Deflection of a Beam using Double Integration

In summary, the conversation discusses the determination of maximum deflection of a simply supported beam with concentrated couples of equal value and opposite direction at a distance of a. The person is struggling with finding the moment function using double integration method and determining the appropriate boundary conditions. The solution is given as (5(C)(a^2))/(8EI).
  • #1
voodoonoodle
2
0

Homework Statement



Need to determine the maximum deflection of the beam, by using double integration, and EI is constant.

...>...<
_______(_______)_______
^........o
|<--a-->|<--a-->|<--a-->|

Tried to draw this the best I could. It is a simply supported beam, with a concentrated couples of equal value (C) but opposite direction, with a distance between of (a).

Homework Equations



I pretty sure that my boundary conditions are: v=0 @ x=0 and v=0 @ x=3a

and my max. deflection will occur at x=(3/2)a

The Attempt at a Solution



My problem is that I don't know how to get started, all the examples show only one concentrated moment. I'm assuming I have no reactions, because when I sum the moments around either side, the moments cancel out.

I have tried using that M=C(x-a) and that didn't come out correct. So basically I'm looking for help in trying to get the moment function.

I have also tried M=C where I get the following

slope as a function of x = Cx/EI + C1
deflection as a function of x = Cx^2/2EI + C1x + C2

boundary conditions:
v=0 @ x=0 so C2 = 0
v=0 @ x=3a so C1 = (-3(C)(a))/(2EI)

so then I come up with an answer of (-9(C)(a^2))/(8EI) for the max deflection using x=(3/2)a
 
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  • #2
Well, it's been way too long since I used the double integration method, but it does bring back some haunting memories. If you're looking at y" =M(x)/EI, my suggestion would be to calcualte M(x) using shear and moment diagrams. Looks to me like you've got no shear at all anywher along the beam, and just a constant moment of c in between the two applied couples, and no moment on either side of those couples to the support points.
 
  • #3
PhantomJay, that is correct. I'm just not sure how to come up with M(x). I have tried several things for it but still can't get the answer they give for Max Deflection. Which I know to be (5(C)(a^2))/(8EI).
 
  • #4
voodoonoodle said:
PhantomJay, that is correct. I'm just not sure how to come up with M(x). I have tried several things for it but still can't get the answer they give for Max Deflection. Which I know to be (5(C)(a^2))/(8EI).
Well, M(x) =0 between 0 and a; M(x) = C between a and 2a; and M(x) = 0 between 2a and 3a. Now double integrate and use proper boundary conditions , which is probably what you're trying to do. I think the key is to deternine the boundary conditons. I've been relying on tables and charts, so my calculus is a bit rusty at this point.
 
  • #5
What a coincidence, I found this post: http://forums.mathalino.com/general-engineering-sciences/strength-of-materials/beam-deflection-by-double-integration-method" [Broken]. It is exactly the same problem as this one.
 
Last edited by a moderator:

What is the concept behind double integration in beam deflection?

Double integration is a mathematical method used to calculate the deflection of a beam under a given load. It involves integrating the bending moment equation twice, first to obtain the slope of the beam and then again to obtain the deflection.

How do you determine the boundary conditions for double integration in beam deflection?

The boundary conditions for double integration are determined by considering the supports and loading of the beam. The boundary conditions are typically expressed as fixed or simply supported, and can be used to solve for the integration constants in the deflection equation.

What are the assumptions made in double integration method for beam deflection?

The assumptions made in double integration method include: the beam is initially straight and uniform, the material is elastic and follows Hooke's law, the beam is subjected to a single concentrated load or distributed load, and the beam is loaded within its elastic limit.

How does the load distribution affect the deflection of a beam using double integration?

The load distribution directly affects the deflection of a beam using double integration. A concentrated load will result in a deflection curve with a single point of maximum deflection, while a distributed load will result in a deflection curve with a parabolic shape and a maximum deflection at the midpoint of the beam.

Can double integration be used for all types of beams?

Double integration can be used for most types of beams, including cantilever, simply supported, and overhanging beams. However, it may not be applicable for beams with complex loading or those that do not follow the assumptions of the method.

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