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What is the formula for bending moment of a beam subjected to UDL and a point load?

by buytree
Tags: beam, bending, formula, load, moment, point, subjected
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buytree
#1
Aug17-11, 03:31 PM
P: 34
1. The problem statement, all variables and given/known data
1.What is the formula to calculate the bending moment of a beam subjected to UDL?
2.What is the formula to calculate the bending moment of a beam subjected to point load?

2. Relevant equations

Bending moment (UDL) = WL^2/8 (Kg-mm or Kg-m)

Bending moment (point load) = Force x Distance ( This is actually for a horizontal beam with load acting is a point load) (Kg-mm or Kg-m)

What is the formula for bending moment of a vertical beam subjected to a point load and a UDL on the top of it (load applied axially)?

3. The attempt at a solution

I come through several formulas, like (W * X)(X/2) when a UDL is acting over the cantilever beam. I am bit confused how do they arrive with these formulas and where should I use what?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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rock.freak667
#2
Aug17-11, 04:46 PM
HW Helper
P: 6,202
Quote Quote by buytree View Post
1. The problem statement, all variables and given/known data
1.What is the formula to calculate the bending moment of a beam subjected to UDL?
2.What is the formula to calculate the bending moment of a beam subjected to point load?

2. Relevant equations

Bending moment (UDL) = WL^2/8 (Kg-mm or Kg-m)

Bending moment (point load) = Force x Distance ( This is actually for a horizontal beam with load acting is a point load) (Kg-mm or Kg-m)

What is the formula for bending moment of a vertical beam subjected to a point load and a UDL on the top of it (load applied axially)?

3. The attempt at a solution

I come through several formulas, like (W * X)(X/2) when a UDL is acting over the cantilever beam. I am bit confused how do they arrive with these formulas and where should I use what?
Well a UDL is like a "rectangularly" distributed load, so the load effectively acts at the center of the beam i.e. at a distance of L/2. Since the entire length is L, the load is WL (since UDL = W N/m)

so BM = Force*distance = (WL)(L/2)
SteamKing
#3
Aug17-11, 08:35 PM
Emeritus
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HW Helper
Thanks
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You also need to know something about how the ends of the beam are supported. Are they fixed, free, or simply supported?

mathmate
#4
Aug17-11, 08:44 PM
P: 366
What is the formula for bending moment of a beam subjected to UDL and a point load?

For a horizontal simply supported beam of length L subject to udl w, the maximum bending moment is at the centre and equal to
wL^2/8 distributed along the span parabolically.
For a horizontal simply supported beam of length L, and subject to a point load P at mid-span, the maximum bending moment is PL/4. If the point load is applied at aL (0<a<L) from one end, the maximum bending moment is Pa(1-a)/L just under the load.
The bending moment at any other point on the span can be found by simple statics.

"What is the formula for bending moment of a vertical beam subjected to a point load and a UDL on the top of it (load applied axially)?"
That would be a centrally loaded column, if I understand correctly.
buytree
#5
Aug18-11, 07:09 AM
P: 34
SteamKing: The ends are fixed, there is no movement in any direction.
buytree
#6
Aug18-11, 07:19 AM
P: 34
Thanks everyone for your effects in helping me.
Quote Quote by rock.freak667 View Post
Well a UDL is like a "rectangularly" distributed load, so the load effectively acts at the center of the beam i.e. at a distance of L/2. Since the entire length is L, the load is WL (since UDL = W N/m)

so BM = Force*distance = (WL)(L/2)
Quote Quote by mathmate View Post
For a horizontal simply supported beam of length L subject to udl w, the maximum bending moment is at the centre and equal to
wL^2/8 distributed along the span parabolically.
Check the above post by rock.freak667, it says BM = (WL)(L/2) and you say wL^2/8. I am guessing wL^2/8 is the right BM for the beam, then whats the other formula for?
For a horizontal simply supported beam of length L, and subject to a point load P at mid-span, the maximum bending moment is PL/4. If the point load is applied at aL (0<a<L) from one end, the maximum bending moment is Pa(1-a)/L just under the load.
The bending moment at any other point on the span can be found by simple statics.

"What is the formula for bending moment of a vertical beam subjected to a point load and a UDL on the top of it (load applied axially)?"
That would be a centrally loaded column, if I understand correctly.
Yes exactly it is a centrally loaded column which is subjected to load on top of it, so whats the BM for UDL and point load?
mathmate
#7
Aug18-11, 09:00 AM
P: 366
rock.freak667 will have to interpret the application of his formula.
wL^2/8 is the maximum BM at the centre of a udl of w kg/m for a simply supported beam.

In fact, all these questions about formulas can be resolved by standard tables available in books or the web, such as:
http://structsource.com/analysis/types/beam.htm

You will find the required formulas for the fixed supports subject to udl in your case at the above link. Bending moments at different lengths along the span has to be obtained by superimposing the simply supported moment (parabolic) with the end moments by statics.

For the vertical "beam", are both loads (point and udl) applied vertically and axially?

A centrally loaded column not subject to lateral loads does not incur first order bending moments. Second and higher order bending moments could be caused by lateral buckling or deflections (P-delta effects).
buytree
#8
Aug18-11, 09:36 AM
P: 34
mathmate:In the above link you gave if I use the formula for the FIXED-FIXED BEAM WITH UNIFORM LOAD (my case), my BM at the center would be for x=L/2 is WL^2/24. Hoping this would help, let me check and get back to you. Thanks mathmate.

And my column is only loaded vertically downward over the top. Sorry I might have confused in the above posts.
mathmate
#9
Aug18-11, 10:16 AM
P: 366
Maximum positive bending moment at the centre of wL^2/24 is correct for a fixed-fixed beam.

If you are doing the design of a beam, do not forget that the negative support moments of wL^2/12 are higher than that at the centre.

Draw the bending moment diagram would make it clear.
buytree
#10
Aug18-11, 10:23 AM
P: 34
Quote Quote by mathmate View Post
Maximum positive bending moment at the centre of wL^2/24 is correct for a fixed-fixed beam.

If you are doing the design of a beam, do not forget that the negative support moments of wL^2/12 are higher than that at the centre.

Draw the bending moment diagram would make it clear.
I assume the above case is only for the horizontal beams, what about the BM for column with UDL and point load?
mathmate
#11
Aug18-11, 10:52 AM
P: 366
As per my previous response:

"A centrally loaded column not subject to lateral loads does not incur first order bending moments. Second and higher order bending moments could be caused by lateral buckling or deflections (P-delta effects)."

Is the column monolithic with other structures? If so, there would have to be bending moments at the junctions and would not be considered as purely centrally loaded.

This bending moment (at the junction of beam/column) has to be calculated with an indeterminate structural analysis taking into account of the loads on the beam, size (stiffness) of the beams and columns, and load on the column, the possible deflections due to lateral (wind, earthquake, etc.) or asymmetrical loading, etc.
buytree
#12
Aug19-11, 12:42 PM
P: 34
Thanks. I got it.


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