Stress, Deflection for a simply supported beam with UDL applied

In summary: I forgot to divide by 8 when I calculated the bending stress. I tried again and it worked.In summary, In post #1, the UDL is 89.83 N/m, while in post #6, the UDL is 5wL^4/384EI. The load 49N is what I used in my old project, I just considered the same load. The 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.
  • #1
buytree
34
1
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:
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  • #2
This sounds like a pretty straightforward question, is it homework? Have you tried coming to a solution yourself?
 
  • #3
Mech_Engineer said:
This sounds like a pretty straightforward question, is it homework? Have you tried coming to a solution yourself?

Yes I did, can you check my result.

Length = 8m
UDL = 8063 N/m

Comp Stress = Load /Area
= (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

So did I do it right? Thanks for your reply.
 
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  • #4
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
Draw a shear and moment diagram and you will find out why the units in your moment equations are not working out.
 
  • #6
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 want to verify it.
 
  • #7
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
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
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
Thanks nvn. The Bending Stress formula you gave makes sense to me.
The compressive stress is Force/Crossection Area, Weight(49)/(Length(8)*Width(0.2036)) which gives unit in Pa.
I am finding the combined stress for beam, which is
=(cal compressive stress/allowable comp stress) +(cal bending stress in x/allow bend stress)+(cal bending stress in y/allow bend stress) which should be less than 1 according to IS800 code book. It is not giving a value less than 1. Am I doing anything wrong. If needed I don't mind attaching my work sheet. Thanks again.
 

Related to Stress, Deflection for a simply supported beam with UDL applied

1. What is stress and deflection in a simply supported beam?

Stress refers to the internal forces that act on a material, causing it to deform or change shape. Deflection, on the other hand, is the amount of bending or displacement that occurs in a beam when a load is applied.

2. What does UDL stand for and how does it affect the beam?

UDL stands for Uniformly Distributed Load. This type of load is evenly distributed along the length of the beam, causing a consistent amount of stress and deflection throughout the beam's span.

3. How do you calculate stress and deflection in a simply supported beam with UDL applied?

The equations for calculating stress and deflection in a simply supported beam with UDL applied are based on the beam's length, material properties, and the magnitude and distribution of the load. These equations can be found in a standard engineering textbook or online resources.

4. What factors can affect the stress and deflection of a simply supported beam?

The stress and deflection of a simply supported beam can be affected by various factors such as the material properties of the beam, the magnitude and distribution of the load, the beam's length and cross-sectional shape, and the supports at each end of the beam.

5. How can the stress and deflection of a simply supported beam be minimized?

To minimize stress and deflection in a simply supported beam, engineers can use stronger and stiffer materials, increase the beam's cross-sectional area, or decrease the magnitude and distribution of the load. Other techniques such as adding supports or using different beam shapes can also help reduce stress and deflection.

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