Beam Deflection - UDL's and Point Loads

[Corsan]

Hello,
I have tried for numerous hours to solve this and I'm getting no where, could one of you put me out of my misery?

Homework Statement

[PLAIN]http://img98.imageshack.us/img98/3240/beamloads.jpg

The cross section of the beam is 20mm tall by 10 deep. 800mm length.

Hopefully you can see the above image which is the question.

Homework Equations

The Attempt at a Solution

Using the cross-section above I have calculated EI to be

I= bd^3 = 10 x (20^3) = 6666.66
...12...12
Sorry about the dots, spaces didnt seem to work.

Multiply this by E (71.7GPa - aluminium 7075 series) to give 477999.52 N/mm^2.

After this I have tried various equations such as (WI^3)/48EI to calculate the point load.
Also using (5wI^4)/384EI to calculate the UDL's but I am getting silly figures in the thousands.
Is it not a case of finding the answers to these deflection formulas, adding them together and that is the resultant maximum deflection?

However, I have also read through my notes and found something about slopes etc and that has totally thrown me.

Can anyone offer any help?
Many thanks for any assistance.

 

[pongo38]

"(WI^3)/48EI to calculate the point load" has a typo in the bracket. I should be L. That will give you the deflection due to the point load.
If you use (5wI^4)/384EI - with L replacing I in the bracket, as before - you will have a slight underestimate of the deflection because the formula is for a udl across the whole of the beam. There is a formula for the case you have, but I suspect you are expected to do an integration exercise. Is that right?

 

[Corsan]

"(WI^3)/48EI to calculate the point load" has a typo in the bracket. I should be L. That will give you the deflection due to the point load.
If you use (5wI^4)/384EI - with L replacing I in the bracket, as before - you will have a slight underestimate of the deflection because the formula is for a udl across the whole of the beam. There is a formula for the case you have, but I suspect you are expected to do an integration exercise. Is that right?

Hello, thanks for your response, you are correct about the type - bad eyes!

You are right about the integration although a formula would be nice, how would you advise dealing with the UDL situation?
I believe one method would be to assume it is over the length of the beam and then create a virtual UDL to cancel the areas that aren't under load (Macauleys method?)

Thanks again

 

[pongo38]

The central deflection due to the udl sections is given in the Steel designers manual as Wa(3L^2-2a^2)/(96EI), where a=300 in this case, L=800, and you could use this to check your integration. You can use Macaulay if you like, but personally I find it unnecessarily tedious, error prone and academic. No practising engineer uses it. I do agree you find the deflection due the udl's, and the point load separately, and then add them together. That is the application of the principle of superposition for linear systems.

 

[DeeNos]

Hello, I am a scientist and I would be happy to assist you with your problem. It seems like you have been working hard on this and I understand your frustration. First, I would like to clarify that the equations you have mentioned are correct for calculating the deflection caused by UDL's and point loads on a beam. However, it is important to make sure that you are using the correct units for your calculations. In this case, the cross-section of the beam is given in millimeters, so all of your calculations should be done in millimeters as well. This could be the reason for your "silly" figures in the thousands.

Also, for the point load, you need to make sure that the distance from the support to the point load is taken into account in your calculation. This distance is given as 200mm in the question.

As for the slopes, they are related to the deflection of the beam and can be calculated using the equations you have mentioned. However, they are not necessary for finding the maximum deflection. You can simply add the deflections caused by the UDL and point load to get the total deflection at the center of the beam.

I hope this helps and if you still have trouble, please let me know and I would be happy to assist you further. Good luck!