Bending Moments

by smell5
Tags: bending, moments
smell5 is offline
Aug9-11, 05:05 PM
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.
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SteamKing is online now
Aug10-11, 03:45 AM
HW Helper
P: 5,586
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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