# Simply Supported Beams Loads and S/BMD's

## Homework Statement

Here is the Simply Supported Beam with a uniform and triangular load.
http://prntscr.com/jd6b2

## The Attempt at a Solution

Current Step: Shear Force and Bending Moment Diagrams.

Last edited:

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

Here is the Simply Supported Beam with a uniform and triangular load.
http://prntscr.com/jd6b2

## The Attempt at a Solution

http://prntscr.com/jd6my
RAy + RBy = 900 + 225 = 1125kN
ƩMA = 4.5(225) + 12(900) - 18(RBy) = 0
RBy = 656.25kN
RAy = 568.75kN

Is what I've done right so far? (Finding Reactions at A and B)
How are you arriving at these numbers? And what happened to the 200 kN concentrated point load? The UDL of 50 kN per m acts over 9 feet, not 18 feet, and its resultant lies at its cg. The triangular distribution has a resultant of 225 kN, yes, but you show it acting in the wrong location. It too acts at the cg of the triangle.

Hmm, I think I have actually interpreted this question wrong.

Is this image showing one 50kN load being distributed uniformly THEN being triangularly distributed. AND a 200kN point load? (I've been viewing it as two separate loads.

I got 225 by (50kN * 9m)/2 (Uniformly distributed load)

PhanthomJay
Homework Helper
Gold Member
The problem is showing a uniformly distributed load of 50 kN/m acting over the first 9 meters of the beam. It's total load (resultant of the distributted load, which is the 'area' of the rectangle, or (50)9 = 450 kN) acts at its cg 4.5 m from the left end. Thus, when determining reactions, it can be treated as a single concentrated point load of 450 kN acting 4.5 m from the left end. You incorrectly divided that total load by 2, but showed its location correctly.
Then the problem shows a separate 200 kN concentrated point load acting at the center of the beam 9 m from the left. Don't ignore it.
Finally, the problem shows a separate triangularly distributed load that varies from 50 kN/m to 0 kN/m acting over the right half of the beam. It's resultant equivalent concentrated load is found by calculating the 'area' of that triangle and placing that load at the cg of the triangle. You correctly located that load as 12 m from the left end, but its value is wrong.

Alright I've redone it, http://prntscr.com/jdu7d

The next step is to draw bending moment and shear force diagrams. I'll post when I've got them.

I've done the shear force diagram, is this right?

Some reason I can't edit previous posts or the thread.

PhanthomJay
Homework Helper
Gold Member
I've done the shear force diagram, is this right?

Some reason I can't edit previous posts or the thread.
No attachment. You can't edit threads after a certain time has passed..

SteamKing
Staff Emeritus