Bending moment query re. uniformly distributed load and concentrated load(s)

[Gratch]

Homework Statement

A horizontal beam 8m long, resting on two supports 1.5m from each end (supports are 5m apart), carries a uniformly distributed load of 25kN/m between the supports, with concentrated loads of 20kN at the left end of the beam, 30kN at the right end, and 40kN in the centre.

1. Draw shear force diagram
2. Draw bending moment diagram
3. Determine magnitude and position of max bending moment
4. Determine size of beam section required to limit the max bending stress to 500MN/m2

Homework Equations

The Attempt at a Solution

I have worked out the support forces to be 99.5kN for the left support and 115.5kN for the right, and have drawn the shear force diagram (vertical lines for forces, horizontal lines for no forces, and diagonal lines for uniformly distributed load).

If there was no concentrated load at the centre i could draw the bending moment diagram too, but i don't know how to incorporate the centre load.

Do i add the centre load to the centre-point of the uniformaly distributed load (resulting in a higher, steeper curve); do i do the normal curve for the uniformly distributed load with a "spike" of sorts on top for the centre load; or do i do something else entirely?

EDIT: Also, does the bending moment diagram start and end at zero, or is it zero at the supports?

Thanks in advance

 

[pongo38]

"Do the normal curve for the uniformly distributed load with a "spike" of sorts on top for the centre load". This is graphical integration, the result of which is a bent base line. Any vertical ordinate should give you the correct BM at that section. You could answer your own final question if you knew the definition of BM which says: The bending moment at a section is the algebraic sum of moments on one side of that section (or the other side - it doesn't matter - they should be the same) This is the same process you probably used to obtain the reactions, but you haven't shown those details.