Saint Venant's principle applied to ring

[L.Richter]

Homework Statement

An originally complete ring made of linear elastic material (Young's modulus, E and Poisson's ratio, v) is cut by a saw. A gap, delta, is generated by a pair of forces, P. Determine this force, P. (Use Saint Venant's principle) Inner radius of ring, a. Outer radius, b.

Homework Equations

forces: integral over the area, A of tsubA dA = f
moments: integral over the area, A of tsubA X = M

A = area
t = traction tsubi = sigmasubij dot nsubj where sigma represents stress
f = force
M = moment

boundary conditions?

The Attempt at a Solution

I am in a solid mechanics/stress analysis course and I'm having a problem applying Saint Venant's principle to this problem. My thoughts are that the forces, P (equal and opposite) that are generated by sawing the ring (which looks like a washer cut through the bottom thickness only) would equal the force, P that is internal in an uncut section of the ring. So I would be able to use an uncut ring to determine the force, P?

 

[nvn]

No, it sounds like the forces P are applied after the ring is cut, not during cutting.

 

[L.Richter]

If the forces are applied after the cut is made then how does Saint Venant's principle apply? As I understand it the forces that produce the displacement, delta, should equal the force in a distant section of the ring. Can I take a cross sectional area of the ring at a distant point and apply the math?

 

[nvn]

Yes, that sounds correct.

 

[L.Richter]

Thank you for your help. I just keep thinking about the problem statement and it asking to apply SVP. All the examples for the principle involve a beam and taking a point load or distributed load and concentrating it to a "far/distant" point and applying the math. I'm picturing the ring as a bent beam in order to apply the principle.

 

[L.Richter]

I took an element at the top of the ring so that:
N = P cos theta
V = P sin theta

psi(r,theta) = 9Ar^2 + B/r + Cr + Drlnr) cos theta

sigma rr = (2Ar - 2B/r^3 + D/r) cos theta
sigma theta theta = (6Ar + 2B/r^3 + D/r) cos theta
sigma r theta = (2Ar - 2B/r^3 + D/r) sin theta

applied BCs such that @ r =a or b: sigma rr = sigma r theta = 0
@ theta = 0 P = integral from a to b of sigma theta theta dr
I solved for A, B and D. They all contain the force P that I am ultimately solving for though. So now I'm stuck!

 

[L.Richter]

@nvn~ can you give me any suggestions on how to proceed?

 

[nvn]

L.Richter: Would you be able to post a dimensioned free-body diagram, showing applied loads, key points labeled with letters, coordinate system, and if possible, infinitesimal element?

(And maybe Mapes or PhanthomJay will see this, too. Both of these members are phenomenal.)

To answer your question about St. Venant's principle, this principle states that a force applied to a point spreads out to become evenly distributed at a distance (from the point of application) equal to the width or thickness of a body.

 

[L.Richter]

The attachment exceeds the size limits of the forum. Any suggestions??

 

[nvn]

L.Richter: See post https://www.physicsforums.com/showpost.php?p=2937012".

 

[L.Richter]

This is the best that I could do! Hopefully you can see it...

 

[nvn]

L.Richter: Excellent diagram. I did not understand the solution in post 6. If this ring is not a thin ring, I am not sure how to solve it; but maybe someone like Mapes might have ideas. However, if we can pretend your ring is a thin ring, you could solve it as follows. I would say, ignore shear deformation and axial deformation, which are negligible.

r = ring mean radius,
θ = cross section location, measured from -x axis, 0 ≤ θ ≤ pi,
M(θ) = bending moment = P*r*[1 - cos(θ)],
phi(θ) = rotation of the ring,
y(θ) = ring horizontal displacement.

phi(θ) = [1/(E*I)]*integral[M(θ)*r*dθ].

y(θ) = integral[phi(θ)*sin(θ)*r*dθ].

Boundary conditions: phi(pi) = 0; y(pi) = 0.

After you obtain y(θ), set y(0) = 0.5*delta, then solve for P.

 

[L.Richter]

Thank you nvn! Basically I'm solving for P the force that it takes to open the ring delta/2. Or the stress that is produced when the ring is cut that causes the displacement. I propose a psi function based on the vector diagram. Mine was f(r)cos theta. From there I get my Airy stress function psi(r,theta) which contains the constants that I need to solve for. I calculate my stress field (the sigma terms) and use the boundary conditions to plug back into my psi(r,theta) equation. Viola! But when I set the equations up from the boundary conditions I get the constants as functions of P and I'm trying to solve for P!

 

[L.Richter]

..."I solved for A, B and D. They all contain the force P that I am ultimately solving for though. So now I'm stuck!"

A = -P/2M
B = Pa^2b^2/2M
D = P(a^2 + b^2)/M

M = a^2 - b^2 + (a^2 + b^2) ln b/a

Can I use delta = integral from a to b P(x)/A(x)E dx ?? where delta is the displacement, P is the force, A is the cross sectional area and E is the modulus?

 

[L.Richter]

I will try post #12 also. Thanks!