4 Point Load Beam Deflection 1

[wsg1]

Hello,

I am looking at designing a pipe support using a universal beam. I believe I have the correct calculations for bending stress, shear stress etc. But I am struggling to find an equation for deflection. At the moment I am currently estimating the combined loads as a single point load (worst case scenario. Does anyone have an equation for this type of loaded beam (see attachment). I believe my other equations are correct but it has been a good few years since I have done these type of calcs.

Any help would be much appreciated

Cheers,

wsg1

 

[rb1957]

the equation is available in lots of places ... Roark, probably even wiki (sigh); but you can superimpose the deflection from the individual point loads. It'll be a little tricky to make sure you have the maximum deflection.

 

[Neilmo]

Hi wsg1,

1) Draw the bending moment diagram for your loaded beam (M)

2) Remove all of your loads and place a unit load (1.0 no units) at the point on your beam where you want to evaluate the deflection - say mid span

3) Draw a second bending moment diagram for this loading (m)

4) Multiply the values of the two diagrams together to obtain a third diagram

5) Calculate the area of this third diagram (Mm dx)

6) And lastly divide by EI

I know this method as the `method of unit loads`. The formula is - deflection = Mm dx/EI

It is easier than it sounds, give it a try and remember to keep your units consistent.

Good luck, Neil

 

[msquared48]

Use superposition.

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

 

[desertfox]

Hi wsg1

The difficulty with your beam problem is calculating exactly where the beam slope is zero, as the beam is unsymmetrical loaded.
Macaulay's or graphical methods can be used to determine zero beam slope if you want to be real accurate, but the maths can get long winded.
So what I have done is loaded your beam symmetrically with the two highest loads ie 33.43kN and 28.8kN repeated on the right hand side, this allowed me to calculate the maximum deflection at the centre of the beam, ie zero slope. I thought it would provide a comparison with your calculation and if the deflection was similar to yours, the beam would be safe because of the additional loading I have imposed.
Interestingly my calculation using Macaulay’s method for beam deflection yields a value of 3.5879mm

 

[IDS]

Looks like I didn't submitt after previewing last time, so I can get it right this time

The deflection calculation using a lumped load at midspan will be conservative, but if you want a more accurate figure you could find the point of zero slope, and maximum deflection, by trial and error using Goal Seek.

Split the beam into two cantilevers, fixed at the estimated point of maximum deflection, which you could take as mid span for a first guess, and apply the upward reaction force at the ends, and downward applied loads at the indicated positions. The deflection at the free end of a cantilever of length L with a load W applied at a from the free end is given by:

=W/(6EI)(2L^3 - 3L^2a + a^3)

Add the three deflections due to the three loads, then adjust the L value (using Goal Seek) so that the deflection of the two cantilevers is equal. This is the maximum deflection of the beam.

I get 2.42 mm deflection at 1.651 m from the left hand end.

By the way, the calculation for the maximum moment in your spreadsheet is very conservative. The maximum moment is under load P2 and is equal to the area under the shear force diagram =

58.38 * 1.295 - (33.43 * (1.295 - 0.533)) = 50.1 kNm




Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[wsg1]

Thank you all for your help.

IDS, I have drawn the shear force diagram and you are correct my maximum moment was incorrect.

I have tried the deflection calculation you mention above and for the same values get 1.42mm, is the 2.42mm a typo or have I done this incorrectly?

Thanks

wsg1

 

[IDS]

I have tried the deflection calculation you mention above and for the same values get 1.42mm, is the 2.42mm a typo or have I done this incorrectly?

I'm pretty sure that 2.42 is right.

I even checked it in a frame analysis program.

I'll post a spreadsheet when I get time.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[wsg1]

Ok i'll have another look.

Thanks Again

 

[Neilmo]

Hi wsg1,

Generally speaking, no matter what set of random loads and spacings you have on a simply supported beam, the maximum deflection is always within a few percent of the mid-span.

Using your figures, I calculate the maximum deflection to be 2.4217mm at a distance of 1651mm from the LH end.

Regards, Neil

 

[IDS]

Using your figures, I calculate the maximum deflection to be 2.4217mm at a distance of 1651mm from the LH end.

Seems about the right order of magnitude


Here's the spreadsheet.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[wsg1]

IDS,

I have looked over the spreadsheet you attached and it seems the equation you are using is as follows:

d = (W/(6EI)) * (2(L^3) - 3a(L^2) + a^3)

Should this not be

d = (W/(6EI)) * (2(L^3) - 3(L^(2a)) + a^3)

Taking a formula from your spread sheet:

=+D9/(6*$E$4)*(2*$B$6^3-3*$B$6^2*E9+E9^3)

Should this not be as follows

=+D9/(6*$E$4)*(2*$B$6^3-3*$B$6^(2*E9)+E9^3)

It does make a huge difference to the calcs just by adding in this brackets.

This maybe where our differences are occuring

Thanks,

wsg1

 

[IDS]

d = (W/(6EI)) * (2(L^3) - 3a(L^2) + a^3)Should this not bed = (W/(6EI)) * (2(L^3) - 3(L^(2a)) + a^3)

No, 3a(L^2) is right.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Neilmo]

wsg1,

Just to confirm IDS. I used a totally different approach to his method and came up with exactly the same figures.

Neil

 

[Clyde38]

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http://www.excelcalcs.com/repository/strength/finite-element/excsta2d/

[link

http://www.linkedin.com/in/clydehancock

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[wsg1]

Thanks for all the help, got the spreadsheet working now.

Wsg1

 

[BAretired]

You could use the Conjugate Beam Method, i.e. load the conjugate beam with the M/EI diagram from the real beam. The Shear Force diagram and Bending Moment diagram for the conjugate beam will be the rotation and deflection respectively for the real beam.

BA

 

[IDS]

Another way, which also works for multi-span beams, is to use cubic splines:

- Apply a unit deflection at each load position
- Fit a cubic spline through the supports and the deflected point (with appropriate end conditions)
- Find the deflection at the other load points, and the shear force at the deflected points
- Scale the deflections by Applied load/load for unit deflection
- Add the deflections for all the loaded points
- Fit a cubic spline through those points.

I have posted a spreadsheet using that procedure, including the example discussed in this thread, at:

http://newtonexcelbach.wordpress.com/2010/06/20/splinebeam-update/

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

You could use the Conjugate Beam Method

If I ever knew the conjugate beam method, I've long since forgotten it, but there is an interesting article about it here:

http://comp.uark.edu/~icjong/docu/icjASEE04ppr.pdf

I'm not sure I agree with everything he says, but it does provide a clear explanation of how to apply the method.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

Doug,

The Conjugate Beam Method is great for simple span beams, particularly with point loads because the bending moment diagram consists of straight lines. I learned it many years ago and have never forgotten it because there is so little to remember.

BA