Harder beam equation

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Hi, I would need an equation for beam deflection shown in this picture (hope you get it, I did it quickly with paint). I´m interested in the maxmum deflection (middle) and how much the deflection is in the part L1. I´m wery happy if somebody can help!
 

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  • #2
Mech_Engineer
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Have you tried splitting the beam into pieces and solving for each part individually with continuity conditions in between them?

I looked through both my Rourke's manual and Mechanics of materials text book and was unable to find a pre-solved solution for this particular loading condition.
 
  • #4
Thank you both! I´m trying to solve it splitting like Mech_Engineer said, but its easy to make mistakes.. :) It seems that topic that Stewartcs found is very helpful!
 
  • #5
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  • #6
Thanks, looks also helpful. I wonder how you would calculate maximum deflection if moment of inertia is I2 in parts L2 and I1 in part L1?

..and also what is the deflection in that structure only in part L2?
 
  • #7
Try this: http://www.aps.anl.gov/asd/me/Calculators/ElasticBeam2.html
It does not show any equations, but gives you quick result
Actually unfortunately this wasn´t really helpful, because it doesn´t calculate the maximun deflection in case 34. Thank you anyway!

Anyway I am stuck with the problem if the moment of inertia varys like I wrote in my last post and also calculating the deflection only in part L2...
 
  • #8
You can work out the solution analytically.

The first portion L2 of the beam is easy enough. You only have a shear force from the support to give moments. Get the moment equation, integrate it to get displacement as a function of x. Obtain also the moment M1 at the point where q starts, ie beginning of L1.

Next obtain the moment equation for length L1. Remember to add the moment M1 at the beginning of L1. Integrate this equation.

Now, the 1st equation tells you the displacement and gradient of the beam at the end of L2 and where L1 starts. This must be matched when you solve the 2nd equation for L1. Essentially this means solving for the constants after integrating twice.

After this you're done.
 
  • #9
You can work out the solution analytically.

The first portion L2 of the beam is easy enough. You only have a shear force from the support to give moments. Get the moment equation, integrate it to get displacement as a function of x. Obtain also the moment M1 at the point where q starts, ie beginning of L1.

Next obtain the moment equation for length L1. Remember to add the moment M1 at the beginning of L1. Integrate this equation.

Now, the 1st equation tells you the displacement and gradient of the beam at the end of L2 and where L1 starts. This must be matched when you solve the 2nd equation for L1. Essentially this means solving for the constants after integrating twice.

After this you're done.
Thank you!

Can you help is the answer before solving constants something like in picture? And then, how do I solve constants? (F is force from support)

Well, in first equation I get atleast this..
v(0)=0 -> C2=0
 

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  • #10
Thank you!

Can you help is the answer before solving constants something like in picture? And then, how do I solve constants? (F is force from support)

Well, in first equation I get atleast this..
v(0)=0 -> C2=0
You've got the gist.
Since v=0 at x=0, C2 = 0
And using symmetry, dy/dx = 0 at x=(L1/2 + L2). Use this to solve for C3.

Now just equate the vertical displacements and gradients at the point where q starts to solve for your remaining constants. 2 unknown constants, 2 equations, and you're done.
 
  • #11
You've got the gist.
Since v=0 at x=0, C2 = 0
And using symmetry, dy/dx = 0 at x=(L1/2 + L2). Use this to solve for C3.

Now just equate the vertical displacements and gradients at the point where q starts to solve for your remaining constants. 2 unknown constants, 2 equations, and you're done.
Thank you a lot!

Did I do it at all like you ment? Somehow I don´t believe it´s correct.. I´m hopeless.
 

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  • #12
Mech_Engineer
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This post has sort of had me transfixed for a while, so I finally decided to try solving the probem using MathCAD. The answers I came up with seem to at least make sense, but as was mentioned before it is VERY easy to make one small mistake that follows you through to the end.

The final solution I came up with is:

Edit- Equation is incorrect, corrected on Pg.2

This is the maximum deflection along the entire beam, which ends up in the middle due to the symmetric loading condition. Suffice to say it was a LOT of work to get there.
 

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  • #13
Mech_Engineer
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Thank you!

Can you help is the answer before solving constants something like in picture? And then, how do I solve constants? (F is force from support)

Well, in first equation I get atleast this..
v(0)=0 -> C2=0
I noticed that you are trying to define the beam in two parts rather than three. The only way this would work would be to cut the beam down the middle and only solve for one half of it (which should in theory work as long as you get your boundary conditions right); keeping statics in mind for this would be a must.

However, the way you're trying to do it right now doesn't work because you can't just say the equation is the same for the unloaded parts on the left and right hand sides of the beam. This is most easily seen in the fact that the shear forces on the left and right hand sides of the beam are NOT equal; in addition to the fact that you can only have one origin to measure x from if you are using one coordinate system.

What you need to do is split the beam into three pieces, describe the shear force in each section of the beam by doing a shear-force diagram, and then integrate each shear force equation THREE times (or alternatively describe the load on each section of the beam, and integrate 4 times, giving you 12 constants to solve for later). This will give you a grand total of nine unknown coefficients of inegration, which must be solved for by using boundary conditions and continuity conditions between each section of the beam (deflections are equal, slopes are equal, etc.)

Integration procedure tree:

[tex]\nu''''=\frac{q(x)}{EI}[/tex]

[tex]\nu'''=\frac{V(x)}{EI}[/tex]

[tex]\nu''=\frac{M(x)}{EI}[/tex]

[tex]\nu'=\theta(x)[/tex]

[tex]\nu=\delta(x)[/tex]


After you do all of that, you will have a whole bunch of equations that describe discrete sections of the beam's deflection, angle, and other properties.
 
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  • #14
FredGarvin
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This post has sort of had me transfixed for a while, so I finally decided to try solving the probem using MathCAD. The answers I came up with seem to at least make sense, but as was mentioned before it is VERY easy to make one small mistake that follows you through to the end.

The final solution I came up with is:

[tex]\delta_{max}=\frac{qL^4+16qa^4}{384EI}[/tex]

This is the maximum deflection along the entire beam, which ends up in the middle due to the symmetric loading condition. Suffice to say it was a LOT of work to get there.
Nice......
 
  • #15
Mech_Engineer
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Nice......
My favorite part is that is solved that system of equations in about 3 seconds :cool: By hand, that might have taken as long as the rest of the solution combined... with or without linear algebra.
 
  • #16
This post has sort of had me transfixed for a while, so I finally decided to try solving the probem using MathCAD. The answers I came up with seem to at least make sense, but as was mentioned before it is VERY easy to make one small mistake that follows you through to the end.

The final solution I came up with is:

[tex]\delta_{max}=\frac{qL^4+16qa^4}{384EI}[/tex]

This is the maximum deflection along the entire beam, which ends up in the middle due to the symmetric loading condition. Suffice to say it was a LOT of work to get there.

Nice indeed!

Thanks alot for your effort!

I must say thoug, I don´t believe it´s correct. If a=0, you get qL^4/384EI wich is five times smaller than known 5qL^4/384EI. Anyway, I think that equation that prex and mrMikee have found in thread http://www.eng-tips.com/viewthread.cfm?qid=158137&page=10 is more correct, maybe I shold buy/borrow Bloddgett and see how it´s evaluated.

I´m still working with the case of different EI in L2 and L1. I hope I get them with your help. Thank you!
 
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  • #17
stewartcs
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Anyway, I think that equation that prex and mrMikee have found in thread http://www.eng-tips.com/viewthread.cfm?qid=158137&page=10 is more correct, maybe I shold buy/borrow Bloddgett and see how it´s evaluated.
The equation they gave appears correct to me...that's why I gave you the link to it in the first place.

From Blodgett "Design of Welded Structures" Case 3c for beam supported at both ends with uniform load partially distributed over span:

When a = c uniform load is length b centered on span and the deflection at center is given by...

[tex]\delta = wb \cdot \frac{8L^3-4b^2L+b^3}{384EI} [/tex]

where,

w = unit load
L = length

Of course that is just the answer and not how it is derived. The book may not show how it is derived as they typically do not show how every formula (if any) is derived. Hand books just typically give the end result, i.e. a formula you can use to find the answer you need.

I´m still working with the case of different EI in L2 and L1. I hope I get them with your help. Thank you!
If the rigidity (EI) is different, that equation won't work. You'll probably have to treat it as a composite structure.

CS
 
  • #18
Pyrrhus
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Hey there,

I'm wondering if you truly need an equation describing the complete deflection of the beam or you just need the maximum deflection of the beam. If is the latter you require, i recommend using an energetic method like virtual work or Castigliano's theorem which is easier to work with paper and pencil.
 
  • #19
Mech_Engineer
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I must say thoug, I don´t believe it´s correct. If a=0, you get qL^4/384EI wich is five times smaller than known 5qL^4/384EI. Anyway, I think that equation that prex and mrMikee have found in thread http://www.eng-tips.com/viewthread.cfm?qid=158137&page=10 is more correct, maybe I shold buy/borrow Bloddgett and see how it´s evaluated.
Yeah, I had noticed that discrepancy too. it turns out my boundary conditions weren't quite right. When I changed the last boundary condition from the slope being zero in the middle to the bending moment being defined in the middle, everything worked out.

[tex]\delta_{2}(x)=-\frac{qL^3x-6qLa^2x-2qLx^3+qa^4+6qa^2x^2+qx^4}{24EI}[/tex]

[tex]\delta_{max}=-\frac{5qL^4-24qL^2a^2+16qa^4}{384EI}[/tex]
 

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  • #20
Mech_Engineer
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I´m still working with the case of different EI in L2 and L1. I hope I get them with your help.
Now THAT gets really ugly...

[tex]E_{1}[/tex] and [tex]I_{1}[/tex] are in section 1 (left hand side)
[tex]E_{2}[/tex] and [tex]I_{2}[/tex] are in section 2 (middle)
[tex]E_{3}[/tex] and [tex]I_{3}[/tex] are in section 3 (right hand side)
 

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  • #21
Yeah, I had noticed that discrepancy too. it turns out my boundary conditions weren't quite right. When I changed the last boundary condition from the slope being zero in the middle to the bending moment being defined in the middle, everything worked out.

[tex]\delta_{2}(x)=-\frac{qL^3x-6qLa^2x-2qLx^3+qa^4+6qa^2x^2+qx^4}{24EI}[/tex]

[tex]\delta_{max}=-\frac{5qL^4-24qL^2a^2+16qa^4}{384EI}[/tex]

You got it!

Great!
 
  • #22
Now THAT gets really ugly...

[tex]E_{1}[/tex] and [tex]I_{1}[/tex] are in section 1 (left hand side)
[tex]E_{2}[/tex] and [tex]I_{2}[/tex] are in section 2 (middle)
[tex]E_{3}[/tex] and [tex]I_{3}[/tex] are in section 3 (right hand side)

Wow,

thank you a lot. I think I could say problem solved, indeed!
 

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