Page 270 - Intro to Tensor Calculus

264

Plane Strain

The plane strain assumption usually is applied in situations where there is a cylindrical shaped body
whose axis is parallel to the z axis and loads are applied along the z−direction. In any x-y plane we assume
that the surface tractions and body forces are independent of z. We set all strains with a subscript z equal
to zero. Further, all solutions for the stresses, strains and displacements are assumed to be only functions
of x and y and independent of z. Note that in plane strain the stress σ zz is different from zero.
In Cartesian coordinates the strain tensor is expressible in terms of its physical components which can
be represented in the matrix form

   
e 11 e 12 e 13 e xx e xy e xz
 e 21 e 22 e 23  =  e yx e yy e yz  .
e 31 e 32 e 33 e zx e zy e zz

If we assume that all strains which contain a subscript z are zero and the remaining strain components are
functions of only x and y, we obtain a state of plane strain. For a state of plane strain, the stress components
are obtained from the constitutive equations. The condition of plane strain reduces the constitutive equations
to the form:
1 E
e xx = [σ xx − ν(σ yy + σ zz )] σ xx = [(1 − ν)e xx + νe yy ]
E (1 + ν)(1 − 2ν)
1
e yy = [σ yy − ν(σ zz + σ xx )] E
E σ yy = (1 + ν)(1 − 2ν) [(1 − ν)e yy + νe xx ]
1
0= [σ zz − ν(σ xx + σ yy )] E
E σ zz = [ν(e yy + e xx )] (2.4.63)
1+ ν (1 + ν)(1 − 2ν)
e xy = e yx = σ xy
E σ xy = E
1+ ν 1+ ν e xy
e zy = e yz = σ yz =0
E σ xz =0
1+ ν
e zx = e xz = σ xz =0 σ yz =0
E
where σ xx , σ yy , σ zz , σ xy , σ xz , σ yz are the physical components of the stress. The above constitutive
equations imply that for a state of plane strain we will have

σ zz = ν(σ xx + σ yy )
1+ ν
e xx = [(1 − ν)σ xx − νσ yy ]
E
1+ ν
e yy = [(1 − ν)σ yy − νσ xx ]
E
1+ ν
e xy = σ xy .
E
Also under these conditions the compatibility equations reduce to

2 2 2
∂ e xx ∂ e yy ∂ e xy
+ =2 .
∂y 2 ∂x 2 ∂x∂y