Roark's Beam Formula Superposition Question

[curiousmechanical]

Hello Everyone.

I have a beam bending problem that I am currently working on (please see the attached .pdf).

I am looking for the max deflection of the beam.

A colleague gave me the direct formula from Constrado, “STEEL DESIGNERS’ MANUAL,” Fourth Edition, pg. 43.

However, I am a Roark’s fan and I would like to know how to solve the problem using Roark’s bending tables. Roark’s doesn’t give the direct formula, but I can’t help but think that you can solve this using superposition or some other neat manipulation.

You can see my attempt on the attached .pdf, but it doesn’t work out (I came out too low). I checked the solution using FEA and confirmed the direct formula result.

Does anyone know how to manipulate Roark’s formulas to get the correct result?

I am going nuts trying to figure it out.

Thank you!

 

[msquared48]

Looks like you are trying to use equation 2b on page 157?

I think you should be using 2d on page 100, for beams.

The one on page 157 also has the axial "P" load added, which is not your case.

Try page 100.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[prex]

The direct formula is incorrect: you should have a length to the 4th power, if w is load per unit length.
Have a look at the first site below under Beams -> Single beam -> Fixed-fixed -> Distr.load : you'll be able to cross check your formulae.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[IDS]

If W in the direct formula is entered as 121 x 56 (that is the total weight, not weight per inch) it gives 0.0604 inches, which is the same as the Roark formula, and I also get that result with my own calculation.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[curiousmechanical]

First of all, thank you for your quick responses!

msquared48

I am using Roark's, 7th ed. and I am using case 2b on pg.192. Which edition are you using? I don't see any other bending tables 50 or so pages earlier. I will try to hunt down your edition around the office.

prex

You are correct. In the direct formula W is total load (i.e. 6,750lb). In Roark's w is unit load (i.e. 121 lb/in). I will look at your recommended websites when I get home.

IDS

The load you are using is correct, but I am not getting the same result. I just checked and doubled checked and I am still getting 0.0722 in max deflection in the middle. The beam is a W8 x 18 and I modeled it and checked it with CosmosWorks and got 0.07... as well.

Am I still missing something?

 

[rb1957]

i get IDS's result too ... 0.0702 must be finger trouble somewhere ... the "w" is w*(L-2a)

take the example of a UDL y = wL^4/(384*EI) ... you combined the distributed load (lb/in) with L^4. since the bracket has only L^3 the load is in lbs (not lb/in)

your superposition is fine (seems to give the right answer !)

 

[trainguy]

curiousmechanical,

Are you certain that you used the correct values of a in the Roark formulas, during superposition? These definitions change in the various configurations, and a spreadsheet approach (like several of my incorrect ones) could result in this error.

tg

 

[trainguy]

curiousmechanical,

Please disregard my last post, now that I have done the calc myself.

I get the same results as you - I am stumped as well...

tg

 

[msquared48]

curiousmech:

I am using the fifth edition...

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[IDS]

I don't know how people are getting 0.0772 out of the direct formula.

My results are attached, (0.0604, the same as Rourke).

(EVAL() is a UDF by the way, but entering an Excel formula in the usual way gioves the same result).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

It looks like the deflection is about 0.07 when you include shear deflections, which the FEA probably does, but neither of the given formulas include shear deflections.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[curiousmechanical]

Gentleman,

Thank you for your support on this one.

IDS/rb1957

I just put the direct formula in MathCAD and got the 0.0604.

trainguy

Just in case you made the same mistake, I was adding that last term instead of subtracting it.

It hurts to discover that this was the problem all along. I am truly sorry for wasting your time.

Here's how it came to this:

I solved the direct formula (incorrectly) and then confirmed this incorrect result using FEA. Then I tried to use Roark's and couldn't get it to agree with the incorrect answer. It never occurred to me that the initial solution was incorrect because it had the support of the FEA solution (what a sick coincidence). IDS thank you for mentioning the shear deflections because I would still be wondering what was up with that FEA solution.

Well I learned a valuable lesson today.

Thanks guys!

 

[IDS]

Well I learned a valuable lesson today.

Then our time wasn't wasted.


Just a tip on using a FEA program to verify beam or frame calculations: - if you switch off shear deflections in the FEA analysis it should give exactly the same results as a formula based on slope deflection theory, as long as the input values are exactly the same. Even a small difference indicates that there is something not quite right somewhere.

(This only applies to beam elements. Plate and brick elements inherently have some degree of approximation compared with beam theory).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[curiousmechanical]

IDS

Thank you for your help and the tip!

 

[StAnalysis]

For a beam on two supports, I have developed a spreadsheet that calculates the shear, moment and deflections for various loadings (Point Load, Point Moment, or Linear varying distributed load) with end restraints of simple, fixed, cantilever beam, or guided/fixed.