Transcript MCE 571 Theory of Elasticity

Prismatic Bar Subjected to End Loadings
y
Semi-Inverse Method
Assume : x   y  xy  0
S
x
ℓ
R
P
z
M
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Equilibriu m Equations 
 xz  yz

0
z
z
Compatibil ity Equations 
 2 z  2 z  2 z  2 z



0
x 2
y 2
z 2
xy
Integrating 
 z  C1 x  C2 y  C3 z  C4 xz  C5 yz  C6
Extension of Cylinders
y
S
x
ℓ
R
z
Pz
Assumptions
- Load Pz is applied at centroid of crosssection so no bending effects
- Using Saint-Venant Principle, exact end
tractions are replaced by statically
equivalent uniform loading
- Thus assume stress z is uniform over any
cross-section throughout the solid
P
 z  z ,  xz   yz  0
A
and  x   y   xy  0
Using stress results into Hooke’s law and combining with the straindisplacement relations gives
P
u z x
u
P v
P w Pz
AE
Integrating and dropping
 z ,
 z ,

x
AE y
AE z AE
P
rigid-body motion terms
v z y
such that displacements
AE
u v
v w
w u
  0,

 0,

 0 vanish at origin
P
y x
z y
x z
w z z
AE
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Torsion of Cylinders
y
S
x
ℓ
R
z
T
Guided by Observations from Mechanics of Materials
• projection of each section on x,y-plane rotates as
rigid-body about central axis
• amount of projected section rotation is linear
function of axial coordinate
• plane cross-sections will not remain plane after
deformation thus leading to a warping displacement
Elasticity
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M.H. Sadd , University of Rhode Island
Torsional Deformations
y
u  r sin   y
v  r cos  x
P'
P
S
r

  z

O
x
R
 = angle of twist per unit length
u  yz
v  xz
w  w( x, y )
w = warping displacement
Now must show assumed displacement form
will satisfy all elasticity field equations
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Stress Function Formulation
e x  e y  e z  e xy  0
u  yz
v  xz
w  w( x, y )
 x   y   z   xy  0
e xz 
1  w

 y 

2  x

 w

 xz  
 y 
 x

e yz 

1  w

 x 
2  y

 w

 yz  
 x 
 y

Equilibrium Equations
 xz  yz

0
x
y
Compatibility Relation
 xz  yz

 2
y
x
Introduce Prandtl Stress Function  = (x,y) :
 xz 


,  yz  
y
x
Equilibrium will be identically satisfied and compatibility relation gives
 2  2
   2  2  2
x
y
2
a Poisson equation that is amenable to several analytical solution techniques
Elasticity
Theory, Applications and Numerics
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Boundary Conditions
Stress Function Formulation
On Lateral Side: S
  
  n  n  n 0  00
  
  n   n  
n 0
Txn   x nx   yx n y   zx nz  0  0  0
y
T
S
x
n
y
Tzn
xy x
y
y
zy
z
xz x
yz
y
z
z
 dx  dy
d

0 
 0    constant  0
x ds y ds
ds
n
ℓ
R
z
T
Unit Normal
dy dx
nx 

ds dn
dx dy
ny  

ds dn
On End: R (z = constant)
Px   Txn dxdy  0
R
Py   Tyn dxdy  0
R
Pz   Tzn dxdy  0
R
M x   yTzn dxdy  0
R
M y   xTzn dxdy  0
R
M z   ( xTyn  yTxn )dxdy  T  T  2  dxdy
R
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
R
Displacement Formulation
 xz  yz

0
x
y
2w 2w
 2 0
2
x
y
Displacement component satisfies Laplace’s equation
On Lateral Side: S
Tzn   xz nx   yz n y   z nz  0 
 w

dw
 w

 y nx  
 x n y  0 or
 ( yn x  xn y )

dn
 x

 y

On End: R
M z   ( xTyn  yTxn )dxdy  T 
R

w
w 
dxdy
T     ( x 2  y 2 )  x
y
R
y
x 

T  J
Elasticity

x w y w 
dxdy. . . TorsionalRigidity
J    x 2  y 2 

R


y


x


Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Formulation Comparison
y
S
x
O
R
Stress Function Formulation
 2  2
   2  2  2  R
x y
0 S
2
Relatively Simple Governing Equation
Very Simple Boundary Condition
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Displacement Formulation
2w 2w

0 R
x 2 y 2
 w

 w


n y  0  S

y

n


x


 x 

x

y




Very Simple Governing Equation
Complicated Boundary Condition
Multiply Connected Cross-Sections
y
So
C
S1
x
R
Boundary conditions of zero tractions on all lateral surfaces
apply to external boundary So and all internal boundaries
S1, . . . Stress function will be a constant and displacement
be specified as per (9.3.20) or (9.3.21) on each boundary Si,
i = 0, 1, . . .
 w

 w

  i  Si or 
 y nx  
 x n y  0  Si
 x

 y

where i are constants. Value of i may be arbitrarily chosen
only on one boundary, commonly taken as zero on So .
Constant stress function values on each interior boundary are found by
requiring displacements w to be single-valued, expressed by

S1
 ds  2A
dw( x, y)  0
S1
1
where A1 is area enclosed by S1
Value of 1 on inner boundary S1 must therefore be chosen so that relation is satisfied. If crosssection has more than one hole, relation must be satisfied for each hole.
Boundary conditions on cylinder ends will be satisfied, and resultant torque condition will give
T  2 dxdy  21 A1
R
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Membrane Analogy
Stress function equations are identical to those governing static deflection of an elastic membrane
under uniform pressure. This creates an analogy between the two problems, and enables particular
features from membrane problem to be used to aid solution of torsion problem. Generally used to
providing insight into qualitative features and to aid in developing approximate solutions.
z
Deflected Membrane
p
z
Ndy
Ndy
Ndx
dy
R
y
dx
Ndy
Ndx
z
x
pdxdy
Ndy
Membrane Element
S
z  2 z

dx
x x 2
x
x
Static Deflection of a Stretched Membrane
Membrane Equations
F
z
0
 z  z
p



N
x 2 y 2
2
2
z  0 on S
V   zdxdy
R
Equilibrium of Membrane Element
Torsion Equations
 2  2

 2
x 2 y 2
  0 on S
T 2

R
dxdy
Equations are same with:  = z , p/N = 2 , T = 2V
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Contour Line : z  constant
t
n
z
 0   zn  0
s
dz
    zt  
dn

Torsion Solutions Derived from
Boundary Equation
y
f ( x, y )  0
S
x
R
If boundary is expressed by relation f(x,y) = 0, this
suggests possible simple solution scheme of
expressing stress function as  = K f(x,y) where K is
arbitrary constant. Form satisfies boundary
condition on S, and for some simple geometric
shapes it will also satisfy the governing equation
with appropriate choice of K. Unfortunately this is
not a general solution method and works only for
special cross-sections of simple geometry.
Boundary - Value Problem
 2  2

 2  R
x 2 y 2
0S
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 9.1 Elliptical Section
y
x2 y2

1
a2 b2
b
a
x
 x2 y2

Look for Stress Function Solution   K  2  2  1
b
a

a 2 b 2 
 satisfies boundary condition and will satisfy governing governing if K   2
a  b2
Since governing equation and boundary condition are satisfied, we have found solution
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Elliptical Section Results
(Displacement Contours)
(Stress Function Contours)
Stress Field
Displacement Field
2a 2 
2Ty
 xz   2
y


a  b2
ab3
2b 2 
2Tx
 yz  2
x
2
a b
ba3
   2xz   2yz 
 max
2
T (b 2  a 2 )
w
xy
a 3b 3
Loading Carrying Capacity
Angle of Twist
2
2T x
y
 4
4
ab a
b
2T
 (0,b) 
ab 2
2a 2b 2  1
T  2

a  b2  a2

R
x 2 dxdy 
1
b2

R
y 2 dxdy 
a 3b 3
T (a 2  b 2 )
T 2
or  
a  b2
a 3b 3
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island

dxdy 
R


Elliptical Section Results
3-D Warping Displacement Contours
T
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 9.2 Equilateral Triangular Section
y
2a
a
x
For stress function try product form of each boundary line equation
  K ( x  3 y  2a)(x  3 y  2a)(x  a)
 satisfies boundary condition and will satisfy governing governing if K  

6a
Since governing equation and boundary condition are satisfied, we have found solution
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Equilateral Triangular Section Results
(Stress Function Contours)
Stress Field

( x  a) y
a
 2

( x  2ax  y 2 )
2a
 xz 
 yz
 max   yz (a,0) 
Elasticity
3
5 3T
a 
2
18a 3
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
(Displacement Contours)
Displacement Field
w

y (3x 2  y 2 )
6a
Loading Carrying Capacity
Angle of Twist
T
27
3
a 4  I p
5
5 3
Additional Examples That Allow Simple
Solution Using Boundary Equation Scheme
y  a 2  cx 2
y
y
r = 2acos
x  a 2  cy 2
a
r
r=b
a
.

x
x
x   a 2  cy 2
y   a 2  cx 2
Section with Higher Order
Polynomial Boundary (Example 9-3)
  K (a 2  x 2  cy 2 )(a 2  cx2  y 2 )

K  2
, c  3 8
4a (1  2 )
max  (a,0)  (0,a)  2a
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Circular Shaft with Circular
Keyway (Exercise 9-22/23)
 2 2
2a cos 
(b  r )(1 
)
2
r
( max ) keyway
2a
As b / a  0

2
( max ) solid shaft
a

 St ress Concent ration of 2
Examples That Do Not Allow Simple
Solution Using Boundary Equation Scheme
y
y
y = m1x
x=a
x
b
a
x
y = -m2x
General Triangular Section
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Rectangular Section
Example 9.4 Rectangular Section
Fourier Method Solution
y
Previous boundary equation scheme will not create a
stress function that satisfies the governing equation.
Thus we must use a more fundamental solution
technique - Fourier method. Thus look for stress
function solution of the standard form
b
a
x
  h   p
with  p ( x, y)  (a 2  x 2 )
homogeneous solution must then satisfy
2h  0 , h (a, y)  0 , h ( x,b)  (a 2  x 2 )
Separation of Variables Method

 h ( x, y)   Bn cos
n 1
nx
ny
cosh
2a
2a
32a 2
  (a  x ) 
3
2
Elasticity
 h ( x, y)  X ( x)Y ( y)
2
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
 2  2

 2
x 2 y 2
nb 

Bn  32a 2 (1) ( n 1) / 2 /  n 3 3 cosh

2a 

(1) ( n 1) / 2
nx
ny
cos
cosh

nb
2a
2a
n 1, 3, 5 3
n cosh
2a

Rectangular Section Results
Stress Field

16a 
(1) ( n 1) / 2
nx
ny
 xz 

cos
sinh

nb
y
2a
2a
 2 n 1,3,5 2
n cosh
2a


16a
(1) ( n 1) / 2
nx
ny
 yz  
 2x 
sin
cosh

nb
x
2a
2a
 2 n 1,3,5 2
n cosh
2a

16a
1
 max   yz (a,0)  2a 

nb
 2 n 1,3,5 2
n cosh
2a
Loading Carrying Capacity/Angle of Twist
16a 3b 1024a 4
T

3
5
1
nb
tanh
 5
2a
n 1,3,5 n

Displacement Field
32a 2
w  xy 
3
Elasticity
(1) ( n 1) / 2
nx
ny
sin
sinh

nb
2a
2a
n 1, 3, 5 3
n cosh
2a
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island

Rectangular Section Results
(Stress Function Contours)
(Displacement Contours, a/b = 1.0)
(Displacement Contours, a/b = 0.9)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
(Displacement Contours, a/b = 0.5)
Torsion of Thin Rectangular Sections (a<<b)
y
Investigate results for special case of a very thin
rectangle with a << b. Under conditions of b/a >> 1
b
cosh
a
nb
nb
  and tanh
1
2a
2a
  (a 2  x 2 )
 max  2a
x
T
Composite Sections
y
Torsion of sections composed of thin
rectangles. Neglecting local regions where
rectangles are joined, we can use thin
rectangular solution over each section.
Stress function contours shown justify these
assumptions. Thus load carrying torque for
such composite section will be given by
3
1
x
2
(Composite Section)
Elasticity
16
a 3b
3
(Stress Function Contours)
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
N
16
T   ai3bi
3
i 1
Example 9.5 Hollow Elliptical Section
x2
y2

1
(ka) 2 (kb) 2
y
x2 y 2

1
a 2 b2
x
For this case lines of constant shear stress
coincide with both inner and outer boundaries,
and so no stress will act on these lateral
surfaces. Therefore, hollow section solution is
found by simply removing inner core from solid
solution. This gives same stress function and
stress distribution in remaining material.

a 2b 2  x 2 y 2


 2


1
2  2
2

a b a
b



a 2b 2 2
Constant value of stress function on inner boundary is i   2 2 k  1
a b
Load carrying capacity is determined by subtracting load carried by the
removed inner cylinder from the torque relation for solid section
a 3b3 (ka )3 (kb)3 
 3 3
T 2


a b (1  k 4 )
2
2
2
2
2
a b
(ka )  (kb)
a b
2T
1
Maximum stress still occurs at x = 0 and y = b  max 
ab 2 1  k 4
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Hollow Thin-Walled Tube Sections
C Membrane
Tube Centerline
a
a
A
o
B
A
B
(Section aa)
t
With t<<1 implies little variation in membrane slope, and BC can be approximated
by a straight line. Since membrane slope equals resultant shear stress

o
t
  
dxdy  2o Ai  2 A o   2o Ai  2o Ac
R
 2 
where A = section area, Ai = area enclosed by inner boundary, Ac = area enclosed by centerline
T

Combining relations
2 Ac t
TSc
Angle of twist: ds  2Ac   
where Sc = length of tube centerline
Sc
4 Ac2t
Load carrying relation: T  2


Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Cut Thin-Walled Tube Sections
Cut
Cut creates an open tube and produces significant changes to stress function,
stress field and load carrying capacity. Open tube solution can be
approximately determined using results from thin rectangular solution.
Stresses for open and closed tubes can be compared and for identical applied
torques, the following relation can be established (see Exercise 9-24)
3 T
τ OpenTube
τ OpenTube
2 aAs
Ac

6
, but since Ac  As 
 1  τ OpenTube  τ ClosedTube
T
τ ClosedTube
As
τ ClosedTube
2 Ac t
 Stresses are higher in open tubeand thusclosed tube is stronger
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Torsion of Circular Shafts of Variable Diameter
x
.

r
y
z
Displacement Assumption
ur = uz = 0
u = u (r,z)
er  e  e z  erz  0
 r      z   rz  0
u 
1  u
1 u 
er       , ez 
2  r
r 
2 z
u 
u
 u
 r       ,  z   
r 
z
 r
  3   u     3   u  
r
r
  
  0
r  r  r  z  z  r 

r2
3   u 
 r
     r

z

r

 r 
 2  3   2 
Stress Function Approach

 2 0
2
2
u
r

r


r

r
z


 r 3      z
r
z  r  
Boundary Condition
Load Carrying Torque
   dr  dz 
d
T  2[( R( z ), z )  (0, z )]


0

 0    constant


2
ds
r  r ds z ds 
Equilibrium Equations
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Conical Shaft Example 9-7
z
r2  z2
 cos  constanton boundary
r
z
2
Stress Function Solution


z
1
z3

  C 

2
2 3/ 2 
2
2
3
(
r

z
)
 r z

Stresses
Cr 2
 r   2
(r  z 2 ) 5 / 2
Crz
 z   2
(r  z 2 ) 5 / 2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
C
T
2
1
2(  cos   cos 3 )
3
3
Displacement
u  
Cr
 r
3(r 2  z 2 ) 3 / 2
r is rigid-body rotation about z-axis and
 can be determined by specifying shaft
rotation at specific z-location
Conical Shaft Example 9-7  = 30o
Comparison with Mechanics of Materials
Max Shear Stress Comparison
0.06
Mechanics of Materials
Elasticity Theory
0.05
( )
z max
/T
0.04
0.03
0.02
0.01
0
Elasticity
4
5
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
6
7
z
8
9
10
Numerical FEA Torsion Solutions
A
(4224 Elements, 2193 Nodes)
(4928 Elements, 2561 Nodes)
(4624 Elements, 2430 Nodes)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island

(Stress Function Contours)
(Stress Function Contours)
(Stress Function Contours)
Flexure of Cylinders
y
z
Consider flexure of cantilever beam of
arbitrary section with fixed end at z = 0
S
and transverse end loadings Px and Py at
x
z = ℓ. Problem is solved in Saint-Venant
sense, so only resultant end loadings Px
Py
and Py will be used to formulate boundary
ℓ
conditions at z = ℓ.
(xo,yo) .
R
Px
From general formulation  x   y   xy  0 , and motivated
from strength of materials choose  z  ( Bx  Cy )(l  z),
where B and C are constants. Stresses xz and yz will be
determined to satisfy equilibrium and compatibility
relations and all boundary conditions.


Remaining equilibrium equation xz  yz  ( Bx  Cy )  0 will be identically
x
y
F 1 2


satisfied if we introduce stress function F(x,y) such that xz y  2 Bx
F 1 2
 yz  
 Cy
x 2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Flexure Formulation
Remaining Beltrami-Michell Compatibility Relations

B
( 2 F ) 
0

y
1 
2F 
(Cx  By )  2
1 

C
 ( 2 F ) 
0
x
1 
Zero Loading Boundary Condition on Lateral Surface S
 xz nx   yz n y  0
dF
1
dy
dx
  ( Bx 2
 Cy 2 )
ds
2
ds
ds
Separate Stress Function F into Torsional Part  and Flexural Part 
F ( x, y)  ( x, y)  ( x, y)
   2 in R
d
 0 on S
ds
2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island

(Cx  By) in R
1 
d
1
dy
dx
  ( Bx 2
 Cy 2 ) on S
ds
2
ds
ds
 2 
Flexure Formulation
General solution to  2  
 ( x, y )  f ( x, y ) 

(Cx  By )
1 
1 
(Cx 3  By 3 ) where 2f = 0
6 1 
Boundary Conditions on end z = ℓ


R
R
 xz dxdy  Px
BI y  CI xy  Px
 yz dxdy  Py
BI xy  CI x  Py
B
C
Px I x  Py I xy
I x I y  I xy2
Py I y  Px I xy
I x I y  I xy2
where x, y and  xy are the area moments of inertia of section R
 [ x
R
yz
 y xz ]dxdy  xo Py  yo Px
1

 
J    (Cxy 2  Bx2 y)  ( x  y ) dxdy  xo Py  yo Px
R 2
x
y 

where J is the torsional rigidity – final relation determines angle of twist 
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Flexure Example - Circular Section with No Twist
a
z
P
x
ℓ
y
2  
Polar Coordinate Formulation
 P
r cos
1  Ix
1  1 P 2

a sin 3  on r  a
a  2 I x
P  3  2 2
1  2
1  2 3 
2

a
x

xy

x
I x  8(1  )
8(1  )
24(1  ) 
P 1  2
 xz  
xy
4I x 1  
Stress Solution:
P 3  2
P 3  2 2
1  2 2
max   yz (0,0)  2
 yz 
[a  y 2 
x ]
I x 8(1  )
3  2
a 2(1  )
P
4 P
 z   y (l  z )
 max 
Strength
of
Materials:
Ix
3 a 2
Solution:  
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island