![694_Curvilinear coordinates.png](https://secure.tutorsglobe.com/CMSImages/694_Curvilinear%20coordinates.png)
Figure 1. (a) Curvilinear coordinates at the mid-plane showing unit vectors. (b) Geometry of change in unit normal vector.
The reference orthogonal curvilinear coordinate system with corresponding unit vectors is shown in Figure 1. It is clear that the unit vectors are functions of the in-plane coordinates, α1, and α2. In order to develop a metric for distance, and subsequently displacement and strain, in terms of the curvilinear coordinates, it is necessary to determine the functional relationships between the unit vectors and their derivatives, and the curvilinear coordinates, α1, and α2. Towards that end, we note the following:
dr = ∂r/∂α1dα1 + ∂r/∂α2dα2 (1)
Letting,
rα1 = ∂r/∂α1 (2)
and
rα2 = ∂r/∂α2 (3)
we have
dr = rα1.dα1 + rα1.dα2 (4)
From Figure 1 b, where to first-order approximation the arclength defined by Aidαi may be approximated by its chord, and where the change in the unit normal in the ith-direction is given by
(Δe3)i = (∂e3/∂αi)dαi (5)
Applying a similar-triangle argument, we can determine the unit normal derivative relations
∂e3/∂α1 = (A1/R1)e1 (6)
and similarly,
∂e3/∂α2 = (A2/R2)e2 (7)
where A1 = |rα1| (8)
and A2 = |rα2| (9)
are the Lame parameters. Given the above, we can now define the curvilinear unit vectors in terms of known quantities as follows:
e1 = rα1/A1 (10)
e2 = rα2/A2 (11)
and
e3 = e1 x e2 (12)
Noting that we may write
rα1 = A1e1 (13)
and
rα2 = A2e2 (14)
we define the following very useful identity
∂rα1/∂rα2 = ∂rα2/∂rα1 (15)
Which, of course, implies
∂(A1e1)/∂α2 = ∂(A2e2)/∂α1 (16)
Equations (15), and (16) are valid since for analytic functions, the order of differentiation does not matter. Likewise, we have
∂2(e3)/∂α2∂α1 = ∂2(e3)/∂α1∂α2 (17)
Given the above development,
Question: 1. Derive expressions for the derivatives of the unit vectors, e1 and e2 with respect to α1, and α2 in terms of A1, A2, R1, and R2 to add to those already developed for e3. You make use of the orthogonality of the unit vectors, and identities like the following
e2.∂e1/∂α1 = ∂/∂α1(e1.e2) - ∂e2/∂α1.e1 = -∂e2/∂α1.e1 (18)
in this derivation. Note that by orthogonality, the first term on the rhs of (18) vanishes. Equation (18) yields the projection of the given derivative in the e2 direction - i.e. the magnitude of its e2 component.
Question: 2. Derive Codazzi's conditions given by:
∂/∂α1(A2/R2) = 1/R1.∂A2/∂α1 (19)
and
∂/∂α2(A1/R1) = 1/R2.∂A1/∂α1 (20)
You will find the derivative relations that you derived and the identity given by Equation (17) useful in this derivation!
Question: 3. Derive Gauss's condition given by
∂/∂α1(1/A1.∂A2/∂α1) + ∂/∂α2(1/A2.∂A1/∂α2) + A1A2/R1R2 = 0 (21)
You will find Equation (16) useful here.