# Question 7.2: Plot shear and moment diagrams for the beams shown: Given: L......

Plot shear and moment diagrams for the beams shown:

Assume: The only assumptions necessary are implicitly made throughout: equilibrium, and Saint-Venant’s principle.

Step-by-Step
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In each case, we will first find the external reaction forces and/or moments, then use the method of sections at points of interest along the beam to construct the diagrams of V(x) and M(x).
Starting with the first beam, to find reactions, we need an FBD of the whole beam:

\begin{aligned} \curvearrowleft & \sum M_A=0=\frac{P L}{3}-\frac{2 P L}{3}+R_B L \rightarrow R_B=\frac{P}{3} \text { (up), } \\ \curvearrowleft & \sum M_B=0=\frac{P L}{3}-\frac{2 P L}{3}-R_A L \rightarrow R_A=-\frac{P}{3} \text { (down). } \end{aligned}

To construct V diagram, look only at the points where the loading conditions change.

• At the left-hand end of the beam, V must balance $R_A$, so V = P/3.

• At the right-hand end, V must balance $R_B$, so V = P/3.

• Just to the right of the upward applied load P, V = −2P/3

To construct M diagram:

• M = 0 at simply supported ends
• M = 0 at center by symmetry
• At upward P, M = $=\frac{P}{3}\left(\frac{L}{3}\right)=\frac{P L}{9}$
• At downward P, M = $\frac{P}{3}\left(\frac{2 L}{3}\right)-P\left(\frac{L}{3}\right)=-\frac{P L}{9}$

Plot the results.

For the second beam, once again, we start with the external reactions, using the equivalent concentrated load in place of the distributed one:

Since the distributed load is linearly distributed, the shear distribution is parabolic, and the moment distribution is cubic. We may proceed either by integrating the distributed load q(x) = kx once for V(x) and twice for M(x), or by making our imaginary cut at some distance x from the end of the beam, and finding the internal shear and moment. Both methods will provide the same results.
Integration:

\begin{aligned} & V(x)=-\int q \mathrm{~d} x=-\int-k x \mathrm{~d} x=\frac{1}{2} k x^2+C_1, \\ & V(0)=-\frac{1}{2} k L^2=C_1, \\ & V(x)=\frac{1}{2} k x^2-\frac{1}{2} k L^2, \end{aligned}
\begin{aligned} & M(x)=\int V \mathrm{~d} x=\frac{1}{6} k x^3-\frac{1}{2} k L^2 x+C_2 ,\\ & M(0)=\frac{1}{3} k L^3=C_2 ,\\ & M(x)=\frac{1}{6} k x^3-\frac{1}{2} k L^2 x+\frac{1}{3} k L^3 . \end{aligned}

Method of sections:

$\sum F_z=0=\frac{1}{2} k L^2-\frac{1}{2} k x^2+V(x) ,$
so $V(x)=\frac{1}{2} k x^2-\frac{1}{2} k L^2 ,$
$\curvearrowleft \sum M_x=0=M(x)+\frac{1}{3} k L^3-\frac{1}{2} k L^2 x+\frac{1}{2} k x^2\left(\frac{x}{3}\right) .$

(We note that the internal shear V(x) does not cause a moment about the cut at x.)

so $M(x)=\frac{1}{6} k x^3-\frac{1}{2} k L^2 x+\frac{1}{3} k L^3$

The resulting shear and moment diagrams are as shown below. We can check our results by evaluating V(x) and M(x) at x = L. Both are zero there, as they must be at a free end.

We now consider the third beam. The fixed support can offer both reaction forces and a moment, which are found using an FBD:

The shear V is zero at the free end of this cantilever beam, and must balance the upward reaction force $q_0$L at the fixed end. Since the distributed load is uniformly distributed, the shear distribution is linear.

The moment M is zero at the free end and must balance the reaction moment $q_0L^2$/2 at the fixed end. Since the shear distribution is linear (proportional to x), the bending moment distribution is parabolic (proportional to x²). The shear and moment diagrams are shown:

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