Verification of stress function

[roldy]

Homework Statement

The stress function \phi(x,y) = \frac{q}{4c^2}\left(c^2xy - cxy^2 - xy^3 + cLy^2 + Ly^3\right) is proposed as giving the solution for a cantilever (0 = x < L, -c \leq y \leq c) loaded by uniform shear along the edge y = +c, and stress/traction free along the edge y = -c and x = L.

Check if the stress function actually solves the problem in question.

Homework Equations

\phi(x,y) = c_1x^2 + c_2xy + c_3y^2 + c_4x^3 + c_5x^2y + c_6xy^2 +c_7y^3 + c_8x^4 + c_9x^3y + c_{10}x^2y^2 + c_{11}xy^3 + c_{12}y^4 + c_{13}x^5 + c_{14}x^4y + c_{15}x^3y^2 + c_{16}x^2y^3 + c_{17}xy^4 + c_{18}y^5

(1) \sigma_{xx}=\frac{\partial^2\phi}{\partial y^2}
(2) \sigma_{yy}=\frac{\partial^2\phi}{\partial x^2}
(3) \sigma_{xy}=-\frac{\partial^2\phi}{\partial x \partial y}

The Attempt at a Solution

I'm letting q represent \sigma_{xy}

First I setup the boundary conditions:
@ x = 0
\sigma_{xx} = unsure
\sigma_{xy} = unsure

@ x = L
\sigma_{xx} = 0
\sigma_{xy} = 0

@ y = c
\sigma_{yy} = 0
\sigma_{xy} = q

@ x = -c
\sigma_{yy} = 0
\sigma_{xy} = 0



Because the loading is symmetric about the y-axis (x = 0), odd powers of x in \phi(x,y) are canceled.

\phi(x,y) = c_1x^2 + c_3y^2 + c_5x^2y + c_7y^3 + c_8x^4 + c_{10}x^2y^2 + c_{12}y^4 + c_{14}x^4y + c_{16}x^2y^3 + c_{18}y^5

Because the loading is not symmetric about the x-axis (y = 0), even powers of y in \phi(x,y) are canceled.

\phi(x,y) = c_5x^2y + c_7y^3 + c_{14}x^4y + c_{16}x^2y^3 + c_{18}y^5


From this new expression for \phi(x,y), I can use (1), (2), and (3) to find the functions for the stresses.

\sigma_{xx} = 6c_7y + 6c_{16}x^2y + 20c_{18}y^3
\sigma_{yy} = 2c_5y + 12c_{14}x^2y + 2c_{16}y^3
\sigma_{xy} = -2c_5x - 4c_{14}x^3 - 6c_{16}xy^2


Now I apply the boundary conditions

@ x = 0
unsure = \sigma_{xx} = 6c_7y + 20c_{18}y^3
unsure = \sigma_{xy} = ?

@ x = L
0 = \sigma_{xx} = 6c_7y + 6c_{16}L^2y + 20c_{18}y^3 \Rightarrow 0 = 3c_7 + 3c_{16}L^2 + 10c_{18}y^2
0 = \sigma_{xy} = -2c_5L - 4c_{14}L^3 - 6c_{16}Ly^3 \Rightarrow 0 = c_5 + 2c_{14}L^2 + 3c_{16}y^3

@ y = c
0 = \sigma_{yy} = 2c_5c + 12c_{14}x^2c + 2c_{16}c^3 \Rightarrow 0 = c_5 + 6c_{14}x^2 + c_{16}c^2

q = \sigma_{xy} = -2c_5x - 4c_{14}x^3 - 6c_{16}xc^2

qL = F_x = \int \limits^L_0\sigma_{xy}(1)dx = \int \limits^L_0(-2c_5x - 4c_{14}x^3 - 6c_{16}xc^2)(1)dx = -c_5L^2 - c_{14}L^4 - 3c_{16}L^2c^2 \Rightarrow q = -c_5L - c_{14}L^3 - 3c_{16}Lc^2

1/2qL^2 = M = \int \limits^L_0\sigma_{xy}x(1)dx = \int \limits^L_0(-2c_5x - 4c_{14}x^3 - 6c_{16}xc^2)x(1)dx = -2/3c_5L^3 - 4/5c_{14}L^5 -2 10c_{16}L^3c^2 \Rightarrow q = -4/3c_5L -8/5c_{14}L^3 - c_{16}Lc^2

@y = -c
0 = \sigma_{yy} = -2c_5c - 12c_{14}x^2c - 2c_{16}c^3 \Rightarrow 0 = c_5 + 6c_{14}x^2 + c_{16}c^2
0 = \sigma_{xy} = -2c_5x - 4c_{14}x^3 - 6c_{16}x(-c)^2 \Rightarrow 0 = c_5 + 2c_{14}x^2 + 3c_{16}c^2


I first need to solve for c₅, c₁₄, and c₁₆[/itex] by picking 3 equations with only those terms in it.

I've tried using different combinations of the 3 system of equations to solve for the constants but I'm not getting desired results. The answers are a lot more complicated than they should be when compared to the constants in phi(x,y) given.

My questions:

1) Which 3 equations would be best to select to find c₅, c₁₄, and c₁₆?

2) What would be the correct boundary conditions for \sigma_{xx} for x = 0

3) Is the setup correct for the resultant force and moment at y = c?

 

[Päällikkö]

You seem to be trying to derive the stress function, which is a much more complicated problem (and one you do not seem to be given enough information for in the description) than the one given in the problem statement. You are given a stress function, and you should see if it is a valid stress function, ie. that it describes the problem it is supposed to describe.

You have identified all the boundary conditions for the given data. Calculate the stresses \sigma_{xx}, \sigma_{yy}, \sigma_{xy} at all the boundaries using the equations (1)-(3), and see if they match. If they do, the stress function solves the problem in question, if they don't, it doesn't.