{ AXISYMMETRIC_STRESS.PDE
******************************************************
This example shows the application of FlexPDE to problems in
axi-symmetric stress.
The equations of Stress/Strain arise from the balance of forces in a
material medium, expressed in cylindrical geometry as
dr(r*Sr)/r - St/r + dz(Trz) + Fr = 0
dr(r*Trz)/r + dz(Sz) + Fz = 0
where Sr, St and Sz are the stresses in the r- theta- and z- directions,
Trz is the shear stress, and Fr and Fz are the body forces in the
r- and z- directions.
The deformation of the material is described by the displacements,
U and V, from which the strains are defined as
er = dr(U)
et = U/r
ez = dz(V)
grz = dz(U) + dr(V).
The quantities U,V,er,et,ez,grz,Sr,St,Sz and Trz are related through the
constitutive relations of the material,
Sr = C11*er + C12*et + C13*ez - b*Temp
St = C12*er + C22*et + C23*ez - b*Temp
Sz = C13*er + C23*et + C33*ez - b*Temp
Trz = C44*grz
In isotropic solids we can write the constitutive relations as
C11 = C22 = C33 = G*(1-nu)/(1-2*nu) = C1
C12 = C13 = C23 = G*nu/(1-2*nu) = C2
b = alpha*G*(1+nu)/(1-2*nu)
C44 = G/2
where G = E/(1+nu) is the Modulus of Rigidity
E is Young's Modulus
nu is Poisson's Ratio
and alpha is the thermal expansion coefficient.
from which
Sr = C1*er + C2*(et + ez) - b*Temp
St = C1*et + C2*(er + ez) - b*Temp
Sz = C1*ez + C2*(er + et) - b*Temp
Trz = C44*grz
Combining all these relations, we get the displacement equations:
dr(r*Sr)/r - St/r + dz(Trz) + Fr = 0
dr(r*Trz)/r + dz(Sz) + Fz = 0
These can be written as
div(P) = St/r - Fr
div(Q) = -Fz
where P = [Sr,Trz]
and Q = [Trz,Sz]
The natural (or "load") boundary condition for the U-equation defines the
outward surface-normal component of P, while the natural boundary condition
for the V-equation defines the surface-normal component of Q. Thus, the
natural boundary conditions for the U- and V- equations together define
the surface load vector.
On a free boundary, both of these vectors are zero, so a free boundary
is simply specified by
load(U) = 0
load(V) = 0.
The problem analyzed here is a steel doughnut of rectangular cross-section,
supported on the inner surface and loaded downward on the outer surface.
***************************************** }
title "Doughnut in Axial Shear"
coordinates
ycylinder('R','Z')
variables
U { declare U and V to be the system variables }
V
definitions
nu = 0.3 { define Poisson's Ratio }
E = 20 { Young's Modulus x 10**-11 }
alpha = 0 { define the thermal expansion coefficient }
G = E/(1+nu)
C1 = G*(1-nu)/(1-2*nu) { define the constitutive relations }
C2 = G*nu/(1-2*nu)
b = alpha*G*(1+nu)/(1-2*nu)
Fr = 0 { define the body forces }
Fz = 0
Temp = 0 { define the temperature }
Sr = C1*dr(U) + C2*(U/r + dz(V)) - b*Temp
St = C1*U/r + C2*(dr(U) + dz(V)) - b*Temp
Sz = C1*dz(V) + C2*(dr(U) + U/r) - b*Temp
Trz = G*(dz(U) + dr(V))/2
r1 = 2 { define the inner and outer radii of a doughnut }
r2 = 5
q21 = r2/r1
L = 1.0 { define the height of the doughnut }
initial values
U = 0
V = 0
equations { define the axi-symmetric displacement equations }
U: dr(r*Sr)/r - St/r + dz(Trz) + Fr = 0
V: dr(r*Trz)/r + dz(Sz) + Fz = 0
boundaries
region 1
start(r1,0)
load(U) = 0 { define a free boundary along bottom }
load(V) = 0
line to (r2,0)
value(U) = 0 { constrain R-displacement on right }
load(V) = -E/100 { apply a downward shear load }
line to (r2,L)
load(U) = 0 { define a free boundary along top }
load(V) = 0
line to (r1,L)
value(U) = 0 { constrain all displacement on inner wall }
value(V) = 0
line to close
monitors
grid(r+U,z+V) { show deformed grid as solution progresses }
plots { hardcopy at to close: }
grid(r+U,z+V) { show final deformed grid }
contour(U) as "X-Displacement" { show displacement field }
contour(V) as "Y-Displacement" { show displacement field }
vector(U,V) as "Displacement" { show displacement field }
contour(Trz) as "Shear Stress"
surface(Sr) as "Radial Stress"
end