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

Reply to
Bret Cahill
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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

Reply to
Mike Yarwood

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

Reply to
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

Reply to
Billy H

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