 Alright, what we're going to look at today is more of what's, we're going to kind of bring everything together because as, I hope somewhat obvious, a bit of what we've been looking at is a little artificial. We have to do that to get started. So we started with axial loading and looked at the normal stresses associated with that and then some of the shear stresses. We looked at torsion, which was entirely shear, if you remember, and then we put in later some bending and that used most of the very type of loading that we happen to look at in statics and it is some of the most common of the structural type loadings that we'll look at and then we looked just on Monday at the transverse shear loading that we can see. Now we're going to put all of these together in some combination and see what they all have to do with each other. Before that we have to take a little bit of a step. This is considered a combined loading. I'm not exactly sure why because it really only is a strict and particular type of loading and that's the type of stresses seen in thin wall pressure vessels. This is a huge part of the mechanical engineering industry. There's the government regulations that need to be observed in terms of thin wall pressure vessels all built on the very type of thing we're going to look at now, it's just immense. You can imagine what the kind of failure if a high pressure vessel fails and the sudden release of that pressure can be catastrophic even in ways you might not anticipate. Back in Boston in the early 1900s I forget exactly when there was a molasses tank that was holding was full of molasses. The pressure due to the hydrostatic pressure of that molasses ruptured the tank and caused a flood and as embarrassing as a way to die a lot of 100 people or so were killed in a molasses flood. I believe if I remember it was in January as well so they couldn't outrun cold molasses which is I imagine you get up to the pearly gates and you got some explaining to do because that's just a terrible way to go. There are better ways. So we're going to look at thin wall pressure vessels so we'll take kind of a look at a spherical cap just to get the picture going so there's some elemental piece in this and we're assuming that these are all high pressure vessels on the inside so this is like a spherical tank and what that high pressure on the inside causes is tension in all directions in the material that makes up the pressure vessel itself and just trying to give it the appearance of a curved surface like that we're going to see then the normal stress throughout. No resistance to bending in this and all forces are tangential in nature. So we'll be a little more specific now and we'll start with a look at a cylindrical pressure vessel. So we've got it cut away. Of course it's got some kind of cap on the end. High pressure vessels don't do well if you just put a flat piece over the end because as I just mentioned there's no resistance to bending of these very thin materials and the end cap would be almost entirely in bending so it's typical that a more rounded piece is put them in and that's exactly what you see with scuba tanks and the like. So we've got this pressure vessel with an associated coordinate system. First though let's make sure we understand what we mean by thin wall. We have R the inside radius of the tank and then there's some wall thickness T and it's the ratio of those two that we used as our definition as arbitraries it might be. We define a thin wall pressure vessel as one where the radius to the thickness is greater than about 10 so if we had a 10 inch inside radius tank any wall thickness less than one is going to be considered a thin wall pressure vessel. Then as we had before on the spherical picture we've got a little elemental piece there and that's going to see two types of normal stresses on it. There's going to be a hoop stress that we call sigma 1. I don't know why we can't call it sigma H but not one of the books I have looks at it as that. Sigma 1 is a hoop stress hoop obviously because it goes around the vessel. As the type of metal hoops you'd see if we're looking at oak barrels in the length they have exactly that type of thing and we keep looking at the stress. In fact we have a couple problems in the book looking at pressurized oak barrels with the metal hoops running around them and then there's a longitudinal stress called sigma 2. of the vessel itself and it's in its main dimension or all right we need to take this apart and see what we can make of these stresses themselves in regards to the load they're carrying which is the internal pressure. This internal pressure is considered a gauge pressure you may have heard the term before gauge. It's precisely the type of reading if you haven't had a pressure gauge and you attach it to some kind of nozzle that could read the pressure in the tank. It's merely the pressure over atmospheric. That's all your pressure gauge reads when you take the pressure in your tires. It's how much pressure there is in the tire in addition to the atmospheric pressure. What that allows us to do is we only need to look at the forces caused by the internal pressure. We don't need to balance that against the outside the atmospheric pressure also pushing on the pressure vessel in towards the center. We can just ignore that because we subtract it off of both of them so it only has the pressure above atmospheric to worry about. If we consider atmospheric pressure and use the absolute pressure on the inside of the tank they just subtract anyway because they're exerting equal and opposite forces on either side of the wall. So we're only going to look at the gauge pressure we'll use in a fit of creativity the symbol P and now we can look at the type of stresses that we've got there. So we'll start with a little segment of the vessel down the x direction so we'll take a little piece of it that runs like this. So what we've done is we've got a little piece of the tank of some length delta x and I cut it right in half and exposed that tank the type of thing we've done before where we're just imagining a cut through the tank to expose the internal force in the material. So when we do that we've got this internal pressure that's pushing out on the tank. Now you have all taken physics two I believe. Is that right? You looked at hydrogen, David have you? I did take physics two but I'm not sure if we covered this. What I'm looking at is the hydrostatic pressure the pressure of a fluid on a solid surface and in this case we have a curved surface. Now it might seem problematic at first because the pressure is a force that can act normally to a surface only. There's no transverse ability of pressure to exert any forces. So the pressure is going to be acting around the surface like that normal to the surface in every place. And if you remember from hydrostatics in physics two if you actually covered it there was a small section you pull out your physics book it'll show it the forces on submerged curved surfaces. It may seem somewhat problematic because you're going to need to integrate over the entire curved surface on the inside. What it turns out though to to end up in result is very very simple. You only need to look at the projected surface and the forces normal to that. This happens because every upward component of force has an equal and opposite downward component and they all cancel. So any any of the let's see this would be our y direction. Any y direction component to the forces cancel each other and the result is that you only need to look at the equivalent pressure acting on a projected area to get the total force acting on that curved surface which makes it a lot easier for us as we balance the forces because we're balancing the forces really in the z direction because of the exposed surface we've got there. So we've got this projected area which is nothing more than the rectangle there with these on my perspective on my drawings off a little bit. We've got pressure the gauge pressure acting on that and that's all we need to consider. That'll give us the equivalent pressure of having done it all the way around the surface normal to the surface would get the same equivalence. So we've got those forces trying to push that that cylindrical that that the hemicellular semi-cylinder the semi-cylinder towards the minus z direction and then of course the the internal forces in the material which is what we're trying to find are acting in the opposite direction to balance those because we have static equilibrium in all of our situations that we work with. Remember this is this is in the the hoop direction that I'm looking at now. So we can balance the forces on this piece here and see what they equal and that'll allow us to find the stress. So the the first part I guess we can do is is we have this pressure acting on that that area that area is twice the radius times that delta x. So that's 2r delta x that's the that's the pink pressure force exerted by the fluid due to the fluid pressure whether this is compressed air or simply liquid storage liquids being so much heavier than gases they tend to exert a lot of pressure on their own right. So that's the the pink pressure that's got to be balanced by the stresses in the material itself which is delta it's a the stress acting on the area of the wall that's exposed which is delta x times t but there's one top and bottom so it'd be 2t delta x and that's the due to internal forces in the wall itself. Alright so everybody comfortable with the picture for the most part we're balancing the fluid pressure forces trying to push this in the minus e direction and the internal forces in the wall itself as it tries to hang on to the other half of the tank that we've taken out of the drawing and so this simplifies quite quickly the delta x is cancel the 2's cancel and we get then the hoop stress sigma 1 is equal to pr over t. Units work out okay always check our units as we go through algebra yeah remember the units on stress are the very same units as pressure in fact we often use Pascal which is a standard unit of pressure so it works out just fine for the units units of pressure in Pascal we're capitalizing a in that sensor area we don't capitalize units and so that's the hoop stress is as simple as that we need to know what the gauge pressure inside the tank is and then actually multiply it by that ratio that we use to define the thin wall limit itself the pretty straightforward straightforward calculation there to find the hoop stress all right to find the longitudinal stress we need to make the cut in the other direction so it's very much like the picture we've got there so I'll just redraw here so it's a little bit bigger kind of exaggerate the wall thickness so that we can see all the parts of it and now we've got the same type of thing on this projected area here we have the fluid pressure acting so that's that's going to the force acting over that area is the force in the I guess would have been minus x direction and then we have this internal wall force as the wall tries to hold itself together acting all the way around on this exposed bit of wall we've cut open in the imaginary way we've been doing this all along and so again we need the two forces to sum to zero otherwise we won't have static equilibrium so we have the fluid pressure acting on that projected area which is just the inside radius inside circular area pi r squared so again that's the force due to the fluid and that must be balanced by the internal stresses in the material itself to withstand those forces as this pressure is trying to rip the tank apart the tank's trying to hang on as best it can so due to the thin wall approximation all we need to do to estimate this area is we'll just call it two pi r times t it's not quite that it's uh you know we the I guess have to do two pi r outside minus r inside then times t or no our average I guess would be a better one but this is this is close enough with the thin wall approximation and so that's the internal wall forces and we can then solve for the longitudinal stress pi cancels and one r cancels and we get then p over two t r that right p p times r over two two notice that that's half the hoop stress so it's the the circular direction the hoop direction that's going to be the greater of the dangers of the two so that's our longitudinal stress okay fairly straightforward these aren't severely difficult calculations so we'll do a couple problems just to make sure we can work oh no what we have to also do is this is for a cylindrical tank but most of these tanks will have a circular end cap on them so we have to do our analysis for that so we'll now look at a a spherical tank there are some tanks that are completely spherical they're more difficult to make but they do use less material as you remember from probably calc two or three the maximum volume for the minimum material is for a spherical shape which is why not makes water drops round so we'll now look at a spherical tank yes it's also probably one reason why so we have space cancels around the spherical yeah well that's a pressurized a thin wall pressurized vessel yeah the atmospheric pressure p is equal close to zero yeah so the problem like that the the absolute the gauge pressure and the internal absolute pressure are essentially equal in this case since it's spherical there is no preferred direction it's going to be symmetric in all direction so we only need to figure out a single stress hoop there is no hoop direction there's no longitudinal direction so there's only this one single stress we need to figure out but we'll call it sigma one since that is the worst case over in the cylindrical tank so we make sure that we're looking at the right one so if we make a cut through the middle to expose this we get the same kind of picture we have before now we have a hemisphere cut open with fluid pressure like that internal wall stresses like that we get essentially the very same picture we just had and we get the very same solution pr over 2t okay so now we can we can look at a quick problem throw into the numbers here so imagine a long cylindrical tank with spherical end caps something like that maybe a little longer than we might see but you see propane tanks outside of summer camps all over the Adirondacks so this very same type of thing tank is 8 feet long the cylindrical part has a wall thickness of 3 inch and made of steel it's 30 inches in diameter and the end caps themselves are five sixteenths inch steel or as we say five teens we would say that is too embarrassing to say we'll give it a gauge pressure of 180 psi g the g means gauge psi gauge pressure you might also see you'll see this when you take thermodynamics that there'll also be pressure readings and psi a which is pounds per square inch absolute that then includes the effect of atmospheric pressure which we don't need to do because the atmospheric pressure in is subtracted off the part of the atmospheric pressure that's also in there we just used the difference between the two okay so we've got two pieces to look at we've got the end cap so look at and see what the stresses are in there make sure that it won't exceed the steel limits and that's the part we just came up with there's no preferred direction to this symmetry of the sphere so you can just put the pieces in and double check to make sure your unit's all right so that's up to you take a second to do that and then in the body of the tank as well we have two types of stresses to calculate the signal on the hoop stress and the longitudinal stress we can compare those against stress limits in in steel these happen to be well under that so I don't have to bother with that but just check the number to check your units this 30 inch is the outside diameter so technically you need the inside radius and it's not 15 well that's not quite what I had 14.69 oh I'm sorry you're looking at the at the slide in 14.84 and there's going to be uh pressure vessels tend to have a very large factor of safety on them anyway so if we're off by a few little bits it hopefully gets swallowed by the rather large factor of safety at your units most people are getting the pieces there pressure 180 psi what do you have for r you had 14.84 what do you look at different things for r you have 13. something we're all over the map no that's the outside diameter 5 16 oh minus okay yeah for the end cap it's 15 minus 5 16 what's that like special okay what are you doing Tommy we're off we're doing different 13 something 14.688 is that right John finally 6875 remember we don't need too many significant figures on those types of things and then that's thickness itself 5 16 and then the same kind of thing same pressure slightly different r and t on the tank body calculations now that is not psig because these are not fluid pressures these are stresses that we're calculating here there's no such thing as a gauge stress it's not that I know Bill you got that number right so we already know what all we need to charge that pressure is the same this time r is 14.625 and then we only have to cut that in half to get the other one agreed Joey okay no what do you got this or that one not this one not yet or not agree with it not yet okay but others agree with it you're okay with the 70 70 20 20 is that one do another real quick one the ultimate stress no well it depends if you remember when we looked at our stress strain diagrams the yield stress is the end of the linear part the ultimate stress is the highest it goes up to the yield stress is what you're worried about in structural materials because when the load is released you want the stress on the strain to be released too ultimate stress would be for other materials for example this problem we're looking at is that of a basketball and you never use a basketball uninflated well sometimes you have to because you can't find the needle that you lost when your dad told you to put it away where you're supposed to and you didn't so you can't find the needle every household has 12 of them nobody knows where one of them is not that i'm speaking from practice the the ultimate stress from rubber is about 2000 psi we don't need to worry about the yield stress because we're not going to pump up a basketball then unpump it and use it we're going to pump it up and you want to make sure that it'll take as much pressure as you can put into it and not burst so typical uh outside diameter for basketball about 9.5 inches wall thickness of the rubber about an eighth of an inch and typical gauge pressure is about nine psi so determine about how much margin of safety there is to that uh ultimate pressure for the rubber and will you let your kids play with a basketball when it's might be on the verge of exploding taking out his eye suing the school district for millions of dollars and you drive nice cars for the rest of your life i won't let kids play with us like that's pretty much dodge pair oh that's just that's that's more uh an issue for uh it's nice hard self esteem right one yeah so there's no need to be exhausting over a mediocrity well it's in dodgeball when you get hit you have to go line up on the side so everybody knows you're out and then you're not playing everybody plays in america come on you've got kids you guys are kids you grew up in that world you guys are all under depression that you can do anything you want the adults in the room me no better so figure out if we've got a pretty good margin on this so we probably could see the next club should uh should do some myth buster type things you know do do tests like that see what will happen joey'll sign up you will get a case of bearer yeah just what i want to do with 18 you know students all right all right now you should be getting an internal pressure internal stress you're looking for the stresses in the wall of the basketball and i notice everybody's getting oh yeah okay we're all getting about the same number what a factor safety 12 give or take a little bit yeah you should be getting the units work out okay we're at nine psi g the radius uh i think comes out to be 4.625 and the inside wall thickness point oh that was given 4.125 and so we had stress in the wall material itself in the rubber of the basketball of well under the expected to break the stress and so it's not common to find the basketballs that will rupture uh on the playground we'll do one more real quick just to give a different illustration of the type of thing that will happen so let's look at a storage tank now in this case uh the entire tank is not under pressure because it's only the hydrostatic pressure of the fluid in the tank itself so you're not going to need a spherical end cap on the top of this because that's not significantly under pressure it's the bottom down here that's going to see the greater pressure since the pressure varies with depth and tank so we'll give it an eight meter outside diameter a 16 meter height but the uh we don't need that 16 meter we'll just use 15.5 meters of fluid that's what matters it's the height of the fluid unless it is a closed tank and the uh space above it is pressurized which could happen if they're uh worrying about you know you want the tank to empty quicker you pressurize the tank to force it out faster so wall thickness of 16 millimeters at the bottom determine at the bottom what the hoop stress is then and we know the longitudinal stress is half that so the one thing you'll need is the density of water is a thousand kilograms per cubic meter so we'll see if you remember how to determine the pressure at the bottom of a column of the bottom of a liquid column and we're assuming the top is open so it's open to atmospheric so this would be the gauge pressure at the bottom let's see who the members have figured out the pressure at the bottom of the fluid column did you guys do some hydrostatics and fluids and physics too? I exactly didn't do that at RIT and you didn't yeah that made me I'm wondering depending upon the sequence some schools put that in in physics three and put other stuff in physics two I don't think we probably would have done this in physics three at RIT it was a lot like physics three here we did electrical stuff one atmosphere for 10 meters yeah something like that yeah I guess it would be pressure at depth there's a function of the density actually the specific weight and the height of the fluid column at the top you can do without the conversion if you just use row gh where rows the density of the fluid g is the gravitational field strength so that together is the specific weight where density is actually specific mass and then h is the the height of the fluid column which is the 15-5 so if you do that then you're not in units of atmospheric pressure atmospheric pressure is about 14.7 psi if I remember I tried my best not to memorize conversion factors because then when you get them wrong you don't know it so you look them up each time and then you know they're right what's my memory sucks if my memory sucks I'll be a medical doctor I'll be driving a Porsche with a lawn on my side like you guys and I'd be like doesn't sound like much punishment for having poor memory wait till you're 57 Porsche and a lawn sound pretty nice the thing my wife doesn't watch these videos she gave up on those long ago figure out the pressure row gh at the bottom at the bottom because the pressure is much greater at the bottom in fact the gauge pressure gauge pressure at the top is zero because it's atmospheric pressure at the top there's no fluid weighing it down and then once you find that pressure you can just figure out the little yeah figure out the radius thickness is given other than that you do have some units to worry about because we've got some meters and some millimeters in there but that's pretty minor so that'll give us newtons per cubic meter whatever that means I guess that yeah that's the specific weight and then times the 15.5 meters so that'll give us newtons per square meter or Pascal Pascal's we need to try being at considerate pressure that is the gauge pressure because you're only worrying about the fluid column weight itself above that next five and there is no designation like kphg at least none that i've ever seen where if you kill a Pascal's gauge yeah weird and had to kill a Pascal's yeah wow that was this machine actually could have been Vermont they're pretty progressive over there ordering a whole little pocket of liberalism all right so we've got all the numbers there I hope you do uh at least then we've got the 152 times 10 to the three Pascal's newtons square meter so the units will all work out the radius 3.984 right remember that's the inside radius that's the outside radius that's given there so have to subtract the 16 millimeters off of that then the wall thickness was just 16 millimeters so get what 37.9 mega Pascal's and then you can check that against the yield stress since this is something that needs to work both stressed and unstressed because the tank's empty and full at different times then you'd probably want to design for the yield stress rather than the ultimate stress like we would with a basketball okay take our next step into remember what we're looking at is a combined loading so far we've got normal or axial stresses that we've looked at that uh that we came up with very early and it's the same type of thing we're looking at here may not use a P there let's that looks too much like the pressure it's a force over some unit area we also just added bending stresses and that was if you remember due to whatever the bending moment was they were a maximum at the farthest away piece of the material with respect to the neutral axis over the moment inertia of the cross section we also had in here shear stresses first was caused by torsion which was the applied torque times the radius of the piece over the polar moment of inertia and then we just had the uh transfer shear stresses that we just added that's the vq over it so we over the term put together those four types of stresses two normal stresses two shear stresses now we're going to look at the possibility of loading such that these things will be combined for example we could have a beam that was loaded in this fashion simply supported like this simply we could possibly be is too many supports and then some kind of load in here what we typically look at is some kind of load like this so we'll put 3.6 kilonewtons there and that'll be it this will be 375 millimeters we'll put 1175 no sorry 1125 millimeters over there nothing new there but remember what we started the term with was axial loading so let's go ahead and throw that into we have not before looked at these two things in combination it turns out it's very very very easy to handle what we get here is what the axial loading that we started with which is just simply this 25 kilonewtons this one happens to be in tension this is the stuff we were doing the first week when we added on this kind of thing oh i'll give you a cross sectional shape and area of the beam of 50 by 75 millimeters so simple rectangular shape and so we figured out back in the first week that the normal stress was the load being carried by a particular cross sectional area and we've got all this for this axial load 25 kilonewtons being held by 55 by 75 square millimeter area that comes out to be 6.75 kilopascals in tension and we have to pay very close attention to whether we're talking about tensile or compressive forces now because in addition to that we also have this bending that we figured out uh just a few weeks ago just before break we were looking at the bending you can figure it out for yourself but this leaves a maximum moment of well we can figure it out i guess yeah the maximum maximum moment comes just just before here and you could you could prove that to yourselves with a quick shear moment diagram if you want comes out to be well it's the the reaction here which is 2.2.7 is the reaction on that side so it's that times the moment on by the time it gets all the way over there which is the full 375 the maximum moment occurs right at the where that load is uh i don't happen to have that number because what we really need then is the stress that that maximum moment causes what is that moment john kill a newton meter that's uh that's causing the bending see well due to the rectangular cross section we know that c is half of the 75 millimeter height which is 37.5 millimeters 0.375 meters and what's i for a rectangular cross section you remember 112 now that's just the bending stress remember that comes out to be uh 21 6 mega pascals because here's the deal with what's going on we've got this beam due to uh under axial loading i mean label this bending we've got the axial loading that puts a stress profile that looks something like this uh it's intention all across the beam and it's well we're looking at the average so it's uniform and that's the 6.75 kilopascals that's just the axial effect alone we have to add to it this bending effect that we've got which looks like this we know that it's linearly distributed about the neutral axis in this case the neutral axis is right down the center and because of the type of bending we've got we've got compression in the top of the beam tension in the bottom what we figure out here remember was the maximum stress now those two effects are happening at the same time on this beam so we need to add those together to get the real effect of the stress on the beam due to the combined loading so that's going to look something like this uh on the top of the beam we have tension stresses combining with compressing stresses so they're going to cancel each other to some measure um we have 6.75 stress and tension on the top half of the beam and subtracting from a maximum of 21.6 on the bottom of the beam so it's going to look something like when i have that number what's uh what's 6.75 minus 21.6 15 something 6.75 and tension and a maximum of 21.6 working compression on the top what so this is we have at the top of the beam we have a compression of 14.2 we'll call 14.9 uh kilopastals so we're essentially subtracting well adding these two uh one constant one linearly distributed so we get a linear distribution of something like this at the bottom of the beam we're adding the 6.75 in tension with the 21.6 in tension at the bottom of the beam so that should be 28.3 what i think we think the arrows in the wrong direction well depends on how you draw this this this i show going in as compression would it's going from outside the solid inside so that's compression it's an mpa mega pascal oh mega pascal no but the other ones in kta so the difference is like no which one's which 21.6 is mega pascal isn't this as well no that's gone i have i have i have this is actually mega pascals and i have the right kilopastals this is mega pascals yeah so we're still we're still adding them let's fix that kilopastals there and i had the written down the other one this is mega pascals all right and so that's mega pascals there as well but we have a lot less of it in a lot less compression and a lot more of it in a lot more tension and notice that the neutral axis has shifted to because of this combined load now and so now our maximum normal stress is this 28.3 at the bottom which neither one of them approached independently but it's as simple as that as simple as adding the stresses together nothing more than a a superposition of the the stresses that we're calculating so i'll set up a problem and we'll get started and we'll finish it very quickly on um monday so imagine we have a bracket fastened to a wall here folded through and there's some kind of uh some kind i don't know what it would be called some kind of thing over there to which we attach a 250 pound load and this slope there is three by four so some of the other dimensions i have to mention there's one and a quarter that's from the bottom up to the top part of the bracket because we want to look at the stresses in the bracket at two places we want to look at it in one place right here and one place right below that so one place in the middle we'll call a and one place right at the bottom that we'll call b so is it an effect we're looking at one place up here in the middle to see what the stress is and we're also looking at one place at the bottom so up to that middle pleat piece it's uh one and a quarter it's two inches from the load over to those places and the cross section of the the piece looks like this it's a half an inch high and three quarters of an inch deep we're looking at the half inch cross section here so what we're going to get is some axial force through the horizontal component of this that's trying to pull this bracket uh in an x direction if you will then we also have some bending component due to both of those components both the horizontal component is trying to bend it one way the vertical component is trying to bend it the other way whichever one of those is dominant tells us which direction the bending is there but we need to add that stress to the to the axial stress that we're also feeling so let me give you some of the things just to help you through this what we call the section properties which are those pieces of the geometry that we will need for the calculation of each of these we'll need the cross sectional area of course that's uh 0.375 inches squared we also need the moment of inertia the 112 bh cubed because it's nice rectangular cross section that's 0078125 inches to the fourth 0078125 and then there's also shear going on for both of these so we're going to need q and we're going to need it from both points now if you look at the cross section the neutral axis of course is right through the center and so the q four point a which is right in the middle is this entire area above there and you calculate this and double check it remember it's just y bar a is 0 234 375 0 234 375 we'll carry a lot of significant figures through tell the end just because some of these numbers are very very small but you need some of the detail until the end now q b is a little bit different remember b is at the very bottom of the piece so what is q for that remember this q is based upon the area beyond the point of interest away from the neutral axis and there is no area beyond the neutral axis we are at the extreme point so that is zero for point b because it is on the outer most bit of material don't think that we always do the area above the neutral axis because that implies that this has something to do with gravity itself which it doesn't it has to do with where the forces are and where the points of interest are that we're concerned with so we'll have axial stress combined with the bending stress and then we'll also have shear due to the the transverse component of the shear on either point so we'll finish that up on uh on uh Monday