{"id":192,"date":"2021-03-01T09:01:51","date_gmt":"2021-03-01T14:01:51","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&p=192"},"modified":"2025-08-01T17:07:53","modified_gmt":"2025-08-01T21:07:53","slug":"shear-moment-diagrams","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/shear-moment-diagrams\/","title":{"raw":"6.2 Shear\/Moment Diagrams","rendered":"6.2 Shear\/Moment Diagrams"},"content":{"raw":"

6.2.1 What are Shear\/Moment Diagrams?<\/h1>\r\nShear\/Moment diagrams are graphical representations of the internal shear force and bending moment along the whole beam.\r\n\r\n \r\n\r\n[caption id=\"attachment_554\" align=\"aligncenter\" width=\"569\"]\"Simply<\/a> Source (image): By XFEM Skier - Own work, CC BY-SA 3.0, https:\/\/commons.wikimedia.org\/w\/index.php?curid=29178249[\/caption]\r\n\r\n
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Shearing Force Diagram<\/span><\/h2>\r\n

This is a graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame. As a convention, the shearing force diagram can be drawn above or below the\u00a0<\/span>x<\/i>-centroidal axis of the structure, but it must be indicated if it is a positive or negative shear force.<\/p>\r\n\r\n

Bending Moment Diagram<\/span><\/h3>\r\n

This is a graphical representation of the variation of the bending moment on a segment or the entire length of a beam or frame. As a convention, the positive bending moments are drawn above the\u00a0<\/span>x<\/i>-centroidal axis of the structure, while the negative bending moments are drawn below the axis.<\/p>\r\nBelow is a simple example of what shear and moment diagrams look like. Afterwards, the relation between the load on the beam and the diagrams will be discussed.\r\n\r\nSource: Internal Forces in Beams and Frames, LibreTexts. https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a>\r\n\r\n<\/div>\r\n \r\n\r\n \r\n

6.2.2 Distributed Loads & Shear\/Moment Diagrams<\/h1>\r\n

There is a relationship between distributed loads and shear\/moment diagrams. Simply put:<\/p>\r\n

[latex]\\frac{dM}{dx}=V(x)[\/latex]<\/p>\r\n

[latex]\\frac{dV}{dx}=-w(x)[\/latex]<\/p>\r\n

[latex]\\frac{d^2M}{dx^2}=-w(x)[\/latex]<\/p>\r\nOr:\r\n

$latex \\Delta M=\\int V(x)dx $<\/p>\r\n

$latex \\Delta V=\\int w(x)dx$<\/p>\r\nSo, if there is a constant distributed load, then the slope of shear will be linear, and the slope of the moment will be parabolic. If the distributed load is 0, then the shear will be constant and the slope of the moment will be linear (as shown in Example 1 in the next section).\r\n

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For the derivation of the relations among\u00a0<\/span>w, V<\/i>, and\u00a0<\/span>M<\/i>, consider a simply supported beam subjected to a uniformly distributed load throughout its length, as shown in the figure below<\/span>. Let the shear force and bending moment at a section located at a distance of\u00a0<\/span>x<\/i>\u00a0<\/span>from the left support be\u00a0<\/span>V<\/i>\u00a0<\/span>and\u00a0<\/span>M<\/i>, respectively, and at a section\u00a0<\/span>x<\/i>\u00a0<\/span>+\u00a0<\/span>dx<\/i>\u00a0<\/span>be\u00a0<\/span>V<\/i>\u00a0<\/span>+\u00a0<\/span>dV<\/i>\u00a0<\/span>and\u00a0<\/span>M<\/i>\u00a0<\/span>+\u00a0<\/span>dM<\/i>, respectively. The total load acting through the center of the infinitesimal length is\u00a0<\/span>wdx<\/i>.<\/p>\r\n\"Beam\r\n\r\nTo compute the bending moment at section x + dx, use the following:\r\n\r\n[latex]\r\n\\begin{aligned}\r\nM_{x+dx} &= M + Vdx - wdx \\cdot \\frac{dx}{2} \\\\\r\n&= M + Vdx \\quad \\text{(neglecting the small second order term } \\frac{wdx^2}{2} \\text{)}\r\n\\end{aligned}\r\n[\/latex]\r\n\r\nor\r\n

$latex \\frac{dM}{dx}=V(x)$\u00a0 (Equation 6.1)<\/p>\r\nEquation 6.1 implies that the first derivative of the bending moment with respect to the distance is equal to the shearing force. The equation also suggests that the slope of the moment diagram at a particular point is equal to the shear force at that same point.\u00a0<\/span>Equation 6.1<\/a>\u00a0suggests the following expression:<\/span>\r\n

$latex \\Delta M=\\int V(x)dx $ (Equation 6.2)<\/p>\r\nEquation 6.2 states that the change in moment equals the area under the shear diagram. Similarly, the shearing force at section\u00a0<\/span>x<\/i>\u00a0+\u00a0<\/span>dx<\/i>\u00a0is as follows:<\/span>\r\n

$latex V_{x+dx}=V-wdx\\\\V+dV=V-wdx $<\/p>\r\nor\r\n

$latex \\frac{dV}{dx}=-w(x) $\u00a0 (Equation 6.3)<\/p>\r\n

Equation 6.3 implies that the first derivative of the shearing force with respect to the distance is equal to the intensity of the distributed load.\u00a0<\/span>Equation 6.3<\/a>\u00a0<\/span>suggests the following expression:<\/p>\r\n

$latex \\Delta V=\\int w(x)dx$ (Equation 6.4)<\/p>\r\nEquation 6.4 states that the change in the shear force is equal to the area under the load diagram.\u00a0<\/span>Equations 6.1<\/a>\u00a0and 6<\/span>.3<\/a>\u00a0suggest the following:<\/span>\r\n

$latex \\frac{d^2M}{dx^2}=-w(x)$ (Equation 6.5)<\/p>\r\n

Equation 6.5 <\/span>implies that the second derivative of the bending moment with respect to the distance is equal to the intensity of the distributed load.<\/p>\r\nSource: Internal Forces in Beams and Frames, LibreTexts. https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a>\r\n\r\n<\/div>\r\n \r\n\r\n \r\n

6.2.3 Producing a Shear\/Moment Diagram<\/h1>\r\nThere are many methods you can use to solve a shear\/moment diagram. First, you can find the equation for each portion and integrate using the above equations.\r\n\r\nSecond, you could use the method shown in the previous section to calculate the internal forces at important points (where loads are applied, the start and end of distributed loads, at reaction points). Plot these points on the V and M plots at the x locations, then connect the dots using the appropriate shape slope (more on this at the bottom of this page).\r\n\r\nThird, you can find the equations by using the equilibrium equations (so there's no integration\/differentiation).\r\n
    \r\n \t
  1. \r\n\r\n[caption id=\"attachment_1501\" align=\"alignright\" width=\"334\"]\"Internal Adapted from original source https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames[\/caption]\r\n\r\nDraw a FBD of the structure<\/li>\r\n \t
  2. Calculate the reactions using the equilibrium equations (may not need to do this if choosing a cantilever beam and using the free side for the FBD).<\/li>\r\n \t
  3. Make a cut and add internal forces N V and M using the positive sign convention. Depending on the number of loads, you may need multiple cuts. Recall the positive convention:<\/li>\r\n \t
  4. For shear, find an equation (expression) of the shear that is x distance from the origin (often the reaction) for each cut.<\/li>\r\n \t
  5. For moment, find an equation (expression) of the shear that is x distance from the origin (often the reaction) for each cut.<\/li>\r\n \t
  6. Plot these equations on a plot on top of each other.<\/li>\r\n<\/ol>\r\n \r\n\r\nThe rest of this section will use this method.\r\n

    Example 1<\/h2>\r\n(adapted from https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a>)<\/em>\r\n

    Draw the shear force and bending moment diagrams for the cantilever beam supporting a concentrated load of 5 lb at the free end 3 ft from the wall.<\/p>\r\n\"Cantilever\r\n\r\n1. Draw a FBD of the structure<\/em>\r\n\r\n\"A\r\n\r\n \r\n\r\n2. Calculate the reactions using the equilibrium equations (may not need to do this if choosing a cantilever beam and using the free side for the FBD).<\/em>\r\n

    First, compute the reactions at the support. Since the support at B<\/i> is fixed, there will be three reactions at that support, namely By<\/sub><\/i>, Bx<\/sub><\/i>, and MB<\/sub><\/i>. Applying the conditions of equilibrium suggests the following:<\/p>\r\n$latex \\sum F_{x}=0: \\quad \\underline{B_{x}=0}$\r\n\r\n$latex \\sum F_{y}=0: \\quad-5 lb+B_{y}=0$\r\n$latex \\qquad \\quad \\underline{B_{y}=5 lb}$\r\n\r\n$latex \\sum M_{B}=0: \\quad(5 lb )(3 \\mathrm{ft})-M=0$\r\n$latex \\qquad \\quad \\underline{M=15 ft \\cdot lb}$\r\n

    \u00a0<\/span><\/p>\r\n3. Make a cut and add internal forces N V and M using the positive sign convention. Depending on the number of loads, you may need multiple cuts<\/em>\r\n\r\nOnly 1 cut needed because only 1 load is added at the end. (If it were in the middle there would be 2 sections to consider). The value x could be 0 to 3 ft.\r\n\r\n\"Diagram\r\n\r\n4. For shear, find an equation (expression) of the shear that is x distance from the origin (often the reaction) for each cut.<\/em>\r\n

    x<\/i> is the distance from the free end of the cantilever beam to the cut. The shearing force at that section is due to the applied load. Using the equilibrium equations,<\/p>\r\n$latex \\sum F_y = -5 lb - V = 0 \\\\ \\qquad \\quad \\underline{V = - 5 lb} \\text{\u00a0 (- indicates V acts in opposite direction)}$\r\n\r\nThe constant number for shear means that it doesn't change or vary by x. (If there were a distributed load, x would be part of the equation).\r\n

    <\/span>The negative sign indicates the shear actually goes the opposite direction. (This is due to the fact that the sign convention for a shearing force states that a downward transverse force on the left of the section under consideration will cause a negative shearing force on that section.)<\/p>\r\n\"Cantilever\r\n\r\n5. For the moment, find an equation (expression) of the shear that is x distance from the origin (often the reaction) for each cut.<\/em>\r\n\r\nHere, x is measured from the left. Using sum of the moments equations, find an expression for M. You could choose to sum the moments about the end point where the load is applied, or you could do it at the moving point x. Both take the same effort for this problem, so let's choose the left hand side where the 5 lb are being applied.\r\n\r\n$latex \\sum M_L = -Vx - M = 0 $\r\n\r\n$latex \\qquad \\quad M = + Vx = (-5 lb) * x $\r\n\r\n$latex \\qquad \\quad \\underline{M = -(5lb)x } \\text{\u00a0 (the negative sign indicates the arrow goes the other direction.}$\r\n\r\n\"Cantilever\r\n

    The obtained expression is valid for the entire beam (the region 0 < x<\/i> < 3 ft). The negative sign indicates a negative moment, which was established from the sign convention for the moment, so the moment actually goes in the opposite direction. The moment due to the 5 lb force tends to cause the segment of the beam on the left side of the section to exhibit a downward concavity, and that corresponds to a negative bending moment, according to the sign convention for bending moment.<\/p>\r\n \r\n\r\n6. Plot these equations on a plot on top of each other.<\/em>\r\n

    Note that because the shearing force is a constant, it must be of the same magnitude at any point along the beam. As a convention, the shearing force diagram is plotted above or below a line corresponding to the neutral axis of the beam, but a plus sign must be indicated if it is a positive shearing force, and a minus sign should be indicated if it is a negative shearing force. A way to check the answer is to ensure the reaction force brings the problem back to 0. The shear is -5 until the last moment when the reaction force of +5lb brings the force to 0.<\/p>\r\n

    Since the function for the bending moment is linear, the bending moment diagram is a straight line. Thus, it is enough to use the two principal values of bending moments determined at x<\/i> = 0 ft and at x<\/i> = 3 ft to plot the bending moment diagram. As a convention, negative bending moment diagrams are plotted below the neutral axis of the beam, while positive bending moment diagrams are plotted above the axis of the beam.<\/p>\r\n\"Shear\r\n\r\nNotice the units are included in the axes.<\/em>\r\n\r\n \r\n\r\nHere is a second explanation for how to create shear\/moment diagrams:\r\n

    \r\n\r\nShear Diagram<\/em>\r\n\r\nTo create the shear force diagram, we will use the following process.\r\n
      \r\n \t
    1. Solve for all external forces<\/strong> acting on the body.<\/li>\r\n \t
    2. Draw out a free-body diagram<\/strong> of the body horizontally<\/strong>. Leave all distributed forces as distributed forces and do not replace them with the equivalent point load.<\/li>\r\n \t
    3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal shear force.<\/li>\r\n \t
    4. Starting at zero at the right side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except...\r\n