# Shear Force Diagrams and Bending Moment Diagrams

• nobodyuknow

## Homework Statement

http://img189.imageshack.us/img189/9131/22465062cfdd4934b51d7f5.png [Broken]

Hopefully the image isn't too difficult to decipher... Basically a beam balanced on to pin jointed pivots with 6 metres in between each one.

ƩFy = 0
ƩFx = 0
ƩMP = 0

## The Attempt at a Solution

I've solved the first section of the part...
The internal moment comes out to be -56kN/m
Axial Force is 0
And Shear Force is -6kN

Calculating the second section...
I'm stuck here, hence, why I don't think I can proceed with Section 3 either
I calculated the Internal Moment to be... 72 + v(6) kN/m (Which I think is wrong)..
The Shear Force came out to be -24kN
Axial Force is... 0?

Need help with completing the rest

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You need to be able to identify the resultant force and location of the uniformly distributed load and the resultant force and location of the triangularly distributed load. Then solve for the reactions next by summing moments. Then draw the shear diagram. Practice with a more simple case first, like a uniform load on a simply supported beam. Triangularly distributed loads add a level of difficulty to the problem.