EM 424: Equilibrium Equations
21
Equations of Equilibrium
Consider the small element of a body whose sides are cut out parallel to the x, y, z axes.
Each of these sides has a traction vector acting on it, and since these tractions vary with
position in a body, we must take into account those variations when we consider the
tractions on each face. We also assume there is a body force f (force/unit volume) acting
on the element.
(e 2)
(e 2)
T + dT /dx 2
x2 , y
-T
(e 3)
(e )
-T
(e )
dx 2
f
(e3 )
T 3 + dT /dx3
x3 , z
(e1 )
T 1 + dT /dx1
(e1 )
dx1
dx 3
-T
(e 2)
x1 , x
EM 424: Equilibrium Equations
22
Summing forces on the element we find
 (e 1 ) ∂T(e 1 )

dx1  dx2 dx 3 − T(e1 )dx2 dx 3
T +
∂x1


e


∂T( 2 )
+ T (e2 ) +
dx 2  dx1dx 3 − T(e2 )dx1 dx3
∂x 2




∂T(e 3 )
+ T (e3 ) +
dx 3  dx1 dx 2 − T(e 3 )dx1 dx2
∂x3


+ f dx1 dx2 dx 3 = 0
which gives, when canceling terms and dividing by the volume dx1 dx2 dx3 of the element
(e )
(e )
(e )
∂T 3
∂T 1 ∂T 2
+
+
+ f =0
∂x1
∂x 2
∂x3
and which can be compactly written as
(e )
∂T i
∑
+ f =0
i =1 ∂x i
3
If we express these traction vectors in terms of their component stresses with respect to
the x, y, z axes and likewise express the body force in terms of its components, i.e.
3
T(e i ) = ∑ σ ij e j
j =1
3
f = ∑ fj e j
j =1
we find

3  3 ∂σ
∑  ∑ ij + f j  e j = 0
j =1 i =1 ∂x i

and each component of this equation must vanish, leading to the three equations of local
equilibrium for the stresses given by
3
∂σ ij
i =1
∂x j
∑
which, when expanded out, become
+ fj = 0
( j = 1,2,3)
EM 424: Equilibrium Equations
23
∂σ xx ∂σ yx ∂σ zx
+
+
+ fx = 0
∂x
∂y
∂z
∂σ xy ∂σ yy ∂σ zy
+
+
+ fy = 0
∂x
∂y
∂z
∂σ xz ∂σ yz ∂σ zz
+
+
+ fz = 0
∂x
∂y
∂z
(1)
Equilibrium equations for plane stress and plane strain
For the special case of plane stress where σ zz = σ xz = σ yz = 0 and f z = 0 these equations
reduce to
∂σ xx ∂σ yx
+
+ fx = 0
∂x
∂y
∂σ xy ∂σ yy
+
+ fy = 0
∂x
∂y
(2)
Similarly, for plane strain problems where ezz = exz = eyz = 0 from the stress strain
relations it follows that for an isotropic elastic solid the shear stresses vanish. Since in
plane strain problems we also assume e xx = exx (x, y), e xy = exy (x, y), e yy = eyy (x, y) , it
follows also from the stress-strain equations that σ zz = σ zz (x, y) so that with f z = 0 the
third equilibrium equation in Eq.(1) is again automatically satisfied and the other two
equations just reduce to that of Eq. (2) for the plane stress case.