ME340A Elasticity Spring 2010 Wei Cai Stanford University
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C:\Documents and Settings\Wei Cai\My Documents\Courses...\S522a.m
%adpated from maple file S522 from J. R. Barber, Elasticity
% http://www-personal.engin.umich.edu/~jbarber/elasticity/maple/S522
%
% adapted by Wei Cai, caiwei@stanford.edu, for ME 340 Elasticity
% Spring 2006, Stanford University
%
%This file gives the solution of the simply-supported beam problem,
%following the strategy of Section 5.2.2.
%
clear all
syms C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18
syms x y a b p
%stress function
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+C13*x^5+C14*x^4*y+ ...
C15*x^3*y^2+C16*x^2*y^3+C17*x*y^4+C18*y
%stress field
sxx = diff(diff(phi,y),y)
syy = diff(diff(phi,x),x)
sxy = -diff(diff(phi,x),y)
%traction force on y=b
%Ty on y=b
t1 = subs(syy,y,b)
%Tx on y=b
t2 = subs(sxy,y,b)
%traction force on y=-b
%Ty on y=-b
t3 = subs(syy,y,-b)
%Tx on y=-b
t4 = subs(sxy,y,-b)
%traction force on x=a
%Tx on x=a
t5 = subs(sxx,x,a)
%Ty on x=a
t6 = subs(sxy,x,a)
%traction force on x=-a
%Tx on x=-a
t7 = subs(sxx,x,-a)
%Ty on x=-a
t8 = subs(sxy,x,-a)
%find coefficients of polynomials t1,t2,t3,t4
s1 = subs(diff(t1,x,3),x,0)/factorial(3)
s2 = subs(diff(t1,x,2),x,0)/factorial(2)
s3 = subs(diff(t1,x,1),x,0)/factorial(1)
s4 = subs(t1,x,0)
s5 = subs(diff(t2,x,3),x,0)/factorial(3)
s6 = subs(diff(t2,x,2),x,0)/factorial(2)
s7 = subs(diff(t2,x,1),x,0)/factorial(1)
s8 = subs(t2,x,0)
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s9 = subs(diff(t3,x,3),x,0)/factorial(3)
s10 = subs(diff(t3,x,2),x,0)/factorial(2)
s11 = subs(diff(t3,x,1),x,0)/factorial(1)
s12 = subs(t3,x,0)
s13 = subs(diff(t4,x,3),x,0)/factorial(3)
s14 = subs(diff(t4,x,2),x,0)/factorial(2)
s15 = subs(diff(t4,x,1),x,0)/factorial(1)
s16 = subs(t4,x,0)
%The biharmonic equation is 4th order, so applying it to a 5th order polynomial
%generates a first order polynomial.
%
biharm = diff(phi,x,4)+diff(phi,y,4)+2*diff(diff(phi,x,2),y,2)
%coefficients of biharm
b1 = subs(diff(biharm,x,1),{x,y},{0,0})
b2 = subs(diff(biharm,y,1),{x,y},{0,0})
b3 = subs(biharm,{x,y},{0,0})
%integrated force and torque on x=a
Fx1 = simplify(int(t5, y, -b, b))
Fy1 = simplify(int(t6, y, -b, b))
M1 = simplify(int(t5*y, y, -b, b))
%
%integrated force and torque on x=-a
Fx2 = simplify(int(t7, y, -b, b))
Fy2 = simplify(int(t8, y, -b, b))
M2 = simplify(int(t7*y, y, -b, b))
%Solve all these equations together
s = solve(s1, s2, s3, s4+p, s5, s6, s7, s8, ...
s9, s10,s11,s12, s13,s14,s15,s16, ...
b1, b2, b3,
Fx1,M1, Fy1-p*a, ...
'C1','C2','C3','C4','C5','C6','C7','C8','C9',...
'C10','C11','C12','C13','C14','C15','C16','C17','C18')
%construct cell arrays for future use
coeffs = {C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C11,C12,C13,C14,C15,C16,C17,C18}
solution = {s.C1,s.C2,s.C3,s.C4,s.C5,s.C6,s.C7,s.C8,s.C9, ...
s.C10,s.C11,s.C12,s.C13,s.C14,s.C15,s.C16,s.C17,s.C18}
%stress function and stress field solution
phi2 = simplify(subs(phi, coeffs, solution))
sxx2 =
syy2 =
sxy2 =
simplify(subs(sxx, coeffs, solution))
simplify(subs(syy, coeffs, solution))
simplify(subs(sxy, coeffs, solution))
%print out results
disp('phi=') pretty(phi2)
disp('sxx=') pretty(sxx2)
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disp('syy=')
disp('sxy=')
C:\Documents and Settings\Wei Cai\My Documents\Courses...\S522a.m
pretty(syy2)
pretty(sxy2)
%plot stress fields
bn =
pn =
an =
xn = [-1:0.05:1]*an
yn = [-1:0.1:1]*bn
xx = ones(size(yn'))*xn
yy = yn'*ones(size(xn))
phin = subs(phi2,{a,b,p},{an,bn,pn})
sxxn = subs(sxx2,{a,b,p},{an,bn,pn})
syyn = subs(syy2,{a,b,p},{an,bn,pn})
sxyn = subs(sxy2,{a,b,p},{an,bn,pn})
phin = subs(phin,{x,y},{xx,yy})
sxxn = subs(sxxn,{x,y},{xx,yy})
syyn = subs(syyn,y,yy)
sxyn = subs(sxyn,{x,y},{xx,yy})
figure(1)
subplot(2,2,1) mesh(xn,yn,phin)
title('\phi') xlabel('x') ylabel('y')
subplot(2,2,2) mesh(xn,yn,sxxn)
title('\sigma_{xx}') xlabel('x') ylabel('y')
subplot(2,2,3) mesh(xn,yn,syyn)
title('\sigma_{yy}') xlabel('x') ylabel('y')
subplot(2,2,4) mesh(xn,yn,sxyn)
title('\sigma_{xy}') xlabel('x') ylabel('y')
% results
%phi=
%
%
2 3
2
2
3 2
3 2
2 3
5
%
p (10 x b + 15 x y b - 2 y b + 5 y a - 5 x y + y )
%
- 1/40 -----------------------------------------------------------%
3
%
b
%sxx=
%
%
2
2
2
2
%
p y (-6 b + 15 a - 15 x + 10 y )
%
- 1/20 ----------------------------------%
3
%
b
%syy=
%
%
3
2
3
%
p (-2 b - 3 y b + y )
%
1/4 ----------------------%
3
%
b
%sxy=
%
%
2
2
%
p x (-b + y )
%
- 3/4 -------------%
3
%
b
3 of 3
σxx
φ
5
5
0
0
−5
1
−5
1
2
0
−1
−2
2
0
0
y
14
0
y
x
−1
−2
x
σxy
σyy
2
0
0
−0.5
−1
1
2
0
0
y
−1 −2
x
−2
1
2
0
y
0
−1
−2
x
Appendix
Potential Field of a Uniformly Charged Ellipsoid
Wei Cai
Department of Mechanical Engineering, Stanford University, CA 94305-4040
May 28, 2007
Contents
1 Problem Statement
1
2 Orthogonal Curvilinear Coordinates
2
3 Elliptic Coordinates
3
4 Alternative Definition of Elliptic Coordinates
4
5 Ellipsoidal Coordinates
5
6 Ellipsoidal Conductor
8
7 Ellipsoidal Shell
9
8 Uniformly Charged Ellipsoid
11
9 Special Cases
12
10 Application to Hertz Contact Problem
14
A Matlab Files for Analytic Derivation
16
1
Problem Statement
The purpose of this document is to discuss the derivation of a very useful result, which states that
the potential field of a uniformly charged ellipsoid is a quadratic function inside the
ellipsoid [1]. Specifically, consider an ellipsoid with uniform charge density ρ inside the following
region,
x2 y 2 z 2
+ 2 + 2 =1
a2
b
c
(1)
Let V0 specify the volume of the ellipsoid. The potential field is defined as
Z
ρ
φ(x) ≡
dV (x0 )
0|
|x
−
x
V0
1
(2)
where x = (x, y, z) and x0 = (x0 , y 0 , z 0 ).
If point x is inside the ellipsoid [2],
Z ∞
x2
ds
y2
z2
p
1− 2
φ(x, y, z) = π abc ρ
− 2
− 2
a +s b +s c +s
ϕ(s)
0
(3)
where
ϕ(s) ≡ (a2 + s)(b2 + s)(c2 + s)
(4)
If point x is outside the ellipsoid [1],
Z ∞
x2
ds
y2
z2
p
1− 2
φ(x, y, z) = π abc ρ
− 2
− 2
a +s b +s c +s
ϕ(s)
λ
(5)
where λ is the greatest root of the equation f (s) = 0, where
f (s) ≡
x2
y2
z2
+
+
−1
a2 + s b2 + s c2 + s
(6)
The physical significance of function f (s) is that, for each s > 0, the equation f (s) = 0 defines
an ellipsoid (larger than the original ellipsoid), which is an isosurface of the potential field φ(x)
generated by the original ellipsoid (with uniform charge density). Therefore, all the value of s ∈
[0, ∞) corresponds to a family of ellipsoids, called the confocal family; the potential field is a
constant on each ellipsoid in the family.
Physical significance: As a consequence of the above result, the second spatial derivatives of the
potential field is a uniform second order tensor inside the ellipsoid. This is closely analogous to
Eshelby’s Theorem, which states that the stress field inside an ellipsoidal inclusion is uniform [3].
Landau and Lifshitz also used the above result to show that normal stress distribution inside the
contact area between two smooth elastic media (Hertzian contact problem) has the ellipsoidal
shape [2].
2
Orthogonal Curvilinear Coordinates
The proof of this result is best discussed using ellipsoidal coordinates, which is an orthogonal curvilinear coordinate system. In this section, we first summarize the major results concerning orthogonal curvilinear coordinates. We will then introduce the elliptic coordinates (2D) and ellipsoidal
coordinates (3D) in the following sections.
Let the Cartesian coordinates be specified by (x1 , x2 , x3 ) = (x, y, z). An arbitrary differential
length in space ds is specified by (ds)2 = (x1 )2 + (x2 )2 + (x3 )2 . Now consider a general orthogonal
curvilinear coordinate system, (q1 , q2 , q3 ), which are related to the Cartesian coordinates by
qm = qm (x1 , x2 , x3 ),
xm = xm (q1 , q2 , q3 )
(7)
An arbitrary differential length in space can be expressed by
(ds)2 = (h1 dq1 )2 + (h2 dq2 )2 + (h3 dq3 )2
(8)
2
x = const
y = const
µ = const
ν = const
x = -a
x=a
Figure 1: Contour lines in Cartesian and elliptic coordinates [5].
where hi are called scale factors and have the following expressions [4].
(h1 )2 =
(h2 )2 =
(h3 )2 =
∂xk ∂xk
∂q1 ∂q1
∂xk ∂xk
∂q2 ∂q2
∂xk ∂xk
∂q3 ∂q3
(9)
The index notation is used here and k is a dummy index that is summed from 1 to 3. Let ek be
the basis vectors of the Cartesian coordinates (x1 , x2 , x3 ) and let êk be the basis vectors of the
curvilinear coordinates (q1 , q2 , q3 ). The gradient of a scalar field φ(q1 , q2 , q3 ) is,
∇φ = ê1
1 ∂φ
1 ∂φ
1 ∂φ
+ ê2
+ ê3
h1 ∂q1
h2 ∂q2
h3 ∂q3
(10)
The Laplacian of a scalar field φ(q1 , q2 , q3 ) is,
∂
h2 h3 ∂φ
∂
h3 h1 ∂φ
∂
h1 h2 ∂φ
1
2
+
+
∇ φ=
h1 h2 h3 ∂q1
h1 ∂q1
∂q2
h2 ∂q2
∂q3
h3 ∂q3
In 2-dimension, the Laplacian of a scalar field φ(q1 , q2 ) reduces to the following.
∂
h2 ∂φ
∂
h1 ∂φ
1
∇2 φ =
+
h1 h2 ∂q1 h1 ∂q1
∂q2 h2 ∂q2
3
Elliptic Coordinates
The most common definition of elliptic coordinate (µ, ν) is [5],
x = a cosh µ cos ν
y = a sinh µ sin ν
3
(11)
(12)
With this definition, we can show that
x2
x2
+
a2 cosh2 µ a2 sinh2 µ
x2
x2
−
a2 cos2 ν
a2 sin2 ν
= cos2 ν + sin2 ν = 1
= cosh2 µ − sinh2 µ = 1
(13)
Therefore, the contour lines of µ = const form a set of ellipses, and the contour lines of ν = const
form a set of hyperbolas, as shown in Fig. 1. This figure also shows that in the limit of a → 0,
or when the distance from the origin is much greater than a, the elliptic coordinates becomes very
close to polar coordinates.
The Jacobian matrix between Cartesian coordinates (x, y) and elliptic coordinates (µ, ν) is
"
# ∂x
∂x
a sinh µ cos ν −a cosh µ sin ν
∂µ
∂ν
J ≡ ∂x ∂x =
a cosh µ sin ν
a sinh µ cos ν
∂µ
∂ν
(14)
The elliptic coordinates is an orthogonal coordinate system because the two columns of matrix J
are orthogonal to each other. The scale factors are
s 2
∂x 2
∂y
hµ =
+
∂µ
∂µ
s ∂x 2
∂y 2
hν =
+
(15)
∂ν
∂ν
It is easy to show that
q
p
hµ = hν = a sinh2 µ + sin2 ν = det(J)
(16)
Therefore, the Laplacian of a scalar field φ is
2
∂
∂2
2
+
φ(x, y)
∇ φ =
∂x2 ∂y 2
2
1
∂2
∂
=
+
φ(µ, ν)
a2 (sinh2 µ + sin2 ν) ∂µ2 ∂ν 2
4
(17)
Alternative Definition of Elliptic Coordinates
An alternative definition of elliptic coordinates makes it more natural to generalize the concept to
ellipsoidal coordinates in 3D. Consider an ellipse
x2 y 2
+ 2 =1
a2
b
(18)
We will assume a > b without loss of generality. Now consider a family of curves defined by
f (s) ≡
x2
y2
+
−1=0
a2 + s b2 + s
(19)
When s > −b2 , it defines an ellipse. When −b2 > s > −a2 , it defines a hyperbola.
4
For a given (x, y), let (µ, ν) be the two largest roots of the equation f (s) = 0. There is a one-to-one
correspondence between (x, y) and (µ, ν), if we assume x > 0, y > 0. Specifically,
x2 =
y2 =
(a2 + µ)(a2 + ν)
a2 − b2
2
(b + µ)(b2 + ν)
b2 − a2
(20)
This relationship can be verified by plugging it into the definition of f (s),
x2
y2
a2 + ν
b2 + ν
+
−
1
=
+
−1=0
a2 + µ b2 + µ
a2 − b2 b2 − a2
b2 + µ
x2
y2
a2 + µ
+
−1=0
f (ν) =
+
−
1
=
a2 + ν
b2 + ν
a2 − b2 b2 − a2
The four components of the Jacobian matrix can be obtained.
1/2
1/2
∂x
1
a2 + ν
∂x
1
a2 + µ
=
=
∂µ
2 (a2 + µ)(a2 − b2 )
∂ν
2 (a2 + ν)(a2 − b2 )
1/2
1/2
1
b2 + ν
∂y
1
b2 + µ
∂y
=
=
∂µ
2 (b2 + µ)(b2 − a2 )
∂ν
2 (b2 + ν)(b2 − a2 )
f (µ) =
(21)
(22)
The (µ, ν) coordinate system is orthogonal because
∂x ∂x ∂y ∂y
+
=0
∂µ ∂ν
∂µ ∂ν
(23)
Define function ϕ(s) ≡ (a2 + s)(b2 + s). The scale factors can be expressed as
r
1 µ−ν
hµ =
2 ϕ(µ)
r
1 ν−µ
hν =
2 ϕ(ν)
Therefore, the Laplacian of a scalar field φ(µ, ν) is
1
∂ hν ∂φ
∂ hµ ∂φ
2
∇ φ =
+
hµ hν ∂µ hµ ∂µ
∂ν hν ∂ν
s
s
"
!
!#
p
4 ϕ(µ)ϕ(ν) ∂
∂
ϕ(µ) ∂φ
ϕ(µ) ∂φ
=
+
µ−ν
∂µ
ϕ(ν) ∂µ
∂ν
ϕ(ν) ∂ν
p
p
4
∂ p
∂φ
∂ p
∂φ
=
ϕ(µ)
ϕ(µ)
+ ϕ(ν)
ϕ(ν)
µ−ν
∂µ
∂µ
∂ν
∂ν
(24)
(25)
For derivation details see elliptic coord.m.
5
Ellipsoidal Coordinates
Generalizing the elliptic coordinates defined above, we obtain the ellipsoidal coordinates [6]. Consider an ellipsoid, Consider an ellipse
x2 y 2 z 2
+ 2 + 2 =1
a2
b
c
(26)
5
ν = const
µ = const
λ = const
Figure 2: Isosurfaces of ellipsoidal coordinates [7].
We will assume a > b > c without loss of generality. Now consider a family of curves defined by
f (s) ≡
x2
y2
z2
+
+
−1=0
a2 + s b2 + s c2 + s
(27)
For λ > −c2 , f (λ) = 0 defines an ellipsoid. When −c2 > µ > −b2 , f (µ) = 0 defines a one-sheet
hyperbola. When −b2 > ν > −a2 , f (ν) = 0 defines a two-sheet hyperbola, as shown in Fig. 2.
For a given (x, y, z), let (λ, µ, ν) be the three largest roots of equation f (s) = 0. There is a
one-to-one correspondence between (x, y, z) and (λ, µ, ν), if we assume x > 0, y > 0, z > 0.
x2 =
y2 =
z2 =
(a2 + λ)(a2 + µ)(a2 + ν)
(a2 − b2 )(a2 − c2 )
(b2 + λ)(b2 + µ)(b2 + ν)
(b2 − a2 )(b2 − c2 )
2
(c + λ)(c2 + µ)(c2 + ν)
(c2 − a2 )(c2 − b2 )
(28)
(29)
The following limit applies,
λ > −c2 > µ > −b2 > ν > −a2
(30)
The nine components of the Jacobian matrix can be obtained.
6
∂x
∂λ
=
∂x
∂µ
=
∂x
∂ν
=
∂y
∂λ
=
∂y
∂µ
=
∂y
∂ν
=
∂z
∂λ
=
∂z
∂µ
=
∂z
∂ν
=
1/2
(a2 + µ)(a2 + ν)
1
2 (a2 + λ)(a2 − b2 )(a2 − c2 )
1/2
1
(a2 + λ)(a2 + ν)
2 (a2 + µ)(a2 − b2 )(a2 − c2 )
1/2
(a2 + λ)(a2 + µ)
1
2 (a2 + ν)(a2 − b2 )(a2 − c2 )
(31)
1/2
(b2 + µ)(b2 + ν)
1
2 (b2 + λ)(b2 − a2 )(b2 − c2 )
1/2
(b2 + λ)(b2 + ν)
1
2 (b2 + µ)(b2 − a2 )(b2 − c2 )
1/2
(b2 + λ)(b2 + µ)
1
2 (b2 + ν)(b2 − a2 )(b2 − c2 )
(32)
1/2
1
(c2 + µ)(c2 + ν)
2 (c2 + λ)(c2 − a2 )(c2 − b2 )
1/2
1
(c2 + λ)(c2 + ν)
2 (c2 + µ)(c2 − a2 )(c2 − b2 )
1/2
(c2 + λ)(c2 + µ)
1
2 (c2 + ν)(c2 − a2 )(c2 − b2 )
(33)
The (λ, µ, ν) coordinate system is orthogonal because
∂x ∂x ∂y ∂y
∂z ∂z
+
+
= 0
∂λ ∂µ ∂λ ∂µ ∂λ ∂µ
∂x ∂x ∂y ∂y
∂z ∂z
+
+
= 0
∂µ ∂ν
∂µ ∂ν
∂µ ∂ν
∂z ∂z
∂x ∂x ∂y ∂y
+
+
= 0
∂λ ∂ν
∂λ ∂ν
∂λ ∂ν
Define function ϕ(s) ≡ (a2 + s)(b2 + s)(c2 + s). The scale factors can be expressed as
s
1 (λ − µ)(λ − ν)
hλ =
2
ϕ(λ)
s
1 (µ − λ)(µ − ν)
hµ =
2
ϕ(µ)
s
1 (ν − λ)(ν − µ)
hν =
2
ϕ(ν)
The Laplacian of a scalar field φ(λ, µ, ν) is,
1
∂ hµ hν ∂φ
∂ hν hλ ∂φ
∂ hλ hµ ∂φ
∇2 φ =
+
+
hλ hµ hν ∂λ
hλ ∂λ
∂µ
hµ ∂µ
∂ν
hν ∂ν
p
p
4 ϕ(λ)
4 ϕ(µ)
∂ p
∂φ
∂ p
∂φ
=
ϕ(λ)
+
ϕ(µ)
(λ − µ)(λ − ν) ∂λ
∂λ
(µ − λ)(µ − ν) ∂µ
∂µ
p
4 ϕ(ν)
∂ p
∂φ
+
ϕ(ν)
(ν − λ)(ν − µ) ∂ν
∂ν
7
(34)
(35)
(36)
For derivation details see ellipsoidal coord.m.
6
Ellipsoidal Conductor
The discussion from here on follows Kellogg’s book closely [1], with some variables renamed to follow
the notation here. Suppose we would like to solve for the potential function in space produced by
an ellipsoidal conductor that contains surface charges [1]. For a perfect conductor, the potential
on its surface (as well as the interior) is a constant. Therefore, we are trying to solve the Poisson’s
equation,
∇2 φ(x) = 0
(37)
subject to the boundary condition that φ(x) = φ0 when point x is on the ellipsoidal surface,
x2 y 2 z 2
+ 2 + 2 =1
a2
b
c
(38)
and that φ(x) = 0 as |x| → ∞.
Introducing the ellipsoidal coordinates (λ, µ, ν) as defined in the previous section. The surface
of the (original) ellipsoid is simply the isosurface of λ = 0. The limit of |x| → ∞ corresponds to
the limit of λ → ∞. Therefore, when φ is expressed in term of the ellipsoidal coordinates, i.e.
φ(λ, µ, ν), the boundary condition is very simple,
φ(λ = 0, µ, ν) = φ0
φ(λ → ∞, µ, ν) = 0
(39)
Notice that the Laplacian in the Possion’s equation (∇2 φ = 0) in the elliptical coordinates is defined
in Eq. (36). A natural trial solution is a function φ(λ) that only depends on λ, but not on µ or ν.
In this case,
p
4 ϕ(λ)
∂ p
∂φ
2
∇ φ=
=0
(40)
ϕ(λ)
(λ − µ)(λ − ν) ∂λ
∂λ
p
∂φ
ϕ(λ)
∂λ
∂φ
∂λ
= −
E
2
E
= − p
2 ϕ(λ)
Z ∞
E ds
p
φ(λ) =
λ 2 ϕ(s)
(41)
where E is a constant. The potential field inside the conductor is a constant and equals to the
potential on the surface (λ = 0), which is,
Z ∞
E ds
p
φ0 =
(42)
2 ϕ(s)
0
The surface charge of the conductor σ(x) can be obtained from the following relationship.
∂φ(x)
= −4πσ(x)
∂n+
(43)
8
where ∂n∂+ is the gradient along the surface normal.
1 ∂φ(x)
4π ∂n+
1 ∂φ(λ)
1
= −
4π hλ ∂λ
λ=0
p
Notice that at λ = 0, hλ = µν/ϕ(λ)/2. Therefore,
!
p
1 2 ϕ(λ) E
E
σ =
= √
√
4π
µν 2ϕ(λ)
4π µν
σ = −
On the surface of the ellipsoid, λ = 0,
2
y2 z2
2 2 2 x
+ 4 + 4
µν = a b c
a4
b
c
(44)
(45)
(46)
The equation of the plane tangent to ellipsoid at point (x, y, z) is
x
y
z
(X − x) 2 + (Y − y) 2 + (Z − z) 2 = 0
a
b
c
(47)
The surface normal of the tangent plane is
r
x
y
z
x2 y 2 z 2
n=
, 2, 2 /
+ 4 + 4
2
a
b
c
a4
b
c
(48)
The shortest distance from the origin to this plane is
x2
p = qa
2
2
2
y2
z2
c4
+ yb2 + zc2
x2
a4
+ b4 +
=q
1
x2
a4
y2
+ b4 +
z2
c4
abc
=√
µν
(49)
Therefore, the surface charge density is
σ=
E
E
q
p=
4π abc
4π abc x2
1
2
(50)
2
+ yb4 + zc4
a4
This result is related to the problem of a uniformly charged ellipsoid. As will be shown in the
following section, the above expression of σ is exactly the amount of charge contained in a thin
shell between two similar ellipsoids, in the limit of shell thickness going to zero.
7
Ellipsoidal Shell
Consider a set of similar ellipsoids,
x2
y2
z2
x2 y 2 z 2
+
+
=
1
,
equivalently
+ 2 + 2 = u2
(au)2 (bu)2 (cu)2
a2
b
c
(51)
whose semi-axes are au, bu and cu. They are simply the original ellipsoid scaled by a factor u in
all three directions. Notice that this family of ellipsoids are different from the family of ellipsoids
defined by f (λ) = 0 (whose shapes are not similar to each other). For 0 < u < 1, these ellipsoids
9
are smaller than the original ellipsoid (while for 0 < λ < ∞ the ellipsoids defined by f (λ) = 0 are
all larger than the original ellipsoid).
Consider an ellipsoidal shell contained between two ellipsoids defined by u1 and u2 = u1 + ∆u.
Let the volume density of the charge distribution inside the shell to be a constant ρ. In the limit
of ∆u → 0, the shell reduces to a surface with a surface charge density σ. Obviously, the surface
charge density is proportional to the local thickness of the shell, ∆h, i.e.,
σ = ρ ∆h
(52)
Let (ux, uy, uz) be a point on the surface of the ellipsoid defined by u. Let p(x, y, z, u) be the
shortest distance from the origin to the plane tangent to the ellipsoid at point (ux, uy, uz). Then,
u
p(x, y, z, u) = q
2
2
2
x
+ yb4 + zc4
a4
∆u
∆h(x, y, z) =
q
σ =
q
2
2
x2
+ yb4 + zc4
a4
ρ ∆u
x2
a4
(53)
2
2
+ yb4 + zc4
Compare this with Eq. (50), we can conclude that the surface charge of the shell is the same as the
equilibrium surface charge of a ellipsoidal conductor. The correspondence is made complete if we
set u1 = 1, u2 = 1 + ∆u, and
E
= ρ∆u
4πabc
E = 4π abc ρ∆u
(54)
This means that the potential field produced by this ellipsoidal shell (u1 = 1, u2 = 1 + ∆u) is
Z ∞
ds
p
φ(x) = φ(λ) = 2π abc ρ ∆u
(55)
ϕ(s)
λ
The potential field inside the ellipsoidal shell is a constant and equals to the potential on the surface
(λ = 0), which is,
Z ∞
ds
p
φ0 = 2π abc ρ ∆u
(56)
ϕ(s)
0
In summary, Eq. (55) describes the potential generated by an ellipsoidal shell with uniform density
ρ, whose boundary is the original ellipsoid and a similar ellipsoid scaled by a factor (1 + ∆u). This
result can be generalized to an ellipsoidal shell between u1 = u and u2 = u + ∆u for arbitrary u.
The potential field at a point x = (x, y, z) outside this shell is,
Z ∞
ds
2
p
(57)
φ(x) = 2π abc ρ u ∆u
ϕ(u, s)
λ(u)
where λ(u) is the greatest root of equation f (u, s) = 0 for given (x, y, z) and u. f (u, s) and ϕ(u, s)
are generalization of the previously defined functions f (s) and ϕ(s).
x2
y2
z2
+
+
−1
a2 u2 + λ b2 u2 + λ c2 u2 + λ
ϕ(u, s) ≡ (a2 u2 + s)(b2 u2 + s)(c2 u2 + s)
f (u, s) ≡
(58)
(59)
The factor u2 in Eq. (57) accounts for the fact that surface area of the scaled shell and hence its
total charge content is u2 times those of the original shell (u = 1).
10
8
Uniformly Charged Ellipsoid
A uniformly charged ellipsoid can be considered as a collection of many layers of ellipsoid shells
considered above. For a point x outside the ellipsoid, its potential value should be an integral of u
from 0 to 1,
Z 1 Z ∞
ds
p
u2
Ue (x) = 2π abc ρ
du
(60)
ϕ(u, s)
λ(u)
0
Define new variables v ≡ λ(u)/u2 and t ≡ s/u2 . Then φ(u, s) = u2 φ(t).
Z 1 Z ∞
dt
p
u
Ue (x) = 2π abc ρ
du
ϕ(t)
v
0
R
Perform integration by parts on du,
" Z
#1
Z 1 Z ∞
Z
dt
u2 ∞ dt
1 1 2 1 dv
p
p
u
u p
du =
du
+
2 v
2 0
ϕ(t)
ϕ(t)
ϕ(v) du
0
v
(61)
(62)
0
Notice that v is the greatest root of equation,
x2
y2
z2
+
+
= u2
a2 + v b2 + v c2 + v
(63)
For u = 1, v = λ, while for u → 0, v → ∞. Hence
#1
" Z
Z
1 ∞ dt
u2 ∞ dt
p
p
=
2 v
2 λ
ϕ(t)
ϕ(t)
(64)
0
1
2
Z 1
1
dv
u p
du =
ϕ(v) du
0
2
1
2
Z λ
1
= −
2
∞
u2 p
Z ∞
λ
1
ϕ(v)
dv
x2
y2
z2
+
+
a2 + v b2 + v c2 + v
Therefore, the potential outside the ellipsoid is
Z ∞
y2
z2
x2
1
p
Ue (x) = π abc ρ
1− 2
+ 2
+ 2
dv
a +v b +v c +v
ϕ(v)
λ
1
p
ϕ(v)
dv
(65)
(66)
where λ is the greatest root of equation
x2
y2
z2
+
+
=1
a2 + λ b2 + λ c2 + λ
(67)
as previously defined.
Let us now consider the potential field at a point x = (x, y, z) inside the ellipsoid. Here we have to
cut the ellipsoid into two parts. Point x is on the outside of part 1, but is on the inside of part 2.
Let u0 correspond to the ellipsoid that pass through point x, i.e.
x2 y 2 z 2
+ 2 + 2 = u20
a2
b
c
(68)
11
Therefore, part 1 contains the ellipsoidal shells with 0 < u < u0 and part 2 contains the ellipsoidal
shells with u0 < u < 1. The potential at point x from an ellipsoidal shell in part 1 has the same
expression as the above. But potential at point x from an ellipsoidal shell in part 2 equals to
the potential value at the shell surface (because point x is inside the shell). Therefore, the total
potential at an interior point x is,
"Z
#
Z 1 Z ∞
u0 Z ∞
dt
dt
p
p
Ui (x) = 2π abc ρ
u
u
du +
du
(69)
ϕ(t)
ϕ(t)
v
0
0
u0
Perform integration by parts on the first integral,
" Z
# u0
Z
Z u0 Z ∞
1 u0 2 1 dv
u2 ∞ dt
dt
p
p
+
du =
u p
u
du
2 v
2 0
ϕ(t)
ϕ(t)
ϕ(v) du
v
0
(70)
0
Notice that when u = 0, v = ∞, and when u = u0 , v = 0. Hence
Z u0 Z ∞
Z
Z
dt
u20 ∞ dt
1 ∞ 2 1
p
p
u
du =
−
dv
u p
2 0
ϕ(t)
ϕ(t) 2 0
ϕ(v)
0
v
Z
Z u20 ∞ dt
1 ∞
x2
y2
z2
1
p
p
=
−
+ 2
+ 2
dv
2
2 0
a +v b +v c +v
ϕ(t) 2 0
ϕ(v)
The second integral in Eq. (69) can be carried out because the inner integral is a constant.
Z 1 Z ∞
Z
dt
1 − u20 ∞ dt
p
p
u
du =
2
ϕ(t)
ϕ(t)
u0
0
0
(71)
Therefore, the total potential at an interior point x is,
Z ∞
y2
z2
1
x2
p
+ 2
+ 2
Ui (x) = π abc ρ
1− 2
dv
a +v b +v c +v
ϕ(v)
0
(72)
Notice that the only difference between Eq. (66) and Eq. (72) is in the lower limit of integration
(λ versus 0). Because the range of integration is constant, the potential field inside the ellipsoid,
Ui (x), is simply a quadratic function of space.
9
Special Cases
9.1
Uniformly Charged Sphere
The situation of a = b = c describes a sphere with radius a. When the sphere has a uniform charge
density ρ, the potential distribution inside the sphere is,
φ(x, y, z) = π a3 ρ (A − B x2 − B y 2 − B z 2 ) = π a3 ρ (A − B r2 )
(73)
where
Z ∞
2
a
Z0 ∞
ds
2
B =
= 3
5/2
2
3a
(a + s)
0
A =
ds
(a2 + s)3/2
=
(74)
(75)
12
Therefore
2πρ
(3a2 − r2 )
3
φ(r) =
(76)
The potential value on the sphere surface is
φ(r = a) =
2πρ
4πρ 2
(3a2 − a2 ) =
a
3
3
(77)
Notice that Q = 4π a3 ρ/3 is the total charge contained in the sphere. Hence
φ(r = a) =
Q
a
(78)
This is equivalent to the potential produced by a point charge at origin. This confirms Newton’s
Theorem, which states that the potential field outside a uniformly charged sphere is equivalent to
that produced by a point charge located at the center of the sphere. The potential field outside the
sphere (r > a) is
Q
r
φ(r) =
9.2
(79)
Charged Metal Disc
Consider the case of a = b and c → 0. In this limit, the region inside the ellipsoid reduces to a
circle of radius a in the x-y plane (z = 0). The potential distribution inside this area is,
ZZZ
ρ dx0 dy 0 dz 0
p
φ(x, y) =
(x − x0 )2 + (y − y 0 )2 + z 02
p
Because the integration limit for z 0 is ± c 1 − (x0 /a)2 − (y 0 /a)2 ,
s
0 2 0 2
ZZ
0
0
dx dy
x
y
p
φ(x, y) = 2ρ c
1−
−
(80)
0
2
0
2
a
a
02
02
2
(x − x ) + (y − y )
x +y ≤a
Using the results obtained above, the potential must be a quadratic function inside the area of
radius a,
φ(x, y) = π a2 c ρ (A − B x2 − B y 2 )
Z ∞
ds
π
√ =
A =
2 + s) s
a
(a
Z0 ∞
ds
π
√ = 3
B =
2
2
2a
(a + s) s
0
(81)
(82)
In other words, we have obtained the following relationship,
s
0 2 0 2
ZZ
dx0 dy 0
x
y
p
1−
−
=
a
a
(x − x0 )2 + (y − y 0 )2
x02 +y 02 ≤a2
=
where r =
π 2
a (A − B x2 − B y 2 )
2
π2
(2a2 − r2 )
4a
(83)
p
x2 + y 2 . The potential value is maximum at the circumference of the circle, r = a,
φ(r = a) = 2ρ c
π2 2 π Q
a =
4a
3a
(84)
13
where Q = 34 π a2 c ρ is the total charge in the ellipsoid.
The results obtained here can be used to model a very thin circular metal (conductor) plate with
radius a. The total charge in the metal plate is Q, whereas the equilibrium charge distribution on
the surface is
s
0 2 0 2
3Q
x
y
σ(x, y) =
1−
−
(85)
2
2π a
a
a
The potential on the metal surface is a constant
φ0 =
10
πQ
3a
(86)
Application to Hertz Contact Problem
For simplicity, let us consider a rigid sphere of radius R indenting an elastic half space. The
discussion here follows closely that in Landau and Lifshitz [2]. Choose the coordinate system such
that the elastic half space occupies the z ≤ 0 domain and the z = 0 plane is the surface of the half
space. The Boussinesq solution tells us about the surface displacement of the elastic half space in
response to a point force of magnitude F acting at the origin in the −z direction.
uz = −
F (1 − ν 2 ) 1
πE
r
(87)
where E = 2µ(1 + ν) is the Young’s modulus of the elastic half space.
The shape of the indentor can be described by function
u0 (x, y) =
x2
y2
+
2R 2R
(88)
Let d be the indentation depth. Hence inside the region of contact (S), the surface displacement
of the half space is
uz (x, y) = −d + u0 (x, y) = −d +
x2
y2
+
2R 2R
(89)
Let pz (x, y) be the normal stress on the surface of the half space. Inside the region of contact,
pz (x, y) < 0; outside the region of contact, pz (x, y) = 0. The total indenting force F is,
ZZ
F =
−pz (x, y) dx dy
(90)
S
Using the Boussinesq solution, the surface stress pz (x, y) and the surface displacement uz (x, y) are
related by,
ZZ
pz (x0 , y 0 )
1 − ν2
p
uz (x, y) =
dx0 dy 0
(91)
0
2
0
2
πE
(x − x ) + (y − y )
S
Therefore, our task is to find a function pz (x, y) that satisfies the following condition,
ZZ
pz (x0 , y 0 )
πE
x2
y2
0
0
p
dx dy =
−d +
+
1 − ν2
2R 2R
(x − x0 )2 + (y − y 0 )2
S
14
(92)
By symmetry, we expect the contact area to be a circle, and let a be its radius. Motivated by
Eq. (83), we postulate the following form for pz (x, y),
s
0 2 0 2
y
x
pz (x, y) = −p0 1 −
−
(93)
a
a
The constant p0 is related with F by
s
0 2 0 2
ZZ
y
x
F = p0
−
dx dy
1−
a
a
S
2 π a2
3
3F
2 π a2
= p0
p0 =
(94)
Plug the expression of pz (x, y) into Eq. (92), and using Eq. (83), we have
π2
πE
r2
−p0
(2a2 − r2 ) =
−d
+
4a
1 − ν2
2R
2
πE
r2
3F π
2
2
(2a − r ) =
−d +
−
2 π a2 4 a
1 − ν2
2R
2
2
3 (1 − ν ) F
r
(−2a2 + r2 ) = −d +
3
8E a
2R
(95)
Therefore,
3 (1 − ν 2 ) F
4 E a3
1/3
3 (1 − ν 2 ) F R
a =
4E
1
R
=
(96)
The indentation depth is
d =
2/3 2/3
3 (1 − ν 2 ) F 2 3 (1 − ν 2 ) F
3 (1 − ν 2 )
F
2a =
=
3
8E a
4E a
4E
R1/3
(97)
Acknowledgement
I want to thank Prof. David Barnett for useful discussions and lending me Kellogg’s book to read.
15
A
Matlab Files for Analytic Derivation
% File: elliptic_coord.m
% Purpose: analytic derivation of the properties of elliptic coordinates
syms x y a b mu nu
x = sqrt( (a^2+mu)*(a^2+nu)/(a^2-b^2) );
y = sqrt( (b^2+mu)*(b^2+nu)/(b^2-a^2) );
dxdmu = simplify(diff(x,mu));
dxdnu = simplify(diff(x,nu));
dydmu = simplify(diff(y,mu));
dydnu = simplify(diff(y,nu));
disp(’dxdmu=’); pretty(dxdmu);
disp(’dxdnu=’); pretty(dxdnu);
disp(’dydmu=’); pretty(dydmu);
disp(’dydnu=’); pretty(dydnu);
%check orthogonality
simplify(dxdmu*dxdnu + dydmu*dydnu)
h_mu = simplify( sqrt( (dxdmu)^2 + (dydmu)^2 ) );
h_nu = simplify( sqrt( (dxdnu)^2 + (dydnu)^2 ) );
disp(’h_mu=’); pretty(h_mu);
disp(’h_nu=’); pretty(h_nu);
16
% File: ellipsoidal_coord.m
% Purpose: analytic derivation of the properties of ellipsoidal coordinates
syms x y z a b c lm mu nu
x = sqrt( (a^2+lm)*(a^2+mu)*(a^2+nu)/(a^2-b^2)/(a^2-c^2) );
y = sqrt( (b^2+lm)*(b^2+mu)*(b^2+nu)/(b^2-a^2)/(b^2-c^2) );
z = sqrt( (c^2+lm)*(c^2+mu)*(c^2+nu)/(c^2-a^2)/(c^2-b^2) );
dxdlm = simplify(diff(x,lm));
dxdmu = simplify(diff(x,mu));
dxdnu = simplify(diff(x,nu));
dydlm = simplify(diff(y,lm));
dydmu = simplify(diff(y,mu));
dydnu = simplify(diff(y,nu));
dzdlm = simplify(diff(z,lm));
dzdmu = simplify(diff(z,mu));
dzdnu = simplify(diff(z,nu));
disp(’dxdlm=’); pretty(dxdlm);
disp(’dxdmu=’); pretty(dxdmu);
disp(’dxdnu=’); pretty(dxdnu);
disp(’dydlm=’); pretty(dydlm);
disp(’dydmu=’); pretty(dydmu);
disp(’dydnu=’); pretty(dydnu);
disp(’dzdlm=’); pretty(dzdlm);
disp(’dzdmu=’); pretty(dzdmu);
disp(’dzdnu=’); pretty(dzdnu);
%check orthogonality
[ simplify(dxdlm*dxdmu + dydlm*dydmu + dzdlm*dzdmu)
simplify(dxdmu*dxdnu + dydmu*dydnu + dzdmu*dzdnu)
simplify(dxdlm*dxdnu + dydlm*dydnu + dzdlm*dzdnu)
]
h_lm = simplify( sqrt( (dxdlm)^2 + (dydlm)^2 + (dzdlm)^2 ) );
h_mu = simplify( sqrt( (dxdmu)^2 + (dydmu)^2 + (dzdmu)^2 ) );
h_nu = simplify( sqrt( (dxdnu)^2 + (dydnu)^2 + (dzdnu)^2 ) );
disp(’h_lm=’); pretty(h_lm);
disp(’h_mu=’); pretty(h_mu);
disp(’h_nu=’); pretty(h_nu);
17
References
[1] O. D. Kellogg, Foundations of Potential Theory, (Dover, New York, 1953).
[2] L. D. Landau and E. M. Lifshitz, Mechanics, (Permagon, New York, 1959).
[3] C. R. Weinberger, W. Cai, D.
http://micro.stanford.edu/∼me340b
M.
Barnett,
Micromechanics
lecture
notes,
[4] M. H. Sadd, Elasticity: Theory, Applications and Numerics, (Elsevier, New York, 2005).
[5] http://en.wikipedia.org/wiki/Elliptic coordinates
[6] http://en.wikipedia.org/wiki/Ellipsoidal coordinates
[7] http://eom.springer.de/E/e035420.htm
18
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4/16/13 11:34 AM
Plane Strain Tension
7
D:\My Documents\Courses\ME342-Inela...\plane_strain_uniaxial.m
1 of 3
% ME342 Theory and Applications of Inelasticity
% Wei Cai, caiwei@stanford.edu
%
% Example: plane strain uniaxial loading
% Material parameters
mu = 100;
nu = 0.3;
sig_Y = 1;
% shear modulus
% Poisson's ratio
% yield stress
K = 2*mu*(1+nu)/3/(1-2*nu); % bulk modulus
E = 2*mu*(1+nu);
% Young's modulus
k = sig_Y/sqrt(3);
% condition at yield
sig_xx = sig_Y/sqrt(nu^2-nu+1); % see HW 1.1(a)
sig_yy = 0;
sig_zz = nu*sig_xx;
% hydrostatic stress
sig_bar
= (sig_xx + sig_yy + sig_zz)/3;
% deviatoric stress
s_xx = sig_xx - sig_bar;
s_yy = sig_yy - sig_bar;
s_zz = sig_zz - sig_bar;
J2 = (s_xx^2 + s_yy^2 + s_zz^2)/2;
% J2 should equal k^2 at yield
% strain (elastic at onset of yield)
eps_xx = sig_bar/(3*K) + s_xx/(2*mu);
eps_yy = sig_bar/(3*K) + s_yy/(2*mu);
eps_zz = sig_bar/(3*K) + s_zz/(2*mu);
% strain array (going beyond yield point)
eps_xx_data = eps_xx + [0:120]*1e-4;
% initialize arrays to store results
sig_xx_data = zeros(size(eps_xx_data));
sig_zz_data = zeros(size(eps_xx_data));
lambda_dt_over_2mu_data = zeros(size(eps_xx_data));
s_xx_data = zeros(size(eps_xx_data));
s_yy_data = zeros(size(eps_xx_data));
s_zz_data = zeros(size(eps_xx_data));
J2_data = zeros(size(eps_xx_data));
% the first data point is the yield point considered above
sig_xx_data(1) = sig_xx;
sig_zz_data(1) = sig_zz;
s_xx_data(1)
= s_xx;
s_yy_data(1)
= s_yy;
s_zz_data(1)
= s_zz;
Plane Strain Tension
4/16/13 11:34 AM
J2_data(1)
D:\My Documents\Courses\ME342-Inela...\plane_strain_uniaxial.m
= J2;
for i = 2:length(eps_xx_data),
param = [mu, nu, k, sig_xx_data(i-1), sig_zz_data(i-1), ...
eps_xx_data(i-1), eps_xx_data(i)];
%options = optimset('Display','iter','TolFun',1e-10);
options = optimset('Display','off','TolFun',1e-10);
trial = [sig_xx_data(i-1), sig_zz_data(i-1), 0];
sol = fsolve('eqns_plane_strain_uniaxial_plast', trial, options, param);
sig_xx = sol(1); sig_zz = sol(2); lambda_dt_over_2mu = sol(3);
% compute hydrostatic stress and deviatoric stress
sig_bar = (sig_xx + sig_zz)/3;
s_xx = sig_xx - sig_bar;
s_yy = sig_yy - sig_bar;
s_zz = sig_zz - sig_bar;
J2 = (s_xx^2 + s_yy^2 + s_zz^2)/2;
% save results to array
sig_xx_data(i) = sig_xx; sig_zz_data(i) = sig_zz;
lambda_dt_over_2mu_data(i) = lambda_dt_over_2mu;
s_xx_data(i) = s_xx; s_yy_data(i) = s_yy; s_zz_data(i) = s_zz;
J2_data(i) = J2;
end
% plot solution
% plot stress
figure(1);
subplot(3,1,1);
plot(eps_xx_data, sig_xx_data, '.-');
xlabel('\epsilon_{xx}');
ylabel('\sigma_{xx}');
subplot(3,1,2);
plot([0 eps_xx_data], [0 sig_zz_data], '.-');
xlabel('\epsilon_{xx}');
ylabel('\sigma_{zz}');
subplot(3,1,3);
plot([0 eps_xx_data], [nu sig_zz_data./sig_xx_data], '.-');
xlabel('\epsilon_{xx}');
ylabel('\sigma_{zz} / \sigma_{xx}');
% plot deviatoric stress
figure(2);
subplot(3,1,1);
plot(eps_xx_data, s_xx_data, '.-');
xlabel('\epsilon_{xx}');
ylabel('s_{xx}');
subplot(3,1,2);
plot(eps_xx_data, s_yy_data, '.-');
xlabel('\epsilon_{xx}');
ylabel('s_{yy}');
8
2 of 3
4/16/13 11:34 AM
Plane Strain Tension
9
D:\My Documents\Courses\ME342-Inela...\plane_strain_uniaxial.m
3 of 3
subplot(3,1,3);
plot(eps_xx_data, s_zz_data, '.-');
xlabel('\epsilon_{xx}');
ylabel('s_{zz}');
% plot lambda_dt_over_2mu and J2 (should equal k^2)
figure(3);
subplot(2,1,1);
plot(eps_xx_data, [NaN lambda_dt_over_2mu_data(2:end)], '.-');
xlabel('\epsilon_{xx}');
ylabel('\lambda \Delta t / (2\mu)');
subplot(2,1,2);
plot(eps_xx_data, J2_data, '.-');
xlabel('\epsilon_{xx}');
ylabel('J_2');
ylim([0.9 1.1]*k^2);
4/15/13 12:42 AM
Plane Strain Tension
10
D:\My Documents\Courses\...\eqns_plane_strain_uniaxial_plast.m
1 of 2
% ME342 Theory and Applications of Inelasticity
% Wei Cai, caiwei@stanford.edu
%
% equations to be solved by plane_strain_uniaxial.m
function F = eqns_plane_strain_uniaxial_plast(vars, param)
% F(1): eps_xx_tot (specified)
% F(2): eps_zz_tot = 0
% F(3): yield condition
sig_xx = vars(1);
sig_zz = vars(2);
lambda_dt_over_2mu = vars(3);
sig_yy = 0;
mu = param(1);
nu = param(2);
k = param(3);
sig_xx_0 = param(4);
sig_zz_0 = param(5);
sig_yy_0 = 0;
eps_xx_0 = param(6); % total strain at previous time step
eps_xx
= param(7); % total strain at current time step
eps_zz_0 = 0;
eps_zz
= 0;
K = 2*mu*(1+nu)/3/(1-2*nu); % bulk modulus
% change of stress
dsig_xx = sig_xx - sig_xx_0;
dsig_yy = sig_yy - sig_yy_0;
dsig_zz = sig_zz - sig_zz_0;
dsig_bar = (dsig_xx + dsig_yy + dsig_zz)/3;
ds_xx = dsig_xx - dsig_bar;
ds_yy = dsig_yy - dsig_bar;
ds_zz = dsig_zz - dsig_bar;
% increment of elastic strain
deps_elast_xx = dsig_bar/(3*K) + ds_xx/(2*mu);
deps_elast_yy = dsig_bar/(3*K) + ds_yy/(2*mu);
deps_elast_zz = dsig_bar/(3*K) + ds_zz/(2*mu);
% hydrostaic stress
sig_bar
= (sig_xx + sig_yy + sig_zz)/3;
sig_bar_0 = (sig_xx_0 + sig_yy_0 + sig_zz_0)/3;
% deviatoric stress
s_xx = sig_xx - sig_bar;
s_yy = sig_yy - sig_bar;
s_zz = sig_zz - sig_bar;
4/15/13 12:42 AM
Plane Strain Tension
11
D:\My Documents\Courses\...\eqns_plane_strain_uniaxial_plast.m
2 of 2
s_xx_0 = sig_xx_0 - sig_bar_0;
s_yy_0 = sig_yy_0 - sig_bar_0;
s_zz_0 = sig_zz_0 - sig_bar_0;
% average deviatoric stress
ave_s_xx = (s_xx + s_xx_0)/2;
ave_s_yy = (s_yy + s_yy_0)/2;
ave_s_zz = (s_zz + s_zz_0)/2;
% increment of plastic strain
deps_plast_xx = lambda_dt_over_2mu * ave_s_xx;
deps_plast_yy = lambda_dt_over_2mu * ave_s_yy;
deps_plast_zz = lambda_dt_over_2mu * ave_s_zz;
% construct equation
F = [0 0 0]';
% total strain in xx
F(1) = (eps_xx_0 + deps_elast_xx + deps_plast_xx) - eps_xx;
% total strain in zz
F(2) = (eps_zz_0 + deps_elast_zz + deps_plast_zz) - eps_zz;
% yield condition
J2 = (s_xx^2 + s_yy^2 + s_zz^2)/2;
F(3) = J2 - k^2;
1.18
Plane Strain Tension
12
σxx
1.16
1.14
2
4
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10
12
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Plane Strain Tension
13
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Plane Strain Tension
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D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 2D Cylinder Example: Torsion with plasticity using triangular elements
%
% Wei Cai caiwei@stanford.edu
% William Kuykendall wpkuyken@stanford.edu
%
% adapted from FeCalc.m written by Peter Pinsky pinsky@stanford.edu
% universal meshing from Michael Hunsweck
%
% First Adapted 01/03/2013
%
% Last Modified 04/22/2013
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Part 1: Set simulation parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%clear;
global nNodes nElements Coord u IEN K F Nxe Nye EBC ID g penalty_factor
nDoF = 1;
% number of degrees of freedom
nDim = 2;
% number of dimensions
% L: [l w pl pw]
%
l, w:
domain length and width [m]
%
pl, pw: density of triangles along length and width
% Circular cylinder [x0 y0 r]--[center coordinates, radius] [m m m]
%shape = 'circle'; L = [0.8 0.8 10 10]; c = [0 0 0.2];
% Elliptical cylinder [x0 y0 a b]--[(x-x0)/a]^2+[(y-y0)/b]^2=1
%shape = 'ellipse'; L = [1.3 1.3 10 10]; c = [0 0 0.6 0.2];
% Rectangular rod [x0 y0 a b]-- [x_center, y_center, length, width]
%shape = 'rectangle'; L = [0.4 0.4 5 5]; c = [0 0 0.4 0.4];
shape = 'rectangle'; L = [0.6 0.4 5 5]; c = [0 0 0.6 0.4];
E = 200e9;
nu = 0.32;
mu = E/(2*(1+nu));% shear modulus [ N/m^2 ]
tau_max = 4.0e6; % maximum shear stress
%beta
= 3.5e-4;
% twist per unit length [1/m]
beta
= 4e-4;
% twist per unit length [1/m]
% options for solving plasticity problem by iteration
Niter = 2000; dt = 5e6; plotfreq = 100; penalty_factor = 2e-14;
% set axis limits for plotting
if strcmp(shape,'circle')
clim = c(3);
elseif strcmp(shape,'ellipse')
clim = max(c(3:4));
elseif strcmp(shape,'rectangle')
clim = max(c(3:4))/2;
else
fprintf('Unrecognized shape for plotting.\n');
clim = 1;
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end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Part 2: Generate Mesh, Initialize Arrays, Set Up Boundary Conditions
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Step 1: Generate solution mesh
[Coord,IEN,c,shape]
= CreateMesh(L,c,shape); % Generates the mesh
IEN = IEN';
nNodes
= size(Coord,1);
% Number of nodes
nElements
= size(IEN,2);
% Number of elements
nNodesElement = size(IEN,1);
% Number of nodes per element
nEdgesElement = 3;
% Number of edges per element
% Step 2: Set Tolerance for detecting boundary nodes
boundary_tol = 1e-8;
onSurface = logical(zeros(1,nNodes));
% Step 3: Allocate Arrays
C
= spalloc(nElements,nDoF,nElements);
f
= spalloc(nElements,nDoF,nElements);
g
= spalloc(nNodes,nDoF,round(nNodes/4));
EBC = spalloc(nNodes,nDoF,round(nNodes/4));
% Step 4: Set Essential Boundary Condition
% Step 4.1: Get coordinates of all nodes
X = Coord(:,1);
% x-position of all nodes
Y = Coord(:,2);
% y-position of all nodes
% Step 4.2: Define essential boundary condition value(s)
T = [ 0 ]; % Traction
% Step 4.3: Set g, EBC for all nodes on the surface
% Step 4.3a: Find all nodes on external surface
if strcmp(shape,'circle')
% distance squared of node from circle center
D2 = (X-c(1)).^2 + (Y-c(2)).^2;
onSurface = (D2 >= c(3)^2-boundary_tol);
elseif strcmp(shape,'ellipse')
% distance squared of node from ellipse center
D2 = ((X-c(1))/c(3)).^2 + ((Y-c(2))/c(4)).^2;
onSurface = (D2 >= 1-boundary_tol);
elseif strcmp(shape,'rectangle')
for ii = 1:nNodes
onSurface(ii) = abs(X(ii)-c(1))>(c(3)/2-boundary_tol) ...
|| abs(Y(ii)-c(2))>(c(4)/2-boundary_tol) ;
end
else
error('Code should not get here. Shape should have been set.\n');
end
% Adjust the interior mesh coords
[newCoord] = AdjustMesh(Coord, IEN, onSurface, 200, 0e-1, 200);
Coord = newCoord;
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D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
% Step 4.3b: Set g, EBC for node on surface
EBC(onSurface) = 1;
g(onSurface)
= T;
% Step 5: Define the Natural Boundary Condition
% Step 5.1: Allocate h array
h = spalloc(nEdgesElement,nElements,round(nElements/4));
% In this problem there is not any natural boundary condition applied
% Step 6: Define the Load Value (often RHS of equation you are solving)
% Step 1: Set f for all elements
f(:) = 0.1;
f(:) = 10;
f(:) = 1;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Part 3: Create Destination Array (ID) and Location Matrix (LM)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Step 1: Create Data Processing Arrays
% Step 1.1: Allocate arrays
ID = zeros(nNodes,nDoF)';
% Destination Array: ID(A) = P if A is not on essential boundary
%
0 if A is on the essential boundary
%
index is global node number (A)
%
result is global equation number (P)
LM = zeros(nNodesElement*nDoF,nElements);
% Location Matrix: P = LM(a,e)
%
row is degree of freedom (a)
%
column is element number (e)
%
result is global equation number (P)
% Step 1.2: Populate ID Array
I = full(EBC' == 0);
% Creates logical indexing array
nEquations = sum(sum(I));
% Count number of DoF's
ID(I) = 1:nEquations;
% Assign Equation numbers
ID
= ID';
% Step 1.3: Populate LM Array
P
= ID(IEN,:)';
LM(:) = P(:);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Part 4: Assembly of K and F Matrices
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Step 1: Allocate K and F
K = spalloc(nEquations,nEquations,20*nDoF*nEquations);
F = spalloc(nEquations,1,nEquations);
% Step 2: Assemble K and F
% Step 2.1: Compute element contributiontions
for e = 1:nElements
% Step 2.2: Get coordinates of element nodes
x = Coord(IEN(:,e),1);
y = Coord(IEN(:,e),2);
% Step 2.3: Define Shorthand
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D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
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x12 = x(1) - x(2); x23 = x(2) - x(3); x31 = x(3) - x(1);
y12 = y(1) - y(2); y23 = y(2) - y(3); y31 = y(3) - y(1);
% Step 2.4: Compute element Jacobian
J_e = [ -x31, x23; -y31 y23 ];
% Step 2.5: Compute element size
elementSize = 0.5*det(J_e);
Ae = elementSize;
%==================================================================
% Subsection 1: Computation of k_e
%==================================================================
% For isotropic case.
k_e = 1/(4*2*mu*beta*Ae)* ...
[y23^2+x23^2,
y23*y31+x23*x31,
y23*y12+x23*x12;
y23*y31+x23*x31,
y31^2+x31^2,
y31*y12+x31*x12;
y23*y12+x23*x12,
y31*y12+x31*x12,
y12^2+x12^2];
%==================================================================
% Subsection 2: Computation of f_e
%==================================================================
ff = f(IEN(1,e));
f_e = ff*Ae/3*[1;1;1];
%==================================================================
% Subsection 3: Computation of f_g
%==================================================================
g_e = g(IEN(:,e));
f_g = -k_e*g_e;
%==================================================================
% Subsection 4: Computation of f_h
%==================================================================
% Step 4.1a: Edge load intensity (homogeneous boundary)
f_h = h(:,e);
% Step 4.1b: Edge load intensity (non-homogeneous boundary)
l12 = sqrt(x12^2 + y12^2);
l23 = sqrt(x23^2 + y23^2);
l31 = sqrt(x31^2 + y31^2);
h_e = h(:,e);
f_h = h_e(1)*l12/2*[1;1;0] + h_e(2)*l23/2*[0;1;1] + h_e(3)*l31/2*[1;0;1];
%==================================================================
% End of Subsections
%==================================================================
% Step 2.6: Get Global equation numbers
P = LM(:,e);
% Step 2.7: Eliminate Essential DOFs
I = (P ~= 0);
P = P(I);
% Step 2.8: Insert k_e, f_e, f_g, f_h
K(P,P) = K(P,P) + k_e(I,I);
F(P)
= F(P) + f_e(I) + f_g(I) + f_h(I);
end
% Global matrices for computing spatial gradients and area of elements
Nxe = zeros(nElements,nNodes); Nye = zeros(nElements,nNodes); Ae = zeros(nElements,1);
4/26/13 1:42 PM
D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
for e = 1:nElements
% Get element coordinates
xe = Coord(IEN(:,e),1);
ye = Coord(IEN(:,e),2);
% Define Shorthand
x12 = xe(1) - xe(2); x23 = xe(2) - xe(3); x31 = xe(3) - xe(1);
y12 = ye(1) - ye(2); y23 = ye(2) - ye(3); y31 = ye(3) - ye(1);
% Area of a triangle:
Ae(e) = (xe(1)*(ye(2)-ye(3))+xe(2)*(ye(3)-ye(1))+xe(3)*(ye(1)-ye(2)))/2;
% Compute element jacobian
J_e = [ -x31, x23; -y31 y23 ];
% Calculate natural derivatives of N1,N2,N3
N1es = 0; N2es = 1; N3es = -1;
N1er = 1; N2er = 0; N3er = -1;
d_Nrs = [N1er N1es; N2er N2es; N3er N3es];
d_Nxy = d_Nrs/J_e;
Nxe(e,IEN(:,e)) = d_Nxy(:,1)';
Nye(e,IEN(:,e)) = d_Nxy(:,2)';
end
% compute volume contribution from each node
for ii = 1:nNodes
[row,col] = find(IEN == ii);
V_ave(ii)= sum(Ae(col))/3; % 3 for triangular elements
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Part 5: Compute Finite Element Solution (including plasticity)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Step 1: Solve elasticity problem
d_slash = K\F;
% Step 2: Solve plasticity problem by
% minimize (0.5*d'*K*d - F'*d)_in_elast_region + (penalty)_in_plast_region
% using steepest descent algorithm
% Plast_region is defined for nodes on elements whose slope exceeds tau_max
% Elast_region is defined for remaining nodes
% For nodes in Plast_region, the (elastic) gradient
% from (0.5*d'*K*d - F'*d) term is set to zero, and the gradient from
% penalty function = (tau^2-tau_max^2)*penalty_factor is added.
% initialize d from elasticity solution (scaled)
if ~exist('d')
d = d_slash*0.25;
end
I = (EBC == 0); u = zeros(size(g)); tau_max2 = tau_max^2;
for iter = 1:Niter,
% grad is the residual of Poisson's equation
grad = K*d - F;
u(I) = d(ID(I)); u(~I) = g(~I);
dxe = Nxe*u; dye = Nye*u;
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D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
% total torque
Torque = 2*dot(V_ave, u);
% find elements whose slope exceeds tau_max
tau2 = (dxe).^2+(dye).^2;
element_factor = (tau2 >= tau_max2);
ind_e = find(element_factor);
id_in_elements = reshape(ID(IEN(:,ind_e)),length(ind_e)*3,1);
non_zero_entries = find(id_in_elements>0);
% for these elements, set the residual term from Poisson's equation to zero
grad(id_in_elements(non_zero_entries)) = 0;
% for elements whose slope exceeds tau_max, add penalty function
penalty = sum( (tau2-tau_max2).*element_factor ) * penalty_factor;
dpenalty_du = zeros(size(u)); dpenalty = zeros(size(d));
for e = 1:nElements,
if element_factor(e)
% derivative of slope wrt nodal value
dpenalty_du = dpenalty_du + ...
(2*(Nxe(e,:)*u)*Nxe(e,:)' + 2*(Nye(e,:)*u)*Nye(e,:)');
end
end
dpenalty_du = dpenalty_du * penalty_factor;
dpenalty(ID(I)) = dpenalty_du(I);
grad = grad + dpenalty;
% move d in the steepest descent direction
d = d - grad*dt;
% plot intermediate results during minimization
if (mod(iter,plotfreq)==0)
disp(sprintf('iter = %d/%d Torque = %e sum(d) = %e norm(grad) = %e', ...
iter,Niter,Torque,sum(d),norm(grad)));
figure(1)
x = Coord;
T = trisurf(IEN',x(:,1),x(:,2),u);
xlim([-clim clim]); ylim([-clim clim]);
figure(3)
for ii = 1:nNodes
[row,col] = find(IEN == ii);
du_ave(ii,1)= mean(dxe(col,:),1);
du_ave(ii,2)= mean(dye(col,:),1);
end;
T = trisurf(IEN',x(:,1),x(:,2),sqrt(du_ave(:,2).^2+du_ave(:,1).^2));
xlim([-clim clim]); ylim([-clim clim]);
drawnow
end
end
% Step 3: Arrange results (d) into solution vector (u)
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D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
u = zeros(nNodes,nDoF);
I
= (EBC == 0);
u(I) = d(ID(I));
u(~I) = g(~I);
Torque = 2*dot(V_ave, u);
% Step 4: Create coordinate array
x = Coord;
% Step 5: compute average slope on nodes
for ii = 1:nNodes
[row,col] = find(IEN == ii);
du_ave(ii,1)= mean(dxe(col,:),1);
du_ave(ii,2)= mean(dye(col,:),1);
end;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Part 6: Plotting
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Step 1: Plot Prandtl's stress function (u, i.e. phi)
figure(1)
T = trisurf(IEN',x(:,1),x(:,2),u);
title('Torsion: Prandtl Stress Function',...
'Interpreter','Latex','FontSize',12);
xlabel('$x$','Interpreter','Latex','FontSize',12);
ylabel('$y$','Interpreter','Latex','FontSize',12);
zlabel('$\phi$','Interpreter','Latex','FontSize',12);
set(gca,'FontSize',8);
axis([-clim clim -clim clim 0 max(u*1.1)])
co = colorbar;
set(co,'FontSize',8);
colormap jet
shading interp
set(T,'EdgeColor','k');
% Step 2: Plot sigma_xz and sigma_yz (gradient of u)
figure(2)
subplot(2,1,1)
T = trisurf(IEN',x(:,1),x(:,2),du_ave(:,2));
axis equal
axis([-clim clim -clim clim])
view([0 90])
title('FE Solution for $\sigma_{xz}$',...
'Interpreter','Latex','FontSize',12);
xlabel('$x$','Interpreter','Latex','FontSize',12);
ylabel('$y$','Interpreter','Latex','FontSize',12);
zlabel('$\sigma_{xz}$','Interpreter','Latex','FontSize',12);
co = colorbar; set(co,'FontSize',8);
colormap jet; shading interp
set(T,'EdgeColor','k');
subplot(2,1,2)
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D:\My Documents\Courses\ME342-Inelas...\torsion_plast_example.m
T = trisurf(IEN',x(:,1),x(:,2),-du_ave(:,1));
axis equal
axis([-clim clim -clim clim])
view([0 90])
title('FE Solution for $\sigma_{yz}$',...
'Interpreter','Latex','FontSize',12);
xlabel('$x$','Interpreter','Latex','FontSize',12);
ylabel('$y$','Interpreter','Latex','FontSize',12);
zlabel('$\sigma_{yz}$','Interpreter','Latex','FontSize',12);
co = colorbar; set(co,'FontSize',8);
colormap jet; shading interp
set(T,'EdgeColor','k');
% Step 3: Plot maximum shear stress (magnitude of gradient of u)
figure(3)
T = trisurf(IEN',x(:,1),x(:,2),sqrt(du_ave(:,2).^2+du_ave(:,1).^2));
xlim([-clim clim]); ylim([-clim clim]);
title('Torsion: Maximum Shear Stress',...
'Interpreter','Latex','FontSize',12);
xlabel('$x$','Interpreter','Latex','FontSize',12);
ylabel('$y$','Interpreter','Latex','FontSize',12);
zlabel('$\sigma_{xz}$','Interpreter','Latex','FontSize',12);
cbar = colorbar; set(cbar,'FontSize',8);
colormap jet; shading interp
set(T,'EdgeColor','k');
% Step 4: Plot contour of stress function with elast-plast boundary
figure(4);
lxi = linspace(-1,1,300);lyi = linspace(-1,1,300);[xi,yi] = meshgrid(lxi,lyi);
ui = griddata(x(:,1),x(:,2),u,xi,yi,'cubic');
dui = griddata(x(:,1),x(:,2),sqrt(du_ave(:,2).^2+du_ave(:,1).^2),xi,yi,'cubic');
[c h]=contour(xi,yi,ui,20);
hold on
c2=contourc(lxi,lyi,dui,[0.95 0.95]*tau_max,'b--');
plot(c2(1,2:end),c2(2,2:end),'r.');
hold off
title('Torsion: Prandtl Stress Function',...
'Interpreter','Latex','FontSize',12);
axis equal; axis([-clim clim -clim clim])
figure(5);
xi = linspace(-1,1,300);yi = linspace(-1,1,300);[xi,yi] = meshgrid(xi,yi);
contour(xi,yi,dui,20);
title('Torsion: Maximum Shear Stress',...
'Interpreter','Latex','FontSize',12);
axis equal; axis([-clim clim -clim clim])
8 of 8
Plastic solution
5
Torsion: Prandtl Stress Function
x 10
6
5
x 10
7
5
6
5
4
φ
4
3
3
2
1
2
0
0.3
0.2
0.3
0.1
0.2
0
1
0.1
0
−0.1
−0.2
y
−0.1
−0.2
0
x
Torsion: Prandtl Stress Function
0.25
0.2
0.15
0.1
0.05
0
−0.05
−0.1
−0.15
−0.2
−0.25
−0.25 −0.2 −0.15 −0.1 −0.05
0
0.05
0.1
0.15
0.2
0.25
Plastic solution
Torsion: Maximum Shear Stress
6
x 10
4
6
x 10
3.5
4.5
4
3
3.5
3
2.5
σxz
2.5
2
2
1.5
1
1.5
0.5
0
0.3
1
0.2
0.3
0.1
0.2
0
0.5
0.1
0
−0.1
−0.1
−0.2
−0.2
y
x
FE Solution for σxz
0.3
6
x 10
4
3
0.2
2
y
0.1
1
0
0
−1
−0.1
−2
−0.2
−3
−4
−0.2 −0.1
0
x
0.1
0.2
FE Solution for σyz
0.3
6
x 10
4
3
0.2
2
y
0.1
1
0
0
−1
−0.1
−2
−0.2
−3
−4
−0.2 −0.1
0
x
0.1
0.2
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0.8
num
εrr
num
θθ
anl
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anl
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0.6
ε
ε μ / p0
0.4
0.2
0
−0.2
−0.4
−0.6
0
0.5
1
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r
2
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num
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num
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1.2
1.4
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2
4/29/13 10:04 AM
D:\My Documents\Courses\ME342-Inelastic...\cylind_tube_elast.m
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 2D Example: solve strain field in pressurized cylindrical tube
%
elastic deformation only
%
% Wei Cai caiwei@stanford.edu
% Created
03/24/2013
% Last Modified 04/29/2013
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
mu = 100;
nu = 0.3;
lambda = 2*mu*nu/(1-2*nu);
a = 1; b = 2;
dr = 0.05;
r = [a:dr:b];
p0 = 1;
eps_rr = zeros(size(r));
eps_qq = zeros(size(r));
Niter = 10000;
plotfreq = 1000; % 50;
param = [a b dr mu nu p0];
F0 = eqns_cylind_tube_elast([eps_rr, eps_qq],param);
%options = optimset('Display','iter','TolFun',1e-8);
options = optimset('Display','off','TolFun',1e-8);
sol = fsolve('eqns_cylind_tube_elast', [eps_rr, eps_qq], options, param);
eps_rr = sol(1:end/2); eps_qq = sol(end/2+1:end);
F1 = eqns_cylind_tube_elast([eps_rr, eps_qq],param);
% (Alternative method using cgrelax, require gradient evaluation)
% [U dU_depsrrqq] = grad_cylind_tube_elast([eps_rr, eps_qq],param);
% acc = 1e-6; maxfn = 40000; dfpred = 1e-5; n = length([eps_rr, eps_qq]);
% [U,eps_rrqq_rlx] = cgrelax('grad_cylind_tube_elast',n,acc,maxfn,dfpred,...
%
[eps_rr,eps_qq],param);
% eps_rr = eps_rrqq_rlx(1:end/2); eps_qq = eps_rrqq_rlx(end/2+1:end);
% stress field
sig_rr = (lambda + 2*mu) * eps_rr + lambda * eps_qq;
sig_qq = (lambda + 2*mu) * eps_qq + lambda * eps_rr;
sig_zz = lambda * (eps_rr + eps_qq);
pp=p0*a^2/(b^2-a^2); E = 2*mu*(1+nu);
sig_rr_anl = pp*(1-b^2./r.^2);
sig_qq_anl = pp*(1+b^2./r.^2);
eps_rr_anl = (1-nu^2)/E*sig_rr_anl - nu*(1+nu)/E*sig_qq_anl;
eps_qq_anl = (1-nu^2)/E*sig_qq_anl - nu*(1+nu)/E*sig_rr_anl;
1 of 2
4/29/13 10:04 AM
D:\My Documents\Courses\ME342-Inelastic...\cylind_tube_elast.m
% plot solution
fs = 17;
figure(1);
plot(r, eps_rr /p0*mu, '.', r, eps_qq/p0*mu, 'd', ...
r, eps_rr_anl/p0*mu, r, eps_qq_anl/p0*mu);
xlim([0 max(r)*1.5]);
set(gca,'FontSize',fs);
xlabel('r'); ylabel('\epsilon \mu / p_0');
legend('\epsilon_{rr}^{num}','\epsilon_{\theta\theta}^{num}',...
'\epsilon_{rr}^{anl}','\epsilon_{\theta\theta}^{anl}');
figure(2);
plot(r, sig_rr/p0, '.', r, sig_qq/p0, 'o', r, sig_rr_anl/p0, r, sig_qq_anl/p0);
xlim([0 max(r)*1.5]);
set(gca,'FontSize',fs);
xlabel('r'); ylabel('\sigma / p_0');
legend('\sigma_{rr}^{num}','\sigma_{\theta\theta}^{num}',...
'\sigma_{rr}^{anl}','\sigma_{\theta\theta}^{anl}');
2 of 2
4/29/13 9:54 AM
D:\My Documents\Courses\ME342-Inela...\eqns_cylind_tube_elast.m
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 2D Example: solve strain field in pressurized cylindrical tube
%
elastic deformation only
% Construct the equations to be solved by cylind_tube_elast.m
%
% Wei Cai caiwei@stanford.edu
% Created
03/24/2013
% Last Modified 04/29/2013
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function F = eqns_cylind_tube_elast(vars, param)
% F(1:N-1): equilibrium equation
% F(N:2N-2): compatibility equation
% F(2N-1:2N): boundary condition
a = param(1);
b = param(2);
dr = param(3);
mu = param(4);
nu = param(5);
p0 = param(6);
N = length(vars)/2; F = zeros(1,2*N);
eps_rr = vars(1:N); eps_qq = vars(N+1:end);
lambda = 2*mu*nu/(1-2*nu);
r = [a:dr:b]; r_avg = (r(2:end) + r(1:end-1)) / 2;
% stress field
sig_rr = (lambda + 2*mu) * eps_rr + lambda * eps_qq;
sig_qq = (lambda + 2*mu) * eps_qq + lambda * eps_rr;
sig_zz = lambda * (eps_rr + eps_qq);
dsigrrdr = (sig_rr(2:end) - sig_rr(1:end-1)) / dr;
sig_rr_avg = (sig_rr(2:end) + sig_rr(1:end-1)) / 2;
sig_qq_avg = (sig_qq(2:end) + sig_qq(1:end-1)) / 2;
% equilibrium equation
F(1:N-1) = dsigrrdr + (sig_rr_avg-sig_qq_avg)./r_avg;
% strain field
depsqqdr = (eps_qq(2:end) - eps_qq(1:end-1)) / dr;
eps_qq_avg = (eps_qq(2:end) + eps_qq(1:end-1)) / 2;
eps_rr_avg = (eps_rr(2:end) + eps_rr(1:end-1)) / 2;
% compatibility condition
F(N:2*N-2) = depsqqdr - (eps_rr_avg-eps_qq_avg)./r_avg;
% boundary condition
F(2*N-1:2*N) = [sig_rr(1) + p0, sig_rr(end)];
1 of 1
5/10/13 8:39 AM
D:\My Documents\Courses\ME342-Inelastici...\cylind_tube_plast.m
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 2D Example: solve strain field in pressurized cylindrical tube
%
pressure large enough to cause plastic deformation
%
% Wei Cai caiwei@stanford.edu
% Created
03/24/2013
% Last Modified 04/29/2013
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
mu = 100;
nu = 0.3;
k = 4;
% yield stress / sqrt(3)
E = 2*mu*(1+nu);
% Young's modulus
K = 2*mu*(1+nu)/3/(1-2*nu); % bulk modulus
a = 1; b = 2;
dr = 0.02; drho = dr;
r = [a:dr:b]; rho = r;
p0 = 1;
S_rr_sol
= zeros(length(r), length(rho));
S_qq_sol
= zeros(length(r), length(rho));
sig_rr_sol = zeros(length(r), length(rho));
sig_qq_sol = zeros(length(r), length(rho));
sig_zz_sol = zeros(length(r), length(rho));
eps_rr_sol = zeros(length(r), length(rho));
eps_qq_sol = zeros(length(r), length(rho));
% analytic solution in elastic region
for i=1:length(r),
for j=1:length(rho),
% limit to elastic region
if r(i) >= rho(j),
ppp=k*((1-2*nu)^2/3 + b^4/rho(j)^4)^(-1/2);
sig_rr_1 = ppp*(1-b^2./r(i).^2);
sig_qq_1 = ppp*(1+b^2./r(i).^2);
sig_zz_1 = nu*(sig_rr_1 + sig_qq_1);
S_rr_1 = sig_rr_1 - (sig_rr_1+sig_qq_1+sig_zz_1)/3;
S_qq_1 = sig_qq_1 - (sig_rr_1+sig_qq_1+sig_zz_1)/3;
eps_rr_1 = (1-nu^2)/E*sig_rr_1 - nu*(1+nu)/E*sig_qq_1;
eps_qq_1 = (1-nu^2)/E*sig_qq_1 - nu*(1+nu)/E*sig_rr_1;
S_rr_sol(i,j)
= S_rr_1;
S_qq_sol(i,j)
= S_qq_1;
sig_rr_sol(i,j) = sig_rr_1;
sig_qq_sol(i,j) = sig_qq_1;
sig_zz_sol(i,j) = sig_zz_1;
eps_rr_sol(i,j) = eps_rr_1;
eps_qq_sol(i,j) = eps_qq_1;
end
end
end
1 of 3
5/10/13 8:39 AM
D:\My Documents\Courses\ME342-Inelastici...\cylind_tube_plast.m
2 of 3
% find solution in plastic region
disp('find solution in plastic region...');
for j=2:length(rho),
disp(sprintf('j = %d / %d',j, length(rho)));
for i=j-1:-1:1,
% find solution at (r, rho+drho)
eps_rr_1 = eps_rr_sol(i,j-1); eps_qq_1 = eps_qq_sol(i,j-1);
S_rr_1
= S_rr_sol(i,j-1);
eps_rr_2 = eps_rr_sol(i+1,j); eps_qq_2 = eps_qq_sol(i+1,j);
S_rr_2
= S_rr_sol(i+1,j);
param = [r(i), dr, mu, nu, k, eps_rr_1, eps_qq_1, ...
S_rr_1, eps_rr_2, eps_qq_2, S_rr_2];
%options = optimset('Display','iter','TolFun',1e-8);
options = optimset('Display','off','TolFun',1e-8);
%F0 = eqns_cylind_tube_plast([eps_rr_1, eps_qq_1, S_rr_1], param);
%sol = fsolve('eqns_cylind_tube_plast', [eps_rr_1, eps_qq_1, S_rr_1], ...
%
options, param);
[sol,Fval,exitflag] = fsolve('eqns_cylind_tube_plast', ...
[eps_rr_1, eps_qq_1, S_rr_1], options, param);
switch exitflag
case 0, fprintf(['Solution incorrect: ' ...
'Maximum number of function evaluations or iterations reached.\n']);
case -1, fprintf(['Solution incorrect: ' ...
'Algorithm terminated by the output function.\n']);
case -2, fprintf(['Solution incorrect: ', ...
'Algorithm seems to be converging to a point that is not a root.\n']);
case -3, fprintf(['Solution incorrect: ', ...
'Trust region radius became too small.\n']);
case -4, fprintf(['Solution incorrect: ' ...
'Line search cannot sufficiently decrease the residual '...
'along the current search direction.\n']);
end
%F1 = eqns_cylind_tube_plast(sol, param);
eps_rr_sol(i,j) = sol(1); eps_qq_sol(i,j) = sol(2); S_rr_sol(i,j) = sol(3);
S_qq_sol(i,j) = (-S_rr_sol(i,j) + sqrt(4*k^2-3*S_rr_sol(i,j)^2))/2;
sig_rr_sol(i,j) = S_rr_sol(i,j) + K*(eps_rr_sol(i,j)+eps_qq_sol(i,j));
sig_qq_sol(i,j) = S_qq_sol(i,j) + K*(eps_rr_sol(i,j)+eps_qq_sol(i,j));
sig_zz_sol(i,j) = -S_rr_sol(i,j)-S_qq_sol(i,j) ...
+ K*(eps_rr_sol(i,j)+eps_qq_sol(i,j));
end
end
% verify solution
disp('verify solution in plastic region...');
for j=2:length(rho),
for i=j-1:-1:1,
% find solution at (r, rho+drho)
eps_rr_1 = eps_rr_sol(i,j-1); eps_qq_1 = eps_qq_sol(i,j-1);
S_rr_1
= S_rr_sol(i,j-1);
eps_rr_2 = eps_rr_sol(i+1,j); eps_qq_2 = eps_qq_sol(i+1,j);
5/10/13 8:39 AM
D:\My Documents\Courses\ME342-Inelastici...\cylind_tube_plast.m
S_rr_2
= S_rr_sol(i+1,j);
param = [r(i), dr, mu, nu, k, eps_rr_1, eps_qq_1, ...
S_rr_1, eps_rr_2, eps_qq_2, S_rr_2];
F = eqns_cylind_tube_plast([eps_rr_sol(i,j), eps_qq_sol(i,j), ...
S_rr_sol(i,j)], param);
disp(sprintf('i = %d j = %d F = %e %e %e',i,j,F(1),F(2),F(3)));
end
end
% plot solution
fs = 17;
figure(1);
skip = find(r/r(1)==1.2,1,'First')-1;
plot(r/a, sig_rr_sol(:,1:skip:end)/(2*k), '.-');
set(gca,'FontSize',fs);
xlabel('r / a');
ylabel('\sigma_{rr} / (2k)');
legend('\rho/a = 1','1.2','1.4','1.6','1.8','2.0','Location','SouthEast');
figure(2);
plot(r, sig_qq_sol(:,1:skip:end)/(2*k), '.-');
set(gca,'FontSize',fs);
xlabel('r / a');
ylabel('\sigma_{\theta\theta} / (2k)');
legend('\rho/a = 1','1.2','1.4','1.6','1.8','2.0','Location','NorthWest');
figure(3);
plot(r, sig_zz_sol(:,1:skip:end)/(2*k), '.-');
set(gca,'FontSize',fs);
xlabel('r / a');
ylabel('\sigma_{zz} / (2k)');
legend('\rho/a = 1','1.2','1.4','1.6','1.8','2.0','Location','SouthEast');
3 of 3
4/29/13 10:07 AM
D:\My Documents\Courses\ME342-Inel...\eqns_cylind_tube_plast.m
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 2D Example: solve strain field in pressurized cylindrical tube
%
pressure large enough to cause plastic deformation
% Construct the equations to be solved by cylind_tube_plast.m
%
% Wei Cai caiwei@stanford.edu
% Created
03/24/2013
% Last Modified 04/29/2013
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function F = eqns_cylind_tube_plast(vars, param)
% F(1): equilibrium equation
% F(2): compatibility equation
% F(3): plastic flow rule
eps_rr = vars(1);
eps_qq = vars(2);
S_rr
= vars(3);
r = param(1);
dr = param(2); drho = dr;
mu = param(3);
nu = param(4);
k = param(5);
eps_rr_1 = param(6);
eps_qq_1 = param(7);
S_rr_1
= param(8);
eps_rr_2 = param(9);
eps_qq_2 = param(10);
S_rr_2
= param(11);
K = 2*mu*(1+nu)/3/(1-2*nu); % bulk modulus
F = [0 0 0]';
S_qq
= (-S_rr
+ sqrt(4*k^2-3*S_rr^2 ))/2;
S_qq_1 = (-S_rr_1 + sqrt(4*k^2-3*S_rr_1^2))/2;
S_qq_2 = (-S_rr_2 + sqrt(4*k^2-3*S_rr_2^2))/2;
dSrr_dr = (S_rr_2-S_rr)/dr;
dSqq_dr = (S_qq_2-S_qq)/dr;
depsrr_dr = (eps_rr_2-eps_rr)/dr;
depsqq_dr = (eps_qq_2-eps_qq)/dr;
depsrr_drho = (eps_rr - eps_rr_1)/drho;
depsqq_drho = (eps_qq - eps_qq_1)/drho;
dSrr_drho
= (S_rr - S_rr_1)/drho;
dSqq_drho
= (S_qq - S_qq_1)/drho;
1 of 2
4/29/13 10:07 AM
D:\My Documents\Courses\ME342-Inel...\eqns_cylind_tube_plast.m
% equilibrium condition
F(1) = dSrr_dr + K*( depsrr_dr + depsqq_dr ) ...
+ ( (S_rr_2+S_rr)/2 - (S_qq_2+S_qq)/2 ) / (r+dr/2);
% compatibility condition
F(2) = depsqq_dr - ( (eps_rr_2+eps_rr)/2 - (eps_qq_2+eps_qq)/2 ) / (r+dr/2);
% plastic flow rule
F(3) = 2*mu*( (2/3*depsrr_drho - 1/3*depsqq_drho)*(S_qq+S_qq_1)/2
...
-(2/3*depsqq_drho - 1/3*depsrr_drho)*(S_rr+S_rr_1)/2 ) ...
- dSrr_drho*(S_qq+S_qq_1)/2 + dSqq_drho*(S_rr+S_rr_1)/2 ;
2 of 2
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OtherSlipLineFieldlSolutions
for Notches
L. M. Ka.chanov,
Fundamentals
of il-'eTheoryof plasticity,Dover 2004
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16
Slip Resistance in Polycrystal
mean dissip = 3.0665 (Taylor Factor)
3.6
0.7
3.4
0.5
3.2
0.4
3
y
0.6
0.3
2.8
0.2
2.6
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x
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0.8
mean dissip = 3.0665 (Taylor Factor)
3.5
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18
Crystal Rotation
rotation of tensile axis
0.7
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6/4/13 12:38 PM
D:\My Documents\Courses\ME342-Inelasti...\fcc_slip_system_map.m
1 of 3
% FCC slip systems for single crystal under uniaxial tension
%
plane
slip vector
slip_system = [ 1 1 1
1 -1 0
1 1 1
1 0 -1
1 1 1
0 1 -1
1 1 -1
1 -1 0
1 1 -1
1 0 1
1 1 -1
0 1 1
1 -1 1
1 1 0
1 -1 1
1 0 -1
1 -1 1
0 -1 -1
-1 1 1
-1 -1 0
-1 1 1
-1 0 -1
-1 1 1
0 1 -1 ];
N_slip_system = length(slip_system(:,1));
%domain_type = 1; % a simple parameterization covering several triangles
domain_type = 2; % parameterization of the standard triangle
% G. Y. Chin and W. L. Mammel,
%
Computer Solutions of the Taylor Analysis for Axisymmetric Flow
%
Trans. Metall. Soc. AIME 239, 1400-1405 (1967).
if domain_type == 1
theta = linspace(1e-4,1-1e-4,50)*acos(1/sqrt(3));
phi
= linspace(-0.2+1e-4,1-1e-4,50)*(pi/2);
elseif domain_type == 2
theta = linspace(1e-4,1-1e-4,50)*(pi/4);
phi
= linspace(1e-4,1-1e-4,50)*(pi/4);
else
disp('unknown domain_type');
end
X = zeros(length(theta), length(phi));
Y = X; x = X; y = X; z = X;
maxS = X; maxSind = X;
secondS = X; secondSind = X;
T_select = [ 0 0 1
1 0 1
1 1 1
0 1 1
1 1 2
2 1 3 ];
for ai = 1:length(theta),
for bi = 1:length(phi),
if domain_type == 1
x(ai,bi) = sin(theta(ai))*cos(phi(bi));
y(ai,bi) = sin(theta(ai))*sin(phi(bi));
z(ai,bi) = cos(theta(ai));
6/4/13 12:38 PM
D:\My Documents\Courses\ME342-Inelasti...\fcc_slip_system_map.m
elseif domain_type == 2
factor = atan(sin(theta(ai)))/(pi/4);
x(ai,bi) = sin(theta(ai))*cos(phi(bi)*factor);
y(ai,bi) =
sin(phi(bi)*factor);
z(ai,bi) = cos(theta(ai))*cos(phi(bi)*factor);
end
X(ai,bi) = 2*x(ai,bi)/(1+z(ai,bi));
Y(ai,bi) = 2*y(ai,bi)/(1+z(ai,bi));
T = [x(ai,bi) y(ai,bi) z(ai,bi)];
S = zeros(N_slip_system,1);
for i = 1:N_slip_system
n = slip_system(i,1:3);
b = slip_system(i,4:6);
S(i) = abs(dot(T,n)/norm(T)/norm(n) * dot(T,b)/norm(T)/norm(b));
end
[sortedS ind] = sort(S,1,'descend');
maxS(ai,bi) = sortedS(1);
maxSind(ai,bi) = ind(1);
secondS(ai,bi) = sortedS(2); secondSind(ai,bi) = ind(2);
end
end
X_select = zeros(length(T_select(:,1)),1);
Y_select = X_select;
maxS_select = X_select;
maxSind_select = X_select;
secondS_select = X_select;
secondSind_select = X_select;
for si=1:length(X_select),
T = T_select(si,:); T = T / norm(T);
X_select(si) = 2*T(1)/(1+T(3));
Y_select(si) = 2*T(2)/(1+T(3));
S = zeros(N_slip_system,1);
for i = 1:N_slip_system,
n = slip_system(i,1:3);
b = slip_system(i,4:6);
S(i) = abs(dot(T,n)/norm(T)/norm(n) * dot(T,b)/norm(T)/norm(b));
end
[sortedS ind] = sort(S,1,'descend');
maxS_select(si) = sortedS(1);
maxSind_select(si) = ind(1);
secondS_select(si) = sortedS(2); secondSind_select(si) = ind(2);
end
%plot results
figure(1);
contourf(X,Y,maxS,20);
hold on
plot3(X_select,Y_select,ones(size(X_select)),'ko','LineWidth',2);
hold off
xlabel('x');
2 of 3
6/4/13 12:38 PM
D:\My Documents\Courses\ME342-Inelasti...\fcc_slip_system_map.m
ylabel('y');
axis equal
title('highest Schmid factor');
figure(2);
mesh(X,Y,maxSind);
hold on
plot3(X_select,Y_select,13*ones(size(X_select)),'ko','LineWidth',2);
hold off
view([0 90]);
xlabel('x');
ylabel('y');
axis equal
title('slip system ID with highest S');
figure(3);
mesh(X,Y,secondSind);
hold on
plot3(X_select,Y_select,13*ones(size(X_select)),'ko','LineWidth',2);
hold off
view([0 90]);
xlabel('x');
ylabel('y');
axis equal
title('slip system ID with second highest S');
figure(4);
contourf(X,Y,maxS./secondS-1,20);
colorbar
hold on
plot3(X_select,Y_select,ones(size(X_select)),'ko','LineWidth',2);
hold off
xlabel('x');
ylabel('y');
axis equal
title('S_2/S_1 - 1');
3 of 3
6/4/13 1:55 PM
D:\My Documents\Courses\ME342-I...\fcc_slip_system_polycrystal.m
% FCC slip systems for poly-crystal under uniaxial tension
% Find the 5 slip systems activated for each tensile axis
% following Taylor 1938
% G. Y. Chin and W. L. Mammel,
%
Computer Solutions of the Taylor Analysis for Axisymmetric Flow
%
Trans. Metall. Soc. AIME 239, 1400-1405 (1967).
%
slip_system = [
plane
slip vector
1 1 1
1 -1 0
1 1 1
1 0 -1
1 1 1
0 1 -1
1 1 -1
1 -1 0
1 1 -1
1 0 1
1 1 -1
0 1 1
1 -1 1
1 1 0
1 -1 1
1 0 -1
1 -1 1
0 -1 -1
-1 1 1
-1 -1 0
-1 1 1
-1 0 -1
-1 1 1
0 1 -1 ];
N_slip_system = length(slip_system(:,1));
theta = [0:0.01:1]*(pi/4);
phi
= [0:0.01:1]*(pi/4);
dtheta = theta(2)-theta(1);
dphi
= phi(2)-phi(1);
Ntrial = 3;
dTfactor = 1e-4;
X = zeros(length(theta), length(phi));
Y = X; x = X; y = X; z = X;
activeS = zeros(length(theta), length(phi), N_slip_system);
dissip
= zeros(length(theta), length(phi)); weight = dissip;
Taxis
= zeros(length(theta), length(phi), 3); netrot = Taxis; dTaxis = Taxis;
Taxisnew = Taxis; Xnew = X; Ynew = Y; dX = X; dY = Y;
strain = zeros(N_slip_system,6);
rotation = zeros(N_slip_system,3);
for i = 1:length(slip_system(:,1)),
n = slip_system(i,1:3); n = n/norm(n);
b = slip_system(i,4:6); b = b/norm(b);
tmp = (n'*b + b'*n)/2;
strain(i,:) = [tmp(1,1) tmp(2,2) tmp(3,3) tmp(2,3) tmp(3,1) tmp(1,2)];
rotation(i,:) = cross(n,b)/2;
end
% compute participation ratio of each slip system
for ai = 1:length(theta),
if mod(ai,10)==0
disp(sprintf('ai = %d / %d ', ai, length(theta)));
end
for bi = 1:length(phi),
1 of 3
6/4/13 1:55 PM
D:\My Documents\Courses\ME342-I...\fcc_slip_system_polycrystal.m
factor = atan(sin(theta(ai)))/(pi/4);
x(ai,bi) = sin(theta(ai))*cos(phi(bi)*factor);
y(ai,bi) =
sin(phi(bi)*factor);
z(ai,bi) = cos(theta(ai))*cos(phi(bi)*factor);
X(ai,bi) = 2*x(ai,bi)/(1+z(ai,bi));
Y(ai,bi) = 2*y(ai,bi)/(1+z(ai,bi));
T = [x(ai,bi) y(ai,bi) z(ai,bi)];
Taxis(ai,bi,:) = T;
tmp = (T'*T - (T*T'/3)*eye(3))*(3/2);
tensile_strain = [tmp(1,1) tmp(2,2) tmp(3,3) tmp(2,3) tmp(3,1) tmp(1,2)];
%options = optimset('Display','iter','TolFun',1e-8);
options = optimset('Display','off','TolFun',1e-8);
% double # of slip systems so that coeff is non-negative
Aeq = [strain', -strain'];
beq = [tensile_strain'];
lb = zeros(length(slip_system(:,1))*2,1);
f= ones(length(slip_system(:,1))*2,1);
coeff = linprog(f, [], [], Aeq, beq, lb, [], [], options);
% half # of coeff to store as positive and negative values
compact_coeff = coeff(1:end/2)-coeff(end/2+1:end);
activeS(ai,bi,:) = compact_coeff;
dissip(ai,bi) = sum(abs(compact_coeff));
netrot(ai,bi,:) = rotation'*compact_coeff;
end
end
% compute change of Taxis
for ai = 1:length(theta),
for bi = 1:length(phi),
dTaxis(ai,bi,:) = cross(netrot(ai,bi,:),Taxis(ai,bi,:));
Taxisnew(ai,bi,:) = Taxis(ai,bi,:) + dTaxis(ai,bi,:)*dTfactor;
Xnew(ai,bi) = 2*Taxisnew(ai,bi,1)/(1+Taxisnew(ai,bi,3));
Ynew(ai,bi) = 2*Taxisnew(ai,bi,2)/(1+Taxisnew(ai,bi,3));
end
end
dX = Xnew - X; dY = Ynew - Y;
% compute area of mesh
area = zeros(size(weight));
for ai = 1:length(theta)-1,
for bi = 1:length(phi)-1,
dr1 = [x(ai+1,bi) y(ai+1,bi) z(ai+1,bi)] - [x(ai,bi) y(ai,bi) z(ai,bi)];
dr2 = [x(ai,bi+1) y(ai,bi+1) z(ai,bi+1)] - [x(ai,bi) y(ai,bi) z(ai,bi)];
dr3 = [x(ai+1,bi) y(ai+1,bi) z(ai+1,bi)] ...
- [x(ai+1,bi+1) y(ai+1,bi+1) z(ai+1,bi+1)];
dr4 = [x(ai,bi+1) y(ai,bi+1) z(ai,bi+1)] ...
- [x(ai+1,bi+1) y(ai+1,bi+1) z(ai+1,bi+1)];
area(ai,bi) = (norm(cross(dr1,dr2))+norm(cross(dr3,dr4)))/2;
end
end
2 of 3
6/4/13 1:55 PM
D:\My Documents\Courses\ME342-I...\fcc_slip_system_polycrystal.m
% compute integration (quadrature) weight
for ai = 1:length(theta),
for bi = 1:length(phi),
weight(ai,bi) = area(ai,bi)/4;
if ai > 1
weight(ai,bi) = weight(ai,bi) + area(ai-1,bi)/4;
end
if bi > 1
weight(ai,bi) = weight(ai,bi) + area(ai,bi-1)/4;
end
if ai > 1 && bi > 1
weight(ai,bi) = weight(ai,bi) + area(ai-1,bi-1)/4;
end
end
end
mean_dissip = sum(sum(dissip.*weight))/(4*pi/48);
% plot results
fs = 17;
figure(1);
contourf(X,Y,dissip,50);
colorbar
set(gca,'FontSize',fs);
xlabel('x'); ylabel('y');
axis equal
title(sprintf('mean dissip = %.4f
(Taylor Factor)',mean_dissip));
figure(2);
mesh(X,Y,dissip);
set(gca,'FontSize',fs);
xlabel('x'); ylabel('y');
axis equal
view([30 80]);
title(sprintf('mean dissip = %.4f
(Taylor Factor)',mean_dissip));
figure(3);
skip = 10;
contour(X,Y,dissip,50); hold on
quiver(X(1:skip:end,1:skip:end), Y(1:skip:end,1:skip:end), ...
dX(1:skip:end,1:skip:end),dY(1:skip:end,1:skip:end));
hold off
set(gca,'FontSize',fs);
xlabel('x'); ylabel('y');
axis equal
title('rotation of tensile axis');
t001 = text(-0.01,0.02,'001');
t101 = text( 0.80,0.02,'101');
t111 = text( 0.70,0.70,'111');
3 of 3
0
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