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Engineering Statics Open and Interactive

Section 8.6 Section Cut Method

In this section we’ll extend the method of Section 8.3 where we found the internal forces at a specific point to to find the internal forces at every point needed to produce s shear and bending moment diagram.
The procedure is identical except that the cut is taken at a position designated by \(x\) instead of at a specified point. The analysis produces equations as functions of \(x\) rather than values at a particular \(x\text{.}\) Shear and bending moment diagrams are plots of these equations, and the internal forces at any particular point can be found by substituting that particular value of \(x\) into the equations.
We will use a cantilevered beam fixed to a wall on the left end and subjected to a vertical force \(P\) on the right end as an example. Global equilibrium requires that the reactions at the fixed support at \(A\) are a vertical force \(A_y = P\text{,}\) and a counter-clockwise moment \(M_A = P L\text{.}\)
Cantilevered beam of length L fixed to a wall at A on its left end and subject to a vertical force P on its right end.
By taking a cut at a distance \(x\) from the left we can draw two free-body diagrams with lengths \(x\) and \((L-x)\text{.}\) This beam has one loading segment, because no matter where \(x\) is chosen, the free-body diagrams shown in Figure 8.6.1 (b) and (c) are correct. The internal loadings are named \(V(x)\) and \(M(x)\) to indicate that they are functions of \(x\) and are drawn with positive sense as defined in Section 8.2.
FBD of the left side of the beam.
(a) Left FBD.
FBD of the right side of the beam.
(b) Right FBD.
Figure 8.6.1.
To find the shear and bending moment functions, we apply the equilibrium to one of the free-body diagrams. Either side will work, so we’ll select the right-hand portion Figure 8.6.1.(b) since it doesn’t require us to find the reactions at \(A\text{.}\) Letting \(L\) be the length of the beam and \((L-x)\) the length of the right portion, we find
\begin{align*} \Sigma F_y \amp = 0 \amp \Sigma M_\text{cut} \amp = 0\\ V(x) \amp = P \amp M(x) \amp = - P(L-x) \end{align*}
The plots of the equations for \(V(x)\) and \(M(x)\) are shown below in Figure 8.6.2. These equations indicate that the shear force \(V(x)\) is constant \(P\) over the length of the beam and the moment \(M(x)\) is a linear function of the position of the cut, \(x\) starting at \(-PL\) at \(x=0\) and linearly increasing to zero at \(x=L\text{.}\) Note that the graphs are only valid from \(0 \le x \le L\text{,}\) so the curves outside this range are shown as dotted lines. These two graphs are usually drawn stacked beneath the load diagram of the beam.
Load, Shear and Moment diagrams for the previous FBD. Shear digram rises to P at x=0, then stays there for the length of the beam. Moment diagram starts at -PL and linearly increases to zero over the length of the beam.
Figure 8.6.2. Shear and Bending Moment Diagrams for Cantilever Beam
The previous example was simple because only one FBD was necessary for any point on the beam, but many beams are more complex. Beams with multiple loads must be divided into loading segments between the points where loads are applied or where distributed loads begin or end.
Consider the simply supported beam \(AD\) with a uniformly distributed load \(w\) over the first segment from \(A\) to \(B\text{,}\) and two vertical loads \(B\) and \(C\) .
Simply supported beam AD of length 3 units with a uniformly distributed load w from A to B (1 unit length). Two vertical loads are at B and at C (1 unit to the right of B).
Figure 8.6.3. Simply Suppported Beam with Uniformly Distributed Load
This beam has three loading segments so you must draw three free-body diagrams and analyze each segment independently. For each, make an imaginary cut through the segment, then draw a new free-body diagram of the portion to the left (or right) of the cut. Always assume that the exposed internal shear force and internal bending moment act in the positive direction according to the sign convention.
FBD of the simply supported beam to determine V(x) and M(x) for 0 less than x less than 1.
(a) \(0 \lt x \lt d\)
FBD of the simply supported beam to determine V(x) and M(x) for 1 less than x less than 2.
(b) \(d \lt x \lt 2d\)
FBD of the simply supported beam to determine V(x) and M(x) for 2 less than x less than 3.
(c) \(2d \lt x \lt 3d\)
Figure 8.6.4.
After the equilibrium equations are applied to each segment, the resulting equations \(V(x)\) and \(M(x)\) from each segment are joined to plot the shear and moment diagrams. These diagrams help us visualize the values of \(V\) and \(M\) throughout the beam.

Example 8.6.5. Section Cut Method.

Draw the shear and bending moment diagrams for the beam of Figure 8.6.3 knowing that \(d\) is \(\m{4}\text{,}\) forces \(B\) and \(C\) are both \(\N{10}\text{,}\) and the uniformly distributed load \(w(x) = \Nperm{15}\)
Answer.
Solution.
  1. Replace the \(\Nperm{15}\) with an equivalent force of \(\N{60}\) acting at the centroid of the the rectangular loading, and draw a free body diagram of the whole beam.
  2. Take moments about \(B\) to find the reaction at \(A\text{.}\)
    \begin{equation*} A = \dfrac{60(10) + 10(8) + 10(4)}{12} \end{equation*}
  3. Draw a free-body diagram of the section of the beam between \(0 < x < \m{4}\text{.}\)
    The loading on this segment of the beam is \(w(x) = \Nperm{15}\text{,}\) acting down and constant. The equivalent force is the area of the rectangle \(W = 15 x\text{,}\) which is a first degree function of \(x\text{.}\)
  4. Apply the equations of equlibrium to find the shear and moment functions for the segment \(0 \lt x \lt 4\text{.}\)
    \begin{align} \Sigma F_y \amp = 0\notag\\ 60 -15 x - V \amp = 0\notag\\ V \amp= 60 - 15 x \N{}\tag{8.6.1} \end{align}
    \begin{align} \Sigma M_\textrm{cut} \amp = 0\notag\\ -60 x +\left(15x\right) \left(\frac{x}{2}\right) + M \amp= 0\notag\\ M \amp= 60 x -15 x^2 \Nm{}\tag{8.6.2} \end{align}
  5. Follow a similar procedure for the next segment \(4 \lt x \lt 8\)
    \begin{align} \Sigma F_y \amp= 0: \amp V \amp= \N{-10}\tag{8.6.3}\\ \Sigma M_\textrm{cut}\amp= 0:\amp M \amp= 160 - 10 x \Nm{}\tag{8.6.4} \end{align}
  6. And again for the third segment \(8 \lt x \lt 12\)
    \begin{align} \Sigma F_y \amp= 0: \amp V \amp= -\N{20}\tag{8.6.5}\\ \Sigma M_\textrm{cut}\amp= 0: \amp M \amp= 240 - 20 x \Nm{}\tag{8.6.6} \end{align}
  7. Neatly plot these functions on appropriately scaled graph paper
Note that this method does not give us shear and moment values at exactly \(x = \m{4}\) and \(x = \m{8}\text{.}\) Discontinuities occur at these locations, and the shear function has different values when approaching 4 and 8 from the left and from the right!