# Simply Supported Beams Loads and S/BMD's

by nobodyuknow
Tags: beams, loads, s or bmd, simply, supported
 P: 64 1. The problem statement, all variables and given/known data Here is the Simply Supported Beam with a uniform and triangular load. http://prntscr.com/jd6b2 2. Relevant equations 3. The attempt at a solution Newest attempt: http://prntscr.com/jdu7d Current Step: Shear Force and Bending Moment Diagrams.
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 Quote by nobodyuknow 1. The problem statement, all variables and given/known data Here is the Simply Supported Beam with a uniform and triangular load. http://prntscr.com/jd6b2 2. Relevant equations 3. 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.
 P: 64 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)
 P: 64 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.
 P: 64 I've done the shear force diagram, is this right? Some reason I can't edit previous posts or the thread.