Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations pierreick on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Deflection of Simple Beam With uniform load partially distributed 1

Status
Not open for further replies.

bylar

Structural
Jan 3, 2002
173
Does anyone have a formula for Deflection of Simple Beam With uniform load partially distributed. All the references that I have ignores the deflection along th beam with this type of loading. There is an old post 2006 by an ME but no good answer. One option seemed to be to take the load as from one end to the end of the load and take the load from the end to the begining of the load and subtract it from the first deflection. I tried that and I got some negative number and I know that cannot be. Help!
 
Replies continue below

Recommended for you

If it is simply supported with uniform load on entire length, you can find the formula here:

Other cases are not too difficult to derive from the Euler Bernoulli beam model:
Shear V is integral of Force
Moment M is integral of V
Slope is integral of M divided by the constant E*I
Displacement is integral of slope.

Divide the beam into enough sections so that you have only discrete forces applied at boundary of each section or forces which are constant over the length of each section.

Use conditions of static equilibrium to solve the reaction forces at each support. Now your boundary conditions at the ends and between the sections provide enough info to solve the unknown integration constants. It is pretty easy to do with a computer algebra program like Maple (Maple's "Heaviside" function is the same the "singularity" function of MaCaulay's method).

=====================================
(2B)+(2B)' ?
 
I prefer the use of Singularity Functions for nonsymmetric distribution of loading across the beam. Usually the equations generated through the various sections lead to closed form solution sets for rotation and deflection.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
 
bylar:

The method you used is correct. The answer was wrong because of a mistake you made in the application of the equations. Do not look at your original calculation and do it again.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
 
why not 1st principles ? M(x) is defined over different parts of the span, and integrate away. check your answer with Roark, superposition of a beam with a UDL and a part-span dist'd load. or check your answer by reducing the unloaded span and your solution should appraoch a full span load, make the unloaded span bigger and it should approach a central point load ...
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor