# Bending of tapered (nonsymmetric) beams

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1. May 6, 2015

### FHbabble

I am trying to solve for the deflection and stress in a tapered beam (constant width b) shown in the attachment, and I am getting confused by at least 2 things: 1) how to deal with the "imposed" coordinate system, 2) the nonsymmetric shape and the corresponding neutral axis. [Assume that the bottom taper, f(x), is linear from h1-> h2 for simplicity]

In solving the symmetric tapered beam, it's straight forward enough to put the coordinate system aligned with the neutral axis, formulate the moment(x) from statics, and solve the 4th-order equation. Here, I think the solution y[x] is the displacement of the neutral axis, and the bending stress at a location (x,y) relative to the coord system at the neutral axis ~M(x)*y/I(x).

1) With regards to moving the coordinate system, I tried to solve for the displacement and stress for a simple, straight cantilever bar (length L, width b, height h) with only a vertical load=-P at the end using a coordinate system at the top of the bar and was not able to reconcile the values with the standard result using a coordinate system at the centroid. I was solving:
y1''[x] = P*(x-L)/(E*I1) , I1=bh^3/12 {coord system at centroid of beam}
y2''[x]=P*(x-L)/(E*I2), I2=bh^3/3 {coord system at top of beam}
y2=y1-h/2

2) For the tapered [nonsymm] case, will the neutral axis be curved (maybe following the centroid of the cross section)? Formulating the moment(x) using statics is kind-of a pain, because the internal forces will also contribute to M(x) since they do not go through the "chosen" coordinate system, but at the end of formulating M(x) with respect to the shown coordinate system I am stuck in formulating the differential equation with respect to the neutral axis, similar to my issue stated in 1):
Do I/(How do I) need to calculate the neutral axis for formulating the differential eq for the displacement?
-> How do I calculate the displacement of the top surface of the beam?
Is the stress still M(x)y/I(x), where y is measured from the centroid/neutral axis?

Thank you for taking a look. Your advice on 1) or 2) or both is really appreciated!

2. May 7, 2015

### SteamKing

Staff Emeritus
You raise some interesting questions about applying Euler beam theory. (Note: once the bending moment is determined, the deflection is the solution to a second-order differential equation, not a fourth-order one. The fourth order equation comes from starting with the loading of the beam, which can be dispensed with for uncomplicated loadings.)

You got different numbers because the standard derivation for beam deflection applies only to the neutral surface of the beam, which is assumed to remain unchanged in length during bending. The strains at points not located on the neutral plane are proportional to the bending moment and their location relative to the neutral plane.

For small deflections, the neutral plane is assumed to have zero normal strain. (Note: the normal plane passes through the centroid at each section of the beam.)

I think you are trying to make this analysis more complicated than the theory will permit.

Whatever vertical deflections you calculate will be for the neutral plane only. If you wish to calculate the deflection at the top or bottom surface of the beam, or even on the sides, some adjustments have to be made to the deflection calculated for the neutral plane.

http://www.assakkaf.com/courses/enes220/lectures/lecture9 .pdf

discusses the derivation of Euler beam theory. For the questions raised in this post, I would direct your attention to slide Nos. 43 and 44 in particular.

The latter slide (No. 44) illustrates how the plane of the cross section of the beam deforms due to bending, and what adjustments should be made to the deflection of the neutral plane when trying to evaluate the deflection of the beam cross section elsewhere in this plane.

With regard to calculating the location of the neutral axis (and the neutral plane) of the beam, this is usually done assuming the beam is unloaded. There are examples included in the linked article illustrating how to find the neutral axis of a cross section.

For all beams, as long as the deflections arising from bending are small, the usual bending stress calculation will apply, namely σ = My / I.

When calculating the location of the centroid of a cross section, usually the calculation is extended to determine the moment of inertia (I) as well.

For a tapered beam, even one which only has a linear taper, the location of the neutral axis and the moment of inertia will both be functions of position along the length of the beam, hence finding the deflection of the neutral surface would probably be accomplished easier by numerical integration of the M/EI curve for the beam to determine values of the slope and deflection of the beam at several different points along the length:

$θ(x) = \int^x_0\frac{M(x)}{EI(x)}dx+C_1$

$δ(x) = \int^x_0θ(x)dx+C_2$

where:

θ(x) - slope of the beam
δ(x) - deflection of the beam
I(x) - moment of inertia of the beam, as a function of the length
E - Young's modulus for the beam material

C1 and C2 - constants of integration; determined by applying the boundary conditions at the fixed end, i.e. θ(0) = δ(0) = 0.

3. May 7, 2015

### FHbabble

Thank you for your response and insight!

I am going to try to use the curved neutral axis as my datum and see if this will work for my problem. I am assuming (or I think elasticity assumes) that the solution for the displacement of the surface will follow the displacement of the neutral axis, so whatever y[x] I get the y_surface[x]=y[x]+thickness/2[x] {assuming a square cross section for the beam}
In looking at the reference you pointed to (which is great by the way). It looks like the issue with moving the coordinate system off of the neutral axis is that the strain is defined with respect to stretching the bar and having the non-stretched length provide the coordinate system. I was hoping there would be a straight forward transformation, but I will see if I can avoid that.

With regards to the 4th Order vs 2nd Order ODE, I have some confusion over this. I convinced myself that a distributed load vs a concentrated load (with same moment) gives the same results using a 4th Order and 2nd Order ODE, respectively. Also, I've looked at statically indeterminate supports and see that you need to use the 4th ODE (basically so you have enough constants of integration to evaluate the redundancy). However I am unclear regarding combining axial loads.

Specifically, the transverse loading I am looking at is going to be distributed and also dependent on y[x], and the axial load follows a linear distribution along the side of the bar (starting at a location "b" below the centroid). The axial load is independent of y[x], but the moment arm is a function of y[x].

Working this out for an fixed compressive axial load (P) and a distributed load w(x,y[x]) I ended up at:
EI(x) D^4y/Dx^4 + P D^2y/Dx^2=-w(x)

(Sorry this looks so ugly, I haven't figured out how to get the latex code to actually end up in the thread).

I am unclear how to modify this for both moving from an axial load P to a distributed axial load and positioning the beginning of the load below the centroid. I think the P*D^2y/Dx^2 comes from the moment introduced by displacing the beam y[x]. Can the axial load be incorporated just into the moment equation to eliminate the D^2/Dx^2 term?

Thanks!

Last edited: May 7, 2015
4. May 7, 2015

### SteamKing

Staff Emeritus