MECH 303 Chapter 2
2.6 Strain at a point
The strain components x, y ,
xy
at a point completely determine the
strain state ( deformation state ) of the material at this point.
The normal strain N of any line element PN
N l 2 x m 2 y lm xy
The change of angle between PN and PN '
cos ll ' mm '
cos 1 cos 1 N N ' 2 ll ' x mm ' y lm ' l ' m xy
principal strain, principal plane.
xy 0
y
n
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MECH 303 Chapter 2
2.7 Boundary conditions
(1) 8 basic equations for plane problems:
x xy
X 0
x y
equilibrium
y yx Y 0
y x
u
v
x x y y ,
geometrical
xy u v ,
y x
1
x y
physical y y x ,
E
xy G xy
(plane stress)
Solve the 8 unknown functions under proper boundary conditions.
(2)
stress
3 kinds of boundary conditions : displacement
mixed
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MECH 303 Chapter 2
displacement boundary problem (displacement boundary
condition)
u s= u
, s= v ,
and
are the prescribed functions of
coordinates on the surface.
Stress boundary problem ( stress boundary condition)
Surface forces acting on the boundary of a body are prescribed,
which can be expressed as a condition about the stress
components at the boundary:
l x
m yx
m y
l xy
YN
(relations between boundary stress components and the surface
force) when the boundary is normal to a coordinate axis, the
above boundary conditions are simplified
Mixed boundary conditions
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MECH 303 Chapter 2
2.8 Saint-Venants principal
If a system of forces acting on a small portion of the surface of an elastic
body is replaced by another statically equivalent system of forces acting
on the same portion of the surface, the redistribution of loading produces
substantial changes in the stresses only in the immediate neighborhood of
the loading, and the stresses are essentially the same in the parts of the
body which are at large distances in comparison with the linear
dimension of the surface on which the forces are changed. By statically
equivalent systems we mean that the two systems have the same
resultant force and the same resultant moment.
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MECH 303 Chapter 2
2.9 Solution of plane problem in terms of displacements
Equilibrium equation in terms of stresses:
x xy
X 0
x y
express stress by strain via (physical equation)
y yx
Y 0
y
x
Equilibrium equation in terms of strain :
express strain by displacement
components (geometrical equations) :
Equilibrium equation in terms of displacements: (plane stress)
E 2u 1 2u 1 u 2 v
2 y 2 2 xy
1 2 x 2
E 2v 1 2v 1 2
Y 0
2
2
2
2 x
2 xy
1 y
Solve the above equations of u(x,y) and v(x,y) with the following
boundary conditions
(1) displacement boundary condition :
(2)
stress boundary condition ( in term of displacements) :
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MECH 303 Chapter 2
E u
v
y
1 2 x
E
1 2
v
u
m
x
y
s m
1 u v
s X
2 y x
1 v u
s Y
s l
2 x y
Thus, to solve a plane stress problem in terms of displacements, we
have to solve two differential equations simultaneously and the
obtained u (x, y) and v ( x, y) must satisfy the displacement boundary
condition or stress boundary condition or mixed boundary condition
at the boundary.
Once we have u (x, y) and v (x, y), then
x (x, y) , y (x, y),
xy (x, y)
x (x, y), y (x, y),
xy (x, y) are completely uniquely determined.
For plane strain problem, the formulation is the same as before, only
replace E by
E
1 2
and by
The above method is suitable to solve any plane problems with any
boundary conditions.
y
Exercise 1
Exercise 2
(fixed boundary)
(plane stress)
x
L
u0
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MECH 303 Chapter 2
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MECH 303 Chapter 2
2.10 Solution of plane problem in terms of stress
Take the 3 stress components x , y , xy (xy ) as the basic unknown
functions.
(1) 2 equilibrium equations contain 3 functions
(2) The 3rd equation is the compatibility equation in terms of strain
(or stress):
The necessary and sufficient condition for the existence of
single-valued continuous functions u ( x, y) and v (x, y) is simple
x
u
x
v
y
xy
u v
y x
2
2 xy
2 x y
xy
y 2
x 2
(by using physical equations x x ,...)
X Y
y
x
2 x y 1
2 x y
1 X Y
1 x
y
(plane stress)
(plane strain)
(3) The obtained stress solution should satisfy stress boundary
condition (B.C.)
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MECH 303 Chapter 2
2.11 Case of constant body forces
In the case of constant body forces:
X
0
x
Y
0
y
, then the 3
stress components x , y , xy are determined by equations
x xy
X 0
y
x
y yx
Y 0
and B. C. on S
2 0
x
y
x xy
X 0
x y
The general solution of
y yx
Y 0
y x
(1)
l x s m xy s X
m y s l xy s Y
consists of:
x xy
0
x y
The general solution of
y yx
0
y x
(homogeneous system)
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MECH 303 Chapter 2
(2)
x xy
X 0
x y
A particular solution of
y yx
Y 0
y x
(non-homogeneous system)
When body forces are constant , the particular solution may be taken
as
x X x ,
or:
x 0
y Y y ,
xy yx 0
y=0 ,
xy X y Y x
The general solution of
x xy
0
x y
y yx
0
y x
is
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MECH 303 Chapter 2
2
x 2
y
2
y 2
x
xy yx
always satisfy the homogeneous equations
xy
Airys stress function for plane problems.
The complete solution of the equilibrium equations are
2
x 2 Xx
y
2
y 2 Yy
x
2
xy yx
xy
depends on
To determine , substituting x, y, xy, into compatibility condition,
we have
4 0
2
2
4
2
y 2
x
2
2
2
y 2
x
Thus, in the solution of plane problems in terms of stress when body
forces are constant, it is necessary only to solve for stress function
from the single differential equation
=0, and x, y, xy
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MECH 303 Chapter 2
satisfy stress boundary conditions.
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MECH 303 Chapter 2
2.12 Airys stress function. Inverse method and semi-inverse
method
Direct method
Inverse method
usually impossible
take some function satisfying the compatibility
equation, then obtain stress components and find the surface force
components, thus we know what problem they can solve.
Semi-inverse method
we assume the solution for stress or
displacements in a given problem, then proceed to show that all the
differential equations and boundary conditions are satisfied. If some
of the boundary conditions are not satisfied, then we have to modify
the assumptions made.
(Numerical method ---- Finite element method, directly solve the
equilibrium equation in terms of displacements.)
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