y" + Ay + B = 0

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

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