## Q. 7.1

Find the reactions and determine the axial force P, the shear V, and the bending moment M caused by the applied loads at the specified sections. Also, draw FBDs indicating the sense (direction) of all forces and moments.

Find: Internal forces and bending moment.
Assume: The only assumptions necessary are implicitly made throughout this textbook: Equilibrium, and Saint-Venant’s principle.

## Verified Solution

Our strategy is to find the reactions at the supports from the whole beam’s equilibrium, and then use the method of sections to find the internal forces and moments at the specified locations.

For the purpose of finding reaction forces, we can replace the distributed load by its equivalent concentrated load. The magnitude of this concentrated load is simply the area under the distributed load, in this case W = $\frac{1}{2}$(8 k/ft)(6 ft) = 24 kips. It acts at the centroid of the triangular area under the distributed load: one-third of the way from its maximum intensity, or 2 ft from the left end of the beam. We use this load in our free-body diagram:

\begin{aligned} \curvearrowleft & \sum M_B=0=-R_A(8 \mathrm{ft})+(24 \mathrm{k})(6 \mathrm{ft}) \rightarrow R_A=18 \mathrm{kips} ,\\ \curvearrowleft & \sum M_A=0=-(24 \mathrm{k})(2 \mathrm{ft})+R_B(8 \mathrm{ft}) \rightarrow R_B=6 \mathrm{kips}, \end{aligned}
$\sum F_z=0$ is then used as a check: $R_A + R_B = 24$ kips.

Next, we consider sections a–a and b–b. We make an imaginary cut at the specified location, and realize that considering the loading to the left, or to the right, of the a–a cut will yield equivalent results:

We choose the simpler side to calculate, in this case the portion of the beam to the right of a–a. We simply apply the equilibrium equations to this section of the beam:

\begin{aligned} & \sum F_x=0=N_a ,\\ & \sum F_z=0=-V_a+R_B-6 \mathrm{k} \rightarrow V_a=0 \mathrm{kips}, \end{aligned}
$\curvearrowleft \sum M_{\mathrm{about} a}=0=M_a-(6 \mathrm{k})(1 \mathrm{ft})+R_B(5 \mathrm{ft}) \rightarrow M_a=-24 \mathrm{ft}-\mathrm{kips} .$

Next comes the cut at b–b. It is clear that using the portion of the beam to the right of b–b will be easier, and so we construct an FBD and apply equilibrium:

\begin{aligned} & \sum F_x=0=P_b, \\ & \sum F_z=0=-V_b+R_B \rightarrow V_b=6 \mathrm{kips}, \end{aligned}
$\curvearrowleft \sum M_{\text {about } b}=0=R_B(2 \mathrm{ft})+M_b \rightarrow M_b=-12 \mathrm{ft} \text {-kips. }$

Note: The negative signs on the moments at cuts a–a and b–b indicate that these moments are opposite from the way they are drawn in our FBDs. It is convenient to assume positive shear and bending moment when constructing FBDs, so that a negative sign will always represent negative shear or negative bending moment. Please refer to Figures 7.4 and 7.5 for a reminder of the sign convention for shear and bending moment.