Elasticity/Polynomial solutions

Using the Airy Stress Function : Polynomial Solutions

Example 1 Given: Template:Center top$\phi =a{x}_1^2+b{x}_1{x}_2+c{x}_2^2$Template:Center bottom Find the problem which fits this solution. Template:Center top${\sigma }_{11}={\phi }_{,22}=2c;{\sigma }_{22}={\phi }_{,11}=2a;{\sigma }_{12}=-{\phi }_{,12}=-b$Template:Center bottom This is a homogeneous stress field. An infinite number of problems can satisfy these conditions.

Example 2 Given: Template:Center top$\phi =a{x}_1^3+b{x}_1^2{x}_2+c{x}_1{x}_2^2+d{x}_2^3$Template:Center bottom Find the problem which fits this solution. Template:Center top${\sigma }_{11}=2c{x}_1+6d{x}_2;{\sigma }_{22}=6a{x}_1+2b{x}_2;{\sigma }_{12}=-2b{x}_1-2c{x}_2$Template:Center bottom An infinite set of problems can have this stress field as a solution. If $a=b=c=0$, then Template:Center top${\sigma }_{11}=6d{x}_2;{\sigma }_{22}=0;{\sigma }_{12}=0$Template:Center bottom which corresponds to a plane stress beam under pure bending. File:Elastic beam bending.png Pure bending of an elastic beam

Example 3 Consider a cantilevered beam that is fixed at one end and has a vertical force F applied at the free end. File:Elastic cantilever beam.png Bending of a cantilevered beam The boundary conditions on the beam are $\begin{array}{cc}{\sigma }_{12}& =0;x_2=\pm b\\ {\sigma }_{22}& =0;x_2=\pm b\\ {\sigma }_{11}& =0;x_1=0\\ \underset{-b}{\overset{b}{\int }}{\sigma }_{12}d{x}_2& =F;x_1=0\\ 𝐮& =\mathrm{𝟎};x_1=a\end{array}$ 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 phi:=C1*x^2+C2*x*y+C3*y^2+C4*x^3+C5*x^2*y+C6*x*y^2+ C7*y^3+C8*x^4+C9*x^3*y+C10*x^2*y^2+C11*x*y^3+C12*y^4; $\begin{array}{cc}\varphi & :=𝐶1x^2+𝐶2xy+𝐶3y^2+𝐶4x^3+𝐶5x^{2}y+𝐶6x{y}^2+𝐶7y^3\\ & +𝐶8x^4+𝐶9x^{3}y+𝐶10x^2{y}^2+𝐶11x{y}^3+𝐶12y^4\end{array}$ Take the derivatives of the stress function to obtain the expressions for the stresses. sxx1:= diff(phi,y,y); syy1:= diff(phi,x,x); sxy1:= -diff(phi,x,y); $\begin{array}{cc}𝑠𝑥𝑥1& :=2𝐶3+2𝐶6x+6𝐶7y+2𝐶10x^2+6𝐶11xy+12𝐶12y^2\\ 𝑠𝑦𝑦1& :=2𝐶1+6𝐶4x+2𝐶5y+12𝐶8x^2+6𝐶9xy+2𝐶10y^2\\ 𝑠𝑥𝑦1& :=-𝐶2-2𝐶5x-2𝐶6y-3𝐶9x^2-4𝐶10xy-3𝐶11y^2\end{array}$ Next, use the command unapply(...,x,y) to configure the stresses as functions of x,y so that we can find the value at various points, e.g., $y=b$. sxx2:=unapply(sxx1,x,y): syy2:=unapply(syy1,x,y): sxy2:=unapply(sxy1,x,y): We now find the tractions on $y=b$ as t1:=syy2(x,b); t2:=sxy2(x,b); $\begin{array}{cc}𝑡1& :=2𝐶1+6𝐶4x+2𝐶5b+12𝐶8x^2+6𝐶9xb+2𝐶10b^2\\ 𝑡2& :=-𝐶2-2𝐶5x-2𝐶6b-3𝐶9x^2-4𝐶10xb-3𝐶11b^2\end{array}$ and on $y=-b$ t3:=syy2(x,-b); t4:=sxy2(x,-b); $\begin{array}{cc}𝑡3& :=2𝐶1+6𝐶4x-2𝐶5b+12𝐶8x^2-6𝐶9xb+2𝐶10b^2\\ 𝑡4& :=-𝐶2-2𝐶5x+2𝐶6b-3𝐶9x^2+4𝐶10xb-3𝐶11b^2\end{array}$ On $x=0$, we have t5:=sxx2(0,y); t6:=sxy2(0,y); $\begin{array}{cc}𝑡5& :=2𝐶3+6𝐶7y+12𝐶12y^2\\ 𝑡6& :=-𝐶2-2𝐶6y-3𝐶11y^2\end{array}$ 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 s1:=coeff(t1,x,2); s2:=coeff(t1,x,1); s3:=coeff(t1,x,0); s4:=coeff(t2,x,2); s5:=coeff(t2,x,1); s6:=coeff(t2,x,0); s7:=coeff(t3,x,2); s8:=coeff(t3,x,1); s9:=coeff(t3,x,0); s10:=coeff(t4,x,2); s11:=coeff(t4,x,1); s12:=coeff(t4,x,0); $\begin{array}{cc}𝑠1& :=12𝐶8\\ 𝑠2& :=6𝐶4+6𝐶9b\\ 𝑠3& :=2𝐶1+2𝐶5b+2𝐶10b^2\\ 𝑠4& :=-3𝐶9\\ 𝑠5& :=-2𝐶5-4𝐶10b\\ 𝑠6& :=-𝐶2-2𝐶6b-3𝐶11b^2\\ 𝑠7& :=12𝐶8\\ 𝑠8& :=6𝐶4-6𝐶9b\\ 𝑠9& :=2𝐶1-2𝐶5b+2𝐶10b^2\\ 𝑠10& :=-3𝐶9\\ 𝑠11& :=-2𝐶5+4𝐶10b\\ 𝑠12& :=-𝐶2+2𝐶6b-3𝐶11b^2\end{array}$ 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. biharm:=diff(phi,x$4)+diff(phi,y$4)+2*diff(phi,x,x,y,y); $𝑏𝑖ℎ𝑎𝑟𝑚:=24𝐶8+24𝐶12+8𝐶10$ We also calculate the three force resultants on x=0 by integrating over y: Fx:=int(t5, y=-b..b): Fy:=int(t6, y=-b..b): M:=int(t5*y, y=-b..b): 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. solution:=solve({s1=0,s2=0,s3=0,s4=0,s5=0,s6=0,s7=0, s8=0,s9=0,s10=0,s11=0,s12=0,biharm=0,Fx=0,M=0,Fy=F}, {C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C11,C12}); $\begin{array}{cc}\text{solution}& :=\{𝐶7=0,𝐶8=0,𝐶9=0,𝐶4=0,𝐶10=0,𝐶5=0,\\ & 𝐶12=0,𝐶1=0,𝐶3=0,𝐶6=0,𝐶11=\frac{F}{4b^3},𝐶2=-\frac{3F}{4b}\}\end{array}$ 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:=subs(solution,phi); sxx3:=diff(phi,y,y); syy3:=diff(phi,x,x); sxy3:=-diff(phi,x,y); $\varphi :=-\frac{3Fxy}{4b}+\frac{Fx{y}^3}{4b^3}$ and $\begin{array}{cc}𝑠𝑥𝑥3& :=\frac{3Fxy}{2b^3}\\ 𝑠𝑦𝑦3& :=0\\ 𝑠𝑥𝑦3& :=\frac{3F}{4b}-\frac{3F{y}^2}{4b^3}\end{array}$ File:Elastic cantilever beam stress.png Stress distribution in an elastic cantilevered beam. Displacement Boundary Condition The displacement potential function must satisfy the relations ${\psi }_{,12}={\mathrm{\nabla }}^2\phi$ and ${\mathrm{\nabla }}^2\psi =0$. In this problem, Template:Center top$\phi =-\frac{3F{x}_1{x}_2}{4b}+\frac{F{x}_1{x}_2^3}{4b^3}$Template:Center bottom Therefore, Template:Center top${\psi }_{,12}=\frac{6F{x}_1{x}_2}{4b^3}$Template:Center bottom Integrating, Template:Center top$\psi =\frac{3F}{8}{x}_1^2{x}_2^2+f(x_1)+g(x_2)$Template:Center bottom ${\mathrm{\nabla }}^2\psi =0$ only if Template:Center top$\frac{3F}{4}(x_1^2+x_2^2)+f^{{\text{'}}^{\prime }}(x_1)+g^{{\text{'}}^{\prime }}(x_2)=0$Template:Center bottom which means that Template:Center top$f^{{\text{'}}^{\prime }}(x_1)=-\frac{3F}{4}{x}_1^2+G;g^{{\text{'}}^{\prime }}(x_2)=-\frac{3F}{4}{x}_2^2-G;$Template:Center bottom 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 Template:Center top$\begin{array}{cc}{u}_1& =-\frac{P}{2EI}(a^2-x_1^2)x_2-\frac{P(2+\nu )}{6EI}{x}_2^3+\frac{P(1+\nu )b^2}{8EI}{x}_2\\ u_2& =-\frac{P{a}^3}{6EI}[2-\frac{3x_1}{a}(1-\frac{\nu x_2^2}{a^2})+\frac{x_1^3}{a^3}+\frac{3b^2(1+\nu )}{4a^2}(1-\frac{x_1}{a})]\end{array}$Template:Center bottom where $I=(1/12)w(2b{)}^3$, and $w$ = thickness of the beam. 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 Template:Center top$u_2(x_1,0)=-\frac{P{a}^3}{6EI}[2-\frac{3x_1}{a}+\frac{x_1^3}{a^3}+\frac{3b^2(1+\nu )}{4a^2}(1-\frac{x_1}{a})]$Template:Center bottom If $b/a\to 0$, this prediction approaches beam theory. The maximum deflection is Template:Center top$u_2(0,0)=-\frac{P{a}^3}{3EI}-\frac{P{a}^3}{6EI}\frac{3b^2(1+\nu )}{4a^2}$Template:Center bottom

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 $5$

$\begin{array}{cc}\phi =& C_1{x}^5+C_2{x}^{4}y+C_3{x}^3{y}^2+C_4{x}^2{y}^3+C_{5}x{y}^4+C_6{y}^5+\\ & C_7{x}^4+C_8{x}^{3}y+C_9{x}^2{y}^2+C_{10}x{y}^3+C_{11}{y}^4+\\ & C_{12}{x}^3+C_{13}{x}^{2}y+C_{14}x{y}^2+C_{15}{y}^3+\\ & C_{16}{x}^2+C_{17}xy+C_{18}{y}^2\end{array}$

• Substitute ($\phi$) 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.