CONTINUUM MECHANICS for ENGINEERS

For the plane strain situation (Figure 6-5b) we assume that u 3 = 0 and that
the remaining two displacement components are functions of only x 1 and x 2,
u i = u i(x 1, x 2) (i = 1,2) (6.5-10)
In this case, the equilibrium equations, the strain-displacement relations, and
the strain compatibility equations all retain the same form as for plane stress,
that is, Eqs 6.5-2, 6.5-3, and 6.5-4, respectively. Here, Hooke’s law (Eq 6.4-3a)
may be written in terms of engineering constants as
along with
σ
E
σ = ( 1−
v ) ε11 +vε
( 1+v) 1−2v ( )
[ ]
11 22
E
σ = ( 1−
v ) ε +vε
( 1+v) 1−2v ( )
[ ]
22 22 11
E E γ 12
σ12 = ε12
= = Gγ
1+v 1+v 2
Eν
( ε + ε )= v σ −σ
1+v 1 2v
=
( ) ( − )
33 11 22 11 22
The first three of these equations may be expressed in matrix form by
⎡σ
11⎤
⎢ ⎥
⎢
σ 22⎥
⎣
⎢σ
12⎦
⎥
(6.5-11a)
(6.5-11b)
(6.5-11c)
(6.5-12)
(6.5-13)
Furthermore, by inverting the same three equations, we may express
Hooke’s law for plane strain by the equations
12
( )
⎡1−v
v 0 ⎤ ⎡ε11⎤
Ev ⎢
⎥ ⎢ ⎥
1−0 ε22
1 ( 1 2 ) ⎢
v v
+v v
⎥ ⎢ ⎥
⎣
⎢ 0 0 1−2v⎦ ⎥
⎣
⎢ε12⎦
⎥
= ( ) −
ε
ε
ε
1+ ν
= ( 1−νσ
) −νσ
E
[ ]
11 11 22
1+ ν
= ( 1−νσ
) −νσ
E
[ ]
22 22 11
1+ ν 2 1+ ν σ12 σ12
= σ = =
E E 2 2G
12 12
( )
(6.5-14a)
(6.5-14b)
(6.5-14c)