Problem with values for a bending moment diagram

AI Thread Summary
The discussion focuses on calculating the bending moment for a simply supported beam with a uniform distributed load (UDL) and point loads. The user correctly calculates the reactions at the supports but struggles with the bending moment values at 1m intervals. Feedback suggests that the user may have incorrectly integrated the shear force diagram to derive the bending moments. The importance of using the area under the shear force curve to determine bending moments is emphasized. Clarification on the method seems to have helped the user understand their mistakes.
PizzaWizza
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Homework Statement



I have a 5m long simply supported beam, with supports at points 0m and 5m. There's a UDL of 10kNm^-1 from 0m to 4m. It has downward forces acting at 2m (10kN) & 4m (15kN). I need to sketch a bending moment diagram to determine the position & value of the max bending moment. I'm asked to show workings using 1m intervals for bend moment calcs.


Homework Equations



Taking moments from one end to determine upward supports.

The Attempt at a Solution



UDL = 4m x 10kN^-1 = 40kNm^-1 @ 2m (centre of)

Taking moments from Support 1 (R1) = 10x2 + 40x2 + 15x4 = R2 x5

R2 = 160/5 = 32kN. ∴ R1 - 33kN (As the sum of downward forces = 65kN)

I've sketch a shear force diagram that I'm happy with, now to where I'm becoming undone.

Calculating the bending moments for the beam @ 1m intervals

M0 = 0

M1 = 33x1 - (10x0.5x1(UDL)) = 33-5 = 28kN @ 1meter

M2 = 33x2 - (10x1x2(UDL)) - (10x1) = 66-30 = 36kN @ 2meters

M3 = 33x3 - (10x1.5x3(UDL)) - (10x2) = 99-65 = 34kN @ 3meters

M4 = 33x4 - (10x2x4(UDL)) - (10x3) - (15x1) = 132-125 = 7kN @ 4meters

I'm assuming M5 has to be zero. This just doesn't feel right to me. I've got 15kN acting @ 4m and it doesn't seem to be having an effect on the diagram if graphed. I have to be doing something wrong?

Thanks for any help.
 
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You should post you SF diagram.

The BM is the area under the SF curve starting at x = 0. I think you have incorrectly done your integrations of the SF curve for x = 1, 2, 3, and 4 m.
 
Thanks for the feedback. Here's the Shear force diagram.
 

Attachments

  • Shear bend diagram 001.jpg
    Shear bend diagram 001.jpg
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So far, so good. And the bending moments?
 
The bending moments I've highlighted in the original message. They're wrong, I know. I just don't know why?

M1 = 33x1 - (10x0.5x1(UDL)) = 33-5 = 28kN @ 1meter

M2 = 33x2 - (10x1x2(UDL)) - (10x1) = 66-30 = 36kN @ 2meters

M3 = 33x3 - (10x1.5x3(UDL)) - (10x2) = 99-65 = 34kN @ 3meters

M4 = 33x4 - (10x2x4(UDL)) - (10x3) - (15x1) = 132-125 = 7kN @ 4meters
 
PizzaWizza said:
The bending moments I've highlighted in the original message. They're wrong, I know. I just don't know why?

M1 = 33x1 - (10x0.5x1(UDL)) = 33-5 = 28kN @ 1meter

M2 = 33x2 - (10x1x2(UDL)) - (10x1) = 66-30 = 36kN @ 2meters

M3 = 33x3 - (10x1.5x3(UDL)) - (10x2) = 99-65 = 34kN @ 3meters

M4 = 33x4 - (10x2x4(UDL)) - (10x3) - (15x1) = 132-125 = 7kN @ 4meters

Perhaps because you didn't use the shear force diagram correctly?

SteamKing said:
The BM is the area under the SF curve starting at x = 0.

The shear force diagram is constructed, in part, to allow the ready creation of the bending moment diagram.
 
Thanks. I appreciate it and I think it's sank in now :)
 
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