Question

a) Replace the loading acting on the beam by an equivalent force and a couple moment at point A b) Show the equivalent force

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Answer #1

- F - CF F = 1.5 sin 30 – 2.5 (4) - -1.25kN = !:25kN [ ] & Fey = {ty e = -1.5Cosso -.5(3) -3 – 5.799 KN - 5.799. kN 1) (125)M = 34.8kN-m 1.25 KN 15.799 KN + M - EMA Me = -2:5 ()() – 1.5. Cos 36(C) – 3 (8) = – 34.8 KN.m; - 34.8 KN-m (clock wire).

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