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Deflection of beam, reinforced in the center section, multiple Ix values. 2

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bsmet95

Mechanical
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Aug 16, 2007
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I'm designing a beam for a point load, and to keep compressive stress down, I have to add a stiffener in the center portion, according to the attachment. Is there a way to calculate deflection for this application?
 
you've got a simple enough beam (two constant Is) that you don't really need FEA for it.

use the double integration method, section the beam where the section changes,

another idea is to solve the beam bending moments (they don't change with I), then analyze the two different inertias as two different beams ...
the outer beams are loaded by the external reaction only ... easy enough to calc displacements,
the inner beams deflects relative to its ends (where there's shear and moment into the smaller beam) ... easy enough to calc displacements,
i think you'll use the slope and the end of the smaller (outer) beam as a boundary condition for the inner (larger) beam.
add the two together.
 
another interesting question on your set-up is what l/rt to use to calculate your bending allowable....had the same problem a few years back and could not find a definitive answer...so I went consevative and used the properties of the smaller bm the full length...I may be wrong here, but I do not think your typical 2D/3D has the capability of calculating this allowable bending stress unless the bm is continously braced...I would be tempted to do the same thing for the deflection, unless it's a critical parameter...
 
Thanks, rb1957. I have tried it that way, but wasn't sure if it was OK to use.I also tried our FEA but was having a terrible time trying to get all the constraints, etc., right.
 
A hand calculation using the Conjugate Beam Method is very easy to apply.

First, draw the M diagram and the M/EI diagram for the real beam.

Then, consider the M/EI diagram as a variable load on the conjugate beam. The resulting shear at any point on the conjugate beam represents the rotation at the same point on the real beam. The resulting moment at any point on the conjugate beam represents the deflection at the same point on the real beam.

The Conjugate Beam Method is simply an extension of Moment-Area principles.


BA
 
BA:
You forgot to mention our other old favorite, and that’s Newmark’s Numerical Integration Methods. Much like the Conjugate Beam Method, in fact.
 
dhengr, I agree that the Newmark Method could be readily implemented to solve the problem. I don't believe these methods (Conjugate Beam and Newmark) are currently taught in most engineering schools, but Moment-Area theorems are probably part of the curriculum in most schools today. If not, they can be Googled.

BA
 
It takes more time to input the data in a program than to calculate the answer.

If moment of inertia is I in the middle and kI in both ends, the center deflection can be calculated as:

[Δ] = PL3/48EI + (Pa3/6EI)(1/k-1)



BA
 
If you have the formula easily accessible in your head, or on a conveniently dog-eared page of a book on an adjacent bookshelf, and you only need a simple load configuration, then BA's approach is undoubtedly the quickest; failing that a simple beam program will do the job very nearly as quickly and with much more flexibility.

On the other hand, if it takes more than about a minute to create and solve a single simply supported beam in an FEA program then this would be a good exercise to use in getting some practice at using the program.

Doug Jenkins
Interactive Design Services
 
geeze guys, this is a simple beam problem. IMHO i think we should all be able to work out these problems by hand ... graphically, algebraically, whatever.

there is a place for canned beam solvers ... really complicated loadings and geometries, or for checking your hand calcs (or using your hand calcs to check the canned answer).
 
BAretired,

In your formula, what is k-1 in the denominator?
 
It is not k-1, it is (1/k - 1). So if k = 0.5, (1/k - 1) = 1. And if k = 2, (1/k - 1) = -0.5.

BA
 
I omitted to state: if k = 1, (1/k -1)= 0 and the deflection is the simple beam deflection of a beam with stiffness EI over its full length under a concentrated load at midspan.

BA
 
Ok, I must be having a brain cramp today. I'm not sure how, if k = 0.5, 1/k-1 = 1?
 
it's not 1/(k-1) but (1/k) -1 ...
 
rb1957:

Point taken.

However, the OP did stipulate a point load, but not the location. That adds a different dimension to the problem, and considering the ramifications with other different loading situations, I would defer to the computer here 95% of the time just for the flexibility of analysis, knowing things can change. Setting the program up in RISA takes very little time.

Mike McCann
MMC Engineering
 
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