2 4 6 8 10 12 14
-200 512 -32 -120 160 -100 -40
312 -200 280 120
  1. First, find the reaction forces and double check that they’re correct.
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  2. Establish the shear graph with a horizontal axis below the beam and a vertical axis to represent shear. Positive shears will be plotted above the \(x\) axis and negative below.
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  3. Divide the beam into segments by adding vertical lines at all the “interesting points,”i.e. points.
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  4. Start with a dot at \((0,0)\) and proceed from left to right until you reach the end of the beam. On the first segment the distributed load causes the shear to slope down at a rate of \(\lbperft{25}\text{,}\) making the total drop over the \(\m{4}\) segment \(\N{100}\text{.}\)
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  5. The reaction force at \(x = 4\) causes a \(\N{260}\) upward jump there.
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  6. The \(\lbperft{25}\) distributed load continues after the jump, causing the shear curve to continue sloping down until it reaches \(\N{-40}\) at \(x=12\text{.}\)
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  7. The concentrated moment at this point has no effect on the shear diagram!
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    There is no load on the final segment, so the curve remains at \(\N{-40}\) until the end of the beam, at which point the reaction returns the curve to zero.
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  8. Add another vertical line where the shear diagram crosses the \(x\)-axis and locate the zero crossing.
    \begin{equation*} \dfrac{\N{-25}}{\m{1}} = \dfrac{\N{-160}}{d} \end{equation*}
    \begin{equation*} d= \m{6.4} \qquad d' = (8 - d) = \m{1.6} \end{equation*}
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  9. Divide the area into rectangles and triangles between the vertical lines.
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  10. Calculate the areas.
    \begin{align*} A_1 \amp = \dfrac{ (\m{4})(\N{-100})}{2} = \Nm{-200}\\ A_2 \amp = \dfrac{ (\m{6.4}) (\N{160})}{2} = \Nm{512}\\ A_3 \amp = \dfrac{ (\m{1.6})(\N{-40})}{2} = \Nm{-32}\\ A_4 \amp = (\m{3}) (\N{40}) = \Nm{120}\\ \Sigma A_i \amp = \Nm{160} \end{align*}
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  11. Did you notice that the sum of the four areas didn’t add up to zero!
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    The counterclockwise \(\Nm{160}\) concentrated moment makes up the difference required for equilibrium.
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  12. Establish the moment graph beneath the shear diagram with a vertical axis to represent moment. Positive moments will be plotted above the \(x\) axis and negative below.
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  13. Draw and label dots on the moment diagram starting with a dot at \((0,0)\) then proceed from left to right placing dots until you reach the end of the beam.
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    As you move across each segment move up or down by the “area” of that segment and place a dot. The concentrated moment causes a \(\Nm{160}\) downward jump on the moment diagram.
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    When you reach the end of the beam you should return to \(M=0\text{.}\) If you didn’t, then recheck your work!
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  14. Connect the dots with correctly shaped curves.
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    Segments under uniformly distributed loads are parabolas. The shape can be determined from
    \begin{equation*} \dfrac{dM}{dx} = V \end{equation*}
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    Over segment one, the shear changes from \(\N{0}\) to \(\N{-100}\text{,}\) so the curve starts with a horizontal tangent, then curves down. At \(x=4\) the shear suddenly changes from \(\N{-100}\) to \(\N{+160}\) which causes a kink in the shear curve as it abruptly changes from negative to positive slope. The curve peaks at the zero crossing, and then turns down due to \(A_3\text{.}\)
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  15. The concentrated moment causes a downward jump at \(x=12\text{,}\) and the final segment is linear since the shear above it is constant.
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  16. This is the final result.
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