# Bending Moments

by smell5
Tags: bending, moments
 P: 9 Question : A beam 10 meters long, with 3 forces acting on it and 2 re-acting forces on it to keep it in equilibrium. The re-acting forces are at 2 meters from either end. The forces onto the beam are at 3 meters from the left end of 10kN, 10 meters from the left end of 20KN and a distributed load over the full 10 meter beam of 5KN. a) work out the 2 re-acting forces b) Calculate the bending moment at 1meter intervals along the beam c) Draw a bending moment diagram of the beam Right so there's the question, my issue I believe; is calculating the re-acting forces, I have tried numerous ways, and believe that both reacting force will be 40KN each. As taking moments from R1, (10 x 1) + (50 x 3) + (20 x 8) = R2 x 8 R2 = 40 As the UDL = 50KN at 5 meters, R1 = 80 - 40 =40KN Then I drew a sheer diagram which seemed fine. Then I calculated each meter individually expecting the 10th meter to equal zero. However I get M10 = 0 - (10 x 5 x5) + (40 x 8 ) - (10 x 7 ) + (40 x 2) = 60 Obviously not right!!! Could someone please give me some guidance whether you believe it's the re-acting forces or whether it's my calculations for working out each meter. Any Help Would Be Appreciated.
 HW Helper Thanks P: 4,399 The applied loads are not symmetrical, so the reacting loads are not symmetrical. Double check the signs of your moment calculations. The way the problem statement is given, a load of 5kN is distributed oven the length of the beam. I read this as a UDL of 5 kN / 10 m = 0.5 kN / m

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