F 1 is greater than F 2 so from these three figures it's apparent that my force required to create an extension L in a rod of length initial length l0 of cross-sectional area a is s is proportional to Delta L s is proportional to R 1 over L 0 and s is

00:14:39 proportional to the area of cross section so if we combine all three of them we can write that s is proportional to Delta L 1 over L 0 and a so I can write it a little little more compact by taking this and so my F is proportional to now Delta L over L 0 multiplied by the area of cross section now Delta L by L 0 can be called as the fractional change in length let's write Delta L by L 0 as the fractional change in length and hence so I can write or change this constant I can change this proportionality using a proportionality

00:15:51 constant which can now be told as Y which is equal to Delta L by L 0 into a and this is a very important equation in this study of mechanical properties of solid or rather the elastic properties of solid in which it says that the force required to create an extension Delta L for a material of length l0 and cross-sectional area a is written as this where Y is known as Young's modulus of this Young's modulus is named after Thomas young 1773 to 1829 and this formula is going to be revisited a

00:17:02 number of time in our discussion or some variant of this formula will be revisited a number of times in our discussion of elastic properties of solids so we shall make this ongoing discussion more understandable in the sense that let us take the example of a spring so the spring is tied at one end and it has an unstretched length l0 and it's acted upon by a force F and it undergoes an expansion which we denote by delta L so this is still l0 and this is under the application of the force it undergoes an extension Delta L if we

00:18:00 double the force the extension will be doubled which means that the extension Delta L is proportional to the force applied on the spring if we triple the force the extension is going to be tripled now take another example in which there are two Springs which are acted upon by a force F each will have assuming that their unstretched length are same they extend by the same amount for each one of them which is Delta L which we can show it in a different so this is Delta L and again so this is Delta L in the application under the

00:19:06 application of this force F so which means that your Delta L the extension is proportional to the original length l0 so as we double the length of the spring under the application of the same force the extension or the elongation becomes twice as much so Delta L is proportional to L 0 and we have also got Delta L is proportional to F now to get the third dependence that is the dependence on the area of cross section we need to talk about a wire so let us take a wire of a finite cross-section and now take an

00:19:51 identical wire of an identical cross-section and length and suppose you apply a force here this is equivalent to the fact that there is a larger area of cross section same length and is acted upon by a force F so this has an area of cross section a this has an area of cross section a this has an area of cross section 2a so under the application of this force F the extension will be Delta L by 2 which means as we double the area of cross section the elongation goes down by a factor of 2 similarly if you take three

00:20:46 such wires put them together and apply a force F and then consider an equivalent wire with a radius or rather an area of cross section to be 3/8 then the extension is going to be Delta L by 3 which tells that Delta L is inversely proportional to a and hence all these three relations put together we can claim that Delta L is proportional to F is proportional to L 0 is inversely proportional to a and again as said earlier the proportionality constant is called as the Young's modulus and we can

00:21:44 write down this very familiar form which is y equal to F by a and a Delta L by M Co so with the help of this spring we can understand the dependencies of this elongation on the force the unstretched length l0 and the area of cross section of the wire and we can again realize this same formula that we have derived earlier Y is called as a young mother less so let's just come to this discussion of tensile stress that we had briefly spoken about but hasn't elaborated on it if I simply divide both

00:22:43 sides by a the area of cross-section of the material the rod in our case then we will get it as and Y can simply be written as f over a divided by Delta L over L 0 and as we have said that Y is called as a Young's modulus so the young's modulus is defined as the tensile stress divided by the tensile strain so the tensile stress is divine defined as the force per unit area that is a given to the to the rod of cross-sectional area a and Delta L by L 0 is a fractional change in length that has

00:23:52 occurred because of this tensile stress being given and why simply appears as the ratio of the tensile stress and tensile strain and so in a we may want to know that what are the values of this Young's modulus for most materials for solids is very high a to quote a few are representative values we can write that we can prepare a table which you will get it in your book the value of y so for a material such as aluminium this is a 6 point 9 into 10 to the power 10 and just to point out what could be the

00:24:52 dimension of this F over a which is Newton per meter square in SI unit where as Delta L by L 0 will is dimensionless so this will simply have the dimension of the enzyme stress which is simply a Newton per meter square and for copper it is slightly over 10 to the power 11 Newton per meter square similarly for steel which is known to be quite strong this is of the order so this has a value which is 10 to the power 11 children per 2 into 10 to the power 11 Newton per meter square whereas an island which is known to be elastic is 3

00:25:49 point 7 into 10 to the power 9 which has a value which is less than any of these other metallic items that are quoted here and a brass has something intermediate which is 9 into 10 to the power 9 Newton per meter square so this shows an interesting fact that steel of which has among these has the maximum value for the Youngs modulus which is 2 into 10 to the power 11 Newton per meter square simply tells us that this the force required of the the tensile stress required to create the same elongation

00:26:45 is a maximum in steel as compared to any of the other materials that are presented be sure so let us take a very a practical example of how do we where do we get the stress and what are their relevances in our daily life so let us consider a circus in which in a particular act act that is usually shown in circuses that a particular performer he supports the weight of six other performers his core performance in the group and each of his four performers are safe so there are six two performers in the group

00:27:44 our each weighing 50 kgs and one particular person is going to support their weight in in a certain act and a considered G to be equal to 10 meter per second square which you have seen that in many cases they're approximated because 9.8 is slightly odd number to calculate so the total weight of all those six performers put together is equal to total weight equal to 300 kg so with G equal to ten we get the total force that this particular performer will support is equal to 300 into 10 which is equal to three thousand digits

00:28:49 all right so now this particular performer has got his femur bone the bone in his in his leg it's approximately say for example 0.5 meter long so femur bone of this equal to 0.5 meter long and it's also equal to a cross-sectional area is 10 to the power minus 3 meter square so this is cross sectional area we want to know how much of compression in his femur bones would cause for supporting six performers on top of him so consider the Young's modulus of the bone is approximately 10 to the power 10 Newton per

00:29:56 metre square thus what will happen is that it will cause an extension in his femur bone which will go like which is equal to because he has a two legs so this will be like three thousand divided by two which will get divided into two of his feet and this will be like 0.5 meters divided by 10 to the power 10 Newton per meter square and this will be Newton here and this will be like 10 to the power minus 3 meter square and this will give us approximately a 10 to the power minus 6 meter so thus for

00:30:46 supporting the weight of six performers are the extension or other compression in his femur bone is of the order of 10 to the power 6 meters which means like 1 micrometer which is fairly small so we have talked about the compression on the bones of a performer in the circus now let us take the same example not with the bones of a human being but rather with the with a steel wire say for example so a steel wire just for our convenience the length which is same as unstretched length let's call that as

00:31:34 simple as L which is equal to 1 meter take that to be a thin wire say the area of cross section to be a 10 to the power minus 5 meter square is subjected to weight which is equal to say for example 500 kg so we really want 500 kg yes maybe we want 500 kg so the force acting on the wire so this mass is hung from a steel wire whose unstretched length L is equal to one meter the area of cross-section is 10 to the power minus 5 meter squared and there is a weight which is M is equal to 500 kg let's denote that by M

00:32:31 and hence the with G equal to 10 meters per second square which is an approximate value which we often use we can take this to be 5000 mutant so let's just try to calculate the extension of the steel wire due to this mass being hung at the bottom let's write down this formula that y is equal to F over a and divided by Delta L over L so you have a Delta L over L R equal to F by a Y and Delta L becomes equal to F R into L divided by K Y all the other quantities are given and for a steel wire

00:33:32 y is equal to 2 into 10 to the power 11 Newton per meter square and one can easily find out that this is equal to 2.5 millimeter 2.5 millimeter of extension is actually detectable by the naked eye and so you can see this extension of the rod now we understand that why in this particular example we have taken a mass which is as large as 500 kg now just to let you know that if it's a nylon wire nylon has a Y which is approximately 50 times Young's modulus 50 times lower than the value of T

00:34:23 which means the extension for a nylon rod for a nylon wire rather is going to be 50 times more than 2.5 millimeter so this is a large extension which of course can be recorded so this distinguishes different materials undergoing different extensions under the application of a given force solo we have talked about either a compression of an object or extension of an object along one dimension or along one direction so to say now we can consider a different kind of deformation of an object which is other than this simple

00:35:16 extension or compression so we just for that let's take this example of a thick book which is kept on so this is a side view of a book which is kept on a rough table and if we want to consider that so so if we apply a force on the top of the book the book has a area of cross section to be e so we apply a force on the top surface of the book which has a area of cross section a there is a force that acts in the opposite direction which is of the same magnitude and let's denote it by a negative sign and this will cause an

00:36:29 extension let's call it as Delta X here and this is the original length L Z so this resulting deformation of the book from its initial configuration as shown here is called as a shear or this force a per unit area is called as shearing strength as you see in this picture there is a shearing stress being developed between two parts of the cube they are sliding against each other so there is a shearing stress that is developed between the two parts okay and just similar line of arguments will show

00:37:24 that my so your s will still be proportional to a Delta X which where Delta X is indicated by this extension is proportional to of the original length here and it is also proportional to the area so combining just like before if all these quantities we can write it is Delta X by l0 into a and similarly we can write that F equal to or F over a is equal to G up by Delta X by l0 just like the Youngs modulus G will be called as the shear modulus and we are and F over a as we said will be called as a shearing stress delta x over

00:38:32 l0 is called as the shearing strain we can understand the shearing strain in a better manner if we are draw up a cylinder so this is my cylinder this is my extension which is there and there is a force which is acting this week and equal force is acting on the other direction and my this length is L so the shearing strain is given by our delta x over l or let's just put it as l not as we have been doing and this is given by tan of the angle of the deformation okay and for this theta to be small we can

00:39:49 approximate find a tan theta as sine theta because just to point out that tan theta equal to sine theta over cosine theta so as theta is small fine theta will increase linearly as theta where cosine theta will become equal to one so tan theta becomes equal to sine theta which becomes equal to I am so sorry this theta and sine theta will be written as theta so so these are the shearing stress and the shearing strain relation and this is very similar in not only in the in looks but also in form accepting that you have not caused any

00:40:54 extension or compression but has simply given a shearing for applied a shearing force because of which this upright cylinder has become a cylinder which is inclined at an angle theta so the next thing that we are going to talk about is volume deformation this is to say that it's not only in length or in area we can also create a deformation to a particular object in all three possible directions and by applying a stress or applying a force which acts uniformly in all directions a familiar example of this can be given

00:41:54 for a body submerged in a liquid so this is a liquid and let's take a body the stick tube in Arcis and the forces acting are from all sides and this will definitely clear compression create a compression in the in the system and this since the force is acting perpendicular to this we can replace the force by the area and which is by the pressure due to the liquid so that tells you that in this particular case it caused a deformation to the body where this pressure let's call it a change in pressure compared to

00:43:02 the original pressure which is equal to which should be proportional to Delta V over V zero and hence this Delta P which is again the volume stretch is can be written as B into Delta V by V zero where B is called as the bulk modulus of the system so this is and we have to put a negative sign here in accordance with the fact that as pressure increases volume decreases so as you submerge this body deeper and deeper into the liquid they'll be increased pressure and that will cause more the volume to compress

00:44:06 more and this is what the underwater divers do experience in their course of journey into the deep into the sea and so we have understood now there are three constants which describe adequately describe the mechanical properties of the body which are why as Young's modulus G as shear modulus and be as bulk modulus so as you can see in this picture cube is completely submerged in a liquid so it's been acted upon by the force from all directions and so there is a on the right there is the underwater swimmer and the forces

00:45:17 are again exerted on him on all from all directions so the bodies are undergoing volume deformation so now we are going to look at Hookes law this is a named after Robert Hooke from 1635 to 1703 so this is summarizing the findings that we have so far and we can write for the first case of this tensile stress and the tensile strength we had a delta l over l0 connecting to F over a and similarly the F over a is connected to Delta x over l0 and finally this Delta P which is connected to the bulk modulus by Delta V and just in

00:46:36 accordance with this l0 will write a v-0 which represents the original volume of the system now as you can see that in this left-hand side all these are stresses and in the right-hand side all these are strains so so there is a stress and this is strain and in each one of them the stress is actually proportional to the strain where the proportional proportionality constants Y G and B are called as Young's modulus shear modulus and the bulk modulus which depend upon the properties of the material and this

00:47:29 is known as this fact that stress is proportional to strain is known as Hookes law and so just to revise that the stress has a unit of Pascal which is equal to one Newton per meter square so one Pascal equal to one Newton per meter square and strain is of course dimensionless and this brings us to the talk where we can plot stress versus strain in a generic manner and we'll get a straight line till certain values of stress and after that this is going to have a nonlinear behavior and till this

00:48:39 point till which the straight line behavior is obeyed and that will be called as ax so this we will term as the elastic limit so the significance of elastic limit is that that the body will regain its original configuration once the force is removed and beyond this the body is not going to regain its original configuration because stress is no longer proportion to strain and you will have a deformation here which is usually refer to as the plastic deformation the stress-strain graph that we have talked

00:49:34 about needs a little more closer look if you look at the stress versus strain relationship let's draw that again stress which has the unit of pressure which is like F over a whereas strain it dimensionless and this behaves like this initially which is a straight line which means that stress is proportional to strain and in this regime all the way from this point O to the point e Hookes law is valid so our stress is proportional to strain and we know that the proportionality constant is called

00:50:34 as the Youngs modulus once you surpass this point a which means the the strain becomes larger than the value that is here and you keep applying the stress via the weight or yr the mass that was that was discussed earlier this grass no longer remains a straight line as before and it becomes like this and it goes and it becomes like this let's discuss this region now it is not a straight line but it's fairly flat here between a and P and which tells you that for a small increase in stress the increase in

00:51:37 strain is enormous in other words this region in this particular region the material has a plastic flow so it almost behaves like a fluid and it flows and it reaches a point B so this region from B to C is quite interesting it's interesting in the following sense that there is a non monotonicity at this point C what I mean by non monotonicity is the following that as here from here to here in this region even though the stress is decreasing the strain is increasing which hasn't been the case so far so

00:52:29 which means that even if it could happen that in a particular circumstances the stress could actually decrease however the strain will continue increasing there are specialized equipment where these kind of behavior can be detected and the detection of this behavior can only happen close to the breaking point and this D is called as a breaking point where the material breaks so the basepart between o and a is of importance to us because this is where all our discussions are mostly concentrated - because this is called as

00:53:22 a elastic limit or as I told earlier Hookes law is strictly valid and beyond that Hookes law begins to pay so before we end this class let us take a look at what's going to be done in the next few classes at least a couple of classes so we'll take a look at things such as different materials so different materials will be talked about and maximum allowable loads so that when construction takes place we know exactly how much of load to put such as the tensile stresses or the compressive stresses so these has to be

00:54:35 known for different materials especially the ones that are used in construction then we'll take a look at a experimental determination of of young modulus 1/3 we'll take a look at some of the technical definitions of certain terms which are used very often to discuss the elastic properties of matter and they are so the technical definition of of space quantities such as called a toughness brittleness hardness of resilience and tipnis and finally we will solve some problems these are the plans for the next class