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.
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