Elasticity/Polynomial solutions

Using the Airy Stress Function : Polynomial Solutions

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.

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. 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. Bending of a cantilevered beam 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 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; ${\begin{aligned}\phi &:={\mathit {C1}}\,x^{2}+{\mathit {C2}}\,x\,y+{\mathit {C3}}\,y^{2}+{\mathit {C4}}\,x^{3}+{\mathit {C5}}\,x^{2}\,y+{\mathit {C6}}\,x\,y^{2}+{\mathit {C7}}\,y^{3}\\&+{\mathit {C8}}\,x^{4}+{\mathit {C9}}\,x^{3}\,y+{\mathit {C10}}\,x^{2}\,y^{2}+{\mathit {C11}}\,x\,y^{3}+{\mathit {C12}}\,y^{4}\end{aligned}}$ 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); ${\begin{aligned}{\mathit {sxx1}}&:=2\,{\mathit {C3}}+2\,{\mathit {C6}}\,x+6\,{\mathit {C7}}\,y+2\,{\mathit {C10}}\,x^{2}+6\,{\mathit {C11}}\,x\,y+12\,{\mathit {C12}}\,y^{2}\\{\mathit {syy1}}&:=2\,{\mathit {C1}}+6\,{\mathit {C4}}\,x+2\,{\mathit {C5}}\,y+12\,{\mathit {C8}}\,x^{2}+6\,{\mathit {C9}}\,x\,y+2\,{\mathit {C10}}\,y^{2}\\{\mathit {sxy1}}&:=-{\mathit {C2}}-2\,{\mathit {C5}}\,x-2\,{\mathit {C6}}\,y-3\,{\mathit {C9}}\,x^{2}-4\,{\mathit {C10}}\,x\,y-3\,{\mathit {C11}}\,y^{2}\end{aligned}}$ 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); ${\begin{aligned}{\mathit {t1}}&:=2\,{\mathit {C1}}+6\,{\mathit {C4}}\,x+2\,{\mathit {C5}}\,b+12\,{\mathit {C8}}\,x^{2}+6\,{\mathit {C9}}\,x\,b+2\,{\mathit {C10}}\,b^{2}\\{\mathit {t2}}&:=-{\mathit {C2}}-2\,{\mathit {C5}}\,x-2\,{\mathit {C6}}\,b-3\,{\mathit {C9}}\,x^{2}-4\,{\mathit {C10}}\,x\,b-3\,{\mathit {C11}}\,b^{2}\end{aligned}}$ and on $y=-b$ t3:=syy2(x,-b); t4:=sxy2(x,-b); ${\begin{aligned}{\mathit {t3}}&:=2\,{\mathit {C1}}+6\,{\mathit {C4}}\,x-2\,{\mathit {C5}}\,b+12\,{\mathit {C8}}\,x^{2}-6\,{\mathit {C9}}\,x\,b+2\,{\mathit {C10}}\,b^{2}\\{\mathit {t4}}&:=-{\mathit {C2}}-2\,{\mathit {C5}}\,x+2\,{\mathit {C6}}\,b-3\,{\mathit {C9}}\,x^{2}+4\,{\mathit {C10}}\,x\,b-3\,{\mathit {C11}}\,b^{2}\end{aligned}}$ On $x=0$, we have t5:=sxx2(0,y); t6:=sxy2(0,y); ${\begin{aligned}{\mathit {t5}}&:=2\,{\mathit {C3}}+6\,{\mathit {C7}}\,y+12\,{\mathit {C12}}\,y^{2}\\{\mathit {t6}}&:=-{\mathit {C2}}-2\,{\mathit {C6}}\,y-3\,{\mathit {C11}}\,y^{2}\end{aligned}}$ 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); ${\begin{aligned}{\mathit {s1}}&:=12\,{\mathit {C8}}\\{\mathit {s2}}&:=6\,{\mathit {C4}}+6\,{\mathit {C9}}\,b\\{\mathit {s3}}&:=2\,{\mathit {C1}}+2\,{\mathit {C5}}\,b+2\,{\mathit {C10}}\,b^{2}\\{\mathit {s4}}&:=-3\,{\mathit {C9}}\\{\mathit {s5}}&:=-2\,{\mathit {C5}}-4\,{\mathit {C10}}\,b\\{\mathit {s6}}&:=-{\mathit {C2}}-2\,{\mathit {C6}}\,b-3\,{\mathit {C11}}\,b^{2}\\{\mathit {s7}}&:=12\,{\mathit {C8}}\\{\mathit {s8}}&:=6\,{\mathit {C4}}-6\,{\mathit {C9}}\,b\\{\mathit {s9}}&:=2\,{\mathit {C1}}-2\,{\mathit {C5}}\,b+2\,{\mathit {C10}}\,b^{2}\\{\mathit {s10}}&:=-3\,{\mathit {C9}}\\{\mathit {s11}}&:=-2\,{\mathit {C5}}+4\,{\mathit {C10}}\,b\\{\mathit {s12}}&:=-{\mathit {C2}}+2\,{\mathit {C6}}\,b-3\,{\mathit {C11}}\,b^{2}\end{aligned}}$ 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); ${\mathit {biharm}}:=24\,{\mathit {C8}}+24\,{\mathit {C12}}+8\,{\mathit {C10}}$ We also calculate the three force resultants on x=0 by integrating over y: 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}); ${\begin{aligned}{\text{solution}}&:=\{{\mathit {C7}}=0,\,{\mathit {C8}}=0,\,{\mathit {C9}}=0,\,{\mathit {C4}}=0,\,{\mathit {C10}}=0,\,{\mathit {C5}}=0,\\&{\mathit {C12}}=0,\,{\mathit {C1}}=0,\,{\mathit {C3}}=0,\,{\mathit {C6}}=0,\,{\mathit {C11}}={\cfrac {F}{4\,b^{3}}},\,{\mathit {C2}}=-{\cfrac {3\,F}{4\,b}}\}\end{aligned}}$ 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); $\phi :=-{\cfrac {3\,F\,x\,y}{4\,b}}+{\cfrac {F\,x\,y^{3}}{4\,b^{3}}}$ and ${\begin{aligned}{\mathit {sxx3}}&:={\cfrac {3\,F\,x\,y}{2\,b^{3}}}\\{\mathit {syy3}}&:=0\\{\mathit {sxy3}}&:={\cfrac {3\,F}{4\,b}}-{\cfrac {3\,F\,y^{2}}{4\,b^{3}}}\end{aligned}}$ Stress distribution in an elastic cantilevered beam. Displacement Boundary Condition The displacement potential function must satisfy the relations $\psi _{,12}=\nabla ^{2}{\varphi }$ and $\nabla ^{2}{\psi }=0$. In this problem, $\varphi =-{\cfrac {3Fx_{1}x_{2}}{4b}}+{\cfrac {Fx_{1}x_{2}^{3}}{4b^{3}}}$ Therefore, $\psi _{,12}={\cfrac {6Fx_{1}x_{2}}{4b^{3}}}$ Integrating, $\psi ={\cfrac {3F}{8}}x_{1}^{2}x_{2}^{2}+f(x_{1})+g(x_{2})$ $\nabla ^{2}{\psi }=0$ 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. Since $u_{1}$ is no a linear function of $x_{2}$, plane sections do not remain plane. $u_{1}(a,x_{2})\neq 0$ and $u_{2}(a,x_{2})\neq 0$, but St. Venant's principle can be applied. The deflection of the neutral axis ($x_{2}=0$) is $u_{2}(x_{1},0)=-{\cfrac {Pa^{3}}{6EI}}\left[2-{\cfrac {3x_{1}}{a}}+{\cfrac {x_{1}^{3}}{a^{3}}}+{\cfrac {3b^{2}(1+\nu )}{4a^{2}}}\left(1-{\cfrac {x_{1}}{a}}\right)\right]$ If $b/a\rightarrow 0$, this prediction approaches beam theory. The maximum deflection is $u_{2}(0,0)=-{\cfrac {Pa^{3}}{3EI}}-{\cfrac {Pa^{3}}{6EI}}{\cfrac {3b^{2}(1+\nu )}{4a^{2}}}$

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 $5$

${\begin{aligned}\varphi =&C_{1}x^{5}+C_{2}x^{4}y+C_{3}x^{3}y^{2}+C_{4}x^{2}y^{3}+C_{5}xy^{4}+C_{6}y^{5}+\\&C_{7}x^{4}+C_{8}x^{3}y+C_{9}x^{2}y^{2}+C_{10}xy^{3}+C_{11}y^{4}+\\&C_{12}x^{3}+C_{13}x^{2}y+C_{14}xy^{2}+C_{15}y^{3}+\\&C_{16}x^{2}+C_{17}xy+C_{18}y^{2}\end{aligned}}$

• 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.

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