Page 1 of 13

Torsional shear stresses in non-circular shafts via T6 finite element

MECH 417, Rice University, J.E. Akin, draft 4 4/06/2020

Introduction: The shear stresses in a circular shaft are easy to determine, and are covered in mechanics of materials, but the shear stresses in a non-circular shaft are quite difficult to determine. A theoretical equilibrium formulation was given in 1903 by Prandtl by introducing a non-physical ‘stress function’, u(x, y), whose partial derivatives define the physical shear stress components.

𝜏𝑧𝑥 =𝜕𝑢

𝜕𝑦 and 𝜏𝑧𝑦 = −

𝜕𝑢

𝜕𝑥, 𝝉 = {

𝜏𝑧𝑥𝜏𝑧𝑦}

Therefore, the total shear stress at a point is τmax = √τzx2 + τzy2 and its direction is negative reciprocal

of the gradient vector.

Prandtl’s equation for the stress function, u(x, y), of an isotropic elastic shaft is

𝜕

𝜕𝑥(1

𝐺𝑥

𝜕𝑢

𝜕𝑥) +

𝜕

𝜕𝑦(1

𝐺𝑦

𝜕𝑢

𝜕𝑦) + 2𝜃 = 0 (12.15-1)

Page 2 of 13

where Gx and Gy are the orthotropic shear moduli of the shaft material and θ is the angle of twist per unit

length about the z-axis parallel to the shaft. The most common case is where the shaft is a homogeneous isotropic material where Gx = Gy = G. Note that if the middle term were omitted this would correspond

to several of the 1D applications considered before. The major change in FEA in going from 1D to 3D is that the Jacobian matrix always has more than one row, and usually is variable in space. In the 1D cases covered before only the single top left corner term was used in 1D applications. When addressing a curve in 2D space, below, then only the two terms in the top row were needed.

Prandtl showed that the Dirichlet essential boundary condition is that the stress function is a constant on each material boundary curve, and is set to zero on the outer boundary. (If there is an inner hole, or different material, then numerical tricks are needed to enforce the inner Dirichlet conditions; see Fig. 12.15-3.) Prandtl also showed that the externally applied torque, T, is defined by twice the integral, over the cross-section, of the stress function: 𝑇 = 2 ∫ 𝑢(𝑥, 𝑦) 𝑑

ΩΩ ≡ (G 𝐽) 𝜃. (12.15-3)

where J is considered to be the effective polar moment of inertia. (For a circular shaft Jcircle = πR4 2⁄ .) In practice however, we generally know the applied torque, T, and not the twist angle that appears in the differential equation. Thus, a unit twist angle (θ = 1) is assumed and the finite element model is solved for the stress function. Then the integral of the stress function is computed to give the corresponding torque, say Tfea. Then, all of the shear stresses and the angle of twist are scaled by the ratio Tactual Tfea⁄ to obtain the results that correspond to the actual applied torque. (Or, the FEA study can be re-run with the scaled angle of twist.) However, 1903 Prandtl’s equation could only be solved for simple shapes including a rectangle, an ellipse, and an equilateral triangle (see Appendix for those exact solutions), and solutions for other shapes were obtained experimentally using an electrical conduction analogy or a soap bubble analogy. The advent of finite element analysis made accurate computational solutions possible for any 2D shape.

Finite element formulation:

Let the stress function in an element be interpolated from its nodal values as u(r, s) = H(r, s) ue. Then the Galerkin and variational calculus solutions give the torsional stiffness matrix (similar to heat conduction) as

𝑺𝒆 = ∫ 𝑩𝒆𝑇𝜿𝑒

Ω𝑒𝑩𝒆 𝑑Ω, 𝑩𝒆 ≡ [

𝜕𝑯(𝑟, 𝑠) 𝜕𝑥⁄

𝜕𝑯(𝑟, 𝑠) 𝜕𝑦⁄], [𝜿] = [

1𝐺𝑥⁄ 0

0 1𝐺𝑦⁄

] , |𝜿| > 0,

and the source term from the angle of twist is

Page 3 of 13

𝐜𝛉𝐞 = ∫ 𝐇𝐞𝐓

Ωe𝜃e dΩ

These will be evaluated by quadratures as

𝐜𝛉𝐞 = ∫ 𝑯𝒆𝑇𝜃e

⊡

( |𝑱𝒆| 𝑑 ⊡) = ∑𝑯𝒆𝑇(𝑟𝑞 , 𝑠𝑞 )

𝑛𝑞

𝑞=1

𝜃e |𝑱𝒆(𝑟𝑞 , 𝑠𝑞)|𝑤𝑞

where the Jacobian matrix

[𝑱𝒆(𝑟𝑞 , 𝑠𝑞)] = [𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄

𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄]𝑞

𝑒

= [𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑟⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑠⁄] [𝒙 𝒚]𝑒

has a determinant that defines the physical differential area,

|𝑱𝒆(𝑟𝑞 , 𝑠𝑞)| = det[𝑱𝒆(𝑟𝑞 , 𝑠𝑞)] , dΩ = |𝑱𝒆(𝑟𝑞 , 𝑠𝑞)| 𝑑𝑟 𝑑𝑠

and whose inverse defines components of the operator matrix: [𝑱𝒆(𝑟𝑞 , 𝑠𝑞)]−1= 𝑖𝑛𝑣 (|𝑱𝒆(𝑟𝑞, 𝑠𝑞)|)

𝑩𝒒𝒆 ≡ [

𝜕𝑯(𝑟𝑞, 𝑠𝑞) 𝜕𝑥⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑦⁄] = [𝑱𝒆(𝑟𝑞 , 𝑠𝑞)]

−1[𝜕𝑯(𝑟𝑞, 𝑠𝑞) 𝜕𝑟⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑠⁄]

and the triple matrix product contributes to the stiffness matrix

𝑺𝒆 =∑𝑩𝒒𝒆𝑻(𝑟𝑞 , 𝑠𝑞 )

𝑛𝑞

𝑞=1

𝜿𝑒 𝑩𝒒𝒆(𝑟𝑞 , 𝑠𝑞 ) |𝑱

𝒆(𝑟𝑞 , 𝑠𝑞)| 𝑤𝑞

Those terms are all valuated numerically, along with the physical location of the quadrature point

[𝑥(𝑟𝑞 , 𝑠𝑞) 𝑦(𝑟𝑞 , 𝑠𝑞)] = 𝑯(𝑟𝑞 , 𝑠𝑞)[𝒙 𝒚]𝑒

that is saved for plotting the shear stress magnitude and direction during post-processing of the stress function. Another item needed in post-processing is the total applied torque due to the specified angle of twist:

𝑇 =∑ 𝑇𝑒 , 𝑇𝑒 =

𝑒2∫ 𝑢(𝑥, 𝑦) 𝑑

ΩeΩ = 2∫ 𝐇𝐞

ΩedΩ 𝒖𝐞

𝑇𝑒 = 2 [∑ 𝑯(𝑟𝑞 , 𝑠𝑞)𝑛𝑞𝑞=1 |𝑱𝒆(𝑟𝑞, 𝑠𝑞)|𝑤𝑞] 𝒖

𝐞 ≡ 2[𝑰_𝒐𝒇_𝑯]𝒆 𝒖𝐞.

The newly named matrix, [I_of_H]e, for obtaining the integral of the solution during post-processing is not needed in most applications, but is always needed for torsion analysis (and viscous flow in channels). It will be evaluated while forming the stiffness matrix and stored for later use. Implementing an element: Implementing torsional stress analysis using a six-node quadratic triangle (T6) element is a simple extension of the prior lecture where the area of a curved T6 element was obtained by numerical integration in the script downloaded as Area_T6_curved.m (or .txt). There the local derivatives of the interpolation functions were evaluated at the quadrature point, q, as

Page 4 of 13

[𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑟⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑠⁄]

They were multiplied by the element coordinates to form the Jacobian matrix at that point

[𝑱𝒆(𝑟𝑞 , 𝑠𝑞)] = [𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄

𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄]𝑞

𝑒

= [𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑟⁄

𝜕𝑯(𝑟𝑞, 𝑠𝑞) 𝜕𝑠⁄] [𝒙 𝒚]𝑒

The Matlab function det was used to get the numerical value of its scalar determinant

|𝑱𝒆(𝑟𝑞, 𝑠𝑞)| = det[𝑱𝒆(𝑟𝑞 , 𝑠𝑞)]

for use in finding the physical differential area. The new terms needed for forming the torsional stiffness matrix is to use the Matlab function inv to invert the Jacobian matrix

[𝑱𝒆(𝑟𝑞 , 𝑠𝑞)]−1= 𝑖𝑛𝑣 (|𝑱𝒆(𝑟𝑞 , 𝑠𝑞)|)

so the inverse matrix can be used to get the physical gradient components of the interpolation functions:

𝑩𝒒𝒆 ≡ [

𝜕𝑯(𝑟𝑞, 𝑠𝑞) 𝜕𝑥⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑦⁄] = [𝑱𝒆(𝑟𝑞 , 𝑠𝑞)]

−1[𝜕𝑯(𝑟𝑞, 𝑠𝑞) 𝜕𝑟⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑠⁄]

to numerically form the triple matrix products that defines the new stiffness matrix part of its summation

𝑺𝒒𝒆 = 𝑩𝒒

𝒆𝑻(𝑟𝑞 , 𝑠𝑞 ) 𝜿𝑒 𝑩𝒒

𝒆(𝑟𝑞 , 𝑠𝑞 ) |𝑱𝒆(𝑟𝑞 , 𝑠𝑞)| 𝑤𝑞 in 𝑺

𝒆 =∑𝑺𝒒𝒆

𝑛𝑞

𝑞=1

In the curved triangular area script the interpolation functions, 𝑯(𝑟𝑞 , 𝑠𝑞), were also evaluated, but not

used. They could be used here to determine the source matrix contribution

𝐜𝛉𝐞 =∑𝑯𝒆𝑇(𝑟𝑞 , 𝑠𝑞 )

𝑛𝑞

𝑞=1

𝜃e |𝑱𝒆(𝑟𝑞 , 𝑠𝑞)|𝑤𝑞

as well as the auxiliary data on the physical location of the point

[𝑥(𝑟𝑞 , 𝑠𝑞) 𝑦(𝑟𝑞 , 𝑠𝑞)] = 𝑯(𝑟𝑞 , 𝑠𝑞)[𝒙 𝒚]𝑒

and the integral of the interpolation functions needed later to find the torque:

[𝑰_𝒐𝒇_𝑯]𝒆 =∑𝑯(𝑟𝑞 , 𝑠𝑞)

𝑛𝑞

𝑞=1

|𝑱𝒆(𝑟𝑞 , 𝑠𝑞)|𝑤𝑞

Making those additions to the area script would yield the element matrices for torsion. Of course, it is necessary to specify the data for G and θ before forming the element matrices. Next, insert a pre-loop over all elements to assemble (scatter) the element matrices into the matrix equilibrium equations for the system. Enforcing the Dirichlet BCs, that the stress function is zero on every node on the shaft outer perimeter, renders the system non-singular and its solution gives the non-physical stress function value

Page 5 of 13

at all nodes. That is not particularly useful. That surface could be plotted and reviewed. Where its slope is zero so is the shear stress. At a point where the slope is the highest that is where the maximum shear stress occurs, in a direction perpendicular to that slope. (In theory, the gradient is infinite at the tip of any sharp re-entrant corner.) Clearly, more precise engineering output data are required. A post-processing loop that needs much of the same data used to form the stiffness matrix and the load matrix. They must either be recomputed, or read from a binary storage file. Here, the choice during formulating forming the element stiffness matrix is to store the arrays used at every quadrature point (via Matlab functions fopen, fwrite, and fclose) on a sequential binary file. Then, in the added post-processing loop those data are accessed again sequentially for reading (via Matlab function fopen). In a quadrature loop, within the element loop, those data are read sequentially (via fread) and used to find the gradient and shear stress components at all quadrature locations, and to save (in text file el_qp_xyz_fluxes.txt) the shear stress values and directions for later plotting (using Matlab function quiver). In addition, each element contribution to the total torque is summed. Finally, when all elements have been post-processed, the applied torque is printed, the two files closed, and then the program exits. Here, the plot files are saved in an inefficient text mode (.txt), rather than in the most efficient binary (.bin) mode, so students will have other chances to easily review their values. Selecting the quadrature rule: First, since the element is interpolated in unit coordinates on an equilateral right triangle (and not in area coordinates) the proper tabulated data table must be selected (here from qp_unit_tri_rule.m). Before the data can be extracted the total polynomial degree of the integrand must be determined. For a six-node quadratic triangle the interpolation functions are degree 2 and its local derivative is of degree 1. For example,

𝐻1(𝑟, 𝑠) = 1 − 3𝑟 + 2𝑟2 − 3𝑠 + 4𝑟𝑠 + 2𝑠2 ,

𝜕𝐻1(𝑟, 𝑠)

𝜕𝑟= −3 + 4𝑟 + 4𝑠

The Jacobian matrix involves these local derivatives times a constant matrix (degree 0) of the element coordinates

[𝑱𝒆(𝑟𝑞 , 𝑠𝑞)] = [𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑟⁄

𝜕𝑯(𝑟𝑞 , 𝑠𝑞) 𝜕𝑠⁄] [𝒙 𝒚]𝑒

so the Jacobian matrix is of degree 1 + 0 = 1, but the nodes can be placed so as to degenerate the interpolation to linear, not quadratic. Then the Jacobian becomes constant (degree 0). The determinant of the Jacobian

(𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑠−𝜕𝑦

𝜕𝑟

𝜕𝑥

𝜕𝑠) → (degree 1)(degree 1) → 1 + 1 = 2

is degree 2, but can be degree 0 for a straight sided triangular element. Usually, the compromise is to assume the determinant is of degree 1. Any finite element has a finite number of degrees of freedom and any continuum (represented by a differential equation) has an infinite number. Therefore, a finite element model always over estimates the stiffness and a slight under integration (less stiff element) helps compensate for that. Consider a typical scalar ‘field problem’ (heat conduction, potential flow, torsion, etc.) which always has a triple matrix product to evaluate the stiffness matrix. Usually, the material matrix, κe, is constant within an element (degree 0) but those properties can be defined at the nodes and interpolated (here, degree 2). But today automatic mesh generators create many small elements and there is not much space for a property to change. On the interior of a region mesh generators usually tend toward forming

Page 6 of 13

nearly straight triangles, so the Jacobian approaches a constant (degree 0). Of course, a crude mesh near a highly curved boundary would give a full degree 2 Jacobian.

𝑺𝒒𝒆 = 𝑩𝒒

𝒆𝑻(𝑟𝑞 , 𝑠𝑞 ) 𝜿𝑒 𝑩𝒒

𝒆(𝑟𝑞 , 𝑠𝑞 ) |𝑱𝒆(𝑟𝑞, 𝑠𝑞)| 𝑤𝑞 → ⟨

1 + 0 + 1 + 0 + 0 = degree 2 constant 𝐽

1 + 0 + 1 + 1 + 0 = degree 3 average 𝐽

1 + 2 + 1 + 0 + 0 = degree 4 variable 𝜿𝑒

Today’s computers typically solve 105 to 106, or more, equations and have a similar number of elements. The operation count (time) to form the element stiffness matrices is proportional to the product of the number of elements, ne, and the number of quadrature points, nq. Most programmers use the minimum

number of quadrature points. (It can be shown that for a non-singular (rank sufficient) system square matrix with nd equations assembled from ne elements and subjected to nEBC Dirichlet boundary conditions requires that the number of quadrature points needs to satisfy

𝑛𝑒(𝑛𝑞𝑛𝑟 − 𝑛𝑐) ≥ (𝑛𝑑 − 𝑛𝐸𝐵𝐶) (3.4-2)

where nr corresponds to the number of rows in the B matrix in the usual triple product integrand of the element square matrix: BTκ B, and where each element is subject to nc ≥ 0 constraint equations. Such relations can be used as guides in selecting a minimal value of integration points, nq, in the element.)

For torsion, there are no element constraints, so assume that the triple product is degree 3. The triangle integration table gives the minimum number of quadrature points depending on the polynomial degree of the integrand:

For degree 𝑑 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

Needed 𝑛𝑞 ≥ 1, 3, 4, 6, 7, 12, 13, 16, 19, 25, 27, 33, 37, 42, 48, 52, 61

Therefore, for torsion nq ≥ 4.

Why use more points? It can be shown that the most accurate gradients (and thus stresses) occur at, or near, the integration points used for the stiffness matrix, Se. The gradient values are always least accurate at the nodes; unless they have been through a second post-processing for a super convergent patch (scp) averaging, which is used here. Some codes use only the average gradient from element surrounding a node. The theory of elasticity proves that the maximum stress occurs on the boundary, if there are no interior holes in the region. Most integration points are not on the element boundary. Using more than the minimum number of quadrature points moves some of those points closer to the boundary (see 4 vs. 7 below). Therefore, the reported maximum shear stress found at any one integration point (see download listing) is going to be closest the actual maximum shear stress on the part boundary. If the gradients (stresses) at the nodes have been obtained with additional scp averaging (a required step for accurately calculating the error in the FEA) then the maximum stress reported there is probably closest true maximum stress, even if the FEA region has interior holes. Another reason for using more quadrature points is that when using a crude mesh a plot of the vectors gives a better ‘feel’ for the result (see below). It can also enhance the accuracy of the additional scp averaging loop.

Implementation: Torsion was chosen to illustrate everything students need to know to understand how straight line 1D analytically integrated get extended to 2D or 3D curved domains using numerical integration (quadratures). First, select the element type(s). Here a single type curved six-node quadratic triangle,

Page 7 of 13

T6, element is picked. (If the mesh was allowed to have mixed element types the second compatible type would be a quadrilateral with three nodes per edge, e.g. Q8 or Q9.) Next select the triangular parametric space to use area-coordinates, or unit coordinates (here); and locate the proper table of quadrature data for that non-dimensional space. Of course, functions should be provided to execute the mundane tasks needed in any FEA: input the coordinates, Dirichlet BC values, element type and connection list, material properties, mesh plotters, BC plotters, solution plotters, etc. Here, the two uniform properties are input as arguments. If the torsional domain contains multiple properties then properties would be input with the mundane function. To allow curved elements, any closed form element matrices must be replaced by numerically integrated ones. That requires inserting a quadrature loop within the element matrix formation loop. That quadrature loop will evaluate the interpolation functions, and their local derivatives (as was done in the 1D cases). Here, that is done with specific equations, but generally an element library would be called to return those numerical values. Example: Torsion of a rectangle 16 units x 8 units (quarter symmetry) shaft with G = 1, θ = 1. (See appendix solution for b a⁄ = 2). The torsion script on the download link can be applied to any 2D shaft shape, with a single material. Here, a one-quarter segment of a rectangular bar was selected so it can be validated by comparing the results to one of the few closed form solutions for the torque and the maximum shear stress (at points touched by the smallest enclosed circle). [That validation has not been completed yet.]

Quarter symmetry mesh Contour of stress function values

Page 8 of 13

Stress function surface from the 40 deg eye point Stress function surface from -38 deg (default)

Shear stress, n_q = 4 Shear stress, n_q = 7

Page 9 of 13

Averaged (scp) shear stresses at nodes [exact = 0 at LLC and URC]

Note: The exact torsional shear stress on the exterior boundary is tangent to that boundary.

A condensed listing of the output for this bar study (Download script 15a) is given below.

addpath /clear/www/htdocs/mech517/Akin_FEA_Lib

Torsion_2D_lib(1,1)

...

Torsion of rectangular bar with quarter symmetry

5--8-13-16-21 (8,4) Mesh shown to left.

: / : / : EBC = 0 at 5,8,13,16-21

4 24 12 25 20

:/ :/ : Symmetry planes are x =0

3--7-11-15-19 and y = 0

: / : / :

2 22 10 23 18 Shear stress max at 5 (small R)

:/ :/ : Shear stress zero at 1 and 21

--6--9-14--17 ->X

Material shear modulus = 1.000e+00

Twist angle per unit length = 1.000e+00 deg

...

Created file plot_input_2d_mesh.png

...

Computed Solution:

Node, 1 results per node % stress function value

1, 1.273e-01

...

Created file color_scalar_result.png

Created file result_surface_38.png

Created file result_surface_40.png

...

Page 10 of 13

Element Post-processing: % at quadrature points

...

El, Q_Pt, Coordinates 6, 3, 8.000e-01 3.600e+00

El, Q_Pt, Gradient Vector 6, 3, -3.254e-04 -5.734e-02

El, Q_Pt, Shear stress 6, 3, -5.734e-02 3.254e-04

El, Q_Pt, Magnitude 6, 3, 5.734e-02

...

Max shear stress = 5.734e-02 in element 6

at qp 3 located at 8.00e-01, 3.60e+00

Created file qp_xyz_flux_vector_mesh.png

Total applied torque = 4.0579e+00

Multiply by the number of mesh symmetries used ****

Begin SCP averaging shear stress at the nodes

Range of averaged flux values at mesh nodes

FLUX #: MAXIMUM, NODE MINIMUM, NODE

1: 6.6121e-03, 1 -6.4584e-02, 5 % Tau_x

2: 4.9356e-02, 17 -3.0748e-03, 1 % Tau_y

Patch averaged fluxes at mesh nodes:

Node, 2 flux components per node

1 6.612e-03 -3.075e-03 % Tau_x, Tau_y

...

Max averaged node shear stress = 6.461e-02 % Tau_max

occurs at node 5

located at 0.000e+00 , 4.000e+00

NOTE: ave fluxes saved to file scp_node_ave_fluxes.txt

Created ave_2d_flux_vectors.png

Page 11 of 13

Appendix 1: Analytic Torsional Solutions (Blue lines are symmetry lines limiting a FEA mesh)

Page 12 of 13

Appendix 2: PDE to Galerkin Integral Form

The scalar form for the orthotropic stress function PDE is

𝜕

𝜕𝑥(1

𝐺𝑥

𝜕𝑢

𝜕𝑥) +

𝜕

𝜕𝑦(1

𝐺𝑦

𝜕𝑢

𝜕𝑦) + 2𝜃 = 0

In matrix notation this scalar equation is

[𝜕

𝜕𝑥

𝜕

𝜕𝑦]

[ 1

𝐺𝑥0

01

𝐺𝑦]

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦}

+ 2𝜃 = 0

Or symbolically

𝛁{[𝜿]𝛁𝑻𝑢} + 2𝜃 = 0 where

𝛁 = [𝜕

𝜕𝑥

𝜕

𝜕𝑦] , and [𝜿] =

[ 1

𝐺𝑥0

01

𝐺𝑦]

, and {𝛁𝑻𝑢} =

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦}

For the fully anisotropic the material property matrix is

[𝜿] =

[ 1

𝐺𝑥𝑥

1

𝐺𝑥𝑦1

𝐺𝑦𝑥

1

𝐺𝑦𝑦]

= [𝜿]𝑇

But for an isotropic material it is simply

[𝜿] =1

𝐺[1 00 1

]

For a Galerkin integral form multiply the residual error by the solution

𝐼 = ∫ 𝑢(𝑥, 𝑦) (𝜕

𝜕𝑥(1

𝐺𝑥

𝜕𝑢

𝜕𝑥) +

𝜕

𝜕𝑦(1

𝐺𝑦

𝜕𝑢

𝜕𝑦) + 2𝜃)

Ω

𝑑Ω = 0

or

𝐼 = ∫𝑢(𝑥, 𝑦)( 𝛁{[𝜿]𝛁𝑻𝑢} + 2𝜃)

Ω

𝑑Ω = 0

To reduce the order of derivatives in the first integral Green’s Theorem. First the terms will be re-written as two

terms using an identity from vector calculus:

𝑢( 𝛁{[𝜿]𝛁𝑻𝑢}) = 𝛁{𝑢{[𝜿]𝛁𝑻𝑢}} − 𝛁𝑢 {[𝜿]𝛁𝑻𝑢}

But for any vector, say 𝐕𝑻, the integral of the gradient over the domain can be converted to a boundary integral

of the vector dotted with the surface unit normal vector

∫𝛁

Ω

𝐕𝑻𝑑Ω = ∫𝒏

Γ

𝐕𝑻𝑑Ω

where 𝒏 is the outward unit vector on boundary Γ. Select 𝐕𝑻 = {𝑢{[𝜿]𝛁𝑻𝑢}}, then the first area integral in the

split terms becomes a boundary integral

Page 13 of 13

∫𝛁

Ω

𝑢{[𝜿]𝛁𝑻𝑢}𝑑Ω = ∫𝒏

Γ

𝑢{[𝜿]𝛁𝑻𝑢}𝑑Ω = ∫𝑢(𝑘𝑛𝑛 𝜕𝑢 𝜕𝑛⁄ )

Γ

𝑑Γ

which brings the Neumann boundary conditions on 𝑘𝑛𝑛 𝜕𝑢 𝜕𝑛⁄ into the integral form. Finally, the governing

integral is

𝐼 = ∫ 𝑢 𝑘𝑛𝑛𝜕𝑢

𝜕𝑛

Γ

𝑑Γ − ∫𝛁u([𝜿]𝛁𝑻𝑢)𝑑Ω

Ω

+∫2𝑢𝜃

Ω

𝑑Ω = 0

or

𝐼 = ∫ 𝑢 𝑘𝑛𝑛𝜕𝑢

𝜕𝑛

Γ

𝑑Γ − ∫ [𝜕𝑢

𝜕𝑥

𝜕𝑢

𝜕𝑦]

[ 1

𝐺𝑥0

01

𝐺𝑦]

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦}

Ω

𝑑Ω +∫2𝑢𝜃

Ω

𝑑Ω = 0

𝐼 = ∫ 𝑢 𝑘𝑛𝑛𝜕𝑢

𝜕𝑛

Γ

𝑑Γ − ∫𝜕𝑢

𝜕𝑥(1

𝐺𝑥

𝜕𝑢

𝜕𝑥) +

𝜕𝑢

𝜕𝑦(1

𝐺𝑦

𝜕𝑢

𝜕𝑦)𝑑Ω

Ω

+∫2𝑢𝜃

Ω

𝑑Ω = 0

If the y-derivatives are set to zero this becomes the 1D cases considered earlier.