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4 Point Load Beam Deflection

4 Point Load Beam Deflection

4 Point Load Beam Deflection

(OP)
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

RE: 4 Point Load Beam Deflection

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.

RE: 4 Point Load Beam Deflection

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

 

RE: 4 Point Load Beam Deflection

Use superposition.

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

RE: 4 Point Load Beam Deflection

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
 

RE: 4 Point Load Beam Deflection

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/
 

RE: 4 Point Load Beam Deflection

(OP)
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

RE: 4 Point Load Beam Deflection

Quote:

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/
 

RE: 4 Point Load Beam Deflection

(OP)
Ok i'll have another look.

Thanks Again

RE: 4 Point Load Beam Deflection

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

RE: 4 Point Load Beam Deflection

(OP)
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

RE: 4 Point Load Beam Deflection

Quote:

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/
 

RE: 4 Point Load Beam Deflection

wsg1,

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

Neil  

RE: 4 Point Load Beam Deflection

(OP)
Thanks for all the help, got the spreadsheet working now.

Wsg1

RE: 4 Point Load Beam Deflection

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

RE: 4 Point Load Beam Deflection

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/
 

RE: 4 Point Load Beam Deflection

Quote:

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/
 

RE: 4 Point Load Beam Deflection

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

RE: 4 Point Load Beam Deflection

Great spreadsheet Doug.  That's a creative approach to use cubic splines.

To put in my own words:

you apply a unit displacement at location of a given force and allow cubic spline to determine the displacement all along the beam (it matches the true function which is a cubic within each region between discrete loads).

Knowing the displacement polynomial on each side, we can differentiate 3 times and multiply by E*I to get shear.  Difference in shear computed at the point using left-side formula and right-side formula is applied force to create the unit displacement.  Multiply unit displacement function by ratio of actual force over force associated with unit displacement to get actual displacement

Use superposition to sum all applied loads (it is not required to sum reaction loads associated with zero displacement).

The existing spreadsheet handles fixed or simply-supported end conditions.  It would be nice if it could handle "free" end condition also which should be a fairly straightforward alternate boundary condition.   If I get a chance I might study your code to see if I can modify it to do that.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: 4 Point Load Beam Deflection

Thanks Pete, and a nice summary of how it works.

Yes, it should be fairly easy to allow end cantilevers.  It's on the "to do" list, but I don't know when I'll get round to it.

If you do do it yourself I'll be interested to see what you come up with.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
 

RE: 4 Point Load Beam Deflection

is fitting cubic splines any easier than applying the general solution ...
M(x) = Pb/Lx  x<a         Pb/Lx-P(x-a)  x>a
EIv(x) = Pb/Lx^2/2+C1     Pax-Pa/Lx^2/2+C2
EId(x) = Pb/L^3/6+C1x+K1  Pax^2/2-Pa/Lx^3/6+C2x+K2

four boundary conditions d(0)=0, d(L)=0, v(0)=0, v(L)=0
C1 = K1 = 0
C2 = Pa/L*L^2/2 - PaL = -PaL/2
K2 = Pa/L*L^3/6-PaL^2/2+PaL^2/2 = PaL^2/6

so that
EIv(x) = Pb/Lx^2/2        Pax-Pa/Lx^2/2-PaL/2
EId(x) = Pb/L^3/6         Pax^2/2-Pa/Lx^3/6-PaLx/2+PaL^2/6

i think ...

RE: 4 Point Load Beam Deflection

rb1957 - I don't know if it's easier than other approaches, but if you have a general purpose cubic spline interpolation routine it is very easy to adapt it to finding deflections and forces in a continuous beam, and I find the concept quite neat.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
 

RE: 4 Point Load Beam Deflection

but there's no need !  you don't need to interpolate anything, you know the solution already

RE: 4 Point Load Beam Deflection

You're right that it is not interpolation... it's exact.  The Euler beam solution for discrete loads is a 3rd order polynomial in each region, so the cubic spline solution matching the specified boundary conditions will be identical to an algebraic solution matching those same boundary conditions

But the spline is also easily adapatable to handle more than 2 supports as shown in examples in the spreadsheet example.  And a change in boundary condition (simple support, fixed, free) will just change 2 equations out of n+1, not the whole solution as it would for a an algebraic solution of a single predefined geometry.
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: 4 Point Load Beam Deflection

i picked up on the word "interpolation" without looking at the method.  if it does match the boundary conditions then it'll be the right solution.  could you generalise the method to 4th order equations (it include distributed loads) ?

let's just leave it as "there are many ways to skin cats"

RE: 4 Point Load Beam Deflection

no - could not generalize to distributed loads.
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: 4 Point Load Beam Deflection

i realise a mistake with the solution i posted ... i've assumed fixed ends, for pinned ends v1(a) = v2(a) and d1(a) = d2(a) instead of the two slope BCs

RE: 4 Point Load Beam Deflection

Interpolation - it is interpolation. The method calculates the deflection at the supports (zero or specified deflection) and the loaded points (using standard slope-deflection methods), then fits a cubic spline through those points to find the deflection at intermediate points.  The cubic spline is an approximation to the actual shape followed by a real beam, but the point is that it is exactly the same approximation as used in standard beam bending theory, so the results will be exactly the same as any other method using the standard theory.

Distributed loads - You would need to fit a fourth order curve to deal with distributed loads directly, but you could convert the distributed load to a point load and a moment at each loaded node position, then use the same method with a cubic spline by applying a unit rotation as well as a unit deflection at the load positions.  This is how standard frame analysis programs do it, so the results will still be exactly the same as comes out of a standard program, or any other equivalent procedure assuming a cubic profile for the beam deflected shape in between node positions.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
 

RE: 4 Point Load Beam Deflection

Attached is my attempt to adapt Doug's approach to my own purposes.

I added capability for free boundary conditions as discussed.
I added the capability to put a spring under each intermediate support (along with existing capability to specify an alignment deviation of the support).   

Altering someone else's program was too tedious for me so I rebuilt it from scratch using matrix algebra (but sticking very close to the algorithm that Doug used).

There is an instructions tab, a math tab (explaining the math approach and defining the conventions/symbols used in the vba program) and 3 example tabs.

To me this approach seems to provide tremendous flexibility over trying to solve the Euler problem algebraically (by hand) for more complex problems including those that are statically indeterminate.  I shudder at the thought of trying to solve the "misalignment" example algebraically.

If any comments, suggestions or corrections, let me know.  Thanks again to Doug for his idea.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: 4 Point Load Beam Deflection

Thanks for posting that Pete.  I have been struggling with springs, so I'll be interested to see what you have come up with.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
 

RE: 4 Point Load Beam Deflection

"springs" ... where did that come from ... i reviewed the thread and couldn't see anything ... is the beam on elastic foundations ??

RE: 4 Point Load Beam Deflection

Quote:

"springs" ... where did that come from ... i reviewed the thread and couldn't see anything ... is the beam on elastic foundations ??


No the springs are an optional extra introduced along the way.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
 

RE: 4 Point Load Beam Deflection

Quote (rb1957):

"springs" ... where did that come from ... i reviewed the thread and couldn't see anything ... is the beam on elastic foundations ??
As I said, it was adapted "to my own purposes" which should signal that it was not a direct response to original question, but does relate to some of the followup discussion.  If you are intrested, my purpose in adding a spring was for attempting to estimate effects of shaft misalignment in rotating machines (dynamic effects neglected).

I felt it was warranted to post the spreadsheet back in this thread because:
1 – Doug asked me to share what I came up with.
2 – You suggested the cubic spline technique was inferior (22 Jun 10 8:52).  After the equivalence was clarified, you suggested it was no better than (22 Jun 10 10:11 cat skinning comment) solving the Euler beam equations in traditional algebraic fashion.   These methods are certainly mathematically equivalent, but it is a heck of a lot easier to cover a wide range of problems very easily by spline method imo.     It is a new useful tool for my toolbox and I think it might be for others also.   Please feel free to use it or ignore it as you see fit.
=====================
Here is one more rev of the spreadsheet with a few very minor improvements corrections.  The changes are again shown in RevHistory tab.  

=====================================
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