EM 424: Equilibrium in Cylindrical Coordinates
Equilibrium Equations in Cylindrical Coordinates
The equilibrium equations in Cartesian coordinates in terms of the traction vectors acting
on the coordinate planes were found to be:
(e )
∂T i
+ f =0
i =1 ∂xi
3
∑
where
T(
ei )
3
= ∑ σ ij e j
j =1
(i = 1,2,3)
We would like to be able to generalize these equations to other coordinate systems, with
cylindrical coordinates in particular in mind. We can do this if we place these equations
in a coordinate- invariant form. This can be done with the aid of the vector identity
∇ ⋅ (φ A) = φ ∇ ⋅ A + A ⋅∇ φ
To see this, first consider
3
∂Bi
= ∇ ⋅ B = ∇ ⋅ ∑ (Bi e i )
i =1 ∂x i
i =1
3
∑
If we let φ = Bi and A = ei then we have
3
∂Bi
= ∑ Bi (∇ ⋅ e i ) + (e i ⋅ ∇)Bi
i =1
i =1 ∂x i
3
∑
= ∑ [(∇ ⋅ e i ) + (e i ⋅ ∇)]Bi
3
i =1
which is a coordinate-invariant form of the divergence. Now, consider
3 ∂  3
∂Bi

=∑
 ∑ Bij e j 

i =1 ∂x i
i−1 ∂x i  j =1
3
∑
3 3  ∂B
∂e j 
ij
e j + Bij
= ∑ ∑

i =1 j =1 ∂x i
∂xi 
3 3  ∂B

= ∑ ∑  ij e j + Bij (e i ⋅ ∇)e j 
i =1 j =1 ∂x

i
But from our previous result on the divergence we can write
EM 424: Equilibrium in Cylindrical Coordinates
3
∂Bij
= ∑ [(∇ ⋅ e i ) + (ei ⋅ ∇ )]Bij
i =1 ∂x i
i =1
3
∑
so that we have
3 3
∂Bi
= ∑ ∑ Bij e j (∇ ⋅ e i ) + e j (ei ⋅ ∇ )Bij + Bij (e i ⋅ ∇)e j
i =1 ∂x i
i=1 j =1
[
3
∑
]
= ∑ [Bi (∇ ⋅ e i ) + (ei ⋅ ∇ )Bi ]
3
i=1
3
= ∑ (∇ ⋅ e i + ei ⋅ ∇ )Bi
i=1
which is the coordinate-invariant form we wanted. If we let Bi = T(
equilibrium equations in coordinate-invariant form are
ei )
3
∑ (∇ ⋅ ei + e i ⋅ ∇)T(
ei )
i =1
+f = 0
To use this result, let
e1 ≡ e r = cosθ e x + sinθ e y
e 2 ≡ e θ = − sinθ e x + cosθ e y
e3 ≡ ez
and
T(e 1 ) = σ rr e r + σ rθ eθ + σ rz ez
T(
e2 )
= σ θre r + σ θθ eθ + σ θ ze z
T(e 3 ) = σ zr e r + σ zθ eθ + σ zz e z
f = fr er + fθ eθ + fr e r
Using these results and the fact that in cylindrical coordinates
∇ = er
we find
∂
1 ∂
∂
+ eθ
+ ez
∂r
r ∂θ
∂z
then the
EM 424: Equilibrium in Cylindrical Coordinates
1 ∂ 
∇ ⋅ e1 + e1 ⋅ ∇ =  + 
 r ∂r 
1 ∂
r ∂θ
∂
∇ ⋅ e 3 + e 3 ⋅∇ =
∂z
∇ ⋅ e 2 + e 2 ⋅∇ =
so that the equilibrium equations are explicitly
(e )
(e )
(e )
(e )
T r
1 ∂T θ
∂T z
∂T r
+
+
+
+f = 0
∂r
r
r ∂θ
∂z
or, in terms of the stress components
 ∂σ
σ
σ
∂σ rθ
∂σ rz   σ rr

 rr er +
eθ +
ez  + 
e r + rθ eθ + rz e z 

 ∂r

∂r
∂r
r
r
r 
∂e
∂e 
∂σ
∂σ
1  ∂σ
+  θr er + θθ e θ + θz e z + σ θr r + σ θθ θ 
∂θ
∂θ
∂θ
∂θ 
r  ∂θ
 ∂σ
∂σ zθ
∂σ zz 
eθ +
e z  + ( fr er + fθ eθ + fz e z ) = 0
+  zr e r +
 ∂z

∂z
∂z
which gives
∂σ rr σ rr − σ θθ 1 ∂σ rθ ∂σ rz
+
+
+
+ fr = 0
∂r
r
∂z
r ∂θ
∂σ rθ 2σ rθ 1 ∂σ θθ ∂σ θ z
+
+
+
+ fθ = 0
∂r
r
∂z
r ∂θ
∂σ zr σ zr 1 ∂σ zθ ∂σ zz
+
+
+
+ fz = 0
∂r
r
r ∂θ
∂z
and, as with the Cartesian components of stress, we have, from moment equilibrium
σ rθ = σ θ r , σ zθ = σ θ z , σ z r = σ r z