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
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
RE: 4 Point Load Beam Deflection
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
Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
RE: 4 Point Load Beam Deflection
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
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
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
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
Thanks Again
RE: 4 Point Load Beam Deflection
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
Seems about the right order of magnitude ;)
Here's the spreadsheet.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: 4 Point Load Beam Deflection
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
No, 3a(L^2) is right.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: 4 Point Load Beam Deflection
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
o00o-(_)-o00o
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RE: 4 Point Load Beam Deflection
Wsg1
RE: 4 Point Load Beam Deflection
BA
RE: 4 Point Load Beam Deflection
- 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:/
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: 4 Point Load Beam Deflection
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
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
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.
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RE: 4 Point Load Beam Deflection
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
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
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: 4 Point Load Beam Deflection
RE: 4 Point Load Beam Deflection
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.
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RE: 4 Point Load Beam Deflection
let's just leave it as "there are many ways to skin cats"
RE: 4 Point Load Beam Deflection
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RE: 4 Point Load Beam Deflection
RE: 4 Point Load Beam Deflection
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
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.
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RE: 4 Point Load Beam Deflection
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RE: 4 Point Load Beam Deflection
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: 4 Point Load Beam Deflection
RE: 4 Point Load Beam Deflection
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
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.
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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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