Need help with Beam Calcs 2

[CajunNGnear]

Hey guys, log time reader, first time posting.

I have a 50' long beam with 5 supports on a 12.5' spacing. (0ft,12.5ft,25ft,37.5ft, and 50ft)

The beam has a uniform 500lb/ft load; 25,000 lbs total.

I'm trying to find the greatest moment force to find the stress in order to select the size of beam. My issue is determining the reation loads. I assummed 5,000 lbs on each of the 5 supports???? but I'm lost at finding the shear and moment forces (Haven't done this in 12 years) I've got my Roark's Eq book and I'm going through the Three-Moment Eg but I can't get the numbers to add up.

Is there a simpler method I'm missing???

HELP!

 

[desertfox]

hi CajunNGnear

Try this calculator:-

http://www.epcserver.com/Structural/analysis/beams/continous1.asp

desertfox

 

[drawoh]

CajunNGnear,

I hope you did not throw out your mechanics of materials books from college.

You have a statically indeterminate beam. You have to assume boundary condtions such as zero deflection at the reaction points, then solve for the integration constants.

JHG

 

[zekeman]

Since you have a symmetric load, the problem is considerably simplified by using the 0 deflection, zero slope at the center support.
Now you have a fairly simple problem of uniform loading and two forces, one at the end and one in the middle of the 1/2 beam.
By superposition, you first find the deflections at those points absent those 2 forces, i.e. a uniformly loaded cantilevered beam giving y1 and y2 deflections. Now you place two forces, F1 and F2 at those points so that they cause deflections -y1 and -y2 at those points thus effectively putting the beam in the zero deflection state at those points.
Now you can write two equations, namely
-y1=a11F1+a12F2
-y2=a12F1+ a22F2
where you can get those influence coefficients a11,a12 and a22.
I got a11=1/3*L^3/EI, a22=1/24*L^3/EI, a12=5/48*L^3/EI
The rest should be obvious.

 

[zekeman]

Fox,
I believe that Mc is not zero, only the slope there.

 

[msquared48]

As an approximation for the conditions specified, you could assume, based on the 12.5 foot span and the 500 plf loading, that 45% of the 12.5 foot span goes to the very end support and 55% of 12.5 foot span to the interior support.

Then, based on the information above, and multiplying the resultqang coefficients by the total span weight of 6.25 Kips, the approximate reactions would be:

2.81, 6.88, 6.88, 6.88, 2.81

The approximate max shear would be 6.88/2 = 3.44 Kips.

Mike McCann
MMC Engineering

 

[FeX32]

Internal moment on C not external. ie. Mc=0.


Fe

 

[desertfox]

Hi CajunNGnear

My apologies to all, my assumption that Mc was zero was incorrect as several people have pointed out.
Cajun please ignore my original file and use this one instead, I have gone right through it now and checked my answers against the calculator link I left earlier and they tally.

desertfox

 

[Ron]

Your reactions are:

Support 1 - 2458 lb
Support 2 - 7136 lb
Support 3 - 5812 lb
Support 4 - 7136 lb
Support 5 - 2458 lb

Max.Moment is about 75,000 in-lb and occurs in sections 2 and 3.

 

[rb1957]

read up on "three moment equation". maybe unit force method (for solving redundnat beams). 5 supports triple redundancy. invoking symmetry reduces this to two, the middle support and the end moment (there is a moment on the CL, yes it is internal, but when you split the beam (for symmetry) it becomes external). ron's solution looks reasonable, the peak moment would be on both sides of the middle support.

 

[desertfox]

Hi rb1957

I tried splitting the beam and assuming that the moment was external at that point, but it didn't work for me when using the three moment equation and as several others have pointed out that you cannot assume that C is external moment(unless you know some other way).
I ended up doing the three moment equation and finding the unknowns by simultaneous equations which you can see in the file I uploaded.
The reactions compare very close with Ron's figures and also exactly match the figures from the calculator link I posted earlier so at least my maths appear to check out.

desertfox

 

[CajunNGnear]

Great replies guys. Its all coming back now. Haven't done this type of calcs in a while so I keep second guessing myself.

 

[desertfox]

Hi CajunNGear

Look at the file I uploaded it gives the working out

desertfox

 

[SWmechE]

Hello everyone

This topic interested me as I solved a simpler version (two span, concentrated loads, symmetric) earlier by using superpostion: there's equations in an old textbook (Blodgett's) describing this that I walked through and thought I had a reasonable answer.

I wanted to see if the web program that desertfox posted gave the same answers as me but for some reason it doesn't... I was hoping that someone else can give me feedback; in my mind this should output symmetric loading over the three support points and the maximum moments located where the loads are. Actually to me it looks like the Shear and Moment Diagrams appear correct in the program but the values do not.

Attached is the output of the web calc as well as a quick sketch of the problem that I was trying to solve.

Thanks

http://files.engineering.com/getfil...b-369fafd0287e&file=continous_beam_output.pdf

http://files.engineering.com/getfil...-f83e-42a0-a405-e112b13a593c&file=problem.PDF

 

[desertfox]

Hi SWmech

It would be better if you had started a new thread on this however I'll take a look and let you know.

desertfox

 

[desertfox]

Hi SW

I think I agree with you, the calculator link seems to give different answers for beams with point loads when compared with hand calcs.
Altough it checked out fine when I did the UDL loading.
I checked the calculator with a text book problem for point load with two spans and the figures were out.
Any chance you can post your calcs so we can compare?

desertfox

 

[PEinc]

Why not just get the reactions, moments, and shears from the AISC Steel Construction Manual, 13th Edition, Page 3-225, Table 3-23, Diagram 42 for a continuous, four equal span beam with a uniform distrubuted load?

R1 = 0.393wl
R2 = 1.14wl
R3 = 0.928wl
R4 = 1.14wl
R5 = 0.393wl
Vmax = 0.607wl
Mmax = 0.107wl^2
Max. Deflection = (0.0065wl^4)/(EI)

http://www.PeirceEngineering.com

 

[msquared48]

If you want to do it by hand a different way, use moment distribution, old reliable, and develop the shears and reactions from there.

Mike McCann
MMC Engineering

 

[SWmechE]

desertfox,

Attached are my hand calculations... they're pretty straight forwad (although a bit messy) as I'm just using super-position and standardized formulas that I've found in a few different sources.

http://files.engineering.com/getfil...8af-44a0-8d7c-32816f7d1afd&file=handcalcs.pdf

Here's a web page that outlines a bunch of common load cases: for my load case I used Figure 28: Continuous Beam- Two Equal Spans - Concentrated load at any point. As with everything Engineering related these formulas should be taken with a grain of salt and sound engineering judgement before assuming they are 100% correct.

http://www.awc.org/pdf/DA6-BeamFormulas.pdf

 

[desertfox]

Hi SWmechE

Thanks for the information, I noticed the hand calculations have two spans with a load on each span but the load case 28 you directed me to is a two span beam but only as one point load on one of the spans, also the problem you posted was slightly different to the hand calcs you posted last, I was really after your hand calc you tried to compare with the online calculator.
Anyway I have played around and I can get the hand calc reactions off the formula sheet to be pretty close to the on line calculator results but the intermediate moment on the support seems way off.
Not sure at the moment whats wrong it might be me! I'll let you know.

desertfox