Question 3.2A: Shear and Moment Diagrams of a Simply Supported Beam Using a...

See Figures 3-22a and 3-24

1    Solve for the reaction forces using equations 3.3 (p. 78). Summing moments about the right end and summing forces in the y direction gives

\begin{array}{lll} \sum F_x=0 & \sum F_y=0 & \sum F_z=0 \\ \sum M_x=0 & \sum M_y=0 & \sum M_z=0 \end{array}       (3.3a)

\sum F_x=0 \quad \sum F_y=0 \quad \sum M_z=0      (3.3b)

\begin{aligned} \sum M_z &=0=R_1 l-\frac{w(l-a)^2}{2} \\ R_1 &=\frac{w(l-a)^2}{2 l}=\frac{10(10-4)^2}{2(10)}=18 \end{aligned}      (a)

\begin{aligned} \sum F_y &=0=R_1-w(l-a)+R_2 \\ R_2 &=w(l-a)-R_1=10(10-4)-18=42 \end{aligned}      (b)

2    The shape of the shear diagram can be sketched by graphically integrating the loading diagram shown in Figure 3-24a. As a “device” to visualize this graphical integration process, imagine that you walk backward across the loading diagram of the beam, starting from the left end and taking small steps of length dx. You will record on the shear diagram (Figure 3-24b) the area (force · dx) of the loading diagram that you can see as you take each step. As you take the first step backward from x = 0, the shear diagram rises immediately to the value of R_{1}. As you walk from x = 0 to x = a, no change occurs, since you see no additional forces. As you step beyond x = a, you begin to see strips of area equal to –w · dx, which subtract from the value of R_{1} on the shear diagram. When you reach x = l, the total area w · (l – a) will have taken the value of the shear diagram to –R_{2}. As you step backward off the beam’s loading diagram (Figure 3-24a) and plummet downward, you can now see the reaction force R_{2} which closes the shear diagram to zero. The largest value of the shear force in this case is then R_{2} at x = l.

3    If your reflexes are quick enough, you should try to catch the shear diagram (Figure 3-24b) as you fall, climb onto it, and repeat this backward-walking trick across it to create the moment diagram which is the integral of the shear diagram. Note in Figure 3-24c that from x = 0 to x = a this moment function is a straight line with slope = R_{1}. Beyond point a, the shear diagram is triangular, and so integrates to a parabola. The peak moment will occur where the shear diagram crosses zero (i.e., zero slope on the moment diagram). The value of x at V = 0 can be found with a little trigonometry, noting that the slope of the triangle is –w:

x_{@ V=0}=a+\frac{R_1}{w}=4+\frac{18}{10}=5.8     (c)

Positive shear area adds to the moment value and negative area subtracts. So the value of the peak moment can be found by adding the areas of the rectangular and triangular portions of the shear diagram from x = 0 to the point of zero shear at x = 5.8:

M_{@ x=5.8}=R_1(a)+R_1 \frac{1.8}{2}=18(4)+18 \frac{1.8}{2}=88.2      (d)