[PPT] Two-Dimensional Elasticity

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... Transforming Stress Tensor. Direction Cosines. ... A. Tx.
Tx. Stress Distribution in
Polar Coordinates. ... Procedure. Use stress transformation
to get stresses in xy system ...
civil.engr.siu.edu/Elasticity/ lecture/PlaneElasticity_1.ppt -

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Two-Dimensional Elasticity

Plane Stress
Plane Strain
 Plane Stress
• No traction on one plane passing through
the body.
• One dimension of the body much smaller
than the other two.
 Plane Stress

Out - of - plane normal stress :

 zz  0
or :
 33  0

Out - of - plane shear stresses :

 xz   yz  0
or :
13   23  0
 Plane Stress

Remaining stresses do not vary with

z, but are functions of only x and y.
Let us let x=1, y=2, z=3 in
considering the tensor form of
elasticity equations.
Equilibrium Equations

 xx , x   xy , y  Fx  0
 xy , x   yy , y  Fy  0
Fz  0
 Potential Function

Specify Body Force vector as :


F  V   V, i ê i
V is a potential function. This corresponc ds to
a body force field that is conservative (an energy concept).
Equilibrium Equations

 xx , x   xy , y  V, x  0
 xy , x   yy , y  V, y  0
V, z  0
Airy’s Stress Function

 xx  V  , yy
 yy  V  , xx
 xy   , xy
Stresses of this form identically
satisfy the equilibrium equations.
Stress-Strain Equations
 xx    xx   yy 
1
E

 yy    yy   xx 
1
E

 zz   xx   yy 
E
1
 xy   xy
2G
 xz   yz  0
Strains depend only on x and y, but
not on z. However, we do have a
nonzero strain component in the z
direction.
Stress-Strain Equations

 xx 
E
2
  xx   yy 
1 

 yy 
E
2
  yy   xx 
1 
 xy  2G xy
 Combining Results
 xx   , yy  , xx   1    V 
1
E

 yy   , xx  , yy   1    V 
1
E

 zz   , xx  , yy  2V 
E
1
 xy  , xy
2G
 xz   yz  0
St. Venant Compatibility Equations
2
 2 xx   yy  2 xy
 2
x 2
y 2
 x y
 2 yy  2 zz  2 yz
 2
z 2 y 2
y z
 2 zz  2 xx  2 xz
 2
x 2
z 2
 x z
 2 xy 2
 2 xz   yz  2 xx
  
xz xy x 2
y z
 2 yz  2 xy 2
 2 xz   yy
  
 x z  y  z y 2
 x z
 2 xy 2
 2 xz   yz  2 zz
  
yz xz z 2
xy
Plane Stress Compatibility Equations

2
 2 xx   yy  2 xy
 2
All partials w.r.t. z are zero. x 2
y 2
xy
ε zz  0  2 zz
0
ε xz  ε yz  0 y 2

 2 zz
2
0
x
 2 zz
0
 x y

Ignore last three equations for now.

 Plane Stress Compatibility Equations

2
 2 xx   yy  2 xy
 2
x 2
y 2
xy
Substitution yields :
1
E
 , yyyy  , xxyy   1    V, xx  , xxxx  , xxyy   1    V, yy 

1
  , xxyy
G
E
 2(1   )
G
 Governing Equation

 4   (1   ) 2 V

4 4 4
  
4  4  2 2 2  4
x x  y y
2 2
 
2  2  2
x y

 4  4  4   2V  2V 
4
 2 2 2  4   (1   ) 2  2 
x x y y  x y 
 Governing Equation

4 2
   (1   ) V
 xx  V  , yy
 yy  V  , xx
 xy   , xy
Ignoring Body Forces

4
 0
 xx  , yy
 yy  , xx
 xy   , xy
 Out-of-plane Stress

 zz     xx   yy 
Equilibrium Equations
Body force vector :
 
F  F x , y 
Equilibrium equations :
 xx , x   xy , y  V, x  0
 xy , x   yy , y  V, y  0
 zz  constant
Airy’s Stress Function

 xx  V  , yy
 yy  V  , xx
 xy   , xy
Stresses of this form identically
satisfy the equilibrium equations.
St. Venant Compatibility Equations
2
 2 xx   yy  2 xy
 2
x 2
y 2
xy
 2 yy 2
  zz  2 yz
 2
z 2 y 2
 y z
 2 zz  2 xx  2 xz
 2
x 2
z 2
 x z
 2 xy 2
 2 xz   yz  2 xx
  
xz xy x 2
 y z
 2 yz  2 xy 2
 2 xz   yy
  
xz yz y 2
xz
 2 xy 2
 2 xz   yz  2 zz
  
yz xz z 2
xy
 Plane Stress Compatibility Equations

All partials w.r.t. z are zero. 2

  xx  2 yy  2 xy
 2
ε zz  ε xz  ε yz  0 x 2
y 2
xy
 Combining Results

 xx 
1
E
 
1   2 , yy    1    , xx  1   1  2   V 
 yy 
1
E
 
1   2 , xx    1    , yy  1    1  2   V 
1
 xy  , xy
2G
 Plane Stress Compatibility Equations

2 2
  xx   yy
2   xy
 2
x 2
y 2
xy

 xx 
1
E
 
1   2 , yy   1    , xx  1    1  2   V 
 yy 
1
E
 
1   , xx   1    , yy  1    1  2   V
2

1
 xy  , xy
2G
Substitution yields :
1
E
  
1   2 , yyyy    1    , xxyy  1   1  2   V, yy


1
E
   1
1   , xxxx    1    , xxyy  1    1  2   V, xx   , xxyy
2
G
 Governing Equation

(1  2 ) 2
4    V
1 

 4  4  4 (1  2 )   2 V  2 V 
2 2 2  4   2  2
x 4
x  y y 1    x y 
Ignoring Body Forces

4
 0
 xx  , yy
 yy  , xx
 xy   , xy
Biharmonic Equation

4
 0
General Solution

    Cmnx y m n

m n
 General Solution

    Cmnx y
m n

m n
m  n  3 satisfy biharmonic
individually and exactly
Cylindrical Coordinates

r
 

r

rr
 Equilibrium Equations

σ rr σ rr  σθθ 1 σθθ
   Fr  0
r r r θ
σ rθ 2σ rθ 1 σθθ
   Fθ  0
r r r θ
 Strain Displacement Equations

u r
rr   ur , r
r
u r 1 u  u r 1
      u , 
r r  r r
1  u  u  1 u r  1  u 1 
 r        u  , r   u r, 
2  r r r   2  r r 
x  r cos θ
y  r sin θ
2 2 2
x  y r
-1 
y
θ  tan  
 x
 
1 2 x

1
2 2
r, x  x  y 2x   cos θ
2 r
y
r, y   sin θ
r
1  y  y sin θ
θ, x   2    2  
2

1  xy  x  r r

1  1  y cos θ
θ, y  2  
 2 
1  y
x
 x r r
     
 r, x  θ, x
x r 
        sin θ 
 cos θ   
x r   r 
     
 r, y  θ, y
y r 
        cos θ 
 sin θ   
y r   r 
Transforming Stress Tensor

rr  ri rj ij

    i  j ij
r  ri  j ij
 Direction Cosines

x y z
r cos q sin q 0
q -sin q cos q 0
z 0 0 1
 Stress Transformations

rr   xx cos 2    yy sin 2   2 xy cos  sin 

    xx sin 2    yy cos 2   2 xy cos  sin 

r    xx cos  sin    yy cos  sin    xy cos 2   sin 2  
 xx  , yy
 yy  , xx
 xy   , xy
 Airy’s Stress Function

rr  , yy cos 2   , xx sin 2   2, xy cos  sin 

   , yy sin 2   , xx cos 2   2, xy cos  sin 

r   , yy cos  sin   , xx cos  sin   , xy cos 2   sin 2  
     
 r, x  θ, x
x r 
        sin θ 
 cos θ   
x r   r 
     
 r, y  θ, y
y r 
        cos θ 
 sin θ   
y r   r 
 Airy’s Stress Function
          sin   
, xx       cos     
 x   x   x  r   r  
  2  2  sin      sin2  
  cos    2  cos        0  
 r r   r   r  r 

sin    2  2  sin      sin  cos  sin  

  cos   2       2 cos   
r  r     r     r r2
2 2
cos  sin  cos  sin  sin  sin 
 , rr cos2   2, r  2,   , r  , 
r r r r2
Airy’s Stress Function

Obtain similar results for :

, xy and , xy
Airy’s Stress Function
1 1
rr  , r  2 , 
r r
   , rr
1 1
 r  ,   , r 
r r
Biharmonic Equation

4
 0
 Biharmonic Equation

 
 4    2  2  0
2 1 1
   , rr  , r  2 , 
r r
4   2
1 1  2
    1  1   
2 2
  2   2 2  2   2 2
 r r  r r     r r r r  
Axisymmetric Problems
No dependence on 
1
rr  , r
r
   ,rr
 r  0
2 1 1   
   , rr  , r  r 
r r  r  , r

4   2
1   1 
  2    , rr  , r 
 r r r   r 
4 3 2
4   2   1   1 
  4   2 2 3
r r r 3
r r r r
Stress Function

2 2
  C1r ln r  C2r  C3 ln r  C4
 Stresses

1
rr  C1  1  2 ln r   2C2  C3 2
r
1
   C1  3  2 ln r   2C2  C3 2
r
 r  0
Plane Stress

1
rr   rr    
E
1
        rr 
E

 r 
1
 r 
 1  
 r
2G E
Strain-Displacement Equations

u r
rr   ur , r
r
ur
  
r
 Displacements

dur 1  C3  1    
rr     2C 2  1     C1  1    2 ln r  1   2  C1
dr E  r 2 
ur 1  C 3  1    
     2
 2C2  1     C1  1    2 ln r  1  2C1 
r E r 
 Displacements

Integrate the first equation :

dur 1  C3  1    
   2C 2  1     C1  1    2 ln r  1   2  C1
dr E r2 
1  C3  1    
ur     2C 2  1    r  C1  1    2r ln r  r   2  C1r  constant 
E r 
Solve second equation
ur 1  C 3  1    
   2C 2  1     C1  1    2 ln r  1   2C1
r E r2 
1  C3  1    
ur     2C 2  1    r  C1  1    2r ln r  r   2C r
1 
E r 
 Displacements

1  C3  1    
ur     2C 2  1    r  C1  1    2r ln r  r   2C1r  constant  
E r 
1  C3  1    
   2C 2  1    r  C1  1    2r ln r  r   2C r
1 
E r 
 C1  0 constant  0

1  C3  1    
ur     2C 2  1    r 
E r 
 Quasi-Axisymmetric Problem

If we assume stresses are independent of θ but

displacements may depend on 
dur 1  C3  1    
rr     2C 2  1     C 1  1    2 ln r  1   2  C 1
dr E  r 2 
u r 1 u  1  C 3  1    
       2 C 2  1     C 1  1    2 ln r  1   2C1
r r  E  r2 
1  u  u  1 u r 
     
2  r r r  
 Displacements

Integrate the first equation :

dur 1  C3  1    
   2C 2  1     C1  1    2 ln r  1   2  C1
dr E r2 
1  C3  1    
ur     2C 2  1    r  C1  1    2r ln r  r   2  C1r  f    
E r 
Substitute into second equation
1 u  1  C 3  1     ur
   2C 2  1     C1  1    2 ln r  1   2C1 
r  E r 2
 r
1 u  1  f   
 4C1  
r  E r 
 Displacements
Integrate this equation :
u  1
  4C1r  f    
 E
1
u   4C1r   f    d  g  r  
E
 Displacements
Substitute into shear strain expression :
g r  1 1
g r     f  θ  dθ  f  θ   0
r r r
 1 1 
rg r   g  r      f  θ  dθ  f  θ  
r r 
 rg r   g  r   C5
  f  θ  dθ  f  θ    C5
Displacements

rg r   g  r   C5
g  r    C 5  C 6r
g r   C6
Displacements

 f  θ  dθ  f  θ   C5
f  θ   C7 sin   C8 cos 
 f  θ  dθ  C5  C7 cos   C8 sin 
 Displacements

 C3  1    
1   2C2  1    r  
ur   r 
E
C1  1     2r ln r  r   2C1r  C7 sin   C8 cos 
1
u    4 C1r   C7 cos   C8 sin   C6r
E
 Thick-Walled Cylinder or Disk

• Plane Strain
– Cylinder constrained along the normal axis
• Plane Stress
– Thin disk with circular hole
Thick-Walled Cylinder or Disk
Lame’s Problem

Ro
po
Ri

pi
po pressure load on outer surface.
pi pressure load on inner surface.
Ro outer radius.
Ri inner radius.
 Stresses

1
rr  C1  1  2 ln r   2C2  C3 2
r
1
   C1  3  2 ln r   2C2  C3 2
r
 r  0
Boundary Conditions

rr  R o    po
rr  R i    p i
 Stresses

1
 po  C1  1  2 ln R o   2C2  C3 2
Ro
1
 p i  C1  1  2 ln R i   2C2  C3 2
Ri
 Displacement Condition

1
u    4 C1r   C7 cos   C8 sin   C6r
E
u   0   u   2 
C1  0
Stresses

1
 po  2C2  C3 2
Ro
1
 p i  2C 2  C 3 2
Ri
 Stresses

2 2
p i R i  po R o
C2 
 2
2 Ro  Ri 2

C3 
2 2
Ri Ro  po  p i 
2
Ro  2
Ri 
 Stresses
p i R i2  po R o2 R i2R o2  po  p i 
rr   2 2
 2
Ro  Ri 2
 r Ro  Ri 
2

p i R i2  po R o2 R i2R o2  po  p i 
    2 2
 2
Ro  Ri 2
 r Ro  Ri  2

and for Plane Strain :

 zz   rr     
2  2 2
pi R i  poR o 
 
R o2  R i2
Displacements

1  C3  1    
ur     2C 2  1    r 
E r 
p i R i2  po R o2
C2 

2 R o2  R i2 
R i2R o2  po  p i 
C3 

R o2  R i2 
1  R i2R o2  po  p i  p i R i2  po R o2 
ur    1    1   r
E 2

Ro  Ri r 2
 
2
Ro  Ri 2
 
Pressure in a small hole in an
infinite plate

pi Ri
pi
 Stresses
R i2
pi 2  po
Ro R i  po  pi 
2
rr  
 Ri  2
 Ri 
2
1  2  r 1  2 
2
 R   R 
 o  o

R i2
pi 2  po
Ro R i  po  p i 
2
   
 Ri  2
 Ri 
2
1  2  r 1  2 
2
 R   R 
 o  o
Stresses
po  0
let R o2  
2
Ri
0
R o2
2
 R i pi
rr  2
r
2
R i pi
   2
r
 Plane Stress

rr 
2
piR i 2
 poR o

2 2

R i R o po  pi 
Ro2
 Ri2
 
2
r Ro  2
Ri2

  
2
piR i 2
 poR o

2 2

R i R o po  pi 
Ro2
 Ri2
 2

r Ro  2
Ri2

and for Plane Stress :
 2  p R 2 2
 pi R i 
 zz   rr    o o
E E R 2
o  Ri2

 Thick-walled pipe with end caps

• Neither Plane Stress nor Plane Strain

• Superposition of 2D solution in cross
section and an axial stress.
• Axial stress is assumed to be uniformly
distributed and maintains overall
equilibrium.
F  pA  p i A i  po A o
Net Force in Axial Direction

Fz    2
p o R o  2
p i R i 
2 2
Fz p i R i  p o R o
 zz   2 2
Az R o  R i
2 2
pi R i  poR o
 zz  2 2
Ro  Ri
 1
rr   rr  (     zz ) 
E
1
       ( rr   zz ) 
E
1
 zz    zz  ( rr    ) 
E
1
 r   r
2G
 Shrink Fit
• Inner diameter of outer ring is slightly smaller
than the outer diameter of the inner ring.
• Expand outer ring by heating, slip it over the
inner ring and allow it to cool.
• Disks exert pressure on each other.
• Difference in diameter is called the
interference.
Shrink Fit

b
a c

a,b,c are radii after the

parts are fitted together
 Definitions
• Let p be the pressure exerted between the two
parts
• Inner cylinder
pi = 0
po = -p
• Outer cylinder
pi=-p
p0=0
 1  R i2R o2  p o  p i  p i R i2  p o R o2 
ur    1    1   r
E 2

Ro  Ri r 2
  2
Ro  Ri 2
 
pi  0
po   p
Ro  b
Ri  a
rb
1  a 2b 2p pb 2 
  2  1  1   2 2  1   1 
ur1

E b a b 2
 b a   b


ur1 

pb
E1 b 2  a 2
  1    a   1    b 
1
2
1
2
 1  R i2R o2  po  p i  p i R i2  po R o2 
ur    1    1   r 
E 2

Ro  Ri r 2
 2
Ro  Ri 
2

pi  p
po  0
Ro  c
Ri  b
rb

ur 2 
pb

E2 c 2  b 2
  1   2  c 2
  1   2  b 2

  2( u1  u 2 )


pb

2 E1 b 2  a 2
  1    a
1
2
  1  1  b 2

2
pb

2
E2 c  b 2
  1   2  c 2
  1   2  b 2

 Same material

3 2 2
4b ( c  a ) p
 2 2 2 2
( b  a )( c  b ) E
 Solid Shaft

a0
2
4 b( c ) p
 2 2
(c  b ) E
Plate with a small circular hole.

Assume infinite plate under uniform

tension.
Tx

A 
y

x Tx
Without Hole

 xx  Tx
 yy  0
 xy  0
 Distribution Varies Only
Near Hole

• St. Venant’s Principle

– stresses and strains reasonably distant from the
point of application of an external force are not
significantly altered if the external force is
replaced by a statically equivalent load.
– principle holds true for geometrical objects
(holes, notches, corners, etc.)
 Distribution Varies Only
Near Hole

• Construct a hypothetical circle with a radius

large enough that the effects of the hole are
not felt. Call this radius B.
• B>>A
• Transform uniform stress distribution into
polar coordinates at r=B
Tx

A
y

x Tx
Stress Distribution in Polar Coordinates

Tx
rr  B,    Tx cos  
2
 1  cos 2
2
Tx
   B,    Tx sin  
2
 1  cos 2
2
Tx
rr  B,     Tx cos  sin     sin 2 
2
Alternative Approach
 xx  Tx
1
  Tx y 2
2
y  r sin 
1
  Tx  r sin  
2
2
1 1
  Tx r sin   Tx r 2  1  cos 2 
2 2
2 4
Airy’s Stress Function
1 1
rr  , r  2 , 
r r
   , rr
1 1
 r  ,   , r 
r r
 Stresses
1
  Tx r  1  cos 2 
2
4
Tx
rr  B,     1  cos 2
2
Tx
   B,     1  cos 2
2
Tx
r  B,      sin 2 
2
 Stress Components

Tx Tx Tx
rr  B,     1  cos 2   cos 2
2 2 2
Tx Tx
r  B,      sin 2   0   sin 2 
2 2
Stress Components
Axisymmetric :
 1 Tx
rr  B,   
2
 1
rr  B,    0

 dependent :
 1 Tx
rr  B,    cos 2
2
 1 Tx
rr  B,      sin 2 
2
Axisymmetric State of Stress

Tx
2

y
Non-axisymmetric State of Stress

Tx
cos 2
2

y
Tx
 sin 2
2

x
 Stresses

rr 
2
pi R i 2
 poR o

2 2
R i R o po   pi 
 Ro2
 Ri2
 2
r Ro  2
 Ri2

  
2
pi R i 2
 poR o

2 2
R i R o po   pi 
Ro2
 Ri2
 2
r Ro  2
 Ri2

and for Plane Strain :

 zz    rr     
2  2 2
p i R i  po R o 
 2
Ro  Ri 2

Axisymmetric Stresses
Ri  A

Ro  B

pi  0

Tx
po  
2

 1 Tx  B 2 A 2B 2 
rr   2  2 2

2  B A 2
  
r B  A2 

 1 Tx  B 2 A 2B 2 
    2  2 2

2  B A 2
  
r B  A2 
 Theta Dependent Stresses
4   2
1  1  2
   2
 1   1  2

  2   2 2  2   2 2
 r r r r    r r r r  
From the B.C.' s we can assume :
  F r  cos 2θ
This seperation of variables leads to :
2 9 9
F, rrrr  F, rrr  2 F, rr  3 F, r  0
r r r
1
F r   C1  C2r  C3r  C4 2
2 4
r
 Theta Dependent Stresses

  F r  cos 2θ
1
F r   C1  C2r  C3r  C4 2
2 4
r
 2 4 1
   C1  C2r  C3r  C4 2  cos 2θ
 r 
Airy’s Stress Function
1 1
rr  , r  2 , 
r r
   , rr
1 1
 r  ,   , r 
r r
 Stresses
 2 4 1
   C1  C2r  C3r  C4 2  cos 2θ
 r 
 2  1 1
rr    4C1 2  2C2  6C4 4  cos 2θ
 r r 
 2  2 1
    2C2  12C3r  6C4 4  cos 2θ
 r 
 2  1 2 1
 r    2C1 2  2C2  6C3r  6C4 4  cos 2θ
 r r 
Boundary Conditions
 2
rr  A ,    0
 2
 r  A ,    0
 2 Tx
rr  B,    cos 2
2
 2 Tx
r  B,     sin 2
2
 4 6 
 A 2 2 0  4   C1   0 
A    
 2 6    0 
 2 2 6A 2  4  C 2   
 A A  
   T 
 4 6    x 
  B2 2 0  4  C3
B    2 

 2 6      Tx 
2 6B 2  4 C4   2 
 B 2 B 
A  B

Look at fourth equation :

 A2  2  A2  2 A 
2
6  A2  Tx  A 2 
  2  2 C1  2 2 C2  6B  2 C3  4  2 C4    2 
B B B  B  B B  2 B 

 A2  2  A2  6  A2  Tx  A 2 
  2  2 C1  2 2 C2  6A C3  4  2 C4    2 
2

B B B  B B  2 B 

 A2 
 2   0  6 A 2C 3  0  C 3  0
B 
 
A  B

Solving the third equation :

24 2 6 2 Tx 2
 A 2 C1  2A C2  4 A C4  A
B B 2

 A2  Tx 2 Tx
 2   0  2A C 2 
2
A  C2  
B  2 4
 
Solving the first two equations :

4 Tx 6
 2 C1   4 C4  0
A 2 A

2 Tx 6
 2 C1   4 C4  0
A 2 A

A2
C1  Tx
2

A4
C4   Tx
4
Coefficients
2
A
C1  Tx
2

Tx
C2  
4

C3  0

4
A
C4   Tx
4
 Stresses

 2  A 2 1 3 A4 
rr    2 2    T cos 2θ
4  x
 r 2 2r 

 2  1 3 A 4

       T cos 2θ
4  x
2 2 r 

 2  A 2 1 3 A4 
 r    2    T sin 2θ
4  x
 r 2 2r 
Axisymmetric Stresses
Ri  A

Ro  B

pi  0

Tx
po  
2

 1 Tx  B 2 A 2B 2 
rr   2  2 2

2  B A 2
  

r B  A2 

 1 Tx  B 2 A 2B 2 
rr   2  2 2

2  B A 2
 
r B A  
2 
Axisymmetric Stresses

A
0
B

 
 
 1 T  1 A 2
 Tx  A 2 
rr  x
   1  2 
2  A  2
2 A  2 
2
r 
1  2  r 1  2 
  B  
 B  


 
 
 1 T  1 A 2
 T  A 2

   x
   1  2 
x
2  A  2
2 A  2 
2
r 
1  2  r 1  2 
  B 


 B 


 Total Stresses
 1  2  Tx  A2   A2 1 3 A4 
rr  rr  rr  1  2    2 2    T cos 2θ
4  x
2  r   r 2 2r 

 1  2  Tx  A 2   1 3 A4 
         1  2      T cos 2θ
4  x
2  r  2 2 r 

 1 
 2  A 2
1 3 A 4

 r   r   r    2    T sin 2θ
4  x
 r 2 2r 
 Total Stresses
Tx  A 2  Tx  4A 2 3A 4 
rr  1  2    2  1  4  cos 2θ
2  r  2  r r 
Tx  A 2  Tx  3A 4 
   1  2    1  4  cos 2θ
2  r  2  r 
Tx  2A 2 3A 4 
r    2  1  4  sin 2θ
2  r r 
 Maximum Stress

Maximum Values occur at r  A

rr  0
Tx Tx
   1  1   1  3  cos 2θ  Tx  2Tx cos 2θ
2 2
 r  0
 Maximum Stress

Maximum Values occur at r  A

 3
Maximum Values occur at   2
or 2

 
   A ,    xx  0, A   Tx  2Tx cos 2 2   3Tx
 2
Stress concentrat ion factor K  3
 Stress Distribution

Look at stresses along y axis, at   2

 A 2  Tx  3A 4 
   r , 2  
Tx
1  2    1  4    1

2  r  2  r 
 A 2  Tx  3A 4 
   r , 2  
Tx
1  2    1  4 

2  r  2  r 
 2 4
   r , 2  
 Tx A 3A
2  2  4 
2  r r 
 3

2.5

K 1.5

0.5

0
0 1 2 3 4 5 6 7 8 9 10

r/A
r/A K
1 3.0000
1.5 1.5185
2 1.2188
2.5 1.1184
Note how quickly the effect
3 1.0741 of the hole decays. Stress is
3.5 1.0508 raised only by 2.24% at 5 hole
4 1.0371 radii and by0.52 % at 10 hole
4.5 1.0283
5 1.0224
radii.
5.5 1.0182
6 1.0150
6.5 1.0127
7 1.0108
7.5 1.0094
8 1.0082
8.5 1.0072
9 1.0064
9.5 1.0057
10 1.0052
Plate Under Shear
 S

S S

S
S

S S

S
 S  A 2  S  4A 2 3A 4 
rr  1  2    2  1  4  cos 2θ
2 r  2 r r 
S  A 2
S  3A 
4
   1  2    1  4  cos 2θ
2 r  2 r 
S  2A 2 3A 4 
r    2  1  4  Tx sin 2θ
2 r r 

Stresses due to tension.

 S  A 2  S  4A 2 3A 4 
rr   1  2    2  1  4  cos 2 θ  2 
2 r  2 r r 
S  A2  S  3A 4 
    1  2    1  4  cos 2 θ  2 
2 r  2 r 
S  2A 2 3A 4 
 r   2  1  4  sin 2 θ  
2

2 r r 

Stresses due to compression.

  3A 4 4 A 2 
rr  S 1  4  2  cos 2 θ 
 r r 
 3A 4 
   S 1  4  cos 2 θ 
 r 
 3A 4 2A 2 
r  S 1  4  2  sin 2 θ 
 r r 

Superimposing
  3A 
4
   S 1  4  cos 2 0   4S
 A 
 Next Examples
• Finite Width Plate with Center Hole
• Concentrated Force at a Point of a Straight
Boundary
• Wedge Under Uniform Side Load
• Forces Acting on the End (Tip) of a Wedge
• Stresses in a Circular Disk
• Curved Beam
• Rotating Cylinder
Finite Width Plate with Center Hole
 xx  


2b r

c 

a
Boundary Conditions
At the edge of the hole :
rr  c,    0
 r  c ,    0
At the outer boundary :
0
 xx  a, y   f ( y )  
 xy  a, y   0


2
 yy  x, b   0
 xy  x, b   0
Boundary Conditions

At the edge of the hole :

rr  c,    0
 r  c ,    0
In terms of the Stress Function :
 c,   0

 c,  0
r
Airy’s Stress Function
1 1
rr  , r  2 , 
r r
   , rr
1 1
 r  ,   , r 
r r
 Airy’s Stress Function

2
r r
  a0  b0 ln   c 0   
b b
   r  n  r 
n 2
 r 
n
 r 
2 n
 
  an    b0    c n    d n    cos(n) 
n  2 , 4 , 6   b  b  b  b  

a0  b0  12  ln  
2
b0  r 
c0   2  
2  b 
n2
a n   c n  n  1   2n
 dnn   2( n  1)

b n   c n n   2 ( n  1 )  d n  n  1   2 n

c

b
Outer Boundaries

0
a
r
cos 


2
b
r
sin 
 Procedure
• Use stress transformation to get stresses in
x-y system.
• Satisfy B.C. in least squares sense.
 2 2

  a    a 
      xx  cos  ,     xy  cos  ,   d 
0      

2 
2 2
  b    b 
   yy  sin  ,    xy  sin  ,   d  minimum
      