Superposition to find deflection of beam with multiple point loads?

[StructureMan44]

I know the moment and shear force can be found via superposition. If I have a simply supported beam with 5 point loads, can the deflection be found using superposition of 5 beams with a single point load? I have been adding all 5 point loads together and treating it as a single center point load but the deflection from this is conservative.

 

[rb1957]

yes you can superimpose deflections. you'll have to calc the deflected shape (for a point load at a distance "a"), calc the deflection at different points along the beam for each of the five loads. start with the moment, and double integrate; something you should be able to do in excel, or solve algebraically.

simplifying to a single point load is extremely conservative. applying an equivalent UDL (sum forces over the length of the beam)probably isn't too bad.

another day in paradise, or is paradise one day closer ?

 

[canwesteng]

UDL is likely unconservative, unless the point loads are near the ends. Instead of doing all that integration in the conjugate beam method, you should be able to find a point load beam diagram with a formula for deflection at point x.

 

[rb1957]

yes, but most formulas show max deflection, which don't superimpose.

the double integration isn't That difficult, I could write it out in an email ...

M(x) = (Pb/L)*x ... 0<x<a
M(x) = (Pb/L)*x-P*(x-a) ... a<x<b
or M = (Pa/L)*y ... 0<y<b

EI*v(x) = ...

another day in paradise, or is paradise one day closer ?

 

[rb1957]

i get ...
EI*d(x) = Pb*x^3/(6L)+Cx, 0<x<a and
EI*d = Pa*y^3/(6L)+Dy, 0<y<b

C = Pab/(6L)*(2b-a)
D = Pab/(6L)*(2a-b)

for 5 different loading ... different P, a, b = L-a
evaluate at different x and y
summ
if you really want to, from the summ you could further refine your x and y to home in on the maximum

another day in paradise, or is paradise one day closer ?

 

[JeanLucPicard]

Be careful, you can add forces, stresses, deflections in anyway you want if it's under Hooks law! If it's in the plastic region, be cautious!

Live long and prosper!

 

[dhengr]

StructureMan44:
Given a simple beam with point loads, it is quite likely that the max. deflection will occur, at one of the point loads. Use a numerical integration method, such as Newmark’s Method, and put nodes at least at each of the loads, and you should come up with a pretty good answer. The Conjugate Beam and Moment-Area Methods would also work well on these types of problems. Of course, superposition will work, but you must calc./plot the deflections caused by each load at each of the other load points (nodes), and then repeat this process for each load.

 

[XR250]

Seems like a nice exercise, but don't you have software for this? If you are a practicing engineering, you need it!

 

[canwesteng]

Look at all the math I didn't have to do...

 

[IFRs]

If you have Mathcad, write each equation with a subscript of each inch of the beam for each load, add them together and find the MAX.
This could be done in Excel also.

 

[rb1957]

ok, look at what it gives you for a point load ... max deflection, deflection at load point ... things you can't summ.

it doesn't give you the generalised equation. ok, with some smarts you can modify the expression for "dx, x<a", i think.

the math is relatively easy, perhaps the trick is setting up the equations to simplify the integration.

another day in paradise, or is paradise one day closer ?

 

[canwesteng]

Deflection at point x is deflection at any point along the beam

 

[structSU10]

or you can use singularity functions to create a formula. for deflection due to a point load at any location the equation can be written as:

(1/E*I)*[(P/6)*<x-a>^3 - (P*(L-a)/L)*(x^3/6) + (P*a*(2*L^2 - 3*L*a + a^2)/6*L)*x]

the <x-a>^3 is the singularity function which equals 0 when the term in the <> is negative, then it acts just like a normal (x-a)^3

Programing it in something like Mathcad isn't bad, probably hard in excel, and a true pain if done by hand.

 

[IFRs]

It's trivial in mathcad

 

[rapt]

Yes, as long as the member is acting in the elastic range.

Definitely not for long term deflections for concrete or other inelastic materials.

 

[rb1957]

"Given a simple beam with point loads, it is quite likely that the max. deflection will occur, at one of the point loads." ... actually the odd thing is that it usually doesn't, it typically falls on the longer "b" side of the beam.

btw, there's an error in my calc, i had one of the boundary conditions wrong (and checked with a = L/2 which didn't find it). not unitll i tried to find max deflection (ie zero slope) ...

now i think ...
C = -Pab/6L*(a+2b)
D = -Pab/6L*(2a+b)

at least this seems to give zero slope where i expect it (on the "b" span)



another day in paradise, or is paradise one day closer ?

 

[rb1957]

"Deflection at point x is deflection at any point along the beam" ... read the fine print ... "when x<a"

another day in paradise, or is paradise one day closer ?

 

[desertfox]

Hi StructureMan44

If you are attempting to find the maximum deflection of the beam and at one point it occurs, then draw the shear diagram for the loads on the beam and where the shear force is zero the maximum bending moment location will be found.
I would then proceed to calculate that moment based on the beam loads, this can be followed by integration to find the maximum deflection value.

 

[mfernando]

I've set up an excel spreadsheet that splits the beam into 100 pieces and runs the each load case (point load, uniform, etc.) separately. Then combines shear, moment, and deflection at each point for each used load case. I just use the equations found in AISC so it's limited on the type of loading you can input, but I use it only for wood beam calculations at home. Results are almost identical to my test calculations using Enercalc. Shear is a little high because I haven't told excel to hold the shear constant from the supports to x=d yet.

 

[canwesteng]

@rb - when x<a is basically anywhere since you define a. If x>a, switch a and b around.