# Verification of stress function

## 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?

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.