y" + Ay + B = 0

Feb 26, 2006 3 Replies

A long thin no stretch no stiffness weightless film is supported by horizontal parallel beams and filled with a liquid. What is the shape of the cross section of the trough?



Define:



D: depth of liquid in the center



T: tension or force in film at any point x



Tx: horizontal component of T



Ty: vertical comp. of T



Doing just the right half of the curve:



Weight of liquid from 0 to x = Ty = area of liquid from surface to curve:



Ty = integral of [D - y(x)] dx



but since



y' = Ty/Tx



then



y' = [int (D-y) dx]/Tx



Tx doesn't change with x so taking the derivative of both sides:



y" + Ay + B = 0



Where D/Tx = -B and 1/Tx = A



Is there an analytical solution?


Bret Cahill



I think it is probably defined by a catenary curve along the bottom edge (assuming that the beams are at the same height) .

Best of Luck - Mike

Cos seems to work.

I also tried deflecting it with horizontal forces on the edges and I got the exact same equation:

Define:

F: horizontal force

D: maximum deflection [at center]

The bending moment = F(D-y)

The curvature = y" = the bending moment = F(D-y)

or

y" + Fy -FD = 0

Bret Cahill

d^2y / (y-D) + F dx^2 =0 yes??

If so integrate twice. (Provided F and D are independent variables this should work).

Billy H

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required