X - University of Rhode Island
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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
xy
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
Py 3 3(1 ) P
y3
x
(
y
) o y uo
4 Ec3 2 Ec
4 Ec3
2cE
3c 2
3Pxy 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 tan1 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 2n a n 3 r n a n 4 r 2n ) cos n
n 2
(bn1r n bn 2 r 2n bn 3 r n bn 4 r 2n ) 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 cosd 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
cos1 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)