Elasticity/Polynomial solutions

From Wikiversity
Jump to navigation Jump to search

Using the Airy Stress Function : Polynomial Solutions[edit | edit source]

Example 1[edit | edit source]

Given:

Find the problem which fits this solution.

This is a homogeneous stress field. An infinite number of problems can satisfy these conditions.

Example 2[edit | edit source]

Given:

Find the problem which fits this solution.

An infinite set of problems can have this stress field as a solution.

If , then

which corresponds to a plane stress beam under pure bending.

Pure bending of an elastic beam

Example 3[edit | edit source]

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

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;

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);

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., .
sxx2:=unapply(sxx1,x,y): syy2:=unapply(syy1,x,y): sxy2:=unapply(sxy1,x,y):

We now find the tractions on as
t1:=syy2(x,b); t2:=sxy2(x,b);

and on
t3:=syy2(x,-b); t4:=sxy2(x,-b);

On , we have
t5:=sxx2(0,y); t6:=sxy2(0,y);

The stress function is order 4, so the stresses are order 2 in x and y. The tractions on or might therefore be polynomials in 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);

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);

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});

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);

and

Stress distribution in an elastic cantilevered beam.

Displacement Boundary Condition

The displacement potential function must satisfy the relations and .

In this problem,

Therefore,

Integrating,

only if

which means that

These can be integrated to find and in terms of , and constants. The constants can be determined from the displacement BCs applied so as to fix rigid body motion.

The displacements are given by

where , and = thickness of the beam.

  • Since is no a linear function of , plane sections do not remain plane.
  • and , but St. Venant's principle can be applied.
  • The deflection of the neutral axis () is
If , this prediction approaches beam theory.
  • The maximum deflection is

General Approach For Beam Problems[edit | edit source]

  • Find the highest order polynomial terms and for the normal and shear tractions on .
  • Use a polynomial of order max() excluding constant and linear terms. For example, for a polynomial of order
  • Substitute () 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.