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 Replace the distributed loading with an equivalent  resultant force, and specify its location on the beam  measured from point  \(A\).

\( F_{1}=\left(\frac{1}{2} * 3 * 8\right)=12 k N \)

 

\( \frac{3}{3}=1 \)

\( F_{2}=(3 * 4)=12 k N \)

\( F_{3}=\left(\frac{1}{2} * 3*4\right)=6 \mathrm{kN} \)

\( F_{R}=12+12+6=30 \mathrm{kN} \downarrow \)

 

\( \sum M_{A}=(12 * 2)+(6 * 4)+(12*4 \cdot 5) \)

\(=102 kN.m\)

\(=F_R*d\)

\(d=3.4m\)

 Determine the equivalent resultant force and couple moment at point  O .

 

\( F=\int_{0}^{3} w d x \)

\( =\int_{0}^{3} \frac{1}{3} x^{2} d x=\left[\frac{x^{3}}{9}\right]_{0}^{5}=3 k N \downarrow \)

\( M=\int_{0}^{3}(w \widetilde{x}) d x \)

\( =\int_{0}^{3}\left(\frac{1}{3} x^{2}\right)(3-x) d x \)

\( =\int_{0}^{3}\left(x^{2}-\frac{x^{3}}{3}\right) d x =\left[\frac{x^{3}}{3}-\frac{x^{4}}{12}\right]_{0}^{3}=2.25 \mathrm{kNm} \)

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