when we were looking at the one-dimensional mechanical system we had equal an equilibrium equation that said that say d r DX plus b equals zero so the rate of change of the internal force plus the body force is equal to zero and we had another version of this also where i said the rate of change of the cross sectional area times of stress plus the body force equals zero and what I'd like to now do is take a look at the question of what should equal room look like in terms of stresses in the full multi-dimensional setting and just to keep things simple I'll just look in the 2d case the 3-d case extension is quite straight forward from that and what I'd like to come to is an expression for equilibrium at each point in the body in terms of the stresses so in 1d case we had this expression over here in the box which was equilibrium point by point along the length of the bar but now in 2d we can we can talk about the points up and down all around in the system so to do this what I want to do is look at an arbitrary body and I'm gonna focus in on a point in that body and I'm going to cut out a square of material whose side lengths are Delta X and Delta Y and I'm gonna try and go through an argument to show us what equilibrium looks like in terms of stresses from this construction so if I look at this little square of material i zoom into it its corner coordinates or XY and acting in the middle is I'm just going to draw any forces in the x-direction is the body force in the X direction on the left side I have Sigma xx as a stress and so I have to multiply that by the area of that side to get a force but on the other side the forces and stresses can vary from point to point so on the other side I'll have Sigma xx from the left side plus some change in the value which I'll call Delta Sigma xx so that's the change moving in the next direction on the bottom I don't have Sigma YX and on the top I'll have Sigma YX plus Delta Sigma YX so that's the change relative to the bottom so these are the stresses and all the erections or actually all the stresses that lead to forces in the x-direction so I can multiply each one of these by the area over which they act and I can have a force equilibrium equation so some of the forces in the x-direction can be written out as follows and each one of these terms technically should be multiplied by the thickness of this two-dimensional body which will just take to be one so I don't include that in the expression okay if we look at this expression you'll notice that there are a number of terms that are the same and that they're going to cancel so we can cancel those out so the Sigma X X Delta Y cancels and the Sigma Y X Delta X also cancels so so now I have a slightly different expression but it's still force equilibrium in the X direction I can now divide this expression through by the area of this little element so Delta X Delta Y and so I received this equation here so Delta Sigma xx over Delta X plus Delta Sigma YX over Delta Y plus the body force in the X Direction equals zero and now what I can do is consider taking the limit as Delta X and Delta Y goes to zero and each one of these ratios of changes simply is the rate of change of the corresponding function so I'll have the derivative Sigma xx with respect to X plus the derivative of Sigma XY with respect to Y plus the body force equals zero and so this is basically nothing other than the sum of the forces in the x-direction equals zero so this is an expression of equilibrium of force equilibrium in the X direction with respect to stresses and so it involves these partial derivatives of the different stress components of the system if I go I can continue with the argument and I can do sum of the forces in the Y direction and I'll get another partial differential equation the rate of change of the X Y shear stress with respect to X plus the rate of change of the normal stress in the Y direction with respect to Y plus the body force in the Y direction equals zero and if I do sum of the moments about the z axis I find out that Sigma XY equals Sigma YX so this correspondence between the shear stresses on orthogonal planes we will see all those holes for orthogonal coordinate systems so we'll always have symmetry of the stress tensor itself in all cases and symmetry the stress tensor is the same thing as moment equilibrium about the three different axes the XY and z axis so in this case the z axis tells us that sigma XY equals Sigma YX if we look at and so these are the three equilibrium equations that we have in 2d if we count them we'll see that we have two partial differential equations plus one algebraic equation so there are three equations in total freak alert rim as we would expect for a two-dimensional problem and there are four variables in these equations there's Sigma xx and Sigma YY so the normal stresses in the X and the y directions and the two shears the Sigma X Y in the Sigma YX so you'll notice that there is one less equation then there are unknowns so when we're talking about point wise stress as so you're trying to determine the point Y stresses in a body it will always be statically indeterminate so that that's a feature of the full multi-dimensional theory of mechanics is that stress determination is always statically indeterminate