Shear and Moment Diagrams from First Principles

Shear and moment diagrams are usually handed to you as a list of rules. Jump here by the size of the load. Slope there. Today we derive every one of them from a single idea: a piece of a beam in equilibrium is still in equilibrium after you cut it. Here is where we are going. Underneath a loaded beam you draw two pictures: the shear diagram, and the bending moment diagram. By the end you will sketch both of these for a beam you have not seen, and know where the biggest moment sits before computing anything. But not by pattern matching. Every value in them falls out of one free body diagram. So, the beam. Six metres between the supports, a pin at A, a roller at B, and one point load of twelve kilonewtons, four metres from the left. Before we cut anything we need the reactions. Take moments about A. The reaction at B acts six metres out, the load pulls down four metres out, and those two have to balance. That gives eight kilonewtons at B. Vertical equilibrium then leaves four kilonewtons at A. Both supports push upward. Now the question that matters. A beam does not fail at its supports. It fails somewhere inside, where the material is carrying an internal force and an internal moment. What are they, and how do they change as you walk along? Here is the one move the whole subject rests on. Cut the beam at a distance x from A, and throw away everything to the right. What is left is a piece with the reaction at A still pushing up on it. That piece cannot be in equilibrium on its own. Four kilonewtons up and nothing down. So the part we threw away must have been holding it, at the cut face, with a force and with a moment. Call them V and M. Those are the internal shear and the internal bending moment at position x. Draw them the standard way every time: V pointing down on the cut face, and M bending the beam into a smile. Then a negative answer simply means it acts the other way. Vertical equilibrium of the piece. Four up, V down, nothing else. So V is four kilonewtons. Moments about the cut. The reaction acts a distance x away, and M is the only other thing on the piece. So M is four x. Look at what x did. Nothing at all to the shear: four kilonewtons wherever you cut, as long as you stay left of the load. But the moment grows in proportion to x. Watch the piece get longer. That is what we mean by shear and moment as functions of position. One cut, two equilibrium equations, and we have both of them for the whole left hand stretch of this beam. But only that stretch. Move the cut past the load, and the free body is a different picture: the twelve kilonewton load now sits on the piece we kept. Vertical equilibrium now has three forces. Four up, twelve down, and V drawn downward as before. So V is four minus twelve, which is minus eight kilonewtons. And notice: the shear did not slide down to minus eight. It jumped, and it jumped by exactly twelve, the size of the load. Hold on to that. Moments about the cut again. The reaction still acts x away, and the load now acts on our piece as well, a lever arm of x minus four, turning it the other way. That tidies to forty eight minus eight x. Check it at the far end. Put x equal to six and the moment comes out zero, which is exactly what a roller has to give you. A roller cannot resist any moment at all. So here is the answer. Two expressions for the shear and two for the moment, one pair on each side of the load. Nothing there was remembered. A cut, a free body, and two equilibrium equations, done twice. Now let us draw them.

We have four expressions and a beam. Drawing the shear and moment diagrams is nothing more than plotting them, underneath the beam, on the same horizontal scale. So: position along the beam across the bottom, internal shear up the side. On the left stretch the shear is a constant four. The diagram starts from zero off the end of the beam, jumps straight up to four where the pin pushes in, and then runs flat all the way to the load. On the right stretch it is minus eight, so a horizontal line eight below the axis, running from the load across to B. And those two do not meet. At the load they differ by twelve, so the diagram falls vertically, right there, by twelve. The jump is the load. Then the roller pushes up eight and closes the diagram back to zero. That is a free check on your arithmetic. A shear diagram that does not return to zero at the far end means a reaction is wrong. Watch the shear as the cut walks along. Nothing happens to it until it reaches the load, and then in one step it is on the other side of the axis. Now the moment, on its own axis underneath the shear. On the left stretch it was four x, a straight line climbing out of the origin. It arrives at sixteen kilonewton metres under the load. On the right, forty eight minus eight x brings it back down, in another straight line, to zero at B. For a point load, then, the moment diagram is a triangle, and its peak sits directly under the load. Put the two side by side and two things stare at you. Where the shear is constant, the moment is straight. And where the shear is bigger, the moment climbs faster. Look at the numbers. On the left the shear is four and the moment gains four units per metre. On the right the shear is minus eight and the moment loses eight per metre. The slope of the moment diagram is the shear. And here is the one that earns its keep. The moment peaks exactly where the shear crosses the axis, which on this beam happens inside the vertical jump at the load. We have not proved any of that. We have noticed it, on one beam. So let us take a completely different kind of load and see whether it survives.

Second beam. The same six metre span and the same supports, but instead of one point load it carries four kilonewtons on every metre of its length. A uniformly distributed load. The total is four times six, twenty four kilonewtons, and by symmetry each support takes half. Twelve up at each end. Cut it at x and keep the left piece, exactly as before. Here is the only new idea in this example. The load sitting on our piece is four kilonewtons per metre times x metres, and it acts through the middle of that stretch. Vertical equilibrium. Twelve up, four x down, V down. So the shear is twelve minus four x. It is not constant any more. It falls off at four per metre, which is exactly the load intensity. Moments about the cut. The reaction gives twelve x. The load on our piece gives four x, times its lever arm of x over two, turning the other way. So the moment is twelve x minus two x squared. A parabola. Watch the free body grow. The resultant of the load grows with it, and its arrow slides out to stay at the middle of the piece. Plot the shear. Twelve minus four x is a straight ramp, starting at plus twelve at the pin and finishing at minus twelve at the roller, crossing the axis at the middle of the span. And the moment is a parabola, zero at both supports. By symmetry its top is at three metres: twelve times three, minus two times nine, eighteen kilonewton metres. Both of the things we noticed on the first beam survived, and they got sharper. The shear was flat when there was no load; now it slopes, at exactly minus the load intensity. The shear is straight, and the moment has gone up one degree to a parabola. And the maximum moment is where the shear passes through zero. Not under the heaviest load, not at a support. Where the shear crosses the axis. There is a third thing hiding here, and it is the useful one. Shade the area under the shear diagram, from the left end out to some position x. Now watch the shaded area and the moment together. At three metres the shaded triangle is half of three times twelve, which is eighteen. And the moment there is eighteen. Keep going. Past the middle the shear is negative, so the new shading counts against you, and by the far end the positive and negative areas have cancelled exactly. The moment is back to zero. So the area under the shear, between two points, is the change in the moment between those two points. We have now seen it twice. Time to prove it once.

Both beams told us the same three things, so let us prove them once, for any beam and any load. Take a slice of beam of length d x, out of a beam carrying a distributed load w. On its left face the rest of the beam pushes with a shear V and a moment M. On its right face, a distance d x along, both have had a chance to change: V plus d V, and M plus d M. Vertical equilibrium of the slice. V up on the left, the load w d x pressing down on the top, and V plus d V down on the right. The two V's cancel. What is left is d V equals minus w d x. Divide through by d x, and the derivative of the shear with respect to position is minus the load intensity. Read that as a picture. Where there is no load, the shear is flat. Where the load is uniform, the shear is a straight ramp sloping downward. And where the load is heavy, the shear falls steeply. Now take moments about the right hand face. The two moments oppose each other and leave d M. The shear on the left face acts a distance d x away. And the load has a lever arm of half d x. That last term has d x squared in it. As the slice shrinks it dies away faster than everything else, so it goes. What survives is d M equals V d x. The derivative of the moment is the shear. The slope of the moment diagram, at any point at all, is the height of the shear diagram at that same point. That is the relation we kept noticing. Integrate the two of them and out comes the other half of the folklore. The change in shear between two sections is minus the area under the load diagram between them. And the change in moment between two sections is the area under the shear diagram between them. Exactly the shaded triangle we were watching a moment ago. Now every rule you were ever handed is a line in this table. No load: the shear is constant and the moment is straight. A uniform load: the shear is straight and the moment is a parabola. Each one is a degree higher than the last. A point force is an enormous load intensity over no length at all. Its area is finite, so the shear jumps by the size of the force, and the moment, whose slope is the shear, simply kinks. And a concentrated couple contributes nothing to vertical equilibrium, so the shear does not notice it at all. It sits in the moment equation instead, so the moment jumps by the size of the couple. That is the line people get wrong most often. And finally the sentence you actually use. The moment is stationary where its derivative vanishes, and its derivative is the shear. So the biggest moment sits where the shear diagram crosses the axis, or at a jump that carries it across. Two derivatives and two areas. Everything else about these diagrams is a consequence of them.

Third beam, and it has both of the things people find awkward. It runs eight metres, but the supports are at zero and at six, so the last two metres hang past the roller. The loads are twelve kilonewtons down at two metres, four kilonewtons down at the free tip, and a couple of eight kilonewton metres applied at five metres, twisting counterclockwise. Reactions first. Moments about A. The twelve acts at two, the four acts at eight, the reaction at B acts at six, and the couple goes straight into the sum as eight. It has no lever arm at all. A couple is the same about every point. That comes to forty eight, so the reaction at B is eight kilonewtons, and vertical equilibrium then leaves eight kilonewtons at A as well. Both supports push upward. Now build both diagrams without cutting anything, using only what we proved. Start at zero off the left end. At A the reaction pushes up eight, so the shear jumps to plus eight, and with no load between there and the twelve, it runs flat. At two metres the twelve kilonewton load drops it by twelve, from plus eight to minus four. Flat again after that. There is no distributed load anywhere on this beam. Here is where the couple acts, at five metres, and the shear passes straight through it without so much as a flinch. There is no vertical force in a couple, so there is nothing for the shear to notice. At B the reaction adds eight, taking it from minus four up to plus four. It runs flat along the overhang, and the four kilonewton load at the tip brings it home to zero. The diagram closes. Now the moment, and every piece of it is an area under that shear. From A out to two metres, a rectangle eight high and two wide: sixteen. So the moment climbs in a straight line to sixteen kilonewton metres. From two to five the shear is minus four, so the moment falls at four per metre. Three metres of that is minus twelve, taking it from sixteen down to four. And at five metres, the couple. The shear ignored it. The moment cannot. It drops vertically by eight, the size of the couple, from plus four to minus four. Below the axis now, and the shear is still minus four, so the moment goes on falling at four per metre for the last metre to the roller. Minus eight. Over the overhang the shear is plus four, so the moment climbs at four per metre for two metres and lands on zero exactly at the free tip. It has to. There is nothing beyond the tip to bend it. Check that by hand if you like. Cut the overhang anywhere and keep the right piece: one four kilonewton load, a lever arm of eight minus x, bending the beam the wrong way up. Negative, and shrinking to nothing at the tip. Two things to carry away from this beam. A couple jumps the moment and leaves the shear alone. And an overhang drives the moment below the axis, which means tension on the top of the beam, which is where the steel has to go.

Last beam, and you have not seen it. Eight metres, simply supported, with six kilonewtons per metre spread over the left half only and nothing on the right half. Here is the whole method, in five lines. Reactions, then jumps and slopes, then areas, then the maximum, then the check that it closes. One: reactions, and these you do have to compute. The load totals twenty four kilonewtons acting through the middle of the left half, which is much nearer A. So A takes eighteen of it, and B takes only six. Everything after this comes out of the two derivatives. Two: the shear. It starts at zero, jumps to plus eighteen at A, and then under the distributed load it slopes down at six per metre for four metres. That is a fall of twenty four, from plus eighteen to minus six. Past the load there is nothing pressing down, so the shear runs flat at minus six all the way to B, where the six kilonewton reaction closes it to zero. Three: the moment. Zero at a simple support. Its slope is the shear, which starts big and positive and is falling, so the moment leaves A steeply and bends over. A parabola, concave down. Four: it peaks where the shear crosses the axis. Eighteen divided by six is three, so the maximum sits three metres in. Not four, and certainly not at midspan. After that the shear is a constant minus six, so the moment comes down as a straight line, and it has to arrive at zero at B. Five: it does, and the sketch is finished. None of that needed the bending moment function. If you do want the number, it is the area under the shear out to three metres: half of three times eighteen, twenty seven kilonewton metres. So: cut, equilibrium, and two derivatives. The jumps, the slopes and the areas are not rules to remember. They are what those two derivatives look like once you draw them.