MECH 303 Chapter 2
2.2
Equation of Equilibrium in Plane Problems
0
x
P
xy
 yx
y
A
x 
X
D
x
 xy 
Y
B
 xy
x
 x
dx
x
dx
C
y
 yx 
y 
 y
y
 yx
y
dy
dy
Fig. 2.2.1
(1) Plane Stress

 x  yx
F

0


X 0
 x
x
y


 y  xy

Y  0
 F y  0 

y

x

 M  0 
xy
yx
 D

(2) Plane Strain
2-2
MECH 303 Chapter 2
 yx
  x
 X 0


y

x



 xy
  y
Y  0


x
 y
  z
Z 0 Z 0

z



2-3
MECH 303 Chapter 2
2.3 Stress at a Point ---- Principal Stress
x
0
y
yx
P
A
xy
XN
N
x
S
N
y
B
N
YN
Fig. 2.3.1
F
F
x
y
 0  X N dA   x ldA   yx mdA


 0  YN dA   y mdA   xy ldA 

X N  l x  m yx
 YN  m y  l xy
Normal stress N, shear stress N are:


2
2
 N  lYN  mX N  lm y   x   l  m  xy 
 N  lX N  mYN  l 2 x  m 2 y  2lm xy


i.e., the stress components x, y, xy at point P completely
2-4
MECH 303 Chapter 2
determines the stress states at this point (in any plane AB)
Principle plane (xy=0), Principal stress 
 X N  l ,

 X N  l x  m xy ,
m   x
l   ,
xy
YN  m
YN  m y  l xy
on _ principal _ plane 
on _ any _ plane 
 xy
m

l   y
 (-x) (-y) = xy2
 Two roots of :
1  x   y
 x  y

 
2
2
2

2

   xy2

Let 1 be the angle between 1 and x axis, then
2-5
MECH 303 Chapter 2
tan  1 
sin  1
cos90   1  m1


cos  1
cos  1
l1
 tan  1 
tan  2 
1  x
,
 xy
 2  x
 xy
 tan  1  tan  2  1   1  2
Maximum and minimum stress at a point:
 Prove that maximum & minimum stress are 1 & 2
respectively,
 maximum & minimum shear stress are

1   2
2
respectively
 and acting on planes inclined at 45o with principal
plane.
2-6
MECH 303 Chapter 2
2.4 Stress-strain relations (Physical equations,
Constitutive equations, Hook’s law)
Hook’s
law
in
3-dimensional
form:
(material
property)

,
 xy 
1
 xy
G

,
 yz 
1
 yz
G

,
 zx 
1
 zx
G
x 
1
 x    y   z 
E
y 
1
 y    x   z 
E
z 
1
 z    x   y 
E

G
E
21   
,
xy = xy
(1) Plane Stress ( z=0, yz=xz=0 )
x 
1
 x   y  ,
E
z  

E

x
  y ,
y 
1
 y   x ,
E
 xy 
1
 xy
G
yz = xz = 0.
2-7
MECH 303 Chapter 2
(x, y, xy) <==> (x, y, xy)
(2) Plane Strain ( z=0, yz = xz=0 )
1  2
z = (x+y),  x  E
1  2
y 
E



  x 
 y  ,
1 





  y 
 x  ,
1  

 xy 
21   
 xy ,
E
xz = yz = 0
(x, y, xy,) <==> (x, y, xy)
(3) Plane stress equation  Plane strain equation
E
If replace E by 1   2

 by 1  
~
 ~
 If replace E by
E 1  2 
1   2

 by 1  
These two simple rules enable us to avoid the
mathematical derivation of some equations required for
2-8
MECH 303 Chapter 2
the solution of a plane problem: by replacement of E
and  as stated above.
2.5 Geometric equations (Strain-displacement
relation.) Rigid-body displacements
2-9
MECH 303 Chapter 2
Geometrical aspect of plane problem (suitable to any
kinds of materials as far as they are continuous)
Displacement components at any point (u=u(x,y); v=v(x,y))
(1) Relations between strains and displacements
the normal strain of line PA, by definition, is
x 
P' A'dx
dx
P' A'
u
v

 

  dx  u 
dx  u    v  dx  v 
x
x

 

2
2
u   v 

  dx 
dx    dx 
x   x 

2
2
2
u v
According to the basic assumption: x, ,
are small
x x
u
v
quantities, i.e.x<<1, 
<<1,  x <<1, we have
x
2-10
MECH 303 Chapter 2
x 
u
x ,
similarly,  y 
v
u v
 xy 

,
y
y x
(2) Rigid body displacement
if u (x, y) and v (x, y) are given, then x, y ,x y are
completely determined. If x, y, x
y
are given, the
displacement u, v are not wholly determinate :  rigid
body displacement  the non-zero displacements
corresponding to zero strains:
x 
u
,
x
y 
v
,
y
 xy 
u v

y x
Let  x   y   xy  0, then
u
 0,
x

v
 0,
y
v u

0
x y

2-11
MECH 303 Chapter 2
u = f1(y) , v=f2(x) ,

df 1  y 
 
dy

df 1  y  df 2  x 

dy
dx
df 2  x 
, dx  
constan
t
constant
’s
thus , we have : f1  y   y  u 0 , f 2  x   x  v0
u  u 0  y ,
displacement
v  v0  x
These
components
are
the
corresponding
to
y
y
x=
x=
x
x
2-12
MECH 303 Chapter 2
 x   y   xy  0 , and we can prove that u0 and v0 are the
rigid body translation and  the rigid body rotation.
2-13