The granular material exerts the distributed loading on the beam as shown in Fig. 4–51a. Determine the magnitude and location of the equivalent resultant of this load.

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Accepted Answer

The area of the loading diagram is a trapezoid , and therefore the solution can be obtained directly from the area and centroid formulas for a trapezoid listed on the inside back cover. Since these formulas are not easily remembered, instead we will solve this problem by using “composite” areas.
Here we will divide the trapezoidal loading into a rectangular and triangular loading as shown in Fig. 4–51b. The magnitude of the force represented by each of these loadings is equal to its associated area ,

[latex] \begin{array}{l} F_1=\frac{1}{2}(9 ft )(50 lb / ft )=225 lb \ F_2=(9 ft )(50 lb / ft )=450 lb \end{array} [/latex]

The lines of action of these parallel forces act through the respective centroids of their associated areas and therefore intersect the beam at

[latex] \begin{array}{l} \bar{x}_1=\frac{1}{3}(9 ft )=3 ft \ \bar{x}_2=\frac{1}{2}(9 ft )=4.5 ft \end{array} [/latex]

The two parallel forces [latex] \mathbf{F}_{1}[/latex] and [latex] \mathbf{F}_{2}[/latex] can be reduced to a single resultant [latex] \mathbf{F}_{R}[/latex]. The magnitude of [latex] \mathbf{F}_{R}[/latex] is

[latex] +\downarrow F_R=\Sigma F ; \quad F_R=225+450=675 lb [/latex]

We can find the location of [latex] \mathbf{F}_{R}[/latex] with reference to point A , Fig. 4–51 b and 4–51 c . We require

[latex] \begin{array}{cc} \circlearrowright +\left(M_R ight)_A=\Sigma M_A ; \quad \bar{x}(675)=3(225)+4.5(450) \ \bar{x}=4 ft \end{array} [/latex]

NOTE: The trapezoidal area in Fig. 4–51 a can also be divided into two triangular areas as shown in Fig. 4–51 d . In this case

[latex] \begin{array}{l} F_3=\frac{1}{2}(9 ft )(100 lb / ft )=450 lb \ F_4=\frac{1}{2}(9 ft )(50 lb / ft )=225 lb \end{array} [/latex]

and

[latex] \begin{array}{l} \bar{x}_3=\frac{1}{3}(9 ft )=3 ft \ \bar{x}_4=9 ft -\frac{1}{3}(9 ft )=6 ft \end{array} [/latex]

Using these results, show that again [latex]F_{R} = 675 lb[/latex] and [latex]\bar{x} = 4 ft.[/latex]

Question 4.23: The granular material exerts the distributed loading on the ...

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