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[edit] Using the Airy Stress Function : Polynomial Solutions

[edit] 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.

[edit] 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

[edit] 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;

$\begin{align} \phi &:= \mathit{C1}\,x^{2} + \mathit{C2}\,x\,y + \mathit{C3}\,y ^{2} + \mathit{C4}\,x^{3} + \mathit{C5}\,x^{2}\,y + \mathit{C6}\, x\,y^{2} + \mathit{C7}\,y^{3} \\ & + \mathit{C8}\,x^{4} +\mathit{C9}\,x^{3}\,y + \mathit{C10}\,x^{2}\,y^{2} + \mathit{C11}\,x\,y^{3} + \mathit{C12}\,y^{4} \end{align}$

Take the derivatives of the stress function to obtain the expressions for the stresses.
sxx1:= diff(phi,y,y); syy1:= diff(phi,x,x); sxy1:= -diff(phi,x,y);

$\begin{align} \mathit{sxx1} & := 2\,\mathit{C3} + 2\,\mathit{C6}\,x + 6\,\mathit{ C7}\,y + 2\,\mathit{C10}\,x^{2} + 6\,\mathit{C11}\,x\,y + 12\, \mathit{C12}\,y^{2} \\ \mathit{syy1} & := 2\,\mathit{C1} + 6\,\mathit{C4}\,x + 2\,\mathit{ C5}\,y + 12\,\mathit{C8}\,x^{2} + 6\,\mathit{C9}\,x\,y + 2\, \mathit{C10}\,y^{2} \\ \mathit{sxy1} & := - \mathit{C2} - 2\,\mathit{C5}\,x - 2\,\mathit{ C6}\,y - 3\,\mathit{C9}\,x^{2} - 4\,\mathit{C10}\,x\,y - 3\, \mathit{C11}\,y^{2} \end{align}$

Next, use the command 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);

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

and on $y = - b$
t3:=syy2(x,-b); t4:=sxy2(x,-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
t5:=sxx2(0,y); t6:=sxy2(0,y);

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

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
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);

$\begin{align} \mathit{s1} &:= 12\,\mathit{C8} \\ \mathit{s2} &:= 6\,\mathit{C4} + 6\,\mathit{C9}\,b \\ \mathit{s3} &:= 2\,\mathit{C1} + 2\,\mathit{C5}\,b + 2\,\mathit{ C10}\,b^{2} \\ \mathit{s4} &:= - 3\,\mathit{C9} \\ \mathit{s5} &:= - 2\,\mathit{C5} - 4\,\mathit{C10}\,b \\ \mathit{s6} &:= - \mathit{C2} - 2\,\mathit{C6}\,b - 3\,\mathit{ C11}\,b^{2} \\ \mathit{s7} &:= 12\,\mathit{C8} \\ \mathit{s8} &:= 6\,\mathit{C4} - 6\,\mathit{C9}\,b \\ \mathit{s9} &:= 2\,\mathit{C1} - 2\,\mathit{C5}\,b + 2\,\mathit{ C10}\,b^{2} \\ \mathit{s10} &:= - 3\,\mathit{C9} \\ \mathit{s11} &:= - 2\,\mathit{C5} + 4\,\mathit{C10}\,b \\ \mathit{s12} &:= - \mathit{C2} + 2\,\mathit{C6}\,b - 3\,\mathit{ C11}\,b^{2} \end{align}$

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.
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});

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

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:=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

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

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,

$\varphi = - \cfrac{3Fx_1x_2}{4b} + \cfrac{Fx_1x_2^{3}}{4b^{3}}$

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 $I = (1/12)wb^3 \,$ , 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

$u_2(x_1,0) = -\cfrac{Pa^3}{6EI}\left[2 -\cfrac{3x_1}{a} + \cfrac{x_1^3}{a^3} + \cfrac{3b^2(1+\nu)}{4a^2}\left(1-\cfrac{x_1}{a}\right)\right]$

If $b/a \rightarrow 0$ , this prediction approaches beam theory.

The maximum deflection is

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

[edit] 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.

Introduction to Elasticity/Polynomial solutions

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Contents

[edit] Using the Airy Stress Function : Polynomial Solutions

[edit] Example 1

[edit] Example 2

[edit] Example 3

[edit] General Approach For Beam Problems

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