Beam Deflection 1

[hydromech]

Can anyone help with a the equation for calculating simple beam deflection.

I have a beam with a point load and a UDL, when I calculate the deflection using the moments to left of x I get a different answer than when I consider the whole beam with the point load.

I differentiate both equations twice using the same value for x and also the constants for each equation and the value when I consider the whole beam is always approx half that of when I consider only half of the beam.

what is the equation that considers the point load and the UDL..?

Thanks

Adrian

 

[MiketheEngineer]

Calculate the defl for the UDL and then the point load.

Add them together. Your are done....

 

[desertfox]

Hi hydromech

The simple way to get this deflection is as MikeTheEngineer said work out the deflections seperately and add them together click on this link and scroll down to the heading BENDING there you will find the formula for the deflections of the point loaded beam and UDL loaded beam.

http://www.roymech.co.uk/Useful_Tables/Form/Stress_Strain.html

If the point load is not at the center of the beam then you will need to use another method I would use Macaulay's method and again I attach a link.

http://www.freestudy.co.uk/statics/beams/beam tut3.pdf

If you need to use this method then give me a shout and I can help with the equations

regards

desertfox

 

[zekeman]

"Can anyone help with a the equation for calculating simple beam deflection.

I have a beam with a point load and a UDL, when I calculate the deflection using the moments to left of x I get a different answer than when I consider the whole beam with the point load.

I differentiate both equations twice using the same value for x and also the constants for each equation and the value when I consider the whole beam is always approx half that of when I consider only half of the beam.

what is the equation that considers the point load and the UDL..?"
--------------------------------------------------------
From your statement, it looks like you haven't properly accounted for the constant of integration after integrating
M/EI. You can't get it without dealing with both halves, each having a different moment equation and by then satisfying the equality of the slope at the point load.
and finally after the second integration, the 0 deflections at x=0 and x=L. Not so straightforward doing it analytically.

 

[hydromech]

Thanks for all the help...it is really appreciated. I have offered lots help in the past, but this is the first time I have needed help.

The beam is 5m long, supported at each end. There is a 70KN point load in the middle and a UDL of 15KN/mtr. I want to find the max deflection which obviously occurs at 2.5 metres, exactly where the point load is.

When I calculate the load using the moments about x, I differentiate twice 72.5X-(15X^2)/2 getting (72.5X^3)/6-(15X^4)/24 + CX + D. Setting the boundary conditions I get a value for C, and D goes to zero.

Substituting 2.5 for X and then multiplying the moments by 10^3/40x10^6 gives approximately -3.97x10-3 mtrs.

When I consider the whole beam with x at 5 mtrs, I use the moments 72.5X-(15X^2)/2-70(X-2.5).

Again integrating twice, I get(72.5X^3)/6-(15X^4)/24-70((X-2.5)^3)/6 + CX + D. Setting the boundary conditions I get values for C and D.

Substituting 2.5 for X and then multiplying the moments by 10^3/40x10^6 gives approximately -7.609x10-3 mtrs.

I would think that the second value for the deflection is the correct one because that uses the point load.

I thinking that if I add the point load deflection to the first answer I will get -7.609x10-3 mtrs...? I am close?

Thanks again for the help

Adrian

 

[desertfox]

hi hydromech

We need the second moment of area of the beam (I) and the (E)modulus of elasticity of the material to check your answer.
I could be wrong here but I think you might be better using the Macauley's method which gives an example very similar to your beam on page 12 or 13 on the second link I gave you


Desertfox

 

[desertfox]

hi hydromech

I have attached a file with the solution using Macauley's method, if you try and use the double intergration method I think the calculus gets long winded.
I need the (E) and (I) for your beam and I could of verified your deflections with my solution.
Hope this helps

regards

desertfox

 

[zekeman]

Neither of your solutions is correct since you have 2 domains because you are dealing with a discontinuous function at the center. You can look up the solution for the UDL, but your problem is with the point load at the center, P
for x<L/2 L/2<x<L
left side right side
M=Px/2 M=Px/2-P(x-L/2)


Integrating left side domain ( leaving out EI denominator for convenience)
dy/dx=Px^2/4+C
one more integration yields
y=Px^3/12+Cx+D
Now since y(0)=0, D=0
eq (1) y=Px^3/12+Cx
Now comes your problem.
You have no easy way of getting C since you're constrained to x<L/2 you now must deal with the right half; integrating the piece -P(x-L/2 from L/2 to x and adding to the first solution
y=Px^3/12+Cx-P(x-L/2)^3/6 for x>L/2
Now satisfying
y(L)=0, I get
C=-PL^2/16
and substituting in eq 1
y=Px^3/12-PxL^2/16
at y=L/2
y(L/2)=PL^3/96-PL^3/32=-PL^3/48
Restoring EI factor
ymax=PL^3/48EI
Add this to the UDL solution.
There are easier ways of doing this including looking it up but as an academic exercise it may be worthwhile.


(
y'(L/2)=0
yielding linear eq in C2 and D2
Finally
y(l)=0
yielding another eq in C2 and D2
solving 2 simultaneous eq
yields
C2 and D2
More generally,in the absence of symmetry,i.e point load located at x=p) the 4 constants are obtained by employing

left side right side
y(p) = y(p)
y'(p) = y'(p)
y(0)=0 y(L)=0

 

[zekeman]

correction
Disregard the "junk" below the worthwhile. which I previously aborted but it got through anyway.

 

[zekeman]

The simple way to get this deflection is as MikeTheEngineer said work out the deflections seperately and add them together click on this link and scroll down to the heading BENDING there you will find the formula for the deflections of the point loaded beam and UDL loaded beam.

http://www.roymech.co.uk/Useful_Tables/Form/Stress_Strain.html

--------------------------------------------------------------
Fox

I doubt that is Mccauley's method as you wrote it.
B=0 does NOT satisfy y(0)=0
and I doubt that Mccauley said that the the moment equation that you show which is correct for the right side has no validity for x<a

 

[desertfox]

Hi hydromech

I decided to test my solution by Macauley's method so I assumed a (E)=200*10^9 N/m^2 and (I)= 29900*10^-8 m^4
I worked out the deflection according to my Macauley method
and got a deflection of y= 5.089mm.

I then worked out seperate deflections for your problem firstly as a point loaded beam using the formula from the Roy Mech link given in my first post

single point load simply supported beam defl = 3.0483mm

UDL loading simply supported beam defl = 2.0413mm

total deflection 2.0413mm + 3.0483mm =5.0896mm

This shows that the Macauley solution I posted earlier appears to be correct.

I have attached my verification of the above for your reference.
The result from the Macauley should be valid for any part of the beam however you need to ensure all values for (x) are put into all the terms in the deflection equation and evaluated correctly.

regards

desertfox

 

[zekeman]

Quote"
I posted the link that you have in your last post and if you look I also said MiketheEngineer's method was the simplest.
I thought B must equal zero for y=0 when x=0 otherwise y would have a value.
If I am mistaken then please show what the value of B actually is.

desertfox "

Fox
Fact is as I mentioned previously, you cannot use an equation that is valid only for x>a to satisfy a boundary at x<a, namely at x=0. So, since I think the equation is incorrectly used, why would B be of any consequence.
Look at my earlier post to see this
If you have the original McCauley reference, I would be most interested, and you should take another look as well.Unless, maybe I am missing something.

 

[desertfox]

Hi zekeman

Actually if you look at page 13 on the second link of my first post it explains the Macauley Method and further if you scroll down to example 7 its a very similiar beam problem to the original post.
After the second intergration of the moment equation the constant B as to be zero for y=0 when x=0.
Now the beam problem in the case of the OP, the point load is central therefore the only place the slope of the simply supported beam can be zero is the point of maximum deflection ie centre of the beam. Therefore if I go back to the first intergration for the slope and make dy/dx = 0 I can find the constant (A)by substituting x= L/2 as in my attached file.
Also and I have calculated it, the constant (A) can be found from the second intergration ie the deflection equation by substituting x=L again B=0 the value of the constant (A) comes out at exactly the same value as I obtained with the slope equation.
There are only two points that the beam can have no deflection ie at x=0 and x=L, the slope can only be zero in this case at the centre of the beam.
Therefore the equation derived from the Macauley's can be used to find the deflection at any point along the beam.
You can prove it by working out the deflection at any point along that beam and then compare it by using standard beam formula for both the UDL and point load and summing them.

desertfox

 

[IFRs]

When faced with superposition of loads whose location of greatest deflection occurs a distinctly different places on a beam, I use MathCAD to calculate the deflection at each inch along the beam and then find the maximum of them when added them together. Confirming with graphs.

 

[kslee1000]

Lazy structural guys' solution to determine deflection of a simply supported beam with concentrate and uniform loads:

For single concentrate load (P) at a distance "a" from left end:

Dca = P*(a^2)*(b^2)/(3*E*I*L); L = a+b

For uniform load (W), the deflection at a distance "a" from left end:

Dua = W*a*[7*(L^4) - 10*(L^2)*(a^2) +3*(a^4)]/(360*E*I*L)

Total deflection at "a" by superposition:

D = Dca + Dua

 

[kslee1000]

Note: Dca is the deflection under the load, not the maximum.

 

[zekeman]

QuoteActually if you look at page 13 on the second link of my first post it explains the Macauley Method and further if you scroll down to example 7 its a very similiar beam problem to the original post.
After the second intergration of the moment equation the constant B as to be zero for y=0 when x=0.

------------------------------------------------------------
Fox,
OK, the mystery is solved. As you suggested, I looked at the Mccaulay method which IGNORES any negative bracket and thus effectively shows the different regimes so now I can accept ( without apology) the solution you presented.
I think the method is neat , though I think it is unnecessary and not much different from the classical method which integrates the moment equation through all of the regimes to the endpoint before determining the constants of integration.
An easier and more powerful method is the energy method where you integrate M^2/2EI (including all other loading) over the length and then take the partial derivative with respect to P (at the point xp), the point load and you are done. This gives the deflection under the load. If you want it at another location, just add a dummy load Q at that point , write the energy integral and take partial with respect to Q and then set Q=0.

 

[desertfox]

hi zekeman

I didn't know Macauley's method was a mystery; I assumed everyone was shown that method. I was introduced to it about 30 years ago; anyway the main thing is we agree it’s a solution.
Regarding the method i.e. Strain Energy or the latter method you mention "Castiglano's" I am not convinced they are any easier then the one I propose, however why don't you post a file with a solution using that method and the OP will have a choice and it will be an independent check for my solution.

Cheers

desertfox

 

[hydromech]

Thanks for all the help...

And the star winner is...desert fox.

that does not mean to say all other contributions are not very much appreciated.

thanks all

regards

Adrian

 

[desertfox]

your welcome hydromech

desertfox