# Stress, Deflection for a simply supported beam with UDL applied

1. Aug 11, 2011

Could anyone please tel me how to calculate the Bending Stress, Compressive stress and the Deflection for any beam (UB 203x102x23) subjected to Uniform Distributed Load of 8063 N/m over a 8 m long beam (simply supported beam).

Thanks all.

Last edited: Aug 11, 2011
2. Aug 11, 2011

### Mech_Engineer

This sounds like a pretty straightforward question, is it homework? Have you tried coming to a solution yourself?

3. Aug 11, 2011

Yes I did, can you check my result.

Length = 8m
UDL = 8063 N/m

= (8063*8)/(8*.1018)
= 79204.32 N/m^2

Bending Stress = (Force * Distance)/(I/(D/2))
= (8063*8*8)/((2.11*10^-5)/(0.2032/2))
= 2484779678 N/m^2 for x-axis

For y-axis
= (8063*8*8)/((2.11*10^-5)/(.1018/2))
= (1244835488 N/m^2

Deflection at the center = 5wL^4/384EI
= (5*8063*(8^4))/(384*(200*10^9)*(2.11*10^-5)
= .1019 m

Last edited: Aug 11, 2011
4. Aug 12, 2011

My question is, if I apply a point load or UDL I should get the stress in N/m^2. When using UDL is N/m or N/mm, calculating Compressive Stress is Load/ Cross-section Area. The load should be given in N, so I converted N/m UDL to N by multiplying it with my length. Then I get the stress in N/m^2.

Calculating Bending Stress is Force x Distance/(Moment of Inertia/Section Modulus). Here again the force should be in N, so my UDL x length gives me load or force. I get the stress in N/m^2.

What I am doing is right or what? Any one please answer.

5. Aug 12, 2011

### SteamKing

Staff Emeritus
Draw a shear and moment diagram and you will find out why the units in your moment equations are not working out.

6. Aug 12, 2011

I don't mean to say my units are wrong. They sound right to me, but the actual problem is I have a weight of 49 N sitting on four beams connected to each other. And the beam ends are supported by columns (fixed firmly to the columns) restraining the moment in all directions and the beams are held at 5 mts. If you imagine there are four beams connected each other to make a rectangular connection and four columns at each ends of 5 mt height. With load of 49 N, I am converting it into N/m (UDL). I used this formula ( I don't know if it is right or not).
W(UDL) = ws/3 * (3-m^2)/2, m=S/L, S= 4 m, L=8 m, w = 49.
W(UDL) = 89.83 N/m This is the weight over all the beams. When I apply it in the formula for Compressive Stress and Bending Stress, I have to convert it into N to get the units in N/m^2.

My question is, what I am doing is right?? This I came up with my own solution, but I just wanna verify it.

7. Aug 12, 2011

### nvn

buytree: I am currently not understanding your explanation. In post 1, you said UDL = 8063 N/m, but in post 6, you said UDL = 89.83 N/m. That is a big difference. Figuratively speaking, a load of 49 N is nothing. Is this a school assignment?

Also, you did not define your parameters in post 6. And you did not explain how you derived the equation in post 6. Hint 1: In your bending stress calculation in post 3, didn't you forget to divide by 8? Try again.

8. Aug 15, 2011

1#The UDL is 89.83 N/m (In post #1 my calculation is wrong). The load 49N is what I used in my old project, I just considered the same load.
2#Parameters for #6, L-Length of the section, S-Width, w-load over the support. The equation W (UDL) was taken from structural hand book which I downloaded from web. I am not sure how did they derive it.
3#I did a project when in school which included similar calculation but with a point load. One of my juniors had some question about my project, then I started thinking how does it change with UDL applied.
4#I remember the formula for bending stress, (WL^2)/8. But I tried to used the actual bending stress formula which is Bending Stress = (Force * Distance)/(I/(D/2)), I-moment of inertia, D-Depth of the beam. I am pretty sure this is the right formula. I really don't know how different these two formulas are.

9. Aug 15, 2011

### nvn

buytree: Bending stress for a simply-supported beam with a UDL (w) would be, sigma = M*c/I = [(w*L^2)/8](0.5*D)/I. By the way, N/m^2 is called Pa. And 10^6 Pa is called MPa.

10. Aug 16, 2011