
Re: Beam Deflection in a Tapered Pipe
WE Jun 18, 2013 5:31 PM (in response to ptc4863626)Which values should be the same? y1 and y2=y3?
The way you had setup your double integral is not desireable as in the inner integral you use x as intgration variable and as upper limit. The upper limit should be changed to another name and this would be the integration variable of the outer intgeral. Doing so does not change the result, though.
y1 and y3 would only be the same if x^2*(3*Lx)=L^3.
If you look at your definition of y3 you get (by symbolical evaluation)
L^3*W/(6*E*I)
and compare this to your definition of y1
W*x^2*(3*Lx)/(6*E*I)
So for L=x you are missing a factor 2 to get the same result with y3 as with y1.

Re: Beam Deflection in a Tapered Pipe
AlanStevens Jun 19, 2013 3:07 PM (in response to ptc4863626)When you set Dw equal to D the integrals in y2 are easy to solve analytically (by hand). You get a result which is a factor of 2 different from y1, which is what the numerical results are telling you. Where you have pi/64 you should probably have pi/128
Alan

Re: Beam Deflection in a Tapered Pipe
FredKohlhepp Jun 19, 2013 12:19 PM (in response to ptc4863626)When in doubt, go bac to basics.

Re: Beam Deflection in a Tapered Pipe
ptc4863626 Jun 21, 2013 8:21 AM (in response to ptc4863626)In an effort to simplify the calc and let mathcad do the work, I defined (x=0) @ the free end and (L=x) at the fixed end and set up functions in terms of x. I also included graphs for the diameter, moment, moment of inertia, deflection angle, and deflection all in terms of x.
I think that the functions for diameter, moment and moment of inertia are correct. The deflection graph seems opposite what I expect in that @ x=0 the angle should be 0deg. Could that be due to the definite integration range? Should this change to something else?
The deflection graph crosses 0 so I know that can't be right.
I still think that I have an error in how I define the integrals.
