slide cơ học vật chất rắn chapter 8 2 new two dimensional problem solution

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Chapter 8: Two-dimensional problem solution
(Part 2)

TDT  University  -­‐  2015
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8.4 Polar Coordinate Formulation

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8.6 Example Polar Coordinate Solutions

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8.5 General Solutions in Polar Coordinates

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Institute for computational science

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8.4 Polar Coordinate Formulation

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8.5 General Solutions in Polar Coordinates

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8.6 Example Polar Coordinate Solutions

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8.4 Polar Coordinate Formulation

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Airy Stress Function Approach φ = φ(r,θ)

Biharmonic
 Governing
 Equa-on
 

ng

Airy
 Representa-on
 
⎧
1 ∂ϕ 1 ∂ 2ϕ
⎪σ r = r ∂r + r 2 ∂θ 2
⎪

∂ 2ϕ
⎪
⎨σ θ = 2

co

⎛ ∂ 2 1 ∂ 1 ∂ 2 ⎞⎛ ∂ 2 1 ∂ 1 ∂ 2 ⎞
∇ ϕ =⎜ 2 +
+ 2
+
+ 2
ϕ =0
2 ⎟⎜
2
2 ⎟
r ∂r r ∂θ ⎠⎝ ∂r
r ∂r r ∂θ ⎠
⎝ ∂r
4

th

an

∂r
⎪
⎪
∂ ⎛ 1 ∂ϕ ⎞
τ
=

−
⎜
⎟
⎪ rθ
∂r ⎝ r ∂θ ⎠
⎩

du
o

ng


 

Trac-on
 Boundary
 Condi-ons
 

cu

u

Tr = f r (r , θ ) , Tθ = fθ (r , θ )

σr
 
τrθ
 

R

y

σθ
 

r

r
θ

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8.4 Polar Coordinate Formulation

Strain-­‐Displacement
 

an

du
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σ r = λ (er + eθ ) + 2µ er
σ θ = λ (er + eθ ) + 2µ eθ
σ z = λ (er + eθ ) = ν (σ r + σ θ )
τ rθ = 2µ erθ , τ θ z = τ rz = 0

er =

1
1
(σ r −νσ θ ) , eθ = (σ θ −νσ r )
E
E

ez = −

ν

(σ r + σ θ ) = −

cu

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ν

E
1 −ν
1 +ν
erθ =
τ rθ , eθ z = erz = 0
E

u

⎧
∂ur
⎪er =
∂r
⎪
⎪
∂uθ ⎞
1⎛
⎨eθ = ⎜ ur +
⎟
r
∂
θ
⎝
⎠
⎪
⎪

1 ⎛ 1 ∂ur ∂uθ uθ ⎞
+
− ⎟
⎪erθ = ⎜
2
r
∂
θ
∂
r
r ⎠
⎝
⎩

Plane
 stress
 

co

Plane
 strain
 

ng

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Plane Elasticity Problem

Hooke’s
 Law
 

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(er + eθ )


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8.4 Polar Coordinate Formulation

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8.5 General Solutions in Polar Coordinates

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8.6 Example Polar Coordinate Solutions

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8.5 General Solutions in Polar Coordinates

ϕ (r ,θ ) = f (r )ebθ

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8.3.1 General Michell Solution

⎛ ∂ 2 1 ∂ 1 ∂ 2 ⎞⎛ ∂ 2 1 ∂ 1 ∂ 2 ⎞
∇ ϕ =⎜ 2 +

+ 2
+
+ 2
ϕ =0
2 ⎟⎜
2
2 ⎟
r ∂r r ∂θ ⎠⎝ ∂r
r ∂r r ∂θ ⎠
⎝ ∂r
4

co

ng

2
1 − 2b2
1 − 2b2
b2 (4 + b2 )
f ′′′′ + f ′′′ −
f ′′ +
f ′+
f =0
2
3
r
r
r
r4

ng

+ (a4 + a5 log r + a6 r 2 + a7 r 2 log r )θ

th

ϕ = a0 + a1 log r + a2 r 2 + a3r 2 log r

an

Solving the equation gives the general Michell solution (restricted to the periodic case)

∞

cu

u

du
o

a
+ (a11r + a12 r log r + 13 + a14 r 3 + a15 rθ + a16 rθ log r ) cos θ
r
b
+ (b11r + b12 r log r + 13 + b14 r 3 + b15 rθ + b16 rθ log r ) sin θ
r
+ ∑ (an1r n + an 2 r 2+ n + an 3r − n + an 4 r 2− n ) cos nθ

We will use various
terms from this general
solution to solve
several plane problems
in polar coordinates

n=2
∞

+ ∑ (bn1r n + bn 2 r 2+ n + bn 3 r − n + bn 4 r 2− n ) sin nθ
n=2

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8.5 General Solutions in Polar Coordinates
Navier Equation Approach
u=ur(r)er
(Plane Stress or Plane Strain)

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8.3.2 Axisymmetric Solutions

Stress Function Approach
a1
+ a3 + 2a2
2
r
a
σ θ = 2a3 log r − 12 + 3a3 + 2a2
r
τ rθ = 0

ng

ϕ = a0 + a1 log r + a2 r 2 + a3r 2 log r

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th

ng

Displacements - Plane Stress Case

co

σ r = 2a3 log r +

d 2ur 1 dur 1
+
− ur = 0
dr 2 r dr r 2
1

ur = C1r + C2
r

Gives Stress Forms

σr =

A
A
+
B
,
σ
=
−
+ B , τ rθ = 0
θ
r2
r2

cu

u

du
o

1 ⎡ (1 +ν )
⎤
−

a
+
2(1
−
ν
)
a
r
log
r
−
(1
+
ν
)
a
r
+
2
a
(1
−
ν
)
r
1
3
3
2
⎥⎦

E ⎢⎣
r
+ A sin θ + B cos θ
Underlined terms represent
4rθ
uθ =
a3 + A cos θ − B sin θ + Cr rigid-body motion
E
ur =

•  a3 term leads to multivalued behavior, and is not found following the displacement
formulation approach
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8.4 Polar Coordinate Formulation

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ng

th

an

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8.6 Example Polar Coordinate Solutions

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8.5 General Solutions in Polar Coordinates

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8.6 Example Polar Coordinate Solutions

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Example
 8.6
 Thick-­‐Walled
 Cylinder
 Under
 Uniform
 Boundary
 Pressure


 p2
 

A
σr = 2 + B
r
A
σθ = − 2 + B
r

r1
 

an

co


 p1

 

ng

General
 Axisymmetric
 
Stress
 Solu-on
 

cu

Using Strain Displacement
Relations and Hooke’s Law for
plane strain gives the radial
displacement
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σ r (r1 ) = − p1 , σ r (r2 ) = − p2
r12 r22 ( p2 − p1 )
A=
r22 − r12
r12 p1 − r22 p2
B=
r22 − r12

r12 r22 ( p2 − p1 ) 1 r12 p1 − r22 p2
σr =
+

r22 − r12
r2
r22 − r12

σθ = −

u

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th

r2
 

Boundary
 Condi-ons
 
 

r12 r22 ( p2 − p1 ) 1 r12 p1 − r22 p2
+
r22 − r12
r2
r22 − r12

1 +ν

A
r[(1 − 2ν ) B − 2 ]
E
r
r12 p1 − r22 p2
1 +ν ⎡ r12 r22 ( p2 − p1 ) 1
=
+ (1 − 2ν )
⎢−
E ⎣
r22 − r12
r
r22 − r12

ur =

/>
⎤
r⎥
⎦


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8.6 Example Polar Coordinate Solutions

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Example
 8.6
 Cylinder
 Problem
 Results
 Internal
 Pressure
 Only
 

ng

σθ
 /p
 

th

an

r2
 

r1/r2
 =
 0.5
 

co

p
 

Dimensionless Stress
 

r1
 

σr
 /p
 

du
o

ng

For the case of only internal pressure
(p2 = 0 and p1 = p) with r1/r2 = 0.5.
Radial stress decays from –p to zero.
Hoop stress is positive with a
maximum value at the inner radius:

r/r2
 
Dimensionless Distance,
r/r2

 

u

cu


 

Thin-Walled Tube Case:

t = r2 − r1 << 1 ro = (r1 + r2 ) / 2

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(σ θ ) max = (r12 + r22 ) / ( r22 − r12 ) p = (5 / 3) p

σθ ≈

pro
t

Matches with Strength
of Materials Theory
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8.6 Example Polar Coordinate Solutions

Pressurized Hole in an Infinite
Medium


 

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Special
 Cases
 of
 Example
 8-­‐6
 

Stress Free Hole in an Infinite Medium
Under Equal Biaxial Loading at Infinity

ng

p2 = 0 and r2 → ∞

co

p1 = 0, p2 = −T , r2 → ∞
T

 

th

an


 

ng

r1
 

r1
 

cu

u

du
o


 p
 

r12
r12

σ r = − p1 2 , σ θ = p1 2 , σ z = 0
r
r
2
1 +ν p1r1
ur =
E r
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⎛ r12 ⎞
⎛ r12 ⎞
σ r = T ⎜1 − 2 ⎟ , σ θ = T ⎜1 + 2 ⎟
⎝ r ⎠
⎝ r ⎠

σ max = (σ θ )max = σ θ (r1 ) = 2T
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8.6 Example Polar Coordinate Solutions

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Example 8.7 Infinite Medium with a Stress Free Hole Under Uniform Far Field Loading

 

Boundary
 Condi-ons
 
 

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a
 

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y
 
T
 

an

T
 

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th

x
 

cu

6a
a1
4a
− (2a21 + 423 + 224 ) cos 2θ
2
r
r
r
6a
a
σ θ = a3 (3 + 2 log r ) + 2a2 − 12 + (2a21 + 12a22 r 4 + 423 ) cos 2θ
r
r
6a
2a
τ rθ = (2a21 + 6a22 r 2 − 423 − 224 ) sin 2θ
r
r

σ r = a3 (1 + 2 log r ) + 2a2 +

σ r (a, θ ) = τ rθ (a, θ ) = 0
T
σ r (∞, θ ) = (1 + cos 2θ )
2
T
σ θ (∞, θ ) = (1 − cos 2θ )
2
T
τ rθ (∞, θ ) = − sin 2θ
2

For finite stresses at infinity => a3 = a22 = 0
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Note: Far-field
condition
derived from
the law in
Exercise 3.3

Try
 Stress
 Func-on
 
 
ϕ = a0 + a1 log r + a2 r 2 + a3r 2 log r
+ (a21r 2 + a22 r 4 + a23r −2 + a24 ) cos 2θ
T ⎛ a 2 ⎞ T ⎛ 3a 4 4a 2 ⎞

σ r = ⎜1 − 2 ⎟ + ⎜1 + 4 − 2 ⎟ cos 2θ
2⎝ r ⎠ 2⎝
r
r ⎠
T ⎛ a 2 ⎞ T ⎛ 3a 4 ⎞
σ θ = ⎜1 + 2 ⎟ − ⎜1 + 4 ⎟ cos 2θ
2⎝ r ⎠ 2⎝
r ⎠

τ rθ

T ⎛ 3a 4 2a 2 ⎞
= − ⎜ 1 − 4 + 2 ⎟ sin 2θ
2⎝
r
r ⎠
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8.6 Example Polar Coordinate Solutions
Example 8.7 Stress Results

 
y
 

T ⎛ a 2 ⎞ T ⎛ 3a 4 4a 2 ⎞
σ r = ⎜1 − 2 ⎟ + ⎜1 + 4 − 2 ⎟ cos 2θ
2⎝ r ⎠ 2⎝
r
r ⎠
T ⎛ a 2 ⎞ T ⎛ 3a 4 ⎞
σ θ = ⎜1 + 2 ⎟ − ⎜1 + 4 ⎟ cos 2θ
2⎝ r ⎠ 2⎝
r ⎠

ng

a
 

T
 

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T
 

co

x

 

90

60
2

30

− σ θ (a, θ) / T

du
o

1

cu
240

0

330

u

180

210

σ θ (a, θ) / T

,
 σθ/T
 

150

σ max = σ θ (a, ±π / 2) = 3T

th

3

120

τ rθ

T ⎛ 3a 4 2a 2 ⎞
= − ⎜ 1 − 4 + 2 ⎟ sin 2θ
2⎝
r
r ⎠

ng

theta=0:pi/24:2*pi;
sigma=1-2*cos(2*theta);
X=sigma.*cos(theta);
Y=sigma.*sin(theta);
figure

plot(X,Y,'-r')
axis equal

an


 

300

r π
σθ ( , ) / T
a 2

270

σ θ (a,θ ) = T (1− 2cos 2θ )
σ θ (a,0) = −T ,σ θ (a,15o ) = (1− 3)T ,
σ θ (a,30o ) = 0,σ θ (a,45o ) = T ,
σ θ (a,60
) = 2T ,σ θ (a,90 ) = 3T
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r/a

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ng


  Biaxial Loading Cases
Superposition of Example 8.:

=
 

T2
 

+
 

an

co

T1
 

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8.6 Example Polar Coordinate Solutions

du
o

th

ng

T2
 
Equal Biaxial Tension Case
T1 = T2 = T

cu

u

⎛ r12 ⎞
⎛ r12 ⎞
σ r = T ⎜1 − 2 ⎟ , σ θ = T ⎜1 + 2 ⎟
⎝ r ⎠
⎝ r ⎠

σ max = (σ θ ) max = σ θ (r1 ) = 2T

T1
 
Tension/Compression Case
T1 = T , T2 = -T

⎛ 3a 4 4a 2 ⎞
σ r = T ⎜1 + 4 − 2 ⎟ cos 2θ
r
r ⎠
⎝
⎛ 3a 4 ⎞
σ θ = −T ⎜1 + 4 ⎟ cos 2θ
r ⎠
⎝

τ rθ

⎛ 3a 4 2a 2 ⎞
= −T ⎜1 − 4 + 2 ⎟ sin 2θ
r
r ⎠
⎝

σ θ (a, 0) = σ θ (a, π ) = − 4T , σ θ (a, π / 2) = σ θ ( a,3π / 2) = 4T
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8.6 Example
  Polar Coordinate Solutions

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Review
 Stress
 Concentra-on
 Factors
 Around
 
Stress
 Free
 Holes
 

 
T
 

r1
 

T
 

an

K
 =
 2

 

u

T
 
x
 

K
 =
 3
 

T
 

45o
 

T
 

T
 

K
 =
 4
 

T
 

T
 

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T
 

T
 

du
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ng

th


 

cu

a
 

co

T
 

ng

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T
 


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8.6 Example Polar Coordinate Solutions

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Stress
 Concentra-on
 Around
 -­‐
 Stress

 Free
 Ellip-cal
 Hole
 –
 Chapter
 10
 
Maximum Stress Field


 

ng

σ =S

φ max

co


 y
 

an
th
ng

du
o

u

b ⎞
⎛
= S ⎜1 + 2 ⎟
a ⎠
⎝

25

x
 

Stress Concentration Factor


  a
 

cu

b
 

(σ )

∞
x

20

15

(σϕ)max/S
 

10
5

Circular
 Case
 
0
0

1

2

3

4

5

6

7

Eccentricity Parameter, b/a

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8

9

10


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8.6 Example Polar Coordinate Solutions

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Stress
 Concentra-on
 Around
 
Stress
 Free
 Hole
 in
 Orthotropic
 Material

 –
 Chapter
 11
 

 

ng


 

an

co

σx(0,y)/S
 

th

y
 
S
 

ng

S
 

Orthotropic
 Case
 Carbon/Epoxy
 

Isotropic
 Case
 

cu

u

du
o

x
 

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8.6 Example Polar Coordinate Solutions

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Three
 Dimensional
 Stress
 Concentra-on
 Problem
 –
 Chapter
 13
 

Normal Stress on the x,y-plane (z = 0)

S
 

⎛
4 − 5ν a 3
9
a5 ⎞
σ z (r , 0) = S ⎜1 +
+
⎟
3
2(7
−
5

ν
)
r
2(7 − 5ν ) r 5 ⎠
⎝

ng

z
 
y
 

co

Stress Field

a
 

an

x
 

th

σ z (a, 0) = (σ z ) max =

ng

S
 

u

2.5

Stress Concentration Factor

3

Two Dimensional Case:
 σθ(r,π/2)/S
 

cu

2

1.5
1

0.5

27 − 15ν
S
2(7 − 5ν )

ν = 0.3 ⇒

(σ z ) max
= 2.04
S

2.2

du
o

3.5
Normalized Stress in Loading
Direction


 

Three Dimensional Case:
 σz(r,0)/S
 ,
 ν
 =
 0.3
 

0

2.15
2.1
2.05

2
1.95
1.9

1

2

3

4

Dimensionless Distance, r/a

5

0

0.1

0.2

Poisson's Ratio


 
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0.3

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0.4

0.5


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8.6 Example Polar Coordinate Solutions

 

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Wedge Domain Problems

 y
 

ng

Use general stress function solution to include
terms that are bounded at origin and give
uniform stresses on the boundaries

co

ϕ = r 2 (a2 + a6θ + a21 cos 2θ + b21 sin 2θ )

an

r
 

β
 

ng

th

α
 

θ
 

σ r = 2a2 + 2a6θ − 2a21 cos 2θ − 2b21 sin 2θ
σ θ = 2a2 + 2a6θ + 2a21 cos 2θ + 2b21 sin 2θ
τ rθ = −a6 − 2b21 cos 2θ + 2a21 sin 2θ

x
 

u

 y
 

Quarter Plane Example (α = 0 and β = π/2)

du
o


 

cu

σ θ (r , π / 2) = 0
τ rθ (r , π / 2) = S

S
 
r
 

S π
π
( − 2θ + cos 2θ − sin 2θ )
2 2
2
S π
π
σ θ = ( − 2θ − cos 2θ + sin 2θ )

2 2
2
S
π
τ rθ = (1 − cos 2θ − sin 2θ )
2
2

σr =

θ
 

x
 

σ θ (r , 0) = τ rθ (r , 0) = 0
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8.6 Example Polar Coordinate Solutions

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Example 6: Half-Space Examples Uniform Normal Stress Over x ≤ 0

 

Boundary Conditions
σ θ (r , 0) = τ rθ (r , 0) = 0
τ rθ (r , π ) = 0, σ θ (r , π ) = −T

co

Try Airy Stress Function


 y
 

du
o

ng

th

an

r
 

u

θ
 

cu

x
 

ng

T
 

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ϕ = a6 r 2θ + b21r 2 sin 2θ
σ θ = 2a6θ + 2b21 sin 2θ
τ rθ = −a6 − 2b21 cos 2θ

Use BC’s To Determine Stress Solution
T
(sin 2θ + 2θ )
2π
T
σθ =
(sin 2θ − 2θ )
2π
T
τ rθ =

(1 − cos 2θ )
2π

σr = −

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8.6 Example Polar Coordinate Solutions
Example 7: Half-Space Under Concentrated Surface Force System (Flamant Problem)

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Y
 

Boundary Conditions
σ θ (r , 0) = τ rθ (r , 0) = 0
τ rθ (r , π ) = 0, σ θ (r , π ) = 0

X
 

x
 

r
 

ng

θ
 

∫ Forces = − ( Xe

co

C
 

1

+ Ye 2 )

C

ϕ = (a12 r log r + a15 rθ ) cos θ
+ (b12 r log r + b15 rθ ) sin θ

du
o


 y
 

ng

th

an

Try Airy Stress Function

cu

u

The
 trac6ons
 on
 any
 semicircular
 arc
 C
 enclosing
  Use BC’s To Determine Stress Solution
the
 origin
 must
 balance
 the
 applied
 concentrated
 

2
σ
=
−
[ X cos θ + Y sin θ ]
r
loadings.
  Because
  the
  area
  of
  such
  an
  arc
  is
 
πr
propor6onal
 to
 the
 radius
 r,
 the
 stresses
 must
 be
 
σ θ = τ rθ = 0
of
  order

  1/r
  to
  allow
  such
  an
  equilibrium
 
statement
 to
 hold
 on
 any
 radius.
 The
 appropriate
 
terms
 in
 the
 general
 Michell
 solu6on
 (8.3.6)
 that
 
will
 give
 stresses
 of
 order

 1/r
 are
 specified
 by
 
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8.6 Example Polar Coordinate Solutions
Example 8: Flamant Solution Stress Results Normal Force Case
2Yx 2 y
σ x = σ r cos θ = −
π ( x 2 + y 2 )2
2

or in Cartesian
components

 

2Yxy 2

τ xy = σ r sin θ cos θ = −
π ( x 2 + y 2 )2

th

Y
 

2

co

ng

2Yy 3
σ y = σ r sin θ = −
π ( x 2 + y 2 )2

an

2Y
sin θ
πr
σ θ = τ rθ = 0

σr = −

ng

x

 

y
 =
 a
 


 y
 

Dimensionless Stress
 

cu

u

du
o

σr
 =
 constant
 

τxy/(Y/a)
 

σy/(Y/a)

 

σ y = 2Y / πa
Dimensionless Distance, x/a
 

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Institute for computational science

8.6 Example Polar Coordinate Solutions

∂ur 1
2Y
= (σ r −νσ θ ) = −
sin θ
∂r E
π Er
u 1 ∂uθ 1
2ν Y
εθ = r +
= (σ θ −νσ r ) =
sin θ
r r ∂θ E
π Er

1 ∂ur ∂uθ uθ 1
γ rθ =
+
− = τ rθ = 0
r ∂θ
∂r
r µ

εr =

Y
π
[(1 −ν )(θ − ) cos θ − 2 log r sin θ ]
πE
2
Y
π
uθ =
[−(1 −ν )(θ − ) sin θ − 2 log r cos θ − (1 + ν ) cos θ ]
πE
2

co

ng

ur =

du
o

ng

0.1

cu

u

Y
(1 −ν )
2E
Y
uθ (r , 0) = −uθ (r , π ) = −
[(1 + ν ) + 2 log r ]
πE
ur (r , 0) = ur (r , π ) = −

Note unpleasant feature of 2-D model
that displacements become unbounded as
r è ∞

an

th


 

On Free Surface y = 0

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Example 8: Flamant Solution Stress Results Normal Force Case

Y
 

0
-0.1
-0.2
-0.3
-0.4
-0.5
-0.6
-0.5

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0
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0.5


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Institute for computational science

8.6 Example Polar Coordinate Solutions

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Comparison of Flamant Results with 3-D Theory-Boussinesq’s Problem
Cartesian Solution
P
 

Px ⎛ z 1 − 2ν
u=
−
⎜
4πµ R ⎝ R 2 R + z

⎛z
P ⎡ 3x 2 z
R
x 2 (2 R + z ) ⎞ ⎤
−
(1
−
2
ν
)
−
+
⎢
⎜

2 ⎟⎥
2π R 2 ⎣ R 3
⎝ R R + z R( R + z ) ⎠ ⎦
⎛z
P ⎡ 3y2 z
R
y 2 (2 R + z ) ⎞ ⎤
σy = −
−
(1
−
2
ν
)
−
+
⎢
⎜
2 ⎟⎥
2π R 2 ⎣ R 3
⎝ R R + z R( R + z ) ⎠⎦
3Pz 3
P ⎡ 3xyz (1 − 2ν )(2 R + z ) xy ⎤
σz = −
, τ xy = −
−
⎥
5
2π R
2π R 2 ⎢⎣ R 3

R( R + z )2
⎦

σx = −

an

z
 

P(1 −ν )
u z ( R, 0) =
2πµ R

3Pyz 2
3Pxz 2
τ yz = −
, τ xz = −
2π R 5
2π R 5

du
o

Free Surface Displacements

ng

th

y
 

P ⎛
z2 ⎞
⎞
⎜ 2(1 −ν ) + 2 ⎟
⎟ , w=
4πµ R ⎝
R ⎠
⎠

co

x
 

Py ⎛ z 1 − 2ν
⎞
−
⎟,v=
⎜
4πµ R ⎝ R 2 R + z
⎠

ng


 

Cylindrical Solution

cu

u

Corresponding 2-D Results
P
uθ ( r , 0 ) = −
⎡(1 + ν ) + 2 log r ⎤⎦
πE ⎣

3-D Solution eliminates the
unbounded far-field behavior
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P ⎡ 3r 2 z (1 − 2ν ) R ⎤
rz
(1
−
2
ν
)
r
⎡
⎤ σr =
−
+
ur =
−

2π R 2 ⎢⎣ R 3
R + z ⎥⎦
2
⎢
⎥
4πµ R ⎣ R
R+z ⎦
(1 − 2ν ) P ⎡ z
R ⎤
P ⎡
z2 ⎤
σθ =
−
2
⎢
uz =
2(1 −ν ) + 2 ⎥
2π R ⎣ R R + z ⎥⎦
⎢
4πµ R ⎣
R ⎦
3Pz 3
3P rz 2
σz = −
, τ rz = −
uθ = 0
2π R 5
2π R 5

P

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