View Full Version : Formula for deflection of beam

09-11-2004, 09:01 PM
Hi all, need some really basic physics help here.

I'm trying to do some really simple calculations on beam deflection for a new machine I'm doodling, maybe using igus plastic bearings on drill rod, and want to calculate the deflections based on rod diameter to determine best size/price.

All my text books are in storage, my maths and differential calculus is rusty, college was a long time ago and I can't find what I want on google so I thought I'd ask here. I want to calculate the deflection of a beam with fixed ends and a load a given but not equidistant distance from one end. Anyone have or point me to the equation for that?

For example, and it's been a while but I know that for a beam with fixed ends and centre loaded the deflection can be calculated as;

Deflection = (F*L^3)/(192*E*I)

F = Load
L = Length of beam
E = Youngs modulus for the material
I = Area moment of inertia

So for example a tool steel rod of 36" length and 1" dia with a 100lb load would give; Def = (100*36^3)/( 192*28,000,000*0.049) = 0.0177 inches

I seem to remember that for an intermediate load on a beam with fixed ends the equation included some content for the distance from one end but I can't figure it out and its bugging me.

For what it's worth;
A fixed end cantilever with load at end; Def=(F x L^3)/(3*E*I)
A simple supported beam with centre load Def=(F*L^3)/(48*e*I)

It's worth noting the value of firmly locating the ends of unsupported rod for linear bearings. For the above example with Fixed, immovable end locations for the rod will give a deflection of 0.0177 inches. A simple, non fixed end location will allow the same beam with the same load to deflect 0.071 inches. This would apply to a rotating beam such as a lead screw and shows why as ridged a location as possible for the ends helps reduce deflection.

So, any ideas on an intermediate load solution? thanks..


09-11-2004, 09:35 PM
College was a long time ago for me also, and since my field is electronics, I don't even have any of my ME and structures books left, and don't remember much about it. I don't remember any of the equations, but I use two programs for deflection calculations. The first is BeamBoy which is quick and easy to use, and is freeware. It will do stress and deflection graphs for multiple point or distributed loads, with either simple or cantilever supports. The other is Engineering Power Tools, which does a lot more than just deflections, it has a lot of mechanical, electrical, and HVAC calculators and tables. It is shareware, but the free version will do just about everything except saving the results of calculations. Links for downloading are below.



The Wizard
09-11-2004, 09:42 PM
I don't know if this page will help......


... but that was the third hit on a Yahoo search for 'loads on a beam'.

The Java applet looks quite interesting but I didn't play with it much.


09-12-2004, 12:05 AM
this site is pretty good. If it tells you to register just delete your cookies and keep using the site for free

09-14-2004, 05:23 PM
Thanks for the replys and some interesting links. While they only deal with cantilever, supported and single fixed end solutions, and not the two fixed ends problem theres enough in there to work it out, I think. I got enough from the java script to figure it out and just need to do the differential math to work it through. Of course I can't remember quite how that goes either...

By now you'd have guessed I've found the answer to the real problem another way ages ago and this is nothing more than being obstinate, knowing I did this all the time once and trying to find the solution in a way I like..

09-14-2004, 08:14 PM
BeamBoy will do calculations with cantilever supports at both ends of a beam. I think this would model the case of a rod which is rigidly supported at each end or a lead screw with two rigidly mounted radial bearings at each end.