Cantilevered tapered rod 1

[TeejT]

I want to calculate the deflection of a cantilevered tapered rod with an applied point load P acting perpendicular to the length of the rod at the free end. Let's say at the supported end r = R and at the free end r = r.

I'm not so familiar with this stuff... I have this for a cylindrical rod.

v (deflection) = - (P*L^3) / (3*E*I)

I = 1/4*(pi)*r^4

P = const.
L = const.
E = const.
I = variable from r = r to r = R

Help please!

 

[GregLocock]

There is not a closed form analytical solution for tapered sections, in general. I solve them using a piecewise linear approach based on M/I=E/R, for small slices all the way along the beam. this has the advantage that I can vary in any way you like.

FEA is an alternative of course.




Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[Lion06]

You can solve this. Look at Castigliano's Theorem. I'll post more when I get into the office.

 

[Ltwine]

Setup deflection formula, perform intergration from free end towards support.

 

[rb1957]

i disagree with greg,

I = i(x) ... what that means is that moment of inertia of the beam is a function of x. you've got M = m(x) (the moment in the beam as a function of x, slope = int(M/EI) dx = 1/E*int(m(x)/i(x)) dx; and displacement = 1/E int(int(m(x)/i(x)dx)dx) ... clear as mud.

if you just looking for the maximum value, that's probably in reference books, like Roark (maybe wiki, maybe xcalc.com).

if you're after the solution, this gets messy Very quickly, so i often resort to greg's solution (piecewise linear approximations)

if you're after the solution for a student problem, you shouldn't be posting here ...

 

[Ltwine]

I is function of r, and M & r are function of x (in direction of length). Messy intergration. "Strength of Material" text will show you how to setup.

 

[Lion06]

Ok, I have a little more time to write now.

I disagree with Greg that there isn't a closed-form solution to the problem. As long as you have a I as a function of x, then it's no problem.

I prefer to use Castigliano's Theorem. It relates the internal strain energy to the external work done.

1. Set x=0 at the free end
2. Write the loading as a function of x
3. Write the MOI as a function of x (this is where the messy integration is going to come in)
4. Place a fake point load at the tip and label it P1
5. Develop an expression for the moment as a function of x (in terms of the distributed load from step 2, AND the fake point load from step 4)
6. Determine the partial derivative of the moment function (from step 5) with respect to P1
7. Set up the integral (from x=L to x=0) of (M/EI)*(partial M/partialP1)dx
8. E will be constant, so pull that out and integrate what is left with respect to x
9. Substitute 0 for P1 everywhere it appears
10. You now have the deflection of the tip of the cantilever!!

 

[Ltwine]

There is a closed form solution for r = constant, a special case.

 

[Lion06]

r doesn't have to be constant to get a closed form solution.
I just posted a procedure above.

 

[Ltwine]

I don't think integration over formula contains variables is a "closed form solution".

Check out from links below.

http://mathforum.org/library/drmath/view/54553.html

http://www.riskglossary.com/link/closed_form_solution.htm

 

[Lion06]

My understanding is that closed form is any solution where you don't need to use a numerical approximation to get an answer. If you can get x=0.3", then you're solution is closed form. If you have to use infinte series or some other numerical approximation, then you don't have a closed form solution.

 

[Ltwine]

for this case, x=0, 0.001, 0.01, 0.1, 0.1001.....1.0, an infinate series.

 

[Lion06]

LTwine-

Your links confirm my idea that any solution which can be "solved" is closed form, it doesn't matter if it is integration with variables. By definition, integration contains variables.

Not having a closed form solution is when you use a numerical approximation, it says so right in the links you provided, and this is what I've always learned. Read the links again.

 

[Lion06]

What are you talking about? Not trying to be an ass here, seriously, but brush up on your math. Integration is NOT an infinite series. By your definition, all integration is an infinite series.

 

[TeejT]

StructuralEIT -
Develop an expression for the moment as a function of x (in terms of the distributed load from step 2, AND the fake point load from step 4)

I don't understand why I'm developing this expression when the one I have doesn't even include M.
I have:
v (deflection) = - (P*L^3) / (3*E*I)

Thanks for the input.

 

[Ltwine]

I missed a line - both r & I vary with x.
As pointed out by the linked material, there are at least two camps on what presents a "closed form solution". Basically I want to point out that there wasn't anything wrong with Greglocock's statement, though you may not agree with it.

Teejt:

v = summation of (P*L[x]^3)/(3*E*I[x]), from x = 0 to 1.
Hope I didn't make mistake this time.

 

[paddingtongreen]

StructuralEIT, when I was young and going to the technical college one day a week, I tried to use methods I learned at school, but I wasn't permitted to use calculus. In those days, almost everyone was convinced that calculus was haarrrd so they shied away from it, my boss said nobody was confident enough to check it, in any case, there were means and methods that could be used with the desired accuracy.

In this case, we would divide the length into segments and calculate the M/EI for the junctions and then use those; the smaller the segments the closer we got to your answer. Perhaps six segments would be close enough for design purposes.



Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[EdR]

Deflection at free end is 4 P L^3/(3 pi E r r0^3) where

r=radius at free end and
r0= radius at fixed end....

obtained by integration of y''=M(x)/(E I(x))

Ed.R.

 

[miecz]

Nice solution EdR. The attached solution arrives at the same result, but mathcad performed the integration for me.

 

[Lion06]

Nice job Ed.

That looks like a closed-form solution to me.

paddington-
I don't disagree, I was just pointing out that there is a way to do it with the varying I that didn't involve stepping it.