Solving a Simple Beam Moment Calculation

AI Thread Summary
The discussion revolves around calculating the maximum mid-span moment for a simply supported beam with two equal point loads at one-third intervals. The initial confusion stems from the relationship between the loads and the moment calculation, with the participant questioning whether the moment should be WL/2 or WL/3. Through logical reasoning and calculations, it is clarified that the moment at mid-span can be derived similarly to a single point load at the center, ultimately leading to the conclusion that the moment is WL/4. The participant also discusses the reactions at the supports and how they contribute to the moment calculation. The conversation emphasizes the importance of understanding the distribution of loads and moments in beam analysis.
iviv
Messages
6
Reaction score
0
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 realized 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!
 
Physics news on Phys.org
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.
 
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?
 
iviv said:
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top