Question 5.9: A beam supports a distributed load as shown. (a) Determine t...

STRATEGY: The magnitude of the resultant of the load is equal to the area under the load curve, and the line of action of the resultant passes through the centroid of the same area. Break down the area into pieces for easier calculation, and determine the resultant load. Then, use the calculated forces or their resultant to determine the reactions.
MODELING and ANALYSIS:
a. Equivalent Concentrated Load. Divide the area under the load curve into two triangles (Fig. 1), and construct the table below. To simplify the computations and tabulation, the given loads per unit length have been converted into kN/m.

Thus, \bar{X}\sum{A}=\sum{\bar{x}A}: \quad \quad \quad \bar{X}(18 \text{ kN})=63 \text{ kN.m} \quad \quad \quad \bar{X}=3.5 \text{ m}

The equivalent concentrated load (Fig. 2) is

W=18 kN↓

Its line of action is located at a distance

\bar{X}=3.5 \text{ m to the right of } A

b. Reactions. The reaction at A is vertical and is denoted by A. Represent the reaction at B by its components \pmb{\text{B}}_x \text{ and } \pmb{\text{B}}_y. Consider the given load to be the sum of two triangular loads (see the free-body diagram, Fig. 3). The resultant of each triangular load is equal to the area of the triangle and acts at its centroid.

Write the following equilibrium equations from the free-body diagram:

REFLECT and THINK: You can replace the given distributed load by its resultant, which you found in part a. Then you can determine the reactions from the equilibrium equations \sum{F_x}=0,\sum{M_A}=0, \text{ and } \sum{M_B}=0.
Again the results are

\pmb{\text{B}}_x=0 \quad \quad \quad \pmb{\text{B}}_y=10.5 \text{ kN}\uparrow \quad \quad \quad \pmb{\text{A}}=7.5 \text{ kN}\uparrow