Elasticity/Polynomial solutions

Using the Airy Stress Function : Polynomial Solutions
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Example 1
Given: $$    \varphi = a~x_1^2 + b~x_1~x_2 + c~x_2^2 $$ Find the problem which fits this solution. $$    \sigma_{11} = \varphi_{,22} = 2c ~; \sigma_{22} = \varphi_{,11} = 2a ~; \sigma_{12} = -\varphi_{,12} = -b $$ This is a homogeneous stress field. An infinite number of problems can satisfy these conditions.
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Example 2
Given: $$    \varphi = a~x_1^3 + b~x_1^2~x_2 + c~x_1~x_2^2 + d~x_2^3 $$ Find the problem which fits this solution. $$    \sigma_{11} = 2cx_1 + 6dx_2 ~; \sigma_{22} = 6ax_1 + 2bx_2 ~; \sigma_{12} = -2bx_1 - 2cx_2 $$ An infinite set of problems can have this stress field as a solution.

If $$a = b = c= 0$$, then $$    \sigma_{11} =  6dx_2 ~; \sigma_{22} = 0 ~; \sigma_{12} = 0 $$ which corresponds to a plane stress beam under pure bending. {| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" 
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Example 3
Consider a cantilevered beam that is fixed at one end and has a vertical force F applied at the free end. The boundary conditions on the beam are
 * $$\begin{matrix}

\sigma_{12} & = 0 ~; x_2 = \pm b \\ \sigma_{22} & = 0 ~; x_2 = \pm b \\ \sigma_{11} & = 0 ~; x_1 = 0 \\ \int_{-b}^{b}\sigma_{12} dx_2 & = F ~; x_1 = 0 \\ \mathbf{u} & = \mathbf{0} ~; x_1 = a \end{matrix}$$

We will use Maple to solve the problem.

First, assume a polynomial Airy stress function that has a high enough order. In this case a fourth order polynomial will suffice

\begin{align} \phi &:= \mathit{C1}\,x^{2} + \mathit{C2}\,x\,y + \mathit{C3}\,y ^{2} + \mathit{C4}\,x^{3} + \mathit{C5}\,x^{2}\,y + \mathit{C6}\, x\,y^{2} + \mathit{C7}\,y^{3} \\ & + \mathit{C8}\,x^{4} +\mathit{C9}\,x^{3}\,y + \mathit{C10}\,x^{2}\,y^{2} + \mathit{C11}\,x\,y^{3} + \mathit{C12}\,y^{4} \end{align} $$ Take the derivatives of the stress function to obtain the expressions for the stresses.

\begin{align} \mathit{sxx1} & := 2\,\mathit{C3} + 2\,\mathit{C6}\,x + 6\,\mathit{ C7}\,y + 2\,\mathit{C10}\,x^{2} + 6\,\mathit{C11}\,x\,y + 12\, \mathit{C12}\,y^{2} \\ \mathit{syy1} & := 2\,\mathit{C1} + 6\,\mathit{C4}\,x + 2\,\mathit{ C5}\,y + 12\,\mathit{C8}\,x^{2} + 6\,\mathit{C9}\,x\,y + 2\, \mathit{C10}\,y^{2} \\ \mathit{sxy1} & := - \mathit{C2} - 2\,\mathit{C5}\,x - 2\,\mathit{ C6}\,y - 3\,\mathit{C9}\,x^{2} - 4\,\mathit{C10}\,x\,y - 3\, \mathit{C11}\,y^{2} \end{align} $$ Next, use the command  to configure the stresses as functions of x,y so that we can find the value at various points, e.g.,  $$y=b$$.

We now find the tractions on $$y=b$$ as

\begin{align} \mathit{t1} &:= 2\,\mathit{C1} + 6\,\mathit{C4}\,x + 2\,\mathit{C5 }\,b + 12\,\mathit{C8}\,x^{2} + 6\,\mathit{C9}\,x\,b + 2\, \mathit{C10}\,b^{2}\\ \mathit{t2} &:= - \mathit{C2} - 2\,\mathit{C5}\,x - 2\,\mathit{C6 }\,b - 3\,\mathit{C9}\,x^{2} - 4\,\mathit{C10}\,x\,b - 3\, \mathit{C11}\,b^{2} \end{align} $$ and on $$y=-b$$

\begin{align} \mathit{t3} &:= 2\,\mathit{C1} + 6\,\mathit{C4}\,x - 2\,\mathit{C5 }\,b + 12\,\mathit{C8}\,x^{2} - 6\,\mathit{C9}\,x\,b + 2\, \mathit{C10}\,b^{2}\\ \mathit{t4} &:= - \mathit{C2} - 2\,\mathit{C5}\,x + 2\,\mathit{C6 }\,b - 3\,\mathit{C9}\,x^{2} + 4\,\mathit{C10}\,x\,b - 3\, \mathit{C11}\,b^{2} \end{align} $$

On $$x=0$$, we have

\begin{align} \mathit{t5} &:= 2\,\mathit{C3} + 6\,\mathit{C7}\,y + 12\,\mathit{C12}\,y^{2} \\ \mathit{t6} &:= - \mathit{C2} - 2\,\mathit{C6}\,y - 3\,\mathit{C11}\,y^{2} \end{align} $$

The stress function is order 4, so the stresses are order 2 in x and y. The tractions on $$y=+b$$ or $$-b$$ might therefore be polynomials in $$x$$ of order 2.

We calculate the coefficients of each power of x in these expressions as

\begin{align} \mathit{s1} &:= 12\,\mathit{C8} \\ \mathit{s2} &:= 6\,\mathit{C4} + 6\,\mathit{C9}\,b \\ \mathit{s3} &:= 2\,\mathit{C1} + 2\,\mathit{C5}\,b + 2\,\mathit{ C10}\,b^{2} \\ \mathit{s4} &:= - 3\,\mathit{C9} \\ \mathit{s5} &:= - 2\,\mathit{C5} - 4\,\mathit{C10}\,b \\ \mathit{s6} &:= - \mathit{C2} - 2\,\mathit{C6}\,b - 3\,\mathit{ C11}\,b^{2} \\ \mathit{s7} &:= 12\,\mathit{C8} \\ \mathit{s8} &:= 6\,\mathit{C4} - 6\,\mathit{C9}\,b \\ \mathit{s9} &:= 2\,\mathit{C1} - 2\,\mathit{C5}\,b + 2\,\mathit{ C10}\,b^{2} \\ \mathit{s10} &:= - 3\,\mathit{C9} \\ \mathit{s11} &:= - 2\,\mathit{C5} + 4\,\mathit{C10}\,b \\ \mathit{s12} &:= - \mathit{C2} + 2\,\mathit{C6}\,b - 3\,\mathit{ C11}\,b^{2} \end{align} $$ The biharmonic equation is 4th order, so applying it to a 4th order polynomial generates a constant. And this constant must be equal to zero.

\mathit{biharm} := 24\,\mathit{C8} + 24\,\mathit{C12} + 8\,\mathit{C10} $$

We also calculate the three force resultants on x=0 by integrating over y:

We now solve for the constants so as to satisfy (i) the strong boundary conditions, (ii) the biharmonic equation and (iii) the weak boundary conditions.

\begin{align} \text{solution} & := \{ \mathit{C7}=0, \,\mathit{C8}=0, \,\mathit{C9}=0, \,\mathit{C4}=0, \,\mathit{C10}=0, \,\mathit{C5}=0, \\ & \mathit{C12}=0, \,\mathit{C1}=0, \,\mathit{C3}=0, \, \mathit{C6}=0, \,\mathit{C11}= \cfrac {F}{4\,b^{3}}, \,\mathit{C2}= - \cfrac {3\,F}{4\,b} \} \end{align} $$

Notice that there are more equations than there are constants. Some of the equations are not linearly independent. However, Maple can handle this if there is a solution.

Substitute the solution into the original stress function and calculate the final stresses.

\phi :=  - { \cfrac {3\,F\,x\,y}{4\,b}}  + { \cfrac {F\,x\,y^{3}}{4\,b^{3}}} $$ and

\begin{align} \mathit{sxx3} & := { \cfrac {3\,F\,x\,y}{2\,b^{3}}} \\ \mathit{syy3} & := 0 \\ \mathit{sxy3} & := { \cfrac {3\,F}{4\,b}} - { \cfrac {3\,F\,y^{2}}{4\,b^{3}}} \end{align} $$

Displacement Boundary Condition

The displacement potential function must satisfy the relations $$\psi_{,12} = \nabla^2{\varphi}$$ and $$\nabla^2{\psi} = 0$$.

In this problem, $$  \varphi =  - \cfrac{3Fx_1x_2}{4b} + \cfrac{Fx_1x_2^{3}}{4b^{3}} $$ Therefore, $$   \psi_{,12} = \cfrac{6Fx_1x_2}{4b^3} $$ Integrating, $$   \psi = \cfrac{3F}{8}x_1^2x_2^2 + f(x_1) + g(x_2) $$ $$\nabla^2{\psi} = 0$$ only if $$ \cfrac{3F}{4}(x_1^2 + x_2^2) + f^{}(x_1) + g^{}(x_2) = 0 $$ which means that $$   f^{''}(x_1) = -\cfrac{3F}{4}x_1^2 + G ~; g^{''}(x_2) = -\cfrac{3F}{4}x_2^2 - G ~; $$ These can be integrated to find $$f(x_1)$$ and $$g(x_2)$$ in terms of $$x_1$$, $$x_2$$ and constants. The constants can be determined from the displacement BCs applied so as to fix rigid body motion.

The displacements are given by $$\begin{matrix} u_1 &= -\cfrac{P}{2EI}(a^2-x_1^2)x_2 -\cfrac{P(2+\nu)}{6EI}x_2^3 +\cfrac{P(1+\nu)b^2}{8EI}x_2\\ u_2 &= -\cfrac{Pa^3}{6EI}\left[2 -\cfrac{3x_1}{a}\left(1-\cfrac{\nu x_2^2}{a^2}\right) + \cfrac{x_1^3}{a^3} + \cfrac{3b^2(1+\nu)}{4a^2}\left(1-\cfrac{x_1}{a}\right)\right] \end{matrix}$$ where $$I = (1/12)w(2b)^3 \,$$, and $$w$$ = thickness of the beam. $$             u_2(x_1,0) = -\cfrac{Pa^3}{6EI}\left[2 -\cfrac{3x_1}{a} + \cfrac{x_1^3}{a^3} +             \cfrac{3b^2(1+\nu)}{4a^2}\left(1-\cfrac{x_1}{a}\right)\right] $$
 * Since $$u_1$$ is no a linear function of $$x_2$$, plane sections do not remain plane.
 * $$u_1(a,x_2) \ne 0$$ and $$u_2(a,x_2) \ne 0$$, but St. Venant's principle can be applied.
 * The deflection of the neutral axis ($$x_2 = 0$$) is
 * If $$b/a \rightarrow 0$$, this prediction approaches beam theory.

$$ u_2(0,0) = -\cfrac{Pa^3}{3EI} -\cfrac{Pa^3}{6EI}\cfrac{3b^2(1+\nu)}{4a^2} $$
 * The maximum deflection is
 * }

General Approach For Beam Problems

 * Find the highest order polynomial terms $$n$$ and $$m$$ for the     normal and shear tractions on $$x_2 \equiv y = \pm b$$.
 * Use a polynomial of order max($$m+4, n+5$$) excluding constant and     linear terms.  For example, for a polynomial of order $$5$$
 * $$\begin{align}       \varphi = & C_1 x^5 + C_2 x^4 y + C_3 x^3 y^2 + C_4 x^2 y^3 + C_5 x y^4 +                 C_6 y^5 + \\        & C_7 x^4 + C_8 x^3 y + C_9 x^2 y^2 + C_{10} x y^3 + C_{11} y^4 +\\        & C_{12} x^3 + C_{13} x^2 y + C_{14} x y^2 + C_{15} y^3 + \\        & C_{16} x^2 + C_{17} x y + C_{18} y^2       \end{align}

$$
 * Substitute ($$\varphi$$) into the biharmonic equation to get a set of constraint equations. Also compute the stresses.
 * Apply boundary conditions to obtain the tractions at the boundary.
 * For the strong BCs, find the coefficients of powers of x and y and      equate with expressions for the tractions.
 * For the weak BCs, find algebraic expressions.
 * Solve the set of equations and back-substitute.