Transcript X - University of Rhode Island

Application Solutions of Plane Elasticity
Professor M. H. Sadd
Solutions to Plane Problems
Cartesian Coordinates
Airy Representation
x 
 2
 2
 2
,


,



y
xy
y 2
x 2
xy
Biharmonic Governing Equation
 4
 4
 4
 2 2 2  4  4   0
4
x
x y
y
Traction Boundary Conditions
y
S
R
x
Tx  f x ( x, y) , Ty  f y ( x, y)
Uniaxial Tension of a Beam
y
T
T
2c
x
2l
Boundary Condit ions
 x (  l , y )  T ,  y ( x,  c )  0
 xy ( l , y )   xy ( x, c)  0
Try   A02 y 2  x  T ,  y  xy  0
Displacements
u
1
T
T
 ex  ( x   y )   u  x  f ( y )
x
E
E
E
v
1
T
T
 e y  ( y   x )    v   y  g ( x)
y
E
E
E
f ( y )  o y  uo
 xy
u v
  2exy 
 0  f ( y)  g ( x)  0 
g ( x )  o x  vo
y x

OverallDisplacement BoundaryConditions
u (0,0)  0  uo  0 , v(0,0)  0  vo  0
v(0,0)
 0  o  0
x
Pure Bending of a Beam
y
M
M
2c
x
2l
Boundary Condtions
 y ( x, c)  0 ,  xy ( x,c)   xy (l , y )  0

c
c
 x (l , y )dy  0 ,

  A03 y 3   x  
c
c
3M
y ,  y   xy  0
2c 3
Displacements
u
3M
3M

y

u


xy

f
(
y
)
f ( y )  o y  uo
x
2 Ec3
2 Ec3
3M 2
v
3M
3M 2
x  o x  vo

y v
y  g ( x) g ( x) 
3
3
4 Ec3
y
2 Ec
4 Ec
T heoryof Elasticity
M
x  
y ,  y   xy  0
I
Mxy
M
u
,v
[ y 2  x 2  l 2 ]
EI
2 EI
Note Integrated
Boundary Conditions
 x (l , y ) ydy   M
v(l ,0)  0 andu(l ,0)  0 
uo  o  0 , vo  3Ml 2 / 4Ec3
Strengthof Materials
M
x  
y ,  y   xy  0
I
M
v  v( x,0) 
[x2  l 2 ]
2 EI
Bending of a Beam by Uniform Transverse Loading
w
wl
wl
2c
x
BoundaryConditions
 xy ( x,c)  0 ,  y ( x, c)  0 ,  y ( x,c)   w

c
c
 x (l , y )dy  0 ,

c
c
 x (l , y ) ydy  0 ,

c
c
 xy (l , y )dy   wl
y
2l
  A20 x 2  A21 x 2 y  A03 y 3  A23 x 2 y 3 
A23 5
y
5
T heoryof Elasticity
w 2
w y3 c2 y
2
x 
(l  x ) y  ( 
)
2I
I 3
5
w  y3
2 
 y     c 2 y  c 3 
2I  3
3 
 xy  
w
x (c 2  y 2 )
2I
St rengt hof Mat erials
My w 2
x 

(l  x 2 ) y
I
2I
y  0
 xy 
VQ
w
  x (c 2  y 2 )
It
2I
l/c = 2
l/c = 3
l/c = 4
x/w - Elasticity
x/w - Strength of Materials
Dimensionless Distance, y/c
Bending of a Beam by Uniform Transverse Loading
w
wl
wl
2c
x
y
2l
Displacements
w
x3
2 y 3 2c 2 y
y3
2c 3
[(l 2 x  ) y  x(

)  x (  c 2 y 
)]
2 EI
3
3
5
3
3
w  y 4 c 2 y 2 2c 3 y
y2 y4 c2 y2
 
v

 [(l 2  x 2 ) 

]
2 EI  12
2
3
2
6
5
u
x4 l 2
4  2 2  5wl 4
12 4  c 2

  [  (  )c ] x  
[1  (  ) 2 ]
12 2
5 2
5 5 2 l
 24EI
Note that according to theory of elasticity, plane
sections do not remain plane
5wl 4
12 4  c 2
[1  (  ) 2 ]
24EI
5 5 2 l
5wl 4
Strengthof Materialsvmax 
24EI
v(0,0)  vmax 
For long beams l >>c, elasticity and strength of materials
deflections will be approximately the same
Cantilever Beam Problem
y

P
x
Boundary Conditions
 y ( x, c)   xy ( x, c )  0
N
2c

c

c
c
L
c
 xy  
3P
y2
(1  2 )
4c
c
Strengthof Materials
3Pxy N
x  

2c 3
2c
y  0
 xy  
3P
y2
(1  2 )
4c
c
 x (0, y )dy  N ,

c
 x ( L, y )dy  N ,

c
c
c
c
 xy (0, y )dy   P ,

 xy ( L, y )dy   P ,

c
 x (0, y ) ydy  0
c
c
 x (0, y ) ydy   PL
Displacement Field
Stress Field
3Pxy N
x  

2c 3
2c
y  0
3P 
xy 3  N 2
 xy  2  
y
4c 
3c  4c
u 1
1 3Pxy N
v 1
 3Pxy N
 (  x   y )  ( 
 ),
 (  y   x )   ( 
 )
3
x E
E
2c
2c y E
E
2c 3
2c
3Px 2 y
N
Py 3 3(1  ) P
y3

x


(
y

)  o y  uo
4 Ec3 2 Ec
4 Ec3
2cE
3c 2
3Pxy 2 N
Px3
v

y

 o x  vo
4 Ec3
2 Ec
4 Ec3
v ( L,0)
3PL2
3PL2
0 



0




o
o
x
4 Ec3
4 Ec3
N
PL3
PL3
u( L,0)  0  uo 
L , v ( L,0)  0  vo  


L

o
2 Ec
4 Ec3
2 Ec3
For t hecase N  0
u
Px3 3PL2
PL3
P

x


( x 3  3L2 x  2 L3 )
3
3
3
3
4 Ec
4 Ec
2 Ec
4 Ec
P
P
FromStrengt hof Materials v ( x ) 
( x 3  3L2 x  2 L3 ) 
( x 3  3L2 x  2 L3 )
3
6 EI
4 Ec
T hereforethe two displacement solutionsare thesame!
v ( x,0) 
Cantilever Tapered Beam
p

x

A
p cot 
y 
 2
 x tan   xy  ( x 2  y 2 )(  tan 1 )

2(1   cot  ) 
x 
Boundary Conditions
 y ( x ,0)   p ,  xy ( x ,0)  0 , Tx ( x , x tan  )  Ty ( x , x tan  )  0
B
L tan 
0
L
 x ( L, y )dy  0 ,  0L tan   x ( L, y ) ydy  
1 2 L tan 
pL ,  0
 xy ( L, y )dy   pL
2
y
Stress Field


y
xy 
xy 
y2
1 y
 x  2 K   tan1  2
,


2
K


tan


tan

,






2
K
y
xy
, xy

x x  y 2 
x x 2  y 2 
x2  y2


x=L
x=L
Solutions to Plane Problems
Polar Coordinates
Airy Representation
r 
1  1  2 
 2
  1  
 2 2 ,   2 ,  r   

r r r 
r
r  r  
Biharmonic Governing Equation
 2 1 
1  2   2 1 
1 2 

  0
    2 



r r r 2 2  r 2 r r r 2 2 
 r
4
S
R
y

r

x
Traction Boundary Conditions
Tr  f r (r, ) , T  f  (r, )
General Solutions in Polar Coordinates
2
 2 1 
1 2 

  2 
 2 2    0

r
r

r
r  

4
  a0  a1 log r  a 2 r 2  a3 r 2 log r
 ( a 4  a5 log r  a6 r 2  a7 r 2 log r )
a13
 a14 r 3  a15 r  a16 r log r ) cos 
r
b
 (b11r  b12 r log r  13  b14 r 3  b15 r  b16 r log r ) sin 
r
 ( a11r  a12 r log r 

  ( a n1r n  a n 2 r 2n  a n 3 r n  a n 4 r 2n ) cos n
n 2

  (bn1r n  bn 2 r 2n  bn 3 r n  bn 4 r 2n ) sin n
n 2
Axisymmetr
ic Case
  a0  a1 log r  a2 r 2  a3r 2 log r
Thick-Walled Cylinder Under Uniform
Boundary Pressure
BoundaryConditions
p2
r1
p1
r2
r (r1 )   p1 , r (r2 )   p2
A
r  2  B
r
A
   2  B
r
r12 r22 ( p2  p1 ) 1 r12 p1  r22 p2
r 

r22  r12
r2
r22  r12
r12 r22 ( p2  p1 ) 1 r12 p1  r22 p2
  

r22  r12
r2
r22  r12
Internal Pressure Case
r1/r2 = 0.5

/p
r
/p
r/r2
Dimensionless Distance,
r/r2
Stress Free Hole in an Infinite Medium
Under Uniform Uniaxial Loading at Infinity
  a0  a1 log r  a 2 r 2  a3 r 2 log r
 (a 21 r 2  a22 r 4  a23 r 2  a24 ) cos 2
BoundaryCondtions
 r (a, )   r (a, )  0
y
a
T
T
x
T
 r (, )  (1  cos 2)
2
T
 (, )  (1  cos 2)
2
T
 r (, )   sin 2
2
r 
T  a2
1 
2  r 2
 T  3a 4 4a 2
  1  4  2
r
r
 2
T  a 2  T  3a 4 
   1  2   1  4  cos 2
2 r  2
r 
 r  
T  3a 4 2a 2
1  4  2
2 
r
r

 sin 2

 max    (a, / 2)  3T
90
3
120
60
2
 (a, ) / T
150
30
  (a, ) / T
1
180
0
r 
 ( , ) / T
a 2
210
330
240
300
270

 cos 2

r/a
Stress Concentrations for Other Loading Cases
T
T
T
T
T
Unaxial Loading
Biaxial Loading
K=3
T
K=2
T
T
T
Biaxial Loading
K=4
Stress Concentration Around
Elliptical Hole
 
 max
x  S
y
a
b
x
Stress Concentration Factor
25
b

  x (0,b)  S 1  2 
a

20
15
()max/S
10
5
Circular Case (K=3)
0
0
1
2
3
4
5
6
7
Eccentricity Parameter, b/a
8
9
10
Half-Space Under Concentrated Surface
Force System (Flamant Problem)
Y
  (a12 r log r  a15 r) cos  (b12 r log r  b15 r) sin 
X
x
1
 r  [(a12  2b15 ) cos  (b12  2a15 ) sin ]
r
1
   [a12 cos  b12 sin ]
r
1
 r  [a12 sin   b12 cos]
r

r
C
Bou n daryC on dition s
2

X     r (a, )a cosd   b15   r   r [ X cos   Y sin ]
y
0

Y     r (a, )a sin d  a15
    r  0
0
Normal Loading Case (X=0)
2Y
r  
sin 
r
    r  0
 x   r cos2   
2Yx 2 y
( x 2  y 2 ) 2
 y   r sin 2   
2Yy 3
( x 2  y 2 ) 2
 xy   r sin  cos   
2Yxy 2
( x 2  y 2 ) 2
y=a
xy/(Y/a)
y/(Y/a)
Dimensionless Distance, x/a
Notch-Crack Problems
y
  r  [ A sin   B cos  C sin(  2)  D cos(  2)]
  (r ,0)  0 ,  r (r ,0)  0
r



x
 = 2 - 
lim r  0  singular stresses,finitedisplacments
3

2
3 A

B

r  
cos (3  cos) 
sin (1  3 cos)
2 r
2
2
2 r
3 A

3B

  
cos (1  cos) 
sin (1  cos)
2 r
2
2
2 r
3 A

B

 r 
sin (1  cos) 
cos (1  3 cos)
2 r
2
2
2 r
Contours of Maximum Shear Stress
Two-Dimensional FEA Code
MATLAB PDE Toolbox
- Simple Application Package
For Two-Dimensional Analysis
Initiated by Typing “pdetool”
in Main MATLAB Window
- Includes a Graphical User
Interface (GUI) to:
- Select Problem Type
- Select Material Constants
- Draw Geometry
- Input Boundary Conditions
- Mesh Domain Under Study
- Solve Problem
- Output Selected Results
FEA Notch-Crack Problem
(vonMises Stress Contours)
Curved Beam Problem
P
BoundaryCondit ions
 r (a, )   r (b, )  0
 r (a, )   r (b, )  0

r

b

b
a
a
b
a
b
  (r ,0)dr     (r ,0)rdr  0
a
b
a

(r ,  / 2)dr   P

(r ,  / 2)rdr  P(a  b) / 2
b

a

b
a
B
 Cr  Dr log r ) sin 
r
P
a 2b2 a 2  b2
(r  3 
) sin 
N
r
r
P
a 2b2 a 2  b2
   (3r  3 
) sin 
N
r
r
P
a 2b2 a 2  b2
 r   ( r  3 
) cos 
N
r
r
r 
 r (r ,0)dr  P

  ( Ar 3 
 r (r ,  / 2)dr  0
a/P
 = /2 b/a = 4
Theory of Elasticity
Strength of Materials
Dimensionless Distance, r/a
b
N  a 2  b 2  (a 2  b 2 ) log( )
a
Disk Under Diametrical Compression
P
D
=
P
Flamant Solution (1)
+
+
Flamant Solution (2)
Radial Tension Solution (3)
Disk Under Diametrical Compression
(x1)  
2P
cos1 sin 2 1
r1
 (x2 )  
2P
cos 2 sin 2  2
r2
(y1)  
2P
cos3 1
r1
 (y2 )  
2P
cos3  2
r2
(xy1)  
2P
cos2 1 sin 1
r1
(xy2 )  
2P
cos2  2 sin  2
r2
+
 (x3)
+

( 3)
y
 (xy3)
x  
2P

D
2P

D
0
=
P
r1
x
2
r2
P
r1, 2 
x 2  (R  y)2
2P  ( R  y)3 ( R  y)3 1 
y  

 
  r14
r24
D
2P  ( R  y)2 x ( R  y)2 x 
 xy 


 
r14
r24

y
1
2P  ( R  y) x 2 ( R  y) x 2 1 

 
 
r14
r24
D
Disk Results
Theoretical, Experimental, Numerical
 x (0, y ) 
2P
D
Theoretical Contours of
Maximum Shear Stress
Photoelastic Contours
(Courtesy of Dynamic Photomechanics
Laboratory, University of Rhode Island)
Finite Element Model
(Distributed Loading)