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Moment Equation

  • Thread starter iviv
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  • #1
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Hi again. First, I'd like to say thanks a ton to tiny-tim for helping me through with my last problem. He was a great help, though I didn't want to bump that thread to say thanks. Anyway, I've got a far simpler issue now, and I'm having a brain fart.

1. A simply supported beam of span, L, carries two equal point loads of magnitude, W, at the third points. The maximum MID-SPAN moment is given by:-

Now, I know that when its a single point load at the centre of a beam, the equation is WL/4
And I'm just having a little trouble getting my head around the doubling of the forces. We've been told its not WL/8 or WL/12.

WL/8 was my first guess, and I was pretty sure it was right, but I've been told its wrong, but I wasn't sure why. However, now I've actually written all this down, I think I've realised what I've been doing wrong. I was thinking 'double W so double the number'. But since we're dividing by the number, obviously it should be halved, not doubled >_>

Now, this is more simple maths than physics, but should it be WL/2? That was my current initial thought, but then the load isn't slap bang on the centre, its offset at third intervals, so now I'm leaning towards it being WL/3. However, this is just through logic and feel, rather than actual maths. And I'm sure there should be a mathmatical way to work it out, but I'm not seeing it at the moment.

Thanks!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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To get the moment at midspan, you need to start at the center and take moments as you go to one end of the beam.
 
  • #3
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Hmm. So, assume 10Kn loads, and a beam 3m long. Since its symmetrical, the moments would be 10Kn at each support.

The force is 10Kn at 1/3 of the beam, so at the centre it would be 15. Using the equation, 15=WL/x

15= 20*3/x

x=4

So the equation would be WL/4, which makes it the same as the equation if there were simply a point load at the mid-span of the beam?
 
  • #4
rock.freak667
Homework Helper
6,230
31
Hmm. So, assume 10Kn loads, and a beam 3m long. Since its symmetrical, the moments would be 10Kn at each support.

The force is 10Kn at 1/3 of the beam, so at the centre it would be 15. Using the equation, 15=WL/x

15= 20*3/x

x=4

So the equation would be WL/4, which makes it the same as the equation if there were simply a point load at the mid-span of the beam?
Your reactions at the supports would be 15kN.

The BM at mid-span would be the 10kN at a distance of 0.5 m (1.5-1) from the mid-point and the 15kN reaction at a distance of 1.5m from the mid-point.

One will produce a clockwise moment and the other an anti-clockwise moment.
 

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