Simply Supported Beams Loads and S/BMD's


by nobodyuknow
Tags: beams, loads, s or bmd, simply, supported
nobodyuknow
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#1
Nov9-12, 10:07 PM
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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PhanthomJay
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#2
Nov9-12, 10:49 PM
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Quote Quote by nobodyuknow View Post
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.
nobodyuknow
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#3
Nov10-12, 12:51 AM
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)

PhanthomJay
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#4
Nov10-12, 05:42 AM
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Simply Supported Beams Loads and S/BMD's


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.
nobodyuknow
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#5
Nov10-12, 05:58 AM
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.
nobodyuknow
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#6
Nov10-12, 08:20 PM
P: 64
I've done the shear force diagram, is this right?

Some reason I can't edit previous posts or the thread.
PhanthomJay
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Nov10-12, 09:17 PM
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Quote Quote by nobodyuknow View Post
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..
nobodyuknow
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#8
Nov10-12, 09:54 PM
P: 64
Oops, here it is: http://prntscr.com/jg37y
SteamKing
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#9
Nov11-12, 12:36 AM
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Your shear force diagram is incorrect. You have not included the concentrated load of 200 kN.
PhanthomJay
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#10
Nov11-12, 04:45 PM
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When you try this again, be sure to show the distributed loads in your loading diagram. Do not represent them as concentrated point loads...that only works when you are calculating end reactions. And remember that the slope of the shear diagram at any point is equal to the negative of the load intensity at that point. You have kinks and valleys where there should be none.


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