Lesson 6 V and M Diagrams

Added: 2026-05-31

[00:00:00]so far today we have talked primarily about external forces and when we discussed axial forces last week we didn't really differentiate between external and internal forces and that's because for axial forces the external and internal forces are the same if I pull on a member with 20 kips on one side that 20 Kip force is going to pass through the entire length of the member and then we reach the 20 Kip reaction on the other side if there is a change in the cross section of the member for example if the member gets thicker at one point there will be a change in the stress because we are changing the cross-sectional area however the force will remain the same throughout the member now with perpendicular forces we need to pay close attention to the difference between external and internal forces because our internal forces are going to change as we saw in the foam demonstration the internal forces that we created with our externally applied force our little matcha tea canister with the coins in it those internal forces could change depending on the location we placed our weight even though we weren't changing the magnitude of the applied force the internal forces were changing so let's dive deeper into quantifying the internal forces acting in our beams we saw in the last video now that we have all our external forces defined both the applied loads and the reactions we can plot shear and moment diagrams these diagrams are a way of visualizing the internal forces inside of our member along its length the x-axis represents the position along the length of the member and the y-axis represents the magnitude of the force inside the member we will use these graphs to identify the maximum amount of internal force in our members so that we can quantify a demand for the member once we have our demand load we can design a beam that has sufficient capacity to meet or exceed that demand so let's start with our first example of a simply supported Beam with a point load at mid-span and by simply supported I mean that we have a pin on one side and a roller on the other drawing the shear diagram is easiest because we simply follow our external forces directly so let's start there starting from the left we have our reaction ry1 pushing up on the member by a magnitude of p over 2. and so our Shear diagram will jump up from zero to a magnitude p over 2.

[00:03:18]now as we move to the right along the member we have no external forces and so our Shear diagram remains constant until we reach mid span where we have our applied load p and this applied load p is pushing down so our Shear diagram will jump down by a magnitude of P and since we started at p over 2 p over 2 minus P gives us negative p over 2.

[00:03:55]and again we move to the right at a constant rate because there are no external forces until we reach the end where we have our second reaction ry2 which is also equal to P over 2 and so p over 2 where we were starting plus p over 2 brings us back to zero a note for this Shear diagram we should always start and end at zero if we don't start and end at zero then we have done something wrong with our reaction calculations and we should go back and review but in this case we started and ended at zero so everything checks out now the moment diagram this is where things get a little tricky since we start with a pin on the left here and we have no externally applied moment we know that moment will be zero to start then as we move to the right along the member moment will be equal to the area underneath the shear diagram or if you've taken calculus moment is the integral of the shear diagram but for this course we'll be dealing mostly with regular geometric shapes and so we can simply calculate based off of geometry and we don't need to use complicated integral equations so if you imagine plotting the area underneath this curve as we move to the right the further to the right we go the more area we collect under the Curve and the area that we collect is going to increase at a constant rate because this line for our Shear diagram is flat and so Our Moment diagram increases at a linear rate as we move to the right until we reach a maximum point here at the mid span so to help you visualize that if we took this location here and calculated that area we would end up somewhere along the curve here then if we moved further to the right and calculate the area that we've accumulated up until that point we will have a little more and that continues until we reach this maximum point at the mid span where the edge of our rectangle occurs and that total area is going to be equal to p over 2 times L over 2.

[00:06:58]so we end up with p l over four for our maximum moment PL over four now as we move further to the right the area that we start to collect is negative and so we will be subtracting from Our Moment curve and that moment Curve will start to go down until we reach 0 on the other end and we know that it will go back down to zero because these two rectangles have the same area they both have a total area of PL over four and I'm going to move this negative p over 2 over here so it's not getting confused with our maximum moment and so that is our shear and moment diagram for this very basic loading condition so the magnitude of our internal shear is p over 2 across the entire length of the member although it changes Direction once we pass our Point load the magnitude of our internal moment changes linearly as we move along the member it starts at zero increases to PL over 4 and then decreases back to zero next let's look at our Point load applied at a distance a from our support on the left as in the previous problem we start with zero on our Shear diagram and jump up by our external reaction ry1 a magnitude of p b over l then as we move to the right we have no external forces so our Shear diagram is constant until we reach our applied load p then our Shear diagram jumps down by a magnitude of P and as we saw in our previous video PB over L minus p is equal to negative p a over l and so here we again move constantly and we're at a magnitude of negative PA over l until we reach this point here where our external reaction ry2 pushes our Shear diagram back to zero and so that is our Shear diagram now to calculate Our Moment diagram as before we know that Our Moment is going to start at zero and it's going to increase at a relatively faster Pace or a steeper slope then it will decrease once we have passed our Point load p and that's because the magnitude PB over L is larger than the magnitude PA over l and so the slope here will be greater on this side and then the slope will be lesser on this side and now to calculate this maximum moment here M Max will be equal to the area underneath the shear curve which is PB over l times a or p a b over l and so that's our maximum moment p a b over l and now just to verify that that is correct we should be able to continue calculating the area under the shear curve and end up at zero on the right for Our Moment diagram so if we take p a b over l and now continue adding this negative area we'll get minus p a over l times B the length of this rectangle times the width and indeed this brings us back to zero and so that is Our Moment diagram for this simply supported Beam with a concentrated load at any point next we have our uniformly distributed load and here we start at zero on our Shear diagram we jump up by our external reaction ry1 WL over two but now we have this external load this external distributed load along the entire length of the member and recall that this is given in kips per foot and so our Shear diagram will actually have a slope to it where it's decreasing at that rate provided of Kips per foot and so let's calculate the shear at Mid span so Shear at Mid span is going to be equal to where we started WL over 2 minus the rate that it's decreasing which is the applied load multiplied by the distance here from where we started to Mid span which is L over two and so our Shear at Mid span is zero then as we continue along to the right our Shear at the right support will be negative WL over 2.

[00:13:29]which is equal to our reaction at the right and so our Shear diagram jumps up and ends at zero now let's consider Our Moment diagram we know that it will start at zero and we know that we have some positive area on this side of the shear diagram until we reach mid span so we know that we'll reach our maximum moment at Mid span where the area under the curve switches from positive to negative but the rate at which we are gaining area under the curve is decreasing so if we were to start plotting area under the curve up until this point we may end up here and then if we were to move to a little further to the right we would have more area but the amount of area that we gained is less than the amount of area that we started with we still have in total more area but we gained a little less and so we would end up somewhere here moving further to the right the same thing would occur the amount that we gain is not as much as we gained previously but the total amount is still more and so if we were to connect these points you would see that we form a parabola until we reach our point of Maximum moment so now let's calculate that maximum moment magnitude M Max is going to be equal to the area underneath the shear diagram and the area of a triangle of course is one half the height WL over 2 times the width L over 2.

[00:15:40]and so we get WL squared over eight for our maximum moment and we know that the area on this side of the curve is equal to the area on this side of the Curve and so we end up back at zero so that's our shear and moment diagram for a simply supported Beam with a uniformly distributed load