Sample Problems > Applications > Stress

{ PRESTUBE.PDE

This example models the stress in a tube with an internal pressure.

- from "Fields of Physics on the PC" by Gunnar Backstrom

}

title

' Tube With Internal Pressure'

variables

definitions

mm = 1e-3

r1 = 3*mm

r2 = 10*mm

q21= r2/r1

mu = 0.3

E = 200e9 {Steel}

c = E/(1-mu^2)

G = E/2/(1+mu)

dabs= sqrt(u^2+ v^2)

ex= dx(u)

ey= dy(v)

exy= dx(v)+ dy(u)

sx= c*(ex+ mu*ey)

sy= c*(mu*ex+ ey)

sxy= G*exy

p1= 1e8 { the internal pressure }

{ Exact expressions }

rad= sqrt(x^2+ y^2)

sr_ex= -p1*((r2/rad)^2 - 1)/(q21^2 - 1)

st_ex= p1*((r2/rad)^2 + 1)/(q21^2 - 1)

dabs_ex= abs( rad/E*(st_ex- mu*sr_ex))

equations { Constant temperature, no volume forces }

u: dx( c*(dx(u) + mu*dy(v)) ) + dy( G*(dx(v)+ dy(u)) )= 0

v: dx( G*( dx(v)+ dy(u)) )+ dy( c*(dy(v) + mu*dx(u)) )= 0

constraints { Since all boundaries are free, it is necessary

to apply constraints to eliminate rigid-body motions }

integral(u) = 0

integral(v) = 0

integral(dx(v)-dy(u)) = 0

boundaries

region 1

start (r2,0)

load(u)= 0 { Outer boundary is free }

load(v)= 0

arc to (0,r2) to (-r2,0) to (0,-r2) to close

start (r1,0) { Cut-out }

load(u)= p1*x/r1 { Normal component of x-stress }

load(v)= p1*y/r1 { Normal component of y-stress }

arc to (0,-r1) to (-r1,0) to (0,r1) to close

monitors

contour(dabs)

plots

grid(x+200*u, y+200*v)

elevation(sx, sr_ex) from (r1,0) to (r2,0)

elevation(sy, st_ex) from (r1,0) to (r2,0)

contour(dabs) contour((dabs-dabs_ex)/dabs_ex)

contour(u) contour(v)

vector(u,v) vector(u/dabs, v/dabs)

contour(sx) contour(sy) contour(sxy)

end

prestube

prestube