{"version":1,"lectureId":"01M14TY448ABTYCPADC1FDHE42","attempt":0,"publication":{"slug":"deriving-shear-moment-diagrams","title":"Shear and Moment Diagrams from First Principles","subject":"engineering","summary":"Shear and moment diagrams, built from equilibrium rather than memorised. We cut a simply supported beam at an arbitrary position, draw the free body of what is left, and read the internal shear and bending moment straight off two equilibrium equations, first for a single point load and then for a uniformly distributed load. Plotting those functions gives the diagrams. A slice of beam of length dx then turns the patterns we noticed into two derivatives, dV/dx equals minus w and dM/dx equals V, and into their integrals: the change in shear is the area under the load, the change in moment is the area under the shear. Every jump, slope and area rule follows from those, including the applied couple that jumps the moment diagram while leaving the shear untouched. The lecture closes by sketching both diagrams for an unseen beam and locating the maximum moment before computing it.","metaDescription":"Cut a beam, take the free body, and derive shear and moment diagrams from equilibrium, then from the two derivatives that govern them.","transcript":"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.","watch":{"version":1,"scenes":[{"title":"Cutting the Beam","start":0,"end":249.64514583333334,"objects":{"beam":"a Figure (x_range=(-1.0, 7.4), y_range=(-5.0, 2.6), aspect=(8.4, 7.6))","card":"a Title that says \"Engineering Statics — Shear and Moment Diagrams from First Principles\"","cut":"a VariableNumber (initial_value=1.3, format_spec='.1f')","cut_line":"a Line [yellow] drawn in beam (start=(<VariableNumber cut = 5.0>, 0.8), end=(<VariableNumber cut = 5.0>, -0.95), dashed=True)","fb_arm":"a Line [gray] labelled \"x - 4\" drawn in beam (start=(4.0, -4.28), end=(<VariableNumber cut = 5.0>, -4.28))","fb_line":"a Line [blue] drawn in beam (start=(0.0, -3.1), end=(<VariableNumber cut = 5.0>, -3.1))","fb_load":"a Vector [red] labelled \"P\" drawn in beam (start=(4.0, -2.34), end=(4.0, -3.04))","fb_m":"a CurvedArrow [yellow] labelled \"M\" drawn in beam (start=((cut - 0.8), -3.62), end=((cut - 0.8), -2.58), bend=0.55)","fb_ra":"a Vector [green] labelled \"R_A\" drawn in beam (start=(0.0, -4.05), end=(0.0, -3.16))","fb_v":"a Vector [yellow] labelled \"V\" drawn in beam (start=(<VariableNumber cut = 5.0>, -3.1), end=(<VariableNumber cut = 5.0>, -3.98))","fb_x":"a Line [gray] labelled \"x\" drawn in beam (start=(0.0, -4.62), end=(<VariableNumber cut = 5.0>, -4.62))","head_answer":"a Heading that says \"Shear and Moment as Functions of $x$\"","head_beam":"a Heading that says \"The Beam, and Its Reactions\"","head_cut":"a Heading that says \"Cut It, and Keep the Left Piece\"","head_goal":"a Heading that says \"Where We Are Going\"","head_right":"a Heading that says \"Move the Cut Past the Load\"","here_a":"a Math [gray] that says \"$4 thin upright(\"m\")$\" drawn in beam","here_b":"a Math [gray] that says \"$2 thin upright(\"m\")$\" drawn in beam","lab_left":"a Tex [text] that says \"For $0 < x < 4$:\"","lab_right":"a Tex [text] that says \"For $4 < x < 6$:\"","left_pair":"a Derivation [text] that says \"$V &= 4 \\ M &= 4 x$\"","load":"a Vector [red] drawn in beam (start=(4.0, 1.7), end=(4.0, 0.08))","load_name":"a Math [red] that says \"$P = 12 thin upright(\"kN\")$\" drawn in beam","moment_work":"a Derivation [text] that says \"$sum M_(upright(\"cut\")) &= 0 \\ M - 4 x &= 0 \\ M(x) &= 4 x$\"","name_a":"a Math [text] that says \"$A$\" drawn in beam","name_b":"a Math [text] that says \"$B$\" drawn in beam","pin":"a Polygon [gray] drawn in beam (vertices=((0.0, 0.0), (-0.32, -0.62), (0.32, -0.62)), fill_opacity=0.35)","prev_m":"an Axes (x_range=(0.0, 6.0), y_range=(-2.0, 20.0), aspect=(7.0, 2.4))","prev_m_left":"a FunctionPlot [magenta] drawn in prev_m (function=<function>, x_range=(0.0, 4.0))","prev_m_right":"a FunctionPlot [magenta] drawn in prev_m (function=<function>, x_range=(4.0, 6.0))","prev_v":"an Axes (x_range=(0.0, 6.0), y_range=(-12.0, 8.0), aspect=(7.0, 2.4))","prev_v_jump":"a Line [yellow] drawn in prev_v (start=(4.0, 4.0), end=(4.0, -8.0))","prev_v_left":"a FunctionPlot [yellow] drawn in prev_v (function=<function>, x_range=(0.0, 4.0))","prev_v_right":"a FunctionPlot [yellow] drawn in prev_v (function=<function>, x_range=(4.0, 6.0))","promise":"a Text [text] that says \"Every number in these two pictures will come out of one equilibrium statement written on a cut piece of beam.\"","r_a":"a Vector [green] drawn in beam (start=(0.0, -1.8), end=(0.0, -0.72))","r_a_name":"a Math [green] that says \"$R_A = 4$\" drawn in beam","r_b":"a Vector [green] drawn in beam (start=(6.0, -1.8), end=(6.0, -0.72))","r_b_name":"a Math [green] that says \"$R_B = 8$\" drawn in beam","reactions":"a Derivation [text] that says \"$sum M_A &= 0 \\ 6 R_B - 12 (4) &= 0 \\ R_B &= 8 thin upright(\"kN\") \\ R_A &= 12 - 8 = 4 thin upright(\"kN\")$\"","right_pair":"a Derivation [text] that says \"$V &= -8 \\ M &= 48 - 8 x$\"","right_work":"a Derivation [text] that says \"$V(x) &= 4 - 12 = -8 thin upright(\"kN\") \\ M(x) &= 4 x - 12 (x - 4) \\ &= 48 - 8 x$\"","roller":"a Polygon [gray] drawn in beam (vertices=((6.0, 0.0), (5.68, -0.62), (6.32, -0.62)), fill_opacity=0.35)","shear_work":"a Derivation [text] that says \"$sum F_y &= 0 \\ 4 - V &= 0 \\ V(x) &= 4 thin upright(\"kN\")$\"","span":"a Line [blue] drawn in beam (end=(6.0, 0.0))"},"beats":[{"start":0,"say":"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.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.5,"card: enter:write-left-to-right."],[16.358,"card is hidden from the screen — left the board."]]},{"start":17.558,"say":"Here is where we are going. Underneath a loaded beam you draw two pictures: the shear diagram, and the bending moment diagram.","live":null,"does":[[17.558,"head_goal is shown on the screen, written out."],[22.654999999999998,"prev_v is shown on the screen, written out."],[22.955,"prev_v_left is shown on the screen, drawn."],[23.455,"prev_v_right is shown on the screen, drawn."],[24.048,"prev_m is shown on the screen, written out."],[24.154999999999998,"prev_v_jump is shown on the screen, drawn."],[24.348,"prev_m_left is shown on the screen, drawn."],[24.848,"prev_m_right is shown on the screen, drawn."]]},{"start":26.482999999999997,"say":"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.","live":["prev_v","prev_m","head_goal","prev_v_left","prev_v_right","prev_v_jump","prev_m_left","prev_m_right"],"does":[[26.482999999999997,"promise is shown on the screen, written out."],[39.625499999999995,"head_goal is hidden from the screen — left the board."],[39.625499999999995,"prev_m is hidden from the screen — left the board."],[39.625499999999995,"prev_m_left is hidden from the screen — prev_m left the board."],[39.625499999999995,"prev_m_right is hidden from the screen — prev_m left the board."],[39.625499999999995,"prev_v is hidden from the screen — left the board."],[39.625499999999995,"prev_v_left is hidden from the screen — prev_v left the board."],[39.625499999999995,"prev_v_right is hidden from the screen — prev_v left the board."],[39.625499999999995,"prev_v_jump is hidden from the screen — prev_v left the board."],[39.625499999999995,"promise is hidden from the screen — left the board."]]},{"start":40.8255,"say":"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.","live":[],"does":[[40.8255,"head_beam is shown on the screen, written out."],[40.8255,"beam is shown on the screen, written out."],[41.882,"span is shown on the screen, written out."],[44.958,"pin is shown on the screen, written out."],[45.058,"name_a is shown on the screen, written out."],[46.085,"roller is shown on the screen, written out."],[46.185,"name_b is shown on the screen, written out."],[47.443000000000005,"load is shown on the screen, written out."],[47.64300000000001,"load_name is shown on the screen, written out."],[49.556000000000004,"here_a is shown on the screen, written out."],[49.75600000000001,"here_b is shown on the screen, written out."]]},{"start":51.7115,"say":"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.","live":["beam","head_beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b"],"does":[[55.067,"beam moves to a new place on the board."],[55.067,"reactions is shown on the screen, written out."],[62.125,"reactions is shown on the screen, written out."]]},{"start":63.538,"say":"That gives eight kilonewtons at B. Vertical equilibrium then leaves four kilonewtons at A. Both supports push upward.","live":null,"does":[[64.49,"reactions is shown on the screen, written out."],[64.69,"r_b is shown on the screen, written out."],[65.09,"r_b_name is shown on the screen, written out."],[67.857,"reactions is shown on the screen, written out."],[68.057,"r_a is shown on the screen, written out."],[68.45700000000001,"r_a_name is shown on the screen, written out."]]},{"start":72.346,"say":"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?","live":["beam","head_beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name"],"does":[[87.149,"head_beam is hidden from the screen — left the board."],[87.149,"reactions is hidden from the screen — left the board."]]},{"start":87.74900000000001,"say":"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.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name"],"does":[[87.74900000000001,"head_cut is shown on the screen, written out."],[90.98800000000001,"cut_line is shown on the screen, written out."],[96.17800000000001,"fb_line is shown on the screen, written out."],[96.37800000000001,"fb_x is shown on the screen, written out."],[98.45400000000002,"fb_ra is shown on the screen, written out."]]},{"start":100.23800000000001,"say":"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.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name","head_cut","cut_line","fb_line","fb_x","fb_ra"],"does":[[110.22200000000002,"fb_v is shown on the screen, written out."],[111.10500000000002,"fb_m is shown on the screen, written out."]]},{"start":114.37450000000001,"say":"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.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name","head_cut","cut_line","fb_line","fb_x","fb_ra","fb_v","fb_m"],"does":[]},{"start":130.60150000000002,"say":"Vertical equilibrium of the piece. Four up, V down, nothing else. So V is four kilonewtons.","live":null,"does":[[130.95000000000005,"shear_work is shown on the screen, written out."],[133.63200000000003,"shear_work is shown on the screen, written out."],[135.31500000000003,"shear_work is shown on the screen, written out."]]},{"start":139.81650000000002,"say":"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.","live":null,"does":[[140.16500000000002,"moment_work is shown on the screen, written out."],[144.84400000000002,"moment_work is shown on the screen, written out."],[147.084,"moment_work is shown on the screen, written out."]]},{"start":148.56050000000002,"say":"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.","live":null,"does":[[159.062,"cut_line is redrawn as the numbers it depends on change."],[159.062,"fb_line is redrawn as the numbers it depends on change."],[159.062,"fb_x is redrawn as the numbers it depends on change."],[159.062,"fb_v is redrawn as the numbers it depends on change."],[159.062,"fb_m is redrawn as the numbers it depends on change."],[159.062,"fb_arm is redrawn as the numbers it depends on change."],[159.062,"cut ticks to 3.6."]]},{"start":161.56600000000003,"say":"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.","live":null,"does":[[171.36500000000004,"head_cut is hidden from the screen — left the board."],[171.36500000000004,"moment_work is hidden from the screen — left the board."],[171.36500000000004,"shear_work is hidden from the screen — left the board."]]},{"start":172.56500000000003,"say":"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.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name","cut_line","fb_line","fb_x","fb_ra","fb_v","fb_m"],"does":[[172.56500000000003,"head_right is shown on the screen, written out."],[174.81700000000004,"cut_line is redrawn as the numbers it depends on change."],[174.81700000000004,"fb_line is redrawn as the numbers it depends on change."],[174.81700000000004,"fb_x is redrawn as the numbers it depends on change."],[174.81700000000004,"fb_v is redrawn as the numbers it depends on change."],[174.81700000000004,"fb_m is redrawn as the numbers it depends on change."],[174.81700000000004,"fb_arm is redrawn as the numbers it depends on change."],[174.81700000000004,"cut ticks to 5.0."],[181.44600000000003,"fb_load is shown on the screen, written out."],[181.74600000000004,"fb_arm is shown on the screen, written out."]]},{"start":182.80100000000002,"say":"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.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name","cut_line","fb_line","fb_x","fb_ra","fb_v","fb_m","head_right","fb_load","fb_arm"],"does":[[190.77700000000002,"right_work is shown on the screen, written out."]]},{"start":195.94000000000003,"say":"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.","live":null,"does":[[200.90900000000005,"right_work (the \"-8\" part) is emphasized."],[205.51800000000006,"right_work (the \"-8\" part) is no longer emphasized."]]},{"start":207.44200000000004,"say":"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.","live":null,"does":[[212.96900000000002,"right_work is shown on the screen, written out."],[219.13300000000004,"right_work is shown on the screen, written out."]]},{"start":222.01450000000003,"say":"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.","live":null,"does":[[226.247,"right_work is indicated — a transient flash."],[232.01700000000002,"head_right is hidden from the screen — left the board."],[232.01700000000002,"right_work is hidden from the screen — left the board."]]},{"start":232.61700000000002,"say":"So here is the answer. Two expressions for the shear and two for the moment, one pair on each side of the load.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name","cut_line","fb_line","fb_x","fb_ra","fb_v","fb_m","fb_load","fb_arm"],"does":[[232.61700000000002,"head_answer is shown on the screen, written out."],[234.788,"lab_left is shown on the screen, written out."],[235.08800000000002,"left_pair is shown on the screen, written out."],[235.58800000000002,"left_pair is shown on the screen, written out."],[238.41000000000003,"lab_right is shown on the screen, written out."],[238.71000000000004,"right_pair is shown on the screen, written out."],[239.21000000000004,"right_pair is shown on the screen, written out."]]},{"start":240.288,"say":"Nothing there was remembered. A cut, a free body, and two equilibrium equations, done twice. Now let us draw them.","live":["beam","span","pin","name_a","roller","name_b","load","load_name","here_a","here_b","r_b","r_b_name","r_a","r_a_name","cut_line","fb_line","fb_x","fb_ra","fb_v","fb_m","fb_load","fb_arm","lab_left","lab_right","head_answer"],"does":[[248.6034791666667,"beam is hidden from the screen — left the board."],[248.6034791666667,"span is hidden from the screen — beam left the board."],[248.6034791666667,"pin is hidden from the screen — beam left the board."],[248.6034791666667,"name_a is hidden from the screen — beam left the board."],[248.6034791666667,"roller is hidden from the screen — beam left the board."],[248.6034791666667,"name_b is hidden from the screen — beam left the board."],[248.6034791666667,"load is hidden from the screen — beam left the board."],[248.6034791666667,"load_name is hidden from the screen — beam left the board."],[248.6034791666667,"here_a is hidden from the screen — beam left the board."],[248.6034791666667,"here_b is hidden from the screen — beam left the board."],[248.6034791666667,"r_b is hidden from the screen — beam left the board."],[248.6034791666667,"r_b_name is hidden from the screen — beam left the board."],[248.6034791666667,"r_a is hidden from the screen — beam left the board."],[248.6034791666667,"r_a_name is hidden from the screen — beam left the board."],[248.6034791666667,"cut_line is hidden from the screen — beam left the board."],[248.6034791666667,"fb_line is hidden from the screen — beam left the board."],[248.6034791666667,"fb_x is hidden from the screen — beam left the board."],[248.6034791666667,"fb_ra is hidden from the screen — beam left the board."],[248.6034791666667,"fb_v is hidden from the screen — beam left the board."],[248.6034791666667,"fb_m is hidden from the screen — beam left the board."],[248.6034791666667,"fb_load is hidden from the screen — beam left the board."],[248.6034791666667,"fb_arm is hidden from the screen — beam left the board."],[248.6034791666667,"head_answer is hidden from the screen — left the board."],[248.6034791666667,"lab_left is hidden from the screen — left the board."],[248.6034791666667,"lab_right is hidden from the screen — left the board."],[248.6034791666667,"left_pair is hidden from the screen — left the board."],[248.6034791666667,"right_pair is hidden from the screen — left the board."]]}]},{"title":"Drawing the Two Diagrams","start":249.64514583333334,"end":394.05110416666673,"objects":{"beam":"a Figure (x_range=(-0.9, 7.0), y_range=(-1.7, 2.5), aspect=(7.9, 4.2))","head_moment":"a Heading that says \"Plotting the Moment\"","head_shear":"a Heading that says \"Plotting the Shear\"","lab_left":"a Tex [text] that says \"For $0 < x < 4$:\"","lab_right":"a Tex [text] that says \"For $4 < x < 6$:\"","left_pair":"a Derivation [text] that says \"$V &= 4 \\ M &= 4 x$\"","load":"a Vector [red] drawn in beam (start=(4.0, 1.7), end=(4.0, 0.08))","load_name":"a Math [red] that says \"$P = 12$\" drawn in beam","m_axes":"an Axes (x_range=(0.0, 6.0), y_range=(-2.0, 20.0), aspect=(7.9, 2.8))","m_dot":"a PlotPoint [green] drawn in m_axes (target='m_left', x=<VariableNumber probe_m = 0.5>)","m_left":"a FunctionPlot [magenta] drawn in m_axes (function=<function>, x_range=(0.0, 4.0))","m_peak":"a Point [green] labelled \"16\" drawn in m_axes (location=(4.0, 16.0))","m_right":"a FunctionPlot [magenta] drawn in m_axes (function=<function>, x_range=(4.0, 6.0))","marker":"a Line [green] drawn in beam (start=(<VariableNumber marker_x = 3.7>, 0.7), end=(<VariableNumber marker_x = 3.7>, -0.7), dashed=True)","marker_x":"a VariableNumber (initial_value=0.6)","note_slope":"a Math [text] that says \"$upright(\"slope of \") M = V$\"","pin":"a Polygon [gray] drawn in beam (vertices=((0.0, 0.0), (-0.3, -0.55), (0.3, -0.55)), fill_opacity=0.35)","probe":"a VariableNumber (initial_value=0.6)","probe_m":"a VariableNumber (initial_value=3.7)","r_a":"a Vector [green] drawn in beam (start=(0.0, -1.5), end=(0.0, -0.65))","r_a_name":"a Math [green] that says \"$4$\" drawn in beam","r_b":"a Vector [green] drawn in beam (start=(6.0, -1.5), end=(6.0, -0.65))","r_b_name":"a Math [green] that says \"$8$\" drawn in beam","right_pair":"a Derivation [text] that says \"$V &= -8 \\ M &= 48 - 8 x$\"","roller":"a Polygon [gray] drawn in beam (vertices=((6.0, 0.0), (5.7, -0.55), (6.3, -0.55)), fill_opacity=0.35)","span":"a Line [blue] drawn in beam (end=(6.0, 0.0))","v_axes":"an Axes (x_range=(0.0, 6.0), y_range=(-12.0, 8.0), aspect=(7.9, 2.8))","v_close":"a Line [yellow] drawn in v_axes (start=(6.0, -8.0), end=(6.0, 0.0))","v_dot":"a PlotPoint [green] drawn in v_axes (target='v_left', x=<VariableNumber probe = 0.5>)","v_drop":"a Line [yellow] drawn in v_axes (start=(4.0, 4.0), end=(4.0, -8.0))","v_left":"a FunctionPlot [yellow] drawn in v_axes (function=<function>, x_range=(0.0, 4.0))","v_right":"a FunctionPlot [yellow] drawn in v_axes (function=<function>, x_range=(4.0, 6.0))","v_start":"a Line [yellow] drawn in v_axes (end=(0.0, 4.0))"},"beats":[{"start":249.64514583333334,"say":"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.","live":[],"does":[[249.64514583333334,"head_shear is shown on the screen, written out."],[249.64514583333334,"beam is shown on the screen, written out."],[250.48114583333336,"lab_left is shown on the screen, written out."],[250.68114583333335,"left_pair is shown on the screen, written out."],[251.08114583333335,"left_pair is shown on the screen, written out."],[251.68114583333335,"lab_right is shown on the screen, written out."],[251.70014583333335,"span is shown on the screen, written out."],[251.80014583333335,"pin is shown on the screen, written out."],[252.00014583333333,"roller is shown on the screen, written out."],[252.30014583333335,"load is shown on the screen, written out."],[252.48114583333336,"right_pair is shown on the screen, written out."],[252.70014583333335,"load_name is shown on the screen, written out."],[253.20014583333335,"r_a is shown on the screen, written out."],[253.48114583333336,"right_pair is shown on the screen, written out."],[253.80014583333335,"r_a_name is shown on the screen, written out."],[254.50014583333333,"r_b is shown on the screen, written out."],[255.30014583333335,"r_b_name is shown on the screen, written out."]]},{"start":260.14814583333333,"say":"So: position along the beam across the bottom, internal shear up the side.","live":["lab_left","lab_right","beam","head_shear","span","pin","roller","load","load_name","r_a","r_a_name","r_b","r_b_name"],"does":[[261.10014583333333,"v_axes is shown on the screen, written out."]]},{"start":266.08914583333336,"say":"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.","live":["lab_left","lab_right","beam","v_axes","head_shear","span","pin","roller","load","load_name","r_a","r_a_name","r_b","r_b_name"],"does":[[272.81114583333334,"v_start is shown on the screen, drawn."],[276.34114583333337,"v_left is shown on the screen, drawn."]]},{"start":278.6821458333333,"say":"On the right stretch it is minus eight, so a horizontal line eight below the axis, running from the load across to B.","live":["lab_left","lab_right","beam","v_axes","head_shear","span","pin","roller","load","load_name","r_a","r_a_name","r_b","r_b_name","v_start","v_left"],"does":[[281.5501458333333,"v_right is shown on the screen, drawn."]]},{"start":286.33514583333334,"say":"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.","live":["lab_left","lab_right","beam","v_axes","head_shear","span","pin","roller","load","load_name","r_a","r_a_name","r_b","r_b_name","v_start","v_left","v_right"],"does":[[291.54214583333334,"v_drop is shown on the screen, drawn."]]},{"start":296.4151458333333,"say":"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.","live":["lab_left","lab_right","beam","v_axes","head_shear","span","pin","roller","load","load_name","r_a","r_a_name","r_b","r_b_name","v_start","v_left","v_right","v_drop"],"does":[[298.5401458333333,"v_close is shown on the screen, drawn."]]},{"start":309.7396458333333,"say":"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.","live":["lab_left","lab_right","beam","v_axes","head_shear","span","pin","roller","load","load_name","r_a","r_a_name","r_b","r_b_name","v_start","v_left","v_right","v_drop","v_close"],"does":[[309.7396458333333,"marker is shown on the screen, written out."],[309.7396458333333,"v_dot is shown on the screen, written out."],[311.38814583333334,"marker is redrawn as the numbers it depends on change."],[311.38814583333334,"v_dot is redrawn as the numbers it depends on change."],[311.38814583333334,"probe ticks to 3.7."],[311.38814583333334,"marker_x ticks to 3.7."],[318.53964583333334,"lab_left moves to a new place on the board."],[318.53964583333334,"lab_right moves to a new place on the board."],[318.53964583333334,"left_pair moves to a new place on the board."],[318.53964583333334,"right_pair moves to a new place on the board."],[318.53964583333334,"v_axes moves to a new place on the board."],[318.53964583333334,"beam is hidden from the screen — left the board."],[318.53964583333334,"span is hidden from the screen — beam left the board."],[318.53964583333334,"pin is hidden from the screen — beam left the board."],[318.53964583333334,"roller is hidden from the screen — beam left the board."],[318.53964583333334,"load is hidden from the screen — beam left the board."],[318.53964583333334,"load_name is hidden from the screen — beam left the board."],[318.53964583333334,"r_a is hidden from the screen — beam left the board."],[318.53964583333334,"r_a_name is hidden from the screen — beam left the board."],[318.53964583333334,"r_b is hidden from the screen — beam left the board."],[318.53964583333334,"r_b_name is hidden from the screen — beam left the board."],[318.53964583333334,"marker is hidden from the screen — beam left the board."],[318.53964583333334,"head_shear is hidden from the screen — left the board."]]},{"start":319.13964583333336,"say":"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.","live":["lab_left","lab_right","v_axes","v_start","v_left","v_right","v_drop","v_close","v_dot"],"does":[[319.13964583333336,"head_moment is shown on the screen, written out."],[319.13964583333336,"m_axes is shown on the screen, written out."],[325.9201458333333,"m_left is shown on the screen, drawn."]]},{"start":329.0161458333333,"say":"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.","live":["lab_left","lab_right","v_axes","v_start","v_left","v_right","v_drop","v_close","v_dot","m_axes","head_moment","m_left"],"does":[[330.1771458333334,"m_peak is shown on the screen, written out."],[337.29414583333335,"m_right is shown on the screen, drawn."]]},{"start":340.42464583333333,"say":"For a point load, then, the moment diagram is a triangle, and its peak sits directly under the load.","live":["lab_left","lab_right","v_axes","v_start","v_left","v_right","v_drop","v_close","v_dot","m_axes","head_moment","m_left","m_peak","m_right"],"does":[[344.82514583333335,"m_peak is indicated — a transient flash."]]},{"start":347.6776458333334,"say":"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.","live":null,"does":[[352.48414583333334,"v_left is emphasized."],[354.0161458333333,"m_left is emphasized."],[355.80414583333334,"v_left is no longer emphasized."],[355.80414583333334,"m_left is no longer emphasized."]]},{"start":358.8311458333334,"say":"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.","live":null,"does":[[370.24314583333336,"note_slope is shown on the screen, written out."],[371.93814583333335,"A box is drawn around note_slope."]]},{"start":373.30514583333337,"say":"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.","live":["lab_left","lab_right","v_axes","v_start","v_left","v_right","v_drop","v_close","v_dot","note_slope","m_axes","head_moment","m_left","m_peak","m_right"],"does":[[373.30514583333337,"m_dot is shown on the screen, written out."],[378.11214583333333,"v_dot is redrawn as the numbers it depends on change."],[378.11214583333333,"m_dot is redrawn as the numbers it depends on change."],[378.11214583333333,"probe ticks to 0.5."],[378.11214583333333,"probe_m ticks to 0.5."],[381.72214583333334,"v_drop is indicated — a transient flash."]]},{"start":383.49514583333337,"say":"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.","live":["lab_left","lab_right","v_axes","v_start","v_left","v_right","v_drop","v_close","v_dot","note_slope","m_axes","head_moment","m_left","m_peak","m_right","m_dot"],"does":[[393.00943750000005,"head_moment is hidden from the screen — left the board."],[393.00943750000005,"lab_left is hidden from the screen — left the board."],[393.00943750000005,"lab_right is hidden from the screen — left the board."],[393.00943750000005,"left_pair is hidden from the screen — left the board."],[393.00943750000005,"m_axes is hidden from the screen — left the board."],[393.00943750000005,"m_left is hidden from the screen — m_axes left the board."],[393.00943750000005,"m_peak is hidden from the screen — m_axes left the board."],[393.00943750000005,"m_right is hidden from the screen — m_axes left the board."],[393.00943750000005,"m_dot is hidden from the screen — m_axes left the board."],[393.00943750000005,"note_slope is hidden from the screen — left the board."],[393.00943750000005,"right_pair is hidden from the screen — left the board."],[393.00943750000005,"v_axes is hidden from the screen — left the board."],[393.00943750000005,"v_start is hidden from the screen — v_axes left the board."],[393.00943750000005,"v_left is hidden from the screen — v_axes left the board."],[393.00943750000005,"v_right is hidden from the screen — v_axes left the board."],[393.00943750000005,"v_drop is hidden from the screen — v_axes left the board."],[393.00943750000005,"v_close is hidden from the screen — v_axes left the board."],[393.00943750000005,"v_dot is hidden from the screen — v_axes left the board."]]}]},{"title":"A Distributed Load","start":394.05110416666673,"end":565.4929791666667,"objects":{"area_claim":"a Text [text] that says \"The shaded area under the shear diagram, out to any position, is the bending moment there.\"","beam":"a Figure (x_range=(-1.0, 7.4), y_range=(-5.0, 2.6), aspect=(8.4, 7.6))","cut":"a VariableNumber (initial_value=2.2, format_spec='.1f')","cut_line":"a Line [yellow] drawn in beam (start=(<VariableNumber cut = 4.6>, 0.8), end=(<VariableNumber cut = 4.6>, -0.95), dashed=True)","fb_half":"a Line [gray] labelled \"frac(x, 2)\" drawn in beam (start=(0.0, -4.3), end=((cut / 2.0), -4.3))","fb_line":"a Line [blue] drawn in beam (start=(0.0, -3.1), end=(<VariableNumber cut = 4.6>, -3.1))","fb_m":"a CurvedArrow [yellow] labelled \"M\" drawn in beam (start=((cut - 0.8), -3.62), end=((cut - 0.8), -2.58), bend=0.55)","fb_ra":"a Vector [green] labelled \"12\" drawn in beam (start=(0.0, -4.05), end=(0.0, -3.16))","fb_res":"a Vector [red] labelled \"w x\" drawn in beam (start=((cut / 2.0), -2.15), end=((cut / 2.0), -3.04))","fb_v":"a Vector [yellow] labelled \"V\" drawn in beam (start=(<VariableNumber cut = 4.6>, -3.1), end=(<VariableNumber cut = 4.6>, -3.98))","fb_x":"a Line [gray] labelled \"x\" drawn in beam (start=(0.0, -4.72), end=(<VariableNumber cut = 4.6>, -4.72))","fun_m":"a Math [text] that says \"$M(x) = 12 x - 2 x^2$\"","fun_v":"a Math [text] that says \"$V(x) = 12 - 4 x$\"","head_beam":"a Heading that says \"Four Kilonewtons on Every Metre\"","head_plot":"a Heading that says \"A Ramp and a Parabola\"","m_axes":"an Axes (x_range=(0.0, 6.0), y_range=(-2.0, 22.0), aspect=(7.9, 2.8))","m_peak":"a Point [green] labelled \"18\" drawn in m_axes (location=(3.0, 18.0))","m_plot":"a FunctionPlot [magenta] drawn in m_axes (function=<function>, x_range=(0.0, 6.0))","m_rider":"a PlotPoint [green] drawn in m_axes (target='m_plot', x=<VariableNumber sweep = 5.9>)","moment_work":"a Derivation [text] that says \"$sum M_(upright(\"cut\")) &= 0 \\ M - 12 x + (4 x) frac(x, 2) &= 0 \\ M(x) &= 12 x - 2 x^2$\"","pin":"a Polygon [gray] drawn in beam (vertices=((0.0, 0.0), (-0.32, -0.62), (0.32, -0.62)), fill_opacity=0.35)","r_a":"a Vector [green] drawn in beam (start=(0.0, -1.8), end=(0.0, -0.72))","r_a_name":"a Math [green] that says \"$12$\" drawn in beam","r_b":"a Vector [green] drawn in beam (start=(6.0, -1.8), end=(6.0, -0.72))","r_b_name":"a Math [green] that says \"$12$\" drawn in beam","roller":"a Polygon [gray] drawn in beam (vertices=((6.0, 0.0), (5.68, -0.62), (6.32, -0.62)), fill_opacity=0.35)","shear_work":"a Derivation [text] that says \"$sum F_y &= 0 \\ 12 - 4 x - V &= 0 \\ V(x) &= 12 - 4 x$\"","span":"a Line [blue] drawn in beam (end=(6.0, 0.0))","sweep":"a VariableNumber (initial_value=0.5)","udl_0":"a Vector [red] drawn in beam (start=(0.0, 1.4), end=(0.0, 0.08))","udl_1":"a Vector [red] drawn in beam (start=(1.0, 1.4), end=(1.0, 0.08))","udl_2":"a Vector [red] drawn in beam (start=(2.0, 1.4), end=(2.0, 0.08))","udl_3":"a Vector [red] drawn in beam (start=(3.0, 1.4), end=(3.0, 0.08))","udl_4":"a Vector [red] drawn in beam (start=(4.0, 1.4), end=(4.0, 0.08))","udl_5":"a Vector [red] drawn in beam (start=(5.0, 1.4), end=(5.0, 0.08))","udl_6":"a Vector [red] drawn in beam (start=(6.0, 1.4), end=(6.0, 0.08))","udl_name":"a Math [red] that says \"$w = 4 thin upright(\"kN/m\")$\" drawn in beam","udl_top":"a Line [red] drawn in beam (start=(0.0, 1.45), end=(6.0, 1.45))","v_area":"an AreaUnder [green] drawn in v_axes (x_range=(0.0, <VariableNumber sweep = 5.9>), target='v_plot')","v_axes":"an Axes (x_range=(0.0, 6.0), y_range=(-15.0, 15.0), aspect=(7.9, 2.8))","v_close":"a Line [yellow] drawn in v_axes (start=(6.0, -12.0), end=(6.0, 0.0))","v_plot":"a FunctionPlot [yellow] drawn in v_axes (function=<function>, x_range=(0.0, 6.0))","v_start":"a Line [yellow] drawn in v_axes (end=(0.0, 12.0))","v_zero":"a Point [green] labelled \"x = 3\" drawn in v_axes (location=(3.0, 0.0))"},"beats":[{"start":394.05110416666673,"say":"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.","live":[],"does":[[394.05110416666673,"head_beam is shown on the screen, written out."],[394.05110416666673,"beam is shown on the screen, written out."],[396.36110416666673,"span is shown on the screen, written out."],[396.46110416666676,"pin is shown on the screen, written out."],[396.66110416666675,"roller is shown on the screen, written out."],[399.64710416666674,"udl_top is shown on the screen, written out."],[399.74710416666676,"udl_0 is shown on the screen, written out."],[399.94710416666675,"udl_1 is shown on the screen, written out."],[400.24710416666676,"udl_2 is shown on the screen, written out."],[400.64710416666674,"udl_3 is shown on the screen, written out."],[401.14710416666674,"udl_4 is shown on the screen, written out."],[401.74710416666676,"udl_5 is shown on the screen, written out."],[402.44710416666675,"udl_6 is shown on the screen, written out."],[403.26910416666675,"udl_name is shown on the screen, written out."]]},{"start":405.77360416666676,"say":"The total is four times six, twenty four kilonewtons, and by symmetry each support takes half. Twelve up at each end.","live":["beam","head_beam","span","pin","roller","udl_top","udl_0","udl_1","udl_2","udl_3","udl_4","udl_5","udl_6","udl_name"],"does":[[412.9481041666667,"r_a is shown on the screen, written out."],[413.1481041666667,"r_a_name is shown on the screen, written out."],[413.54810416666675,"r_b is shown on the screen, written out."],[414.1481041666667,"r_b_name is shown on the screen, written out."]]},{"start":415.12710416666675,"say":"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.","live":["beam","head_beam","span","pin","roller","udl_top","udl_0","udl_1","udl_2","udl_3","udl_4","udl_5","udl_6","udl_name","r_a","r_a_name","r_b","r_b_name"],"does":[[415.5331041666667,"cut_line is shown on the screen, written out."],[417.4601041666667,"fb_line is shown on the screen, written out."],[417.6601041666667,"fb_x is shown on the screen, written out."],[418.06010416666675,"fb_ra is shown on the screen, written out."],[426.58610416666676,"fb_res is shown on the screen, written out."],[428.92010416666676,"fb_half is shown on the screen, written out."],[429.2201041666667,"fb_v is shown on the screen, written out."],[429.7201041666667,"fb_m is shown on the screen, written out."]]},{"start":431.0061041666667,"say":"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.","live":["beam","head_beam","span","pin","roller","udl_top","udl_0","udl_1","udl_2","udl_3","udl_4","udl_5","udl_6","udl_name","r_a","r_a_name","r_b","r_b_name","cut_line","fb_line","fb_x","fb_ra","fb_res","fb_half","fb_v","fb_m"],"does":[[431.35410416666673,"beam moves to a new place on the board."],[431.35410416666673,"shear_work is shown on the screen, written out."],[435.58010416666673,"shear_work is shown on the screen, written out."],[437.5421041666667,"shear_work is shown on the screen, written out."]]},{"start":446.68760416666674,"say":"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.","live":null,"does":[[446.8501041666667,"moment_work is shown on the screen, written out."],[453.1551041666667,"moment_work is shown on the screen, written out."],[460.41110416666675,"moment_work is shown on the screen, written out."]]},{"start":461.88710416666675,"say":"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.","live":null,"does":[[462.0091041666667,"cut_line is redrawn as the numbers it depends on change."],[462.0091041666667,"fb_line is redrawn as the numbers it depends on change."],[462.0091041666667,"fb_x is redrawn as the numbers it depends on change."],[462.0091041666667,"fb_res is redrawn as the numbers it depends on change."],[462.0091041666667,"fb_half is redrawn as the numbers it depends on change."],[462.0091041666667,"fb_v is redrawn as the numbers it depends on change."],[462.0091041666667,"fb_m is redrawn as the numbers it depends on change."],[462.0091041666667,"cut ticks to 4.6."],[468.59210416666673,"beam is hidden from the screen — left the board."],[468.59210416666673,"span is hidden from the screen — beam left the board."],[468.59210416666673,"pin is hidden from the screen — beam left the board."],[468.59210416666673,"roller is hidden from the screen — beam left the board."],[468.59210416666673,"udl_top is hidden from the screen — beam left the board."],[468.59210416666673,"udl_0 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_1 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_2 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_3 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_4 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_5 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_6 is hidden from the screen — beam left the board."],[468.59210416666673,"udl_name is hidden from the screen — beam left the board."],[468.59210416666673,"r_a is hidden from the screen — beam left the board."],[468.59210416666673,"r_a_name is hidden from the screen — beam left the board."],[468.59210416666673,"r_b is hidden from the screen — beam left the board."],[468.59210416666673,"r_b_name is hidden from the screen — beam left the board."],[468.59210416666673,"cut_line is hidden from the screen — beam left the board."],[468.59210416666673,"fb_line is hidden from the screen — beam left the board."],[468.59210416666673,"fb_x is hidden from the screen — beam left the board."],[468.59210416666673,"fb_ra is hidden from the screen — beam left the board."],[468.59210416666673,"fb_res is hidden from the screen — beam left the board."],[468.59210416666673,"fb_half is hidden from the screen — beam left the board."],[468.59210416666673,"fb_v is hidden from the screen — beam left the board."],[468.59210416666673,"fb_m is hidden from the screen — beam left the board."],[468.59210416666673,"head_beam is hidden from the screen — left the board."],[468.59210416666673,"moment_work is hidden from the screen — left the board."],[468.59210416666673,"shear_work is hidden from the screen — left the board."]]},{"start":469.7921041666667,"say":"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.","live":[],"does":[[469.7921041666667,"head_plot is shown on the screen, written out."],[469.7921041666667,"v_axes is shown on the screen, written out."],[470.09410416666674,"fun_v is shown on the screen, written out."],[473.57710416666674,"v_plot is shown on the screen, drawn."],[474.25010416666674,"v_start is shown on the screen, drawn."],[478.06910416666676,"v_close is shown on the screen, drawn."],[478.6501041666667,"v_zero is shown on the screen, written out."]]},{"start":481.6356041666667,"say":"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.","live":["fun_v","v_axes","head_plot","v_start","v_plot","v_close","v_zero"],"does":[[481.6356041666667,"m_axes is shown on the screen, written out."],[482.16410416666673,"fun_m is shown on the screen, written out."],[482.72110416666675,"m_plot is shown on the screen, drawn."],[491.34810416666676,"m_peak is shown on the screen, written out."]]},{"start":494.0141041666667,"say":"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.","live":["fun_v","fun_m","v_axes","m_axes","head_plot","v_start","v_plot","v_close","v_zero","m_plot","m_peak"],"does":[]},{"start":509.7536041666667,"say":"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.","live":null,"does":[[511.97110416666675,"v_zero is indicated — a transient flash."],[517.4161041666667,"m_peak is indicated — a transient flash."]]},{"start":519.7926041666667,"say":"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.","live":null,"does":[[524.9011041666668,"v_area is shown on the screen, faded in."],[525.3011041666667,"m_rider is shown on the screen, written out."],[528.5931041666668,"area_claim is shown on the screen, written out."]]},{"start":530.4816041666668,"say":"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.","live":["fun_v","fun_m","area_claim","v_axes","m_axes","head_plot","v_start","v_plot","v_close","v_zero","m_plot","m_peak","v_area","m_rider"],"does":[[531.1321041666667,"v_area is redrawn as the numbers it depends on change."],[531.1321041666667,"m_rider is redrawn as the numbers it depends on change."],[531.1321041666667,"sweep ticks to 3.0."]]},{"start":542.0646041666668,"say":"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.","live":null,"does":[[542.4131041666667,"v_area is redrawn as the numbers it depends on change."],[542.4131041666667,"m_rider is redrawn as the numbers it depends on change."],[542.4131041666667,"sweep ticks to 5.9."]]},{"start":555.3201041666667,"say":"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.","live":null,"does":[[564.4513125000001,"area_claim is hidden from the screen — left the board."],[564.4513125000001,"fun_m is hidden from the screen — left the board."],[564.4513125000001,"fun_v is hidden from the screen — left the board."],[564.4513125000001,"head_plot is hidden from the screen — left the board."],[564.4513125000001,"m_axes is hidden from the screen — left the board."],[564.4513125000001,"m_plot is hidden from the screen — m_axes left the board."],[564.4513125000001,"m_peak is hidden from the screen — m_axes left the board."],[564.4513125000001,"m_rider is hidden from the screen — m_axes left the board."],[564.4513125000001,"v_axes is hidden from the screen — left the board."],[564.4513125000001,"v_start is hidden from the screen — v_axes left the board."],[564.4513125000001,"v_plot is hidden from the screen — v_axes left the board."],[564.4513125000001,"v_close is hidden from the screen — v_axes left the board."],[564.4513125000001,"v_zero is hidden from the screen — v_axes left the board."],[564.4513125000001,"v_area is hidden from the screen — v_axes left the board."]]}]},{"title":"Load, Shear and Moment","start":565.4929791666667,"end":756.0034166666667,"objects":{"consequences":"a Table [text] that says \"On a stretch Shear Moment no load constant straight uniform $w$ straight parabola point force $P$ jumps by $P$ kink couple $M_0$ unchanged jumps by $M_0$\" (rows=(('On a stretch', 'Shear', 'Moment'), ('no load', 'constant', '…, header=True)","dx_dim":"a Line [gray] labelled \"dif x\" drawn in elem (start=(1.0, -2.0), end=(3.0, -2.0))","elem":"a Figure (x_range=(-1.4, 4.8), y_range=(-2.7, 3.2), aspect=(6.2, 5.9))","force_work":"a Derivation [text] that says \"$sum F_y &= 0 \\ V - w thin dif x - (V + dif V) &= 0 \\ dif V &= -w thin dif x \\ frac(dif V, dif x) &= -w$\"","head_int":"a Heading that says \"The Same Two Statements, Integrated\"","head_max":"a Heading that says \"Where the Maximum Moment Is\"","head_slice":"a Heading that says \"A Slice of Beam, of Length $d x$\"","integrals":"a Derivation [text] that says \"$V_2 - V_1 &= - integral_(x_1)^(x_2) w thin dif x \\ M_2 - M_1 &= integral_(x_1)^(x_2) V thin dif x$\"","load_a":"a Vector [red] drawn in elem (start=(1.0, 1.95), end=(1.0, 0.88))","load_b":"a Vector [red] drawn in elem (start=(2.0, 1.95), end=(2.0, 0.88))","load_c":"a Vector [red] drawn in elem (start=(3.0, 1.95), end=(3.0, 0.88))","load_name":"a Math [red] that says \"$w$\" drawn in elem","load_top":"a Line [red] drawn in elem (start=(1.0, 2.0), end=(3.0, 2.0))","m_in":"a CurvedArrow [yellow] labelled \"M\" drawn in elem (start=(0.3, 1.1), end=(0.3, -0.4))","m_out":"a CurvedArrow [yellow] labelled \"M + dif M\" drawn in elem (start=(3.7, -0.4), end=(3.7, 1.1))","moment_work":"a Derivation [text] that says \"$sum M &= 0 \\ dif M - V thin dif x + frac(w (dif x)^2, 2) &= 0 \\ dif M &= V thin dif x \\ frac(dif M, dif x) &= V$\"","note_max":"a Text [text] that says \"The bending moment is stationary wherever the shear crosses the axis. Check the ends and every jump in the shear as well.\"","rule_max":"a Math [text] that says \"$frac(dif M, dif x) = V = 0$\"","slab":"a Polygon [blue] drawn in elem (vertices=((1.0, 0.0), (3.0, 0.0), (3.0, 0.8), (1.0, 0.8)), fill_opacity=0.25)","v_in":"a Vector [yellow] labelled \"V\" drawn in elem (start=(1.0, -1.2), end=(1.0, -0.05))","v_out":"a Vector [yellow] labelled \"V + dif V\" drawn in elem (start=(3.0, -0.05), end=(3.0, -1.2))"},"beats":[{"start":565.4929791666667,"say":"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.","live":[],"does":[[565.4929791666667,"head_slice is shown on the screen, written out."],[565.4929791666667,"elem is shown on the screen, written out."],[570.8339791666667,"load_top is shown on the screen, written out."],[570.9339791666667,"load_a is shown on the screen, written out."],[571.1339791666667,"load_b is shown on the screen, written out."],[571.4339791666667,"load_c is shown on the screen, written out."],[571.8339791666667,"load_name is shown on the screen, written out."],[572.2149791666667,"slab is shown on the screen, written out."],[572.4149791666667,"dx_dim is shown on the screen, written out."]]},{"start":578.7359791666668,"say":"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.","live":["elem","head_slice","slab","dx_dim","load_top","load_a","load_b","load_c","load_name"],"does":[[581.5349791666667,"v_in is shown on the screen, written out."],[582.9629791666667,"m_in is shown on the screen, written out."],[588.2679791666667,"v_out is shown on the screen, written out."],[588.6679791666667,"m_out is shown on the screen, written out."]]},{"start":592.8914791666667,"say":"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.","live":["elem","head_slice","slab","dx_dim","load_top","load_a","load_b","load_c","load_name","v_in","m_in","v_out","m_out"],"does":[[593.0479791666667,"elem moves to a new place on the board."],[593.0479791666667,"force_work is shown on the screen, written out."],[598.5739791666667,"force_work is shown on the screen, written out."],[603.2059791666667,"force_work (the \"V\" part) is slashed through — it cancels."],[603.2059791666667,"force_work (the \"V#2\" part) is slashed through — it cancels."]]},{"start":604.7584791666667,"say":"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.","live":null,"does":[[605.4429791666668,"force_work is shown on the screen, written out."],[611.5969791666668,"force_work is shown on the screen, written out."]]},{"start":616.7014791666667,"say":"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.","live":null,"does":[[617.7229791666667,"force_work (the \"-w\" part) is emphasized."],[629.0199791666666,"force_work (the \"-w\" part) is no longer emphasized."]]},{"start":630.7579791666667,"say":"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.","live":null,"does":[[631.6979791666668,"moment_work is shown on the screen, written out."],[641.2179791666667,"moment_work is shown on the screen, written out."]]},{"start":643.6984791666667,"say":"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.","live":null,"does":[[650.2819791666667,"moment_work (the \"frac(w (dif x)^2, 2)\" part) is slashed through — it cancels."],[651.6749791666667,"moment_work is shown on the screen, written out."]]},{"start":655.4794791666667,"say":"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.","live":null,"does":[[655.9899791666667,"moment_work is shown on the screen, written out."],[662.0849791666667,"A box is drawn around moment_work."],[667.4839791666667,"elem is hidden from the screen — left the board."],[667.4839791666667,"slab is hidden from the screen — elem left the board."],[667.4839791666667,"dx_dim is hidden from the screen — elem left the board."],[667.4839791666667,"load_top is hidden from the screen — elem left the board."],[667.4839791666667,"load_a is hidden from the screen — elem left the board."],[667.4839791666667,"load_b is hidden from the screen — elem left the board."],[667.4839791666667,"load_c is hidden from the screen — elem left the board."],[667.4839791666667,"load_name is hidden from the screen — elem left the board."],[667.4839791666667,"v_in is hidden from the screen — elem left the board."],[667.4839791666667,"m_in is hidden from the screen — elem left the board."],[667.4839791666667,"v_out is hidden from the screen — elem left the board."],[667.4839791666667,"m_out is hidden from the screen — elem left the board."],[667.4839791666667,"force_work is hidden from the screen — left the board."],[667.4839791666667,"head_slice is hidden from the screen — left the board."],[667.4839791666667,"moment_work is hidden from the screen — left the board."]]},{"start":668.6839791666666,"say":"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.","live":[],"does":[[668.6839791666666,"head_int is shown on the screen, written out."],[673.2239791666667,"integrals is shown on the screen, written out."]]},{"start":679.2104791666667,"say":"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.","live":["head_int"],"does":[[679.7209791666667,"integrals is shown on the screen, written out."]]},{"start":689.1104791666667,"say":"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.","live":null,"does":[[692.2799791666667,"consequences is shown on the screen, written out."],[693.0929791666667,"consequences is shown on the screen, written out."],[697.1329791666667,"consequences is shown on the screen, written out."]]},{"start":704.7339791666667,"say":"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.","live":null,"does":[[711.5139791666668,"consequences is shown on the screen, written out."]]},{"start":717.3149791666667,"say":"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.","live":null,"does":[[726.8929791666667,"consequences is shown on the screen, written out."],[727.3929791666667,"consequences (the \"row=5\" part) is emphasized."],[729.4239791666666,"consequences (the \"row=5\" part) is no longer emphasized."],[731.5719791666667,"consequences is hidden from the screen — left the board."],[731.5719791666667,"head_int is hidden from the screen — left the board."],[731.5719791666667,"integrals is hidden from the screen — left the board."]]},{"start":732.1719791666667,"say":"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.","live":[],"does":[[732.1719791666667,"head_max is shown on the screen, written out."],[736.3399791666667,"rule_max is shown on the screen, written out."],[743.7589791666667,"note_max is shown on the screen, written out."],[745.7209791666667,"A box is drawn around rule_max."]]},{"start":748.3064791666667,"say":"Two derivatives and two areas. Everything else about these diagrams is a consequence of them.","live":["rule_max","note_max","head_max"],"does":[[754.96175,"head_max is hidden from the screen — left the board."],[754.96175,"note_max is hidden from the screen — left the board."],[754.96175,"rule_max is hidden from the screen — left the board."]]}]},{"title":"An Overhang and a Couple","start":756.0034166666667,"end":944.9995833333334,"objects":{"beam":"a Figure (x_range=(-1.0, 9.4), y_range=(-2.6, 2.7), aspect=(10.4, 5.3))","couple":"a CurvedArrow [red] drawn in beam (start=(5.55, 0.15), end=(4.45, 0.15), bend=0.9)","couple_mark":"a Line [gray] labelled \"M_0\" drawn in v_axes (start=(5.0, -6.5), end=(5.0, 6.5), dashed=True)","couple_name":"a Math [red] that says \"$M_0 = 8$\" drawn in beam","fact_1":"a Text [text] that says \"A couple leaves the shear untouched and jumps the moment.\"","fact_2":"a Text [text] that says \"Over the overhang the moment is negative, so the beam is in tension along its top face.\"","head_beam":"a Heading that says \"An Overhang, and a Couple at Five Metres\"","head_build":"a Heading that says \"Built From Jumps and Areas\"","load_1":"a Vector [red] drawn in beam (start=(2.0, 1.7), end=(2.0, 0.08))","load_1_name":"a Math [red] that says \"$12$\" drawn in beam","load_2":"a Vector [red] drawn in beam (start=(8.0, 1.7), end=(8.0, 0.08))","load_2_name":"a Math [red] that says \"$4$\" drawn in beam","m_axes":"an Axes (x_range=(0.0, 8.0), y_range=(-11.0, 20.0), aspect=(9.0, 2.7))","m_jump":"a Line [magenta] drawn in m_axes (start=(5.0, 4.0), end=(5.0, -4.0))","m_s1":"a Line [magenta] drawn in m_axes (end=(2.0, 16.0))","m_s2":"a Line [magenta] drawn in m_axes (start=(2.0, 16.0), end=(5.0, 4.0))","m_s3":"a Line [magenta] drawn in m_axes (start=(5.0, -4.0), end=(6.0, -8.0))","m_s4":"a Line [magenta] drawn in m_axes (start=(6.0, -8.0), end=(8.0, 0.0))","name_a":"a Math [text] that says \"$A$\" drawn in beam","name_b":"a Math [text] that says \"$B$\" drawn in beam","pin":"a Polygon [gray] drawn in beam (vertices=((0.0, 0.0), (-0.34, -0.66), (0.34, -0.66)), fill_opacity=0.35)","r_a":"a Vector [green] drawn in beam (start=(0.0, -1.9), end=(0.0, -0.76))","r_a_name":"a Math [green] that says \"$8$\" drawn in beam","r_b":"a Vector [green] drawn in beam (start=(6.0, -1.9), end=(6.0, -0.76))","r_b_name":"a Math [green] that says \"$8$\" drawn in beam","reactions":"a Derivation [text] that says \"$sum M_A &= 0 \\ 6 R_B &= 12 (2) + 4 (8) - 8 \\ &= 48 \\ R_B &= 8 thin upright(\"kN\") \\ R_A &= 16 - 8 = 8 thin upright(\"kN\")$\"","roller":"a Polygon [gray] drawn in beam (vertices=((6.0, 0.0), (5.66, -0.66), (6.34, -0.66)), fill_opacity=0.35)","span":"a Line [blue] drawn in beam (end=(8.0, 0.0))","v_axes":"an Axes (x_range=(0.0, 8.0), y_range=(-8.0, 11.0), aspect=(9.0, 2.7))","v_close":"a Line [yellow] drawn in v_axes (start=(8.0, 4.0), end=(8.0, 0.0))","v_drop":"a Line [yellow] drawn in v_axes (start=(2.0, 8.0), end=(2.0, -4.0))","v_s1":"a Line [yellow] drawn in v_axes (start=(0.0, 8.0), end=(2.0, 8.0))","v_s2":"a Line [yellow] drawn in v_axes (start=(2.0, -4.0), end=(6.0, -4.0))","v_s3":"a Line [yellow] drawn in v_axes (start=(6.0, 4.0), end=(8.0, 4.0))","v_up_a":"a Line [yellow] drawn in v_axes (end=(0.0, 8.0))","v_up_b":"a Line [yellow] drawn in v_axes (start=(6.0, -4.0), end=(6.0, 4.0))"},"beats":[{"start":756.0034166666667,"say":"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.","live":[],"does":[[756.0034166666667,"head_beam is shown on the screen, written out."],[756.0034166666667,"beam is shown on the screen, written out."],[760.0904166666667,"span is shown on the screen, written out."],[761.4484166666667,"pin is shown on the screen, written out."],[761.5484166666666,"name_a is shown on the screen, written out."],[761.7484166666667,"roller is shown on the screen, written out."],[762.0484166666666,"name_b is shown on the screen, written out."]]},{"start":766.8319166666666,"say":"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.","live":["beam","head_beam","span","pin","name_a","roller","name_b"],"does":[[767.7374166666667,"load_1 is shown on the screen, written out."],[767.9374166666666,"load_1_name is shown on the screen, written out."],[770.0364166666667,"load_2 is shown on the screen, written out."],[770.2364166666666,"load_2_name is shown on the screen, written out."],[772.3814166666667,"couple is shown on the screen, written out."],[772.6814166666667,"couple_name is shown on the screen, written out."]]},{"start":777.3004166666667,"say":"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.","live":["beam","head_beam","span","pin","name_a","roller","name_b","load_1","load_1_name","load_2","load_2_name","couple","couple_name"],"does":[[779.5294166666666,"beam moves to a new place on the board."],[779.5294166666666,"reactions is shown on the screen, written out."],[786.4954166666666,"reactions is shown on the screen, written out."],[789.2464166666666,"reactions (the \"- 8\" part) is emphasized."],[791.8244166666667,"reactions (the \"- 8\" part) is no longer emphasized."]]},{"start":793.5039166666667,"say":"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.","live":null,"does":[[794.4324166666667,"reactions is shown on the screen, written out."],[795.5934166666667,"reactions is shown on the screen, written out."],[795.8934166666667,"r_b is shown on the screen, written out."],[796.4934166666667,"r_b_name is shown on the screen, written out."],[798.0314166666667,"reactions is shown on the screen, written out."],[798.3314166666667,"r_a is shown on the screen, written out."],[798.9314166666667,"r_a_name is shown on the screen, written out."],[803.9764166666666,"beam is hidden from the screen — left the board."],[803.9764166666666,"span is hidden from the screen — beam left the board."],[803.9764166666666,"pin is hidden from the screen — beam left the board."],[803.9764166666666,"name_a is hidden from the screen — beam left the board."],[803.9764166666666,"roller is hidden from the screen — beam left the board."],[803.9764166666666,"name_b is hidden from the screen — beam left the board."],[803.9764166666666,"load_1 is hidden from the screen — beam left the board."],[803.9764166666666,"load_1_name is hidden from the screen — beam left the board."],[803.9764166666666,"load_2 is hidden from the screen — beam left the board."],[803.9764166666666,"load_2_name is hidden from the screen — beam left the board."],[803.9764166666666,"couple is hidden from the screen — beam left the board."],[803.9764166666666,"couple_name is hidden from the screen — beam left the board."],[803.9764166666666,"r_b is hidden from the screen — beam left the board."],[803.9764166666666,"r_b_name is hidden from the screen — beam left the board."],[803.9764166666666,"r_a is hidden from the screen — beam left the board."],[803.9764166666666,"r_a_name is hidden from the screen — beam left the board."],[803.9764166666666,"head_beam is hidden from the screen — left the board."],[803.9764166666666,"reactions is hidden from the screen — left the board."]]},{"start":804.5764166666667,"say":"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.","live":[],"does":[[804.5764166666667,"head_build is shown on the screen, written out."],[804.5764166666667,"v_axes is shown on the screen, written out."],[815.6174166666667,"v_up_a is shown on the screen, drawn."],[819.5534166666666,"v_s1 is shown on the screen, drawn."]]},{"start":820.9424166666666,"say":"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.","live":["v_axes","head_build","v_up_a","v_s1"],"does":[[823.4734166666667,"v_drop is shown on the screen, drawn."],[827.2464166666666,"v_s2 is shown on the screen, drawn."]]},{"start":832.4904166666666,"say":"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.","live":["v_axes","head_build","v_up_a","v_s1","v_drop","v_s2"],"does":[[834.5224166666667,"couple_mark is shown on the screen, drawn."],[841.8484166666667,"v_s2 is indicated — a transient flash."]]},{"start":844.1664166666667,"say":"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.","live":["v_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark"],"does":[[845.5714166666667,"v_up_b is shown on the screen, drawn."],[850.3434166666667,"v_s3 is shown on the screen, drawn."],[854.4874166666667,"v_close is shown on the screen, drawn."]]},{"start":858.3389166666667,"say":"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.","live":["v_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark","v_up_b","v_s3","v_close"],"does":[[858.3389166666667,"v_axes moves to a new place on the board."],[858.3389166666667,"m_axes is shown on the screen, written out."],[869.8434166666667,"m_s1 is shown on the screen, drawn."]]},{"start":873.8104166666667,"say":"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.","live":["v_axes","m_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark","v_up_b","v_s3","v_close","m_s1"],"does":[[877.4554166666667,"m_s2 is shown on the screen, drawn."]]},{"start":884.2439166666667,"say":"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.","live":["v_axes","m_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark","v_up_b","v_s3","v_close","m_s1","m_s2"],"does":[[890.2924166666667,"m_jump is shown on the screen, drawn."],[892.0694166666667,"fact_1 is shown on the screen, written out."]]},{"start":895.7574166666667,"say":"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.","live":["fact_1","v_axes","m_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark","v_up_b","v_s3","v_close","m_s1","m_s2","m_jump"],"does":[[900.5754166666667,"m_s3 is shown on the screen, drawn."]]},{"start":906.0284166666667,"say":"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.","live":["fact_1","v_axes","m_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark","v_up_b","v_s3","v_close","m_s1","m_s2","m_jump","m_s3"],"does":[[909.2434166666667,"m_s4 is shown on the screen, drawn."]]},{"start":917.5989166666667,"say":"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.","live":["fact_1","v_axes","m_axes","head_build","v_up_a","v_s1","v_drop","v_s2","couple_mark","v_up_b","v_s3","v_close","m_s1","m_s2","m_jump","m_s3","m_s4"],"does":[]},{"start":931.4349166666667,"say":"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.","live":null,"does":[[937.2754166666667,"fact_2 is shown on the screen, written out."],[940.3984166666667,"fact_2 (the \"tension\" part) is emphasized."],[942.2674166666667,"fact_2 (the \"tension\" part) is no longer emphasized."],[943.9579166666667,"fact_1 is hidden from the screen — left the board."],[943.9579166666667,"fact_2 is hidden from the screen — left the board."],[943.9579166666667,"head_build is hidden from the screen — left the board."],[943.9579166666667,"m_axes is hidden from the screen — left the board."],[943.9579166666667,"m_s1 is hidden from the screen — m_axes left the board."],[943.9579166666667,"m_s2 is hidden from the screen — m_axes left the board."],[943.9579166666667,"m_jump is hidden from the screen — m_axes left the board."],[943.9579166666667,"m_s3 is hidden from the screen — m_axes left the board."],[943.9579166666667,"m_s4 is hidden from the screen — m_axes left the board."],[943.9579166666667,"v_axes is hidden from the screen — left the board."],[943.9579166666667,"v_up_a is hidden from the screen — v_axes left the board."],[943.9579166666667,"v_s1 is hidden from the screen — v_axes left the board."],[943.9579166666667,"v_drop is hidden from the screen — v_axes left the board."],[943.9579166666667,"v_s2 is hidden from the screen — v_axes left the board."],[943.9579166666667,"couple_mark is hidden from the screen — v_axes left the board."],[943.9579166666667,"v_up_b is hidden from the screen — v_axes left the board."],[943.9579166666667,"v_s3 is hidden from the screen — v_axes left the board."],[943.9579166666667,"v_close is hidden from the screen — v_axes left the board."]]}]},{"title":"Sketching a Beam You Have Not Seen","start":944.9995833333334,"end":1074.7540625000001,"objects":{"beam":"a Figure (x_range=(-1.0, 9.4), y_range=(-2.2, 2.7), aspect=(10.4, 4.9))","head":"a Heading that says \"Half Loaded, and Sketched by Eye\"","m_axes":"an Axes (x_range=(0.0, 8.0), y_range=(-2.0, 30.0), aspect=(9.0, 2.7))","m_curve":"a FunctionPlot [magenta] drawn in m_axes (function=<function>, x_range=(0.0, 4.0))","m_line":"a FunctionPlot [magenta] drawn in m_axes (function=<function>, x_range=(4.0, 8.0))","m_peak":"a Point [green] labelled \"27\" drawn in m_axes (location=(3.0, 27.0))","pin":"a Polygon [gray] drawn in beam (vertices=((0.0, 0.0), (-0.34, -0.66), (0.34, -0.66)), fill_opacity=0.35)","r_a":"a Vector [green] drawn in beam (start=(0.0, -1.7), end=(0.0, -0.76))","r_a_name":"a Math [green] that says \"$18$\" drawn in beam","r_b":"a Vector [green] drawn in beam (start=(8.0, -1.7), end=(8.0, -0.76))","r_b_name":"a Math [green] that says \"$6$\" drawn in beam","roller":"a Polygon [gray] drawn in beam (vertices=((8.0, 0.0), (7.66, -0.66), (8.34, -0.66)), fill_opacity=0.35)","span":"a Line [blue] drawn in beam (end=(8.0, 0.0))","steps":"a Block [text] that says \"Reactions, from equilibrium. Shear: start at zero, jump by each force, slope $-w$ between them. Moment: start at zero, add the area under the shear, jump at each couple. Maximum moment where the shear crosses the axis. Both diagrams must c…\"","udl_0":"a Vector [red] drawn in beam (start=(0.0, 1.4), end=(0.0, 0.08))","udl_1":"a Vector [red] drawn in beam (start=(1.0, 1.4), end=(1.0, 0.08))","udl_2":"a Vector [red] drawn in beam (start=(2.0, 1.4), end=(2.0, 0.08))","udl_3":"a Vector [red] drawn in beam (start=(3.0, 1.4), end=(3.0, 0.08))","udl_4":"a Vector [red] drawn in beam (start=(4.0, 1.4), end=(4.0, 0.08))","udl_name":"a Math [red] that says \"$w = 6 thin upright(\"kN/m\")$\" drawn in beam","udl_top":"a Line [red] drawn in beam (start=(0.0, 1.45), end=(4.0, 1.45))","v_axes":"an Axes (x_range=(0.0, 8.0), y_range=(-9.0, 22.0), aspect=(9.0, 2.7))","v_close":"a Line [yellow] drawn in v_axes (start=(8.0, -6.0), end=(8.0, 0.0))","v_flat":"a FunctionPlot [yellow] drawn in v_axes (function=<function>, x_range=(4.0, 8.0))","v_ramp":"a FunctionPlot [yellow] drawn in v_axes (function=<function>, x_range=(0.0, 4.0))","v_up":"a Line [yellow] drawn in v_axes (end=(0.0, 18.0))","v_zero":"a Point [green] labelled \"x = 3\" drawn in v_axes (location=(3.0, 0.0))"},"beats":[{"start":944.9995833333334,"say":"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.","live":[],"does":[[944.9995833333334,"head is shown on the screen, written out."],[944.9995833333334,"beam is shown on the screen, written out."],[948.1575833333334,"span is shown on the screen, written out."],[948.2575833333334,"pin is shown on the screen, written out."],[948.4575833333333,"roller is shown on the screen, written out."],[950.0615833333334,"udl_top is shown on the screen, written out."],[950.1615833333334,"udl_0 is shown on the screen, written out."],[950.3615833333333,"udl_1 is shown on the screen, written out."],[950.6615833333334,"udl_2 is shown on the screen, written out."],[951.0615833333334,"udl_3 is shown on the screen, written out."],[951.5615833333334,"udl_4 is shown on the screen, written out."],[952.3135833333333,"udl_name is shown on the screen, written out."]]},{"start":955.6070833333333,"say":"Here is the whole method, in five lines. Reactions, then jumps and slopes, then areas, then the maximum, then the check that it closes.","live":["beam","head","span","pin","roller","udl_top","udl_0","udl_1","udl_2","udl_3","udl_4","udl_name"],"does":[[957.5225833333334,"beam moves to a new place on the board."],[957.5225833333334,"steps is shown on the screen, written out."]]},{"start":966.8765833333333,"say":"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.","live":["steps","beam","head","span","pin","roller","udl_top","udl_0","udl_1","udl_2","udl_3","udl_4","udl_name"],"does":[[967.9215833333334,"steps (the \"Reactions\" part) is emphasized."],[977.8715833333333,"r_a is shown on the screen, written out."],[978.0715833333334,"r_a_name is shown on the screen, written out."],[979.8105833333334,"r_b is shown on the screen, written out."],[980.0105833333333,"r_b_name is shown on the screen, written out."],[983.9435833333333,"beam is hidden from the screen — left the board."],[983.9435833333333,"span is hidden from the screen — beam left the board."],[983.9435833333333,"pin is hidden from the screen — beam left the board."],[983.9435833333333,"roller is hidden from the screen — beam left the board."],[983.9435833333333,"udl_top is hidden from the screen — beam left the board."],[983.9435833333333,"udl_0 is hidden from the screen — beam left the board."],[983.9435833333333,"udl_1 is hidden from the screen — beam left the board."],[983.9435833333333,"udl_2 is hidden from the screen — beam left the board."],[983.9435833333333,"udl_3 is hidden from the screen — beam left the board."],[983.9435833333333,"udl_4 is hidden from the screen — beam left the board."],[983.9435833333333,"udl_name is hidden from the screen — beam left the board."],[983.9435833333333,"r_a is hidden from the screen — beam left the board."],[983.9435833333333,"r_a_name is hidden from the screen — beam left the board."],[983.9435833333333,"r_b is hidden from the screen — beam left the board."],[983.9435833333333,"r_b_name is hidden from the screen — beam left the board."]]},{"start":984.5435833333333,"say":"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.","live":["steps","head"],"does":[[984.5435833333333,"v_axes is shown on the screen, written out."],[985.9365833333334,"steps (the \"Reactions\" part) is no longer emphasized."],[985.9365833333334,"steps (the \"Shear\" part) is emphasized."],[988.7225833333333,"v_up is shown on the screen, drawn."],[992.3455833333334,"v_ramp is shown on the screen, drawn."]]},{"start":1000.2130833333333,"say":"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.","live":["steps","head","v_axes","v_up","v_ramp"],"does":[[1003.7885833333333,"v_flat is shown on the screen, drawn."],[1007.7945833333333,"v_close is shown on the screen, drawn."]]},{"start":1009.8685833333334,"say":"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.","live":["steps","head","v_axes","v_up","v_ramp","v_flat","v_close"],"does":[[1009.8685833333334,"m_axes is shown on the screen, written out."],[1011.2735833333334,"steps (the \"Moment\" part) is emphasized."],[1011.2735833333334,"steps (the \"Shear\" part) is no longer emphasized."],[1020.9905833333333,"m_curve is shown on the screen, drawn."]]},{"start":1025.3295833333334,"say":"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.","live":["steps","head","v_axes","m_axes","v_up","v_ramp","v_flat","v_close","m_curve"],"does":[[1026.7695833333335,"steps (the \"Maximum moment\" part) is emphasized."],[1026.7695833333335,"steps (the \"Moment\" part) is no longer emphasized."],[1027.7795833333334,"v_zero is shown on the screen, written out."],[1033.0275833333333,"m_peak is shown on the screen, written out."]]},{"start":1038.1670833333333,"say":"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.","live":["steps","head","v_axes","m_axes","v_up","v_ramp","v_flat","v_close","m_curve","v_zero","m_peak"],"does":[[1042.4855833333334,"m_line is shown on the screen, drawn."],[1046.8975833333334,"steps (the \"Maximum moment\" part) is no longer emphasized."],[1046.8975833333334,"steps (the \"close at the far end\" part) is emphasized."],[1048.1745833333334,"steps (the \"close at the far end\" part) is no longer emphasized."]]},{"start":1049.6100833333335,"say":"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.","live":["steps","head","v_axes","m_axes","v_up","v_ramp","v_flat","v_close","m_curve","v_zero","m_peak","m_line"],"does":[[1059.1545833333334,"m_peak is indicated — a transient flash."]]},{"start":1061.8905833333333,"say":"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.","live":null,"does":[[1073.7123958333334,"head is hidden from the screen — left the board."],[1073.7123958333334,"m_axes is hidden from the screen — left the board."],[1073.7123958333334,"m_curve is hidden from the screen — m_axes left the board."],[1073.7123958333334,"m_peak is hidden from the screen — m_axes left the board."],[1073.7123958333334,"m_line is hidden from the screen — m_axes left the board."],[1073.7123958333334,"steps is hidden from the screen — left the board."],[1073.7123958333334,"v_axes is hidden from the screen — left the board."],[1073.7123958333334,"v_up is hidden from the screen — v_axes left the board."],[1073.7123958333334,"v_ramp is hidden from the screen — v_axes left the board."],[1073.7123958333334,"v_flat is hidden from the screen — v_axes left the board."],[1073.7123958333334,"v_close is hidden from the screen — v_axes left the board."],[1073.7123958333334,"v_zero is hidden from the screen — v_axes left the board."]]}]}]},"durationSeconds":1075,"chapters":[{"title":"Cutting the Beam","startSeconds":0,"narration":"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."},{"title":"Drawing the Two Diagrams","startSeconds":249.64514583333334,"narration":"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."},{"title":"A Distributed Load","startSeconds":394.05110416666673,"narration":"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."},{"title":"Load, Shear and Moment","startSeconds":565.4929791666667,"narration":"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."},{"title":"An Overhang and a Couple","startSeconds":756.0034166666667,"narration":"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."},{"title":"Sketching a Beam You Have Not Seen","startSeconds":944.9995833333334,"narration":"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."}]}}