EM 424: Equilibrium in Cylindrical Coordinatese_m.424/Equi_Cylind_Coord.pdf · EM 424: Equilibrium...
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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: ∂T e i ( ) ∂ x i i =1 3 ∑ + f = 0 where T e i ( ) = σ ij e j j =1 3 ∑ 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 ∂B i ∂x i i =1 3 ∑ =∇⋅ B =∇⋅ B i e i ( ) i =1 3 ∑ If we let φ = B i and A = e i then we have ∂B i ∂x i = B i ∇⋅ e i ( ) + e i ⋅∇ ( ) B i i =1 3 ∑ i =1 3 ∑ = ∇⋅ e i ( ) + e i ⋅∇ ( ) [ ] B i i =1 3 ∑ which is a coordinate-invariant form of the divergence. Now, consider ∂B i ∂x i i =1 3 ∑ = ∂ ∂x i B ij e j j =1 3 ∑ i−1 3 ∑ = ∂B ij ∂x i e j + B ij ∂e j ∂x i j =1 3 ∑ i =1 3 ∑ = ∂B ij ∂x i e j + B ij e i ⋅∇ ( ) e j j =1 3 ∑ i =1 3 ∑ But from our previous result on the divergence we can write
Transcript of EM 424: Equilibrium in Cylindrical Coordinatese_m.424/Equi_Cylind_Coord.pdf · EM 424: Equilibrium...
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:
∂T e i( )
∂xii=1
3∑ + f = 0
where T e i( ) = σ ij e j
j=1
3∑ 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 ∇⋅ φA( )= φ ∇⋅A +A ⋅∇φ To see this, first consider
∂Bi∂xii=1
3∑ = ∇⋅B = ∇ ⋅ Bi e i( )
i=1
3∑
If we let φ = Bi and A = ei then we have
∂Bi∂xi
= Bi ∇⋅ e i( )+ e i ⋅ ∇( )Bii=1
3∑
i=1
3∑
= ∇⋅ e i( )+ e i ⋅ ∇( )[ ]Bii=1
3∑
which is a coordinate-invariant form of the divergence. Now, consider
∂Bi∂xii=1
3∑ =
∂∂xi
Bije jj=1
3∑
i−1
3∑
=∂Bij∂xi
e j + Bij∂e j∂xi
j=1
3∑
i=1
3∑
=∂Bij∂xie j + Bij e i ⋅ ∇( )e j
j=1
3∑
i=1
3∑
But from our previous result on the divergence we can write
EM 424: Equilibrium in Cylindrical Coordinates
∂Bij∂xii=1
3∑ = ∇⋅ e i( )+ ei ⋅ ∇( )[ ]Bij
i=1
3∑
so that we have
∂Bi∂xii=1
3∑ = Bije j ∇⋅ e i( )+ e j ei ⋅ ∇( )Bij + Bij e i ⋅ ∇( )e j[ ]
j=1
3∑
i=1
3∑
= Bi ∇⋅ e i( )+ ei ⋅ ∇( )Bi[ ]i=1
3∑
= ∇⋅ e i + ei ⋅ ∇( )i=1
3∑ Bi
which is the coordinate-invariant form we wanted. If we let Bi = T e i( ) then the equilibrium equations in coordinate-invariant form are
∇⋅ ei + e i ⋅ ∇( )T e i( )
i=1
3∑ + f = 0
To use this result, let
e1 ≡ er = cosθ ex + sinθ e ye2 ≡ eθ = −sinθ ex + cosθ eye3 ≡ ez
and
T e 1( ) = σrrer +σ rθeθ + σrzezT e 2( ) = σθrer +σθθeθ +σθzez
T e 3( ) = σ zrer + σ zθeθ + σzzezf = frer + fθeθ + frer
Using these results and the fact that in cylindrical coordinates
∇ = er∂∂r
+ eθ1r
∂∂θ
+ ez∂∂z
we find
EM 424: Equilibrium in Cylindrical Coordinates
∇⋅ e1 + e1 ⋅ ∇ =1r
+∂∂r
∇ ⋅ e2 + e2 ⋅∇ =1r
∂∂θ
∇⋅ e3 + e3 ⋅∇ =∂∂z
so that the equilibrium equations are explicitly
∂T e r( )
∂r+T er( )
r+
1r
∂T eθ( )
∂θ+
∂T ez( )
∂z+ f = 0
or, in terms of the stress components
∂σrr
∂rer +
∂σ rθ
∂reθ +
∂σ rz
∂rez
+
σ rr
rer +
σrθ
reθ +
σrz
rez
+1r
∂σθr
∂θer +
∂σθθ
∂θeθ +
∂σθz
∂θez + σθr
∂er∂θ
+ σθθ∂eθ
∂θ
+∂σ zr
∂zer +
∂σ zθ
∂zeθ +
∂σ zz
∂zez
+ frer + fθeθ + fzez( )= 0
which gives
∂σ rr
∂r+ σrr − σθθ
r+ 1r
∂σrθ
∂θ+ ∂σ rz
∂z+ fr = 0
∂σ rθ
∂r+
2σ rθ
r+
1r
∂σθθ
∂θ+
∂σθz
∂z+ fθ = 0
∂σ zr
∂r+
σzr
r+
1r
∂σ zθ
∂θ+
∂σzz
∂z+ fz = 0
and, as with the Cartesian components of stress, we have, from moment equilibrium σrθ = σθ r , σ zθ = σθ z , σz r = σr z
