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Problem 4.150 K) 2 of 10 Replace the loading by an equivalent foroe Express your answer to three sigaificant figures and incl
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Basically the entire loading consist of 2 loadings which are triangular.

the 1st triangle extends from the left support to 7.5 m

The concentrated load will act at a distance (2/3)*7.5 = 5m from the left end

w = 5.8 kN/m

F1 = (1/2) * 5.8 * 7.5

= 21.75 kN

For the 2nd triangular loading from the right end, it will act at a distance (2/3)*4.5 = 3 m from the right end.

F2 = (1/2)*5.8 * 4.5

= 13.05 kN

and you have the concentrated load of 15 kN

So equivalent force = 21.75 + 13.05 + 15 = 49.8 kN

the force is downward so - 49.8 kN

For equivalent moment we take

\sum M about point O = 0

500 + F1* 5 + F2*9 + 15*12 = M at point O

500 + 21.75*5 + 13.05*9 + 180

= 906.2 kN-m clockwise

So - 906.2 kN-m

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