2001, W. E. Haisler
1
Chapter 10: Analysis of Linear Elastic Solids
Chapter 10 -Analysis of Linear Elastic Solids
For a linear elastic solid under static equilibrium, we can
now write the following equations for any 3-D body:
1. Static Equilibrium Equations (Conservation of Linear
Momentum)
   g  0
 xx  yx  zx


  gx  0
x
y
z
 xy  yy  zy


 gy 0
x
y
z
 xz  yz  zz


  gz  0
x
y
z
 3 equations
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
2
2. Constitutive Equations (Material Response to Stress or Strain)
E
 
[(1  )     ]
xx (1   )(1  2 )
xx
yy
zz
E
 
[  (1  )   ]
yy (1   )(1  2 ) xx
yy
zz
E
 
[    (1  ) ]
zz (1   )(1  2 ) xx
yy
zz
E
 

xy (1   ) xy
E
 

 6 equations
xz (1   ) xz
E
 

yz (1   ) yz
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
Alternately, for the strain-stress equations
1
 xx  [  (   )]
yy
zz
E xx
1
 yy  [  (   )]
xx
zz
E yy
1
 zz  [  (   )]
xx
yy
E zz
1 
 xy  (
)
xy
E
1 
 xz  (
)
xz
E
1 
 yz  (
)
yz
E
3
2001, W. E. Haisler
4
Chapter 10: Analysis of Linear Elastic Solids
3. Geometric Relations (Kinematics) for small strain
 xx  xy  xz 


    yx  yy  yz 
 zx  zy  

zz 
 ux
1  u y  ux

(

)

x
2

x

y

 uy
1  ux  u y


(

)
2  y  x
y
 1  ux  uz 1  u y  uz
 2 (  z   x ) 2 (  z   y )
1  uz  ux 
(

)
2 x
z 
1  uz  u y 
(

)
2 y
z 

 uz

z
 6 equations
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
For the general 3-D linear elasticity problem, we have:
the 15 governing equations
3 equilibrium (linear momentum) equations for []
6 constitutive equations ( [] = [C] {} )
6 kinematic (strain) equations ({} = function of
displacement gradients)
which relate the 15 unknown variables
stresses  xx , yy , zz , xy , xz , yz (6),
strains  xx ,  yy ,  zz ,  xy ,  xz ,  yz (6),
displacements ux , u y , uz (3).
Must specify boundary conditions to solve this boundary
value problem. These must be either displacements or
stresses (tractions) on every point of the boundary.
5
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
6
There are many special cases of the general elasticity
problem that are of practical interest (and can be solved!!!).
These include Plane Stress and Plane Strain problems.
2-D Plane Stress
Plane stress occurs in thin bodies which have predominately
in-plane stress (out-of-plane stresses are zero).
For a geometry which is plane
z y
 xy stress in the x-y plane, we
x
assume  zz   xz   yz  0.
 xx
The stress tensor reduces to
 xx  xy 
 yx
[ ]  

 yy
 xy  yy 
2001, W. E. Haisler
7
Chapter 10: Analysis of Linear Elastic Solids
1. Static Equilibrium becomes (for no body forces)
 xx  yx

0
x
y
 xy  yy

0
x
y
2. Constitutive Equations for a linear elastic Hookean material
become (substitute plane stress requirement of
 zz   xz   yz  0 into general strain equations and solve for
stress components):
 xx

E [   ]
yy
(1 2 ) xx
 yy

E [   ]
yy
(1 2 ) xx
 xy

E
(1   )

xy
1
[Txx   yy ]
E
1
 yy  [Tyy   xx ]
E
1
 xy 
 xy
E
 xx 
or
 zz  

E
( xx   yy )
2001, W. E. Haisler
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Chapter 10: Analysis of Linear Elastic Solids
3. Strain-displacement becomes
 xx
   
 yx
  ux
 xy  
y

 yy   1  u x  u y
2(  y   x )

1  u y  ux 
(

)
2 x y 

 uy
y


The three in-plane strain components must satisfy a relation
which is referred to as compatibility (insures compatibility
of displacements defining the strains):
 2 xy
 2 xx  2 yy

2
 x y
 y2
 x2
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
9
Necessary Equations for a 1-D Stress State
For a 1-D state of stress (function of x only), we have:
1. Static Equilibrium becomes
d xx
x-component:
  gx  0
dx
2. Constitutive Equation for a linear elastic isotropic material
becomes
 xx  E xx
3. Kinematics (Strain-Displacement) equation becomes
 ux
 xx 
x
For the 1-D case, we have 3 unknowns (1 stress, 1 strain and 1
displacement).
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
1-D Special Cases
All of the general 3-D equations relating stress, strain and
deformation may be specialized to some important 1-D cases
which involve long, slender geometries like beams, rods, bars,
tubes, etc.:
1. Bar with axial force only
A1, E1, L1 1
2
P
A2 , E2 , L2
A rigid horizontal bar is
attached to the bottom of each
vertical bar. The horizontal
bar remains horizontal when
pulled downward.
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2001, W. E. Haisler
11
Chapter 10: Analysis of Linear Elastic Solids
2. Bar (or pipe) in torsion
25,000 in-lbs
4”
20”
6
G = 5.5 x 10 psi
2001, W. E. Haisler
Chapter 10: Analysis of Linear Elastic Solids
12
3. Beam of bending (about z-axis)
P
y
y
x
x
a
L
All of the above special cases may be considered separately or
combined. We will look at each case by itself (chapters 11, 12
and 13).