 Hello everyone. In this video I'd like to talk about the basics of stress and strain. So to talk about stress and strain, first we need to review what are the primary loads that we're thinking about. So the types of loads that we could expect to find in a rigid body. So I'm going to go ahead and start writing that down. So typically we have several types of loads and we might be looking at axial, shear, bending moment, and torsion. And recall that we had kind of some basic variables that we used for each of these. So p for axial, v for shear, m for bending moment, and t for torsion. So then for each of these we can think about well what stresses would result from the application of these loads. So if I take each of these four loads I can say well an axial load is going to result in a normal stress due to that axial load. And use my notation Greek letter sigma for normal stress and typically give it a little subscript a. We have sometimes what we call direct shear for the application of a shear force, shear stress then using the letter tau subscript v to refer to the the application of direct shear. Bending moment also results in a normal stress and we can give that a label sigma sub b. And finally we have torsional shear tau sub t. So we have these four possible stresses, two of them being in the normal direction, normal meaning normal to the to the cross sectional surface. And then the shear stress is being parallel to the surface. So when we make that imaginary cut in a rigid body if we have this cut plane normal is going to be stresses that are into or out of the surface and shear stresses are going to be in the plane of that cut. So let's start with our axial stress. So axial stress as you might expect if we have an arbitrary rigid body and on this rigid body I have forces in equal and opposite directions. And then I'm going to make a cut internal to that rigid body and I'm going to pick a side of that doesn't really matter which and say okay I have a force that's still applied to this end. And of course we know then that there has to be an equal and opposite internal force in which case would be that axial and we can represent that because really how it's represented then as as a stress which is distributed across that surface. So this axial stress sigma sub a you know given that we know what axial stress looks like is going to look like something like this force f which is that applied load or whatever you know label we've given it divided by that area. So it's a pressure it's a stress it's a force distributed over an area and that's pretty straightforward. And again it's drawn in this direction which is normal to the surface meaning it points into and out of that that into or out of that surface. And we would consider this to be like an average stress across that surface right which falls under the assumption that it's uniformly distributed in that it's equivalent basically at any point along that surface we would assume that it's an equivalent stress. So next if I can scroll down here a little bit talk about direct shear. So shearing at least for me the easiest way to think about shearing is to think about it like a scissors cutting a piece of paper right if I have a scissors which I don't have in front of me and I have a piece of paper I can imagine what's happening here right I have the two sides to my to my scissors and when those blades pass next to each other basically touching each other but very very close next to each other it cuts through the paper right and that is the best example I can give of of a shearing type motion. So if I have my some arbitrary rigid body and I have two other rigid bodies which I say I might apply a load to and this one is offset on this side basically at the same interface point where these two right hand side of the top one left hand side of the bottom rectangle and we're applying a force to each of those to kind of push them past each other then if we take a look at this location make a cut imaginary cut through here uh and represent that as a free body diagram I could say okay I'm looking at the left hand side of this rigid body I have a force P still applied right and again because it's in uh equilibrium I need an equal and opposite force so the equal and opposite force has to be on that surface because that's the only thing left of an equivalent value of of P. Now this is again a stress distributed over an area you know we know that that the units of stress are you know pounds per square inch or or um uh newtons per meter squared so we distribute that stress over that surface and as a result we have a stressed house of V equal to in this case P over that area that that cross sectional area of my rigid body where I've made that cut and now I've labeled it P here just because that's what I labeled the load don't get that confused with axial load you know we might call this V for for um shear in most circumstances but one thing we need to be careful about uh and you may recall from you know if you've seen this before is that this shear stress is not distributed uniformly across the surface uh what we would actually find and if I do another drawing here of my rigid body without putting my loads on it I would actually find that my that's a bad drawing but my shear stress is distributed as a parabolic function so so greater in the center uh and going to zero at the outer edges so in reality I have a a parabolic curve on that surface which shows what my um stress distribution is so on the top and bottom of my my rectangular rigid body that I have in my picture I would have zero shear stress and I would have greatest shear stress right in the middle so if I wanted to know my actual highest shear stress I would say it's four thirds V over a or in this case P over a whatever I want to call it the shear stress I originally wrote which was just P over a I would call this an average shear stress and this four thirds would be more like a max so it's four thirds so it's 33 percent greater at that center than what it would be on average across that so it's something that's pretty important to remember typically if we're interested in the shear stress as our failure mechanism we need to remember that it's four thirds V over a at that maximum center point now we also need to be careful that it's only at this this central axis um if I kind of dash in a neutral axis here and you'll get why I call that a neutral axis in just a second that's where we would expect to find that max to occur if we're looking at the only the top and the bottom then we have no shear stress and so we need to be careful about where we locate our points now one other thing to think about when we talk about direct shear because it comes up a lot when we're talking about things that are subject to peer for shear which would be like pins or bolts we can have what we might call single shear which is say I have a bolt and I have something that's loaded so this is a pin perhaps such that I have a force on this side and an equal and opposite opposing force on this side so in this example this bolt that's in the middle is holding these two plates together um you know could be like a cotter pin or a pin that holds like a trailer to a lawnmower or something like that and when I look at this example my shear stress as a result occurs where these two plates are together right on that bolt there's one one shear surface one cross-sectional area where that force is applied so then I can you know do the same thing I did before and say that my shear stress is that average as an average is f over a however I might also have the same example of a bolt and I have one plate here with a force applied to it and then I have I don't know what you want to call it like a double a clove clove I can't remember there's some there's a name for this clove or something like that where I have two surfaces so I have my force still applied here but now this has a split in such that there's two sides to it top and bottom around this other one which is pulling this way so that means that there's two surfaces which I'm putting way too many lines on the screen here but you hopefully get what I'm saying there's two surfaces where this stress is being applied to my bolt now what that means is when I go ahead and write my shear stress I have a force over an area but there's kind of two ways I can think about this I have two cross-sectional areas so I could say force over 2a or I could also say well if I take this force back and cut this here the force actually gets distributed one half of over each area so f over 2 divided by one area so of course both of these equations result in the same thing right it's f over a one half multiplied by one half so it ends up being half of the amount of shear stress when it's split in this way so sometimes we call this first example single shear and we might call this second version double shear so nothing's really changing other than in the double shear case my shear load is applied over two different locations on the bolt so the shear stress is in effect halved as it's applied to the bolt all right next I want to talk about bending stress so using the the simplest example I take a rigid body and I apply a moment to it so a bending moment which is basically a torsion applied on equal and opposite sides I have my moment I have my equal and opposite moment on the other end and the result of this is that if I you know exaggerate the deformation I get something that looks like this and you may have heard of this before I have a neutral axis through the middle of this and the reason we call this a neutral axis is because as I bend something and curve it in this way you could see if you looked at the say the top line and compared it to the bottom line my top line is getting shorter while my bottom line is getting longer so I'm compressing on the top and putting it in tension on the bottom and of course that implies that somewhere in the middle it has to be zero right if I have a positive or excuse me a negative on the top because we typically say compression is negative and I have a tension on the bottom which we typically say is positive then somewhere in the middle it has to be zero and that's what we call the neutral axis so it's the axis on which the stress is is zero now when we think about that then we can go ahead and consider how that means our stress is applied so if I again make a cut any random place through the middle of here and look at that as a free body diagram and I have my neutral axis drawn in I could look at the stress on this and say well on the top it's going to be negative in compression zero at the neutral axis positive at the opposite end and I can kind of draw in what my stress vectors might look like on here so in this case we what we see is that the stress is a maximum at the in the positive direction at the bottom and in the negative direction at the top and it's zero at the middle at that neutral axis stress for bending then then has the equation my by i now getting our notation y in this case is considered to be distance from the neutral axis so as I walk away my position from that neutral axis towards the upper or lower position I would say that that's increasing my distance y so it's a maximum at that top or bottom position and therefore you know based on the equation we can see well as y increases the stress increases so it all matches with our result or with our equation with our math I remember is the area moment of inertia and in effect what area moment of inertia is is resistance to bending so it's the resistance of of that part to to bending it's a it's a geometric property so it's dependent on the shape or the cross-sectional shape of my rigid body so if I have a rectangle that's going to be one thing if I have an i-beam that's going to be something else and those are going to have different area moment of moments of inertia and I can calculate those usually the equations are something we look up right back of the textbook style thing or somewhere online so we can look up what those area moment of inertia values would be and and plug that in if you're following along the last one that I need to talk about is our torsional shear stress so torsional shear again just picking an arbitrary rigid body torsional shear is is the stress that results from the application of a torque to something so in this case I have a torque applied to either end of a shaft and that torque results in a stress in between those two now if I again make a cut somewhere in the middle and in this case I'm going to pull out a little stress element so I've drawn a little square on here and pull that out it might have a curved surface on top because I cut it from the outer edge and on the right hand side of my shaft I've got you know looking at it from this direction or from the direction of the end looking in I have a clockwise torque and as I can imagine how this would be applied to to my little stress element that my cut if my stress element is on the left hand side of the cut then my torque is going to go across the face of it this way and it's going to be really hard for me to draw but you can imagine I have a back side to this cube so I have a torque equal and opposite torque on the opposite side and now the interesting thing about this this torsional stress is that if you looked at this little 3d stress element that I've pulled out here I've got one on the front one on the back but it's not an equilibrium right and why is that because it can rotate these are both pointing in the same direction now it's in force equilibrium one this way one this way so equal and opposite my hands aren't on the screen but you hopefully get what I'm saying but it's not in rotational equilibrium so to put it in rotational equilibrium I would need to draw equal and opposite shear stresses on the front and back and that gives me equilibrium so now my torsional stress has the equation tr by j r is the radius of my shaft typically when we do these things we're applying torque to round shafts because they tend to do better it's kind of an energy minimization type thing if I apply a torque to a square shaft or a square thing there's a calculation and I don't have the equation in front of me but basically you can find an equivalent radius which is the equivalent circular shaft that would act the same as that square geometry and that's normally what we would do for that but I'm just going to stick to round shafts for this example so r is my radius j is polar moment of inertia and just like area moment of inertia is resistance to bending polar moment of inertia j is a geometric geometric property that's resistance to twisting so it's how how well the geometry is going to resist being deformed in a twisting fashion and you can probably already guess from this equation how my stress is distributed but if I look at that circular cross section and go out from there my stress increases linearly with radius as I increase r or radius I'm linearly increasing my stress so I get a plot that looks like this where my stress is a maximum at the outer surface of my shaft and zero at the very you know dead center of the shaft so that gives us um our torsional stress and of course any component that we might look at might have any one of these four combinations of stress