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2
4
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16
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24
65
35
-65
20
-211. 25
-260
61.25
80
70
260
3.5'
-10
1
First, determine the reaction forces and moments by drawing a free-body diagram of the entire beam and applying the equilibrium equations. Double check that they are are correct.
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.
Make vertical lines at all the “interesting points,”i.e. points where concentrated forces or moments act on the beam and at the beginning and end of any distributed loads. This divides the beam into segments between vertical lines.
Start the shear diagram with a dot at
\((0,0)\text{.}\)
Draw the shear diagram by proceding from left to right until you reach the end of the beam. Whenever you encounter a concentrated force, jump up or down by that value.
Draw a
\(\lb{65}\) upward jump at
\(x=0\) due to
\(R_A\text{.}\)
Since there is no load acting on the segment from \(x=0\) to \(x=4\text{,}\) the slope of this segment is
\begin{equation*}
\dfrac{dV}{dx} = w = 0
\end{equation*}
Place a dot at \((4,65)\) and draw a horizontal segment.
The
\(\lb{30}\) force
\(x=4\) causes a downward jump to
\(\lb{35}\text{,}\) and there is no load above this segment so place a dot at
\((4,35)\) and draw a horizontal segment to
\((6, 35)\text{.}\)
Whenever you encounter a distributed load, move up or down by the “area” of the distributed load above \(\Delta V\text{.}\)
\begin{equation*}
\Delta V = \int_a^b w(x)\,dx
\end{equation*}
\begin{equation*}
\Delta V = \lbperft{-10}\times\ft{10} = \lb{-100}
\end{equation*}
So go down
\(\lb{100}\) as you move from
\(x=6\) to 16 and place a dot at
\((16,-65)\text{.}\)
The slope of the curve at each point \(x\) is given by
\begin{equation*}
\dfrac{dV}{dx} = w(x)
\end{equation*}
Underneath uniformly distributed loads, the slope is constant, so connect these points with a straight line with slope, \(m = \lbperft{-10}\text{.}\)
Over the next segment, the load is zero, so draw a horizontal segment to
\((20,-65)\)
The
\(\lb{85}\) reaction force causes an upward jump, so draw a jump to
\((20,20)\text{.}\)
The load on the final segment is zero, so draw a horizontal segment to
\((24,20)\text{.}\)
The
\(\lb{-20}\) force at
\(D\) causes a downward jump, so draw a jump to
\((24,0)\text{.}\)
Shear diagram should always end at
\(V =0\text{.}\) If yours doesn’t, recheck your work!
Add another interesting point at any points where the shear diagram crosses the
\(x\) -axis. We will need to find this point.
Determine the
\(x\) position of the zero crossing.
Since we know the slope of this segment, use similar triangles to find the distance \(d\text{.}\)
\begin{equation*}
\dfrac{\lb{-10}}{\ft{1}} = \dfrac{\lb{-35}}{d}
\end{equation*}
\begin{equation*}
d= \ft{3.5} \qquad d' = (10 - d) = \ft{6.5}
\end{equation*}
After you have completed the shear diagram, divide the area under the curve into rectangles and triangles between the vertical lines.
These areas are in fact, moments.
Areas above the axis are positive moments, areas below the axis are negative moments.
Calculate the areas. Notice that the areas have moment units.
\begin{align*}
A_1 \amp = \ft{4} \times \lb{65} = \ftlb{260} \amp A_4 \amp = \dfrac{ \ft{6.5} \times \lb{-65}}{2} = \ftlb{-211.25}\\
A_2 \amp = \ft{2} \times \lb{35} = \ftlb{70}\amp A_5 \amp = \ft{4} \times \lb{-65} = \ftlb{-260}\\
A_3 \amp = \dfrac{ \ft{3.5} \times \lb{35}}{2} = \ftlb{61.25}\amp A_6 \amp = \ft{4} \times \lb{20} = \ftlb{80}
\end{align*}
As a reflection of \(\Sigma M = 0\text{,}\) the positive and negative areas must add up to zero. If they don’t, recheck your work!
\begin{equation*}
\Sigma A_i = 260 + 70 + 61.25 - 211.25 - 260 + 80 = 0~~ \checkmark
\end{equation*}
If the load includes any concentrated moments, they must also be included in the tally.
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.
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. As you move across each segment move up or down by the “area” of that segment and place a dot. If you pass a concentrated moments, jump by that amount.
When you reach the end of the beam you should return to
\(M=0\text{.}\) If you didn’t, then recheck your work!
Connect the dots with correctly shaped lines. Segments under constant shear are straight lines.
Segments under uniformly distributed loads are parabolas. The shape can be determined from
\begin{equation*}
\dfrac{dM}{dx} = V
\end{equation*}
Between
\(x=6\) and
\(x=16\text{,}\) the shear changes from
\(\lb{+35}\) through 0 to
\(\lb{-65}\text{,}\) so the curve starts with a positive slope, reaches a maximum where the shear crosses the
\(x\) -axis, and then turns down to end with negative slope.
This is the final result.