4. calculus review · edge at node 3 ( (1), (1))=( 3, 3). therefore, the edge goes exactly through...

Lecture Notes: Finite Element Analysis, J.E. Akin, Rice University, Copyright. 2017-20. All rights reserved.

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4. Calculus review

4. CALCULUS REVIEW ..................................................................................................................................... 1

4.1 PARAMETRIC GEOMETRY ...................................................................................................................................................... 1

4.2 JACOBIAN MATRIX ............................................................................................................................................................... 4

4.3 INVERSE JACOBIAN ............................................................................................................................................................ 11

4.4 PARAMETRIC SUBSTITUTION IN INTEGRALS ............................................................................................................................. 11

4.5 INTEGRATION BY PARTS ...................................................................................................................................................... 13

4.6 INTEGRAL OF POINT SOURCES .............................................................................................................................................. 14

4.7 AXISYMMETRIC INTEGRALS .................................................................................................................................................. 15

4.8 SUMMARY ....................................................................................................................................................................... 17

4.9 EXERCISES........................................................................................................................................................................ 21

4.1 Parametric geometry: Probably the most common parametric coordinates used in

calculus are polar coordinates, or spherical coordinates. Many integrals in the x-y plane are more

easily evaluated in polar coordinates. Student experiences with such parametric transformations

are not new. For example, consider the integral over the first quadrant of

𝐼 = ∫ ∫ 𝑒−(𝑥2+𝑦2)∞

0

∞

0 𝑑𝑥 𝑑𝑦.

Convert this integral to a parametric form using polar coordinates: 𝑥(𝑅, 𝜃) = 𝑅 cos 𝜃, 𝑦(𝑅, 𝜃) =𝑅 sin 𝜃. For this problem the parameter 𝜃 varies from 0 to 𝜋 2⁄ and the parameter R varies from

0 to ∞. Then 𝑥2 + 𝑦2 = 𝑅2 and (see below) 𝑑𝑥 𝑑𝑦 = 𝑟 𝑑𝑟 𝑑𝜃 and the integral simplifies to

𝐼 = ∫ ∫ 𝑒−𝑅2𝜋 2⁄

0

∞

0 𝑅 𝑑𝑅 𝑑𝜃 =

𝜋

2∫ 𝑒−𝑅

2 𝑅 𝑑𝑅 =

∞

0

𝜋

2[𝑒−𝑅

2

−2]∞ 0=

𝜋

2 1

2=

𝜋

4.

In finite element analysis the ability to handle curved shapes rather than straight sided

triangles and rectangles, and their three-dimensional equivalents, requires the use of non-

dimensional parametric interpolation to transform or map the element geometry from a straight

sided parametric space to its curved physical shape. The use of parametric coordinates

automatically allows the elements to have unique curved edges and curved non-flat faces on

solid elements. By way of contrast, the edge interpolations of an element generated by

interpolating using the physical coordinates are not unique (see Fig. 4.1-1).

If the lack of uniqueness associated with interpolations over a quadrilateral using physical

coordinates was not spotted initially, it would become clear in the third step given in Section 2.2.

Evaluating the proposed interpolation polynomial using the physical coordinates at each of the

four nodes gives the matrix identity 𝒖𝑒 = 𝑮𝒆𝒄 where the square matrix, 𝑮𝒆, now depends on the

physical position of each of the four nodes. Generally, the non-uniqueness of this solution

becomes clear because the inverse, 𝑮𝒆−1, does not exist for a general quadrilateral shape and the


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four constants cannot be found from the theoretical relation 𝒄 = 𝑮𝒆−1𝒖𝑒. The physical

coordinate matrix inverse exists only when the quadrilateral degenerates to a rectangle with its

sides parallel to the x- and y-axes. In contrast, a proper parametric coordinate interpolation will

be unique when evaluated on any edge of a general quadrilateral located in physical space.

Figure 4.1-1 Element uniqueness is lost with physical coordinate interpolations

The simplest curved edge is one defined by three nodal points. In parametric geometry that is

a segment of a parabola. It is not a circular arc, but is a good approximation of one if the change

in slope is less than 25 degrees. On such an edge a two- or three-dimensional parametric

interpolation, with uniform parametric node spacing, degenerates to the unique quadratic line

interpolation given in (2.2-1). As expected, that interpolation depends only on the function

values at those three nodes. Therefore, the function will be continuous across adjacent curved

element edges of a similar type. That is also true for each of the physical coordinates of those

nodes. In other words, no geometric gaps or overlaps occur at adjacent elements with curved

edges of the same polynomial degree.

Example 4.1-1 Given: Use the physical x- and y-coordinates to attempt to interpolate the

straight-sided two-dimensional four node quadrilateral shown in Fig. 4.1-1. Solution: The

derivation of any set of interpolation functions would follow the process given in Section 2.2.

Begin with the incomplete quadratic polynomial assumption to interpolate between the four

nodal values of a solution, u (x, y), in terms of the physical coordinates:

𝑢(𝑥, 𝑦) = 𝑐1 + 𝑐2𝑥 + 𝑐3𝑦 + 𝑐4𝑥𝑦 = 𝑷(𝑥, 𝑦) 𝒄

The failure of this approach is quickly foreseen. All edges have been given to be straight.

Thus, on the edge between nodes 2 and 3 the y-coordinate is described by the slope-intercept

equation: 𝑦23(𝑥) = 𝑏23 + 𝑚23𝑥 where the slope, 𝑚23, and the y-intercept, 𝑏23, are geometric

constants. The solution for u along that edge (that is, the boundary interpolation) degenerates to a

function of x alone:

𝑢23(𝑥) = 𝑢(𝑥, 𝑦23(𝑥)) = 𝑐1 + 𝑐2𝑥 + 𝑐3(𝑏23 + 𝑚23𝑥) + 𝑐4𝑥(𝑏23 + 𝑚23𝑥)

𝑢23(𝑥) = (𝑐1 + 𝑐3𝑏23) + 𝑥 (𝑐2 + 𝑐3𝑚23 + 𝑐4𝑏23) + 𝑥2(𝑐4𝑚23)

𝑢23(𝑥) ≡ 𝑑1 + 𝑑2𝑥 + 𝑑3𝑥2.


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Three values of u are needed to uniquely determine these three constants, but only the two 𝑢2

and 𝑢3 values are available on that edge. The same conclusion is reached on the other three

edges. In other words, global coordinate interpolation for general quadrilaterals fails to give

unique results, unless all of the sides are parallel to the global axes. Global coordinate

interpolation is valid only for straight edged triangles and tetrahedra, and were used with those

elements in the earliest finite element literature.

Example 4.1-2 Given: Consider the element as in Fig. 4.1-1. For a four node quadrilateral in

unit coordinates it can be shown that the interpolation is

𝑢(𝑟, 𝑠) = (1 − 𝑟 − 𝑠 + 𝑟𝑠) 𝑢1 + (𝑟 − 𝑟𝑠) 𝑢2 + 𝑟𝑠 𝑢3 + (𝑠 − 𝑟𝑠) 𝑢4

Evaluate the bi-linear parametric form on edge 2-3. Solution: On edge 2-3, the r coordinate is

constant, 𝑟23 = 1, so the interpolation degenerates to (the edge interpolation)

𝑢23(𝑠, 𝑟 = 1) = (1 − 1 − 𝑠 + 1𝑠) 𝑢1 + (1 − 1𝑠) 𝑢2 + 𝑠 𝑢3 + (𝑠 − 1𝑠) 𝑢4 , or

𝑢23(𝑠) = 0 + (1 − 𝑠)𝑢2 + 𝑠 𝑢3 + 0 = (1 − 𝑠) 𝑢2 + 𝑠 𝑢3

which is a unique linear function dependent only on the two nodal values, 𝑢2 and 𝑢3, on that

edge. An adjacent element of the same (linear) type will share those two values. That guaranties

that an interpolated solution will be continuous across the edges of adjacent elements which

degenerate to the same complete polynomial on their edges.

Example 4.1-3 Given: Determine the geometric shape of an edge of the four node quadrilateral

in Fig. 4.1-1. Solution: The parametric interpolations can be used for any data, including

physical coordinates. From Ex. 4.2-2 the edge interpolation for the element is known. Applying

it to the x- and y-positions gives

𝑥23(𝑠) = (1 − 𝑠) 𝑥2 + 𝑠 𝑥3 = 𝑥2 + 𝑠(𝑥3 − 𝑥2)

𝑦23(𝑠) = (1 − 𝑠) 𝑦2 + 𝑠 𝑦3 = 𝑦2 + 𝑠(𝑦3 − 𝑦2)

At node 2, s = 0, and the interpolation gives (𝑥(0), 𝑦(0)) = (𝑥2, 𝑦2) and at the other end of the

edge at node 3 (𝑥(1), 𝑦(1)) = (𝑥3, 𝑦3). Therefore, the edge goes exactly through the node points

at its ends. In addition, x23(s) and y23(s) both change linearly between those two end points.

Thus, the edge has the shape of a straight line between nodes 2 and 3.

Example 4.1-4 Given: An eight-node quadratic quadrilateral element has three nodes on each

edge. Along the edge where r = 1 the interpolation degenerates to 𝑯𝒃(𝑠) = [(1 − 3𝑠 +2𝑠2) (4𝑠 − 4𝑠2) (−𝑠 + 2𝑠2)]. The system node numbers on that edge are 2, 7, and 9,

respectively. Determine the shape of that edge. Solution: Interpolate the x- and y-positions

along the edge;

𝑥(𝑠) = 𝑯𝒃(𝑠) 𝒙𝒃 = (1 − 3𝑠 + 2𝑠2)𝑥2 + (4𝑠 − 4𝑠2)𝑥7 + (−𝑠 + 2𝑠

2)𝑥9

𝑥(𝑠) = 𝑥2 + 𝑠(4𝑥7 − 3𝑥2 − 𝑥9) + 𝑠2(2𝑥2 − 4𝑥7 + 2𝑥9)

Likewise, the y-coordinate has the same form. Evaluating at the local nodes of s = 0, s = 1/2, and

s = 1 shows that the edge shape goes exactly through the input positions of nodes 2, 7, and 9,

respectively. The quadratic change in both the x- and y-coordinates means that in general the


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edge shape is a segment of a parabola going through the three nodes. If the user input the

coordinate data such that node 7 was exactly the midpoint between nodes 2 and 9 then a special

case occurs with 𝑥7 = (𝑥2 + 𝑥9)/2. Then the 𝑠2 term drops out and

𝑥(𝑠) = 𝑥2 + 𝑠(2𝑥2 + 2𝑥9 − 3𝑥2 − 𝑥9) = 𝑥2 + 𝑠(𝑥9 − 𝑥2)

etc. for y(s) and the edge becomes a straight line between the two end nodes.

4.2 Jacobian matrix: The differentiation of composite functions is required in every

parametric finite element study. Let the dependent variable be u(x, y, z) in a domain with the

coordinate transformations x = f(r, s, t), y = g(r, s, t) and z = h(r, s, t). From the chain rule of

calculus the relationship between the parametric derivatives and the physical derivatives are:

𝝏𝒖

𝝏𝒓=

𝝏𝒖

𝝏𝒙

𝝏𝒙

𝝏𝒓+𝝏𝒖

𝝏𝒚

𝝏𝒚

𝝏𝒓+𝝏𝒖

𝝏𝒛

𝝏𝒛

𝝏𝒓

𝜕𝑢

𝜕𝑠=

𝜕𝑢

𝜕𝑥

𝜕𝑥

𝜕𝑠+𝜕𝑢

𝜕𝑦

𝜕𝑦

𝜕𝑠+𝜕𝑢

𝜕𝑧

𝜕𝑧

𝜕𝑠 (4.2-1)

𝜕𝑢

𝜕𝑡=

𝜕𝑢

𝜕𝑥

𝜕𝑥

𝜕𝑡+𝜕𝑢

𝜕𝑦

𝜕𝑦

𝜕𝑡+𝜕𝑢

𝜕𝑧

𝜕𝑧

𝜕𝑡

or in a matrix form this defines the Jacobian matrix, J(r, s, t),

{

𝜕𝑢

𝜕𝑟𝜕𝑢

𝜕𝑠𝜕𝑢

𝜕𝑡}

=

[ 𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝑠

𝜕𝑦

𝜕𝑠

𝜕𝑧

𝜕𝑠𝜕𝑥

𝜕𝑡

𝜕𝑦

𝜕𝑡

𝜕𝑧

𝜕𝑡]

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦

𝜕𝑢

𝜕𝑧}

≡ [𝑱(𝑟, 𝑠, 𝑡)]

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦

𝜕𝑢

𝜕𝑧}

.

The Jacobian matrix is not always square; it has as may rows as parametric coordinates and

as many columns as physical coordinates. The relevance of different partitions of the full three-

dimensional Jacobian matrix is denoted in the left of Fig. 4.2-1, and its various rectangular sub-

sections are illustrated on the right. In shorthand matrix notation, the parametric partial

derivatives, in a parametric space ⊡, are defined in terms of the physical partial derivatives, in a

space Ω, given in the full equation above are symbolically represented as :

𝜕𝑢

𝜕 ⊡= [𝑱(𝑟, 𝑠, 𝑡)]

𝜕𝑢

∂Ω

or since the derivatives can be applied to any function, and not just u(x, y, z) there is a general

relationship between parametric derivatives and physical derivatives: 𝝏⊡( ) = 𝑱(⊡) 𝝏𝛀( ).


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Figure 4.2-1 Partitions of the transformation Jacobian matrix

In practice, the element Jacobian matrix, 𝑱𝒆(⊡), is usually evaluated numerically as the

product of the local derivative of the geometry interpolation, at a specific local point, times the

input element physical coordinates:

𝑱𝒆(⊡) =𝝏𝑯(⊡ )

𝝏⊡𝒙𝒆 (4.2-2)

which for one-dimensional problems is simply

𝑱𝒆(𝑟) =𝜕𝑯(𝒓 )

𝜕𝑟𝒙𝒆. (4.2-3)

The determinant of the Jacobian of the transformation,

𝐽 = | 𝑱 | =|

|

𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝑠

𝜕𝑦

𝜕𝑠

𝜕𝑧

𝜕𝑠𝜕𝑥

𝜕𝑡

𝜕𝑦

𝜕𝑡

𝜕𝑧

𝜕𝑡

|

| ≡

𝜕(𝑥, 𝑦, 𝑧)

𝜕(𝑟, 𝑠, 𝑡)

relates the physical differential volume to the differential parametric volume:

𝑑Ω = |𝐽(⊡)| 𝑑 ⊡. (4.2-4)

For straight a line 𝑑𝑥 = |𝐽(𝑟)|𝑑𝑟 = |𝜕𝑥(𝑟)

𝜕𝑟| 𝑑𝑟.

For a flat area 𝑑𝐴 = 𝑑𝑥 𝑑𝑦 = |𝐽(𝑟, 𝑠)| 𝑑𝑟 𝑑𝑠 = |𝜕𝑥

𝜕𝑟

𝜕𝑦(𝑟,𝑠)

𝜕𝑠−𝜕𝑦

𝜕𝑟

𝜕𝑥(𝑟,𝑠)

𝜕𝑠| 𝑑𝑟 𝑑𝑠.

For a volume 𝑑𝑉 = 𝑑𝑥 𝑑𝑦 𝑑𝑧 = |𝐽(𝑟, 𝑠, 𝑡)| 𝑑𝑟 𝑑𝑠 𝑑𝑡.

For a non-flat surface, such as a solid element face, it is necessary to use the mathematical topic

of differential geometry to define the vector normal to the surface and to show that the surface

area is related to a Lame’ constant:

𝑑𝑆 = √(𝑥, 𝑟 𝑦, 𝑠 − 𝑦, 𝑟 𝑥, 𝑠)2 + (𝑦, 𝑟 𝑧, 𝑠 − 𝑧, 𝑟 𝑦, 𝑠)2 + (𝑧, 𝑟 𝑥, 𝑠 − 𝑥, 𝑟 𝑧, 𝑠)2 𝑑𝑟 𝑑𝑠

In practical FEA studies the Jacobian matrix and its (generalized) determinant are evaluated

numerically. That is also true for the determinant and the inverse of the Jacobian matrix.

When the Jacobian of a transformation is constant, it is possible to develop closed form exact

integrals of parametric polynomials. Several examples of this will be presented in a later chapter


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for lines, triangles, tetrahedrons, and rectangles. They are useful for simplifying lectures, but for

practical finite element analysis, and its automation, it is usually necessary to employ numerical

integration. For line integrals of polynomials in unit coordinates, with a constant Jacobian, it is

easy to prove that typical terms in the matrix integrals are

∫ 𝑟𝑚 𝑑𝑥𝐿𝑒

0= 𝐿𝑒 (𝑚 + 1)⁄ (4.2-5)

Example 4.2-1 Given: Evaluate the Jacobian for a straight quadratic line element with unequal

spaces between its three nodes. Solution: Since only the relative spacing matters let 𝑥1 = 0, 𝑥2 =𝛼 𝐿𝑒 , 𝑥3 = 𝐿

𝑒 for 0.1 ≤ 𝛼 ≤ 0.5. The quadratic line interpolation is

𝑥(𝑟) = 𝑯 (𝑟) 𝒙𝒆 = (1 − 3𝑟 + 2𝑟2)𝑥1 + (4𝑟 − 4𝑟2)𝑥2 + (2𝑟

2 − 𝑟)𝑥3.

For the given node locations

𝑥(𝑟) = (4𝑟 − 4𝑟2)𝛼 𝐿𝑒 + (2𝑟2 − 𝑟)𝐿𝑒 = 𝑟(4𝛼 − 1)𝐿𝑒 + 𝑟2(2 − 4𝛼)𝐿𝑒

Thus, the variable Jacobian is 𝜕𝑥 𝜕𝑟⁄ = (4𝛼 − 1)𝐿𝑒 + 2𝑟(2 − 4𝛼)𝐿𝑒 = 𝑱𝒆(𝑟). Note that the

Jacobian is only constant within the element when 𝛼 = 0.5, that is when the interior node is at

the middle of the element. Then 𝑱𝒆 = 𝐿𝑒, as expected. For 0.25 < 𝛼 ≤ 0.5 the Jacobian is

positive (which is good). At 𝛼 = 1 4⁄ the Jacobian is zero (which is usually bad). For 𝛼 < 1 4⁄

the Jacobian is negative, which is always bad. For distorted elements (𝛼 ≠ 1 2⁄ ) the guideline is

that the interior node should be in the center quarter of the element length, 0.4 ≤ 𝛼 ≤ 0.6.

Ideally, the element will have the interior node should be exactly at the midpoint. Some

programs force that condition by using the average on the end coordinates, 𝑥2 = (𝑥1 + 𝑥3) 2⁄ .

Likewise, it is easy to show that for a cubic line element the Jacobian will be constant when the

interior nodes are placed at the third-points along the length. For a unit coordinate interpolation

the constant is again the element length, 𝐿𝑒.

Example 4.2-2 Given: For the cubic line element graphed in Fig. 2.2-9 with equally spaced

nodes the nodal data are 𝒙𝒆 = [2 4 6 8]𝑐𝑚, 𝒖𝒆 = [2 3 0 − 1]℃. Evaluate the integral of

the nodal data. In other words find the area under the curve at the top of Fig. 2.2-9. Solution:

The integral is

𝐼𝑢 = ∫ 𝑢(𝑟) 𝑑𝑥

𝐿𝑒= ∫ 𝑢(𝑟) |𝐽(𝑟)| 𝑑𝑟

1

0= ∫ 𝑯(𝑟) 𝒖𝒆 |𝐽(𝑟)𝑒| 𝑑𝑟 = {∫ 𝑯(𝑟) |𝐽(𝑟)𝑒| 𝑑𝑟

1

0} 𝒖𝒆

1

0

Ex. 4.2-1 noted that the Jacobian is constant in a cubic element when the interior nodes are at the

third-points of the length. In this case 𝐽(𝑟)𝑒 = |𝐽(𝑟)𝑒| = 𝐿𝑒/1 = 6 𝑐𝑚 and the can Jacobian be

taken outside the integral, as the nodal constants were:

𝐼𝑢 = ∫ 𝑢(𝑟) 𝑑𝑥

𝐿𝑒= 𝐿𝑒 {∫ 𝑯(𝑟) 𝑑𝑟

1

0} 𝒖𝒆.

1𝑥1 = (1𝑥4) (4𝑥1)


7

𝐼𝑢 = 𝐿𝑒 {∫ [(2 − 11𝑟 + 18𝑟2 − 9𝑟3)

1

0

(18𝑟 − 45𝑟2 + 27𝑟3)

(−9𝑟 + 36𝑟2 − 27𝑟3) (2𝑟 − 9𝑟2 + 9𝑟3)]/2} 𝒖𝒆. The first entry in the integral of the interpolation function is

∫ 𝐻(𝑟)1 𝑑𝑟 1

0= ∫ (2 − 11𝑟 + 18𝑟2 − 9𝑟3)/2 𝑑𝑟

1

0= | (2𝑟 −

11𝑟2

2+18𝑟3

3−9𝑟4

4) /2|0

1 =1

8.

Evaluating the other three terms gives the array, ∫ 𝑯(𝑟) 𝑑𝑟 1

0= [1/8, 3/8, 3/8, 1/8] 𝑐𝑚 and

the final integral is 𝐼𝑢 = ∫ 𝑢(𝑟) 𝑑𝑥

𝐿𝑒= 𝐿𝑒[1/8, 3/8, 3/8, 1/8] 𝒖𝒆. Now, substituting the given

data 𝒖𝒆𝑇 = [2 3 0 −1] ℃ and multiplying the two arrays the final integral value is

𝐼𝑢 = (6 𝑐𝑚) [1

8,

3

8,

3

8,

1

8] {

2 3 0−1

}℃ =60 ℃ 𝑐𝑚

8= 7.5 ℃ 𝑐𝑚

Example 4.2-3 Given: A two-node line segment from 𝑥1 = 20 m to 𝑥2 = 30 m is subjected to

a force per unit length of p(x) that varies linearly from 𝑝1 = 40 N/m to 𝑝2 = 34 N/m. Determine

the resultant moment of that line load, with respect to the origin at x=0:

𝑀 = ∫ 𝑥 [𝑝(𝑥)𝑑𝑥]𝑥2

𝑥1

Solution: The position and line load can be interpolated with a two-node linear element:

𝑥(𝑟) = 𝑯(𝑟) 𝒙𝒆 = (1 − 𝑟)𝑥1 + 𝑟 𝑥2 = 𝒙𝒆𝑇𝑯(𝑟)𝑇

𝑝(𝑟) = 𝑯(𝑟) 𝒑𝒆 = (1 − 𝑟)𝑝1 + 𝑟

Substitute the interpolations into the moment integral, and pull the node constants outside that

integral and note that for a single element, 𝑑𝑥 𝑑𝑟 =⁄ 𝑥2 − 𝑥1 = 𝐿 = 𝐿𝑒:

𝑀 = ∫ 𝒙𝒆𝑇𝑯(𝑟)𝑇 𝑥2

𝑥1

𝑯(𝑟) 𝒑𝒆 𝒅𝒙 = 𝒙𝒆𝑇 [∫ 𝑯(𝑟)𝑇 1

0

𝑯(𝑟) 𝑑𝑥

𝑑𝑟𝒅𝒓] 𝒑𝒆

𝑀 = 𝒙𝒆𝑇 [∫ {(1 − 𝑟)𝑟

}1

0

[(1 − 𝑟) 𝑟] 𝐿 𝑑𝑟] 𝒑𝒆 = 𝒙𝒆𝑇𝐿𝑒∫ [(1 − 2𝑟 + 𝑟2) (𝑟 − 𝑟2)

(𝑟 − 𝑟2) 𝑟2]

1

0

𝑑𝑟 𝒑𝒆

𝑀 = 𝐿𝑒 𝒙𝒆𝑇1

6[2 11 2

] 𝒑𝒆

This can be applied to any such line segment. The specific moment is:

𝑀 =(𝑥2 − 𝑥1)

6[𝑥1 𝑥2] [

2 11 2

] {𝑝1𝑝2} =

(30 − 20)𝑚

6[20 30]𝑚 [

2 11 2

] {4034}𝑁

𝑚= 9,200 𝑁𝑚


8

Of course, if the interpolations are done using natural coordinate interpolations the exact same

matrix integrals and results are obtained.

Example 4.2-4 Given: Calculate the differential area 𝑑𝐴 = 𝑑𝑥 𝑑𝑦 when converted to polar

coordinates using the parametric transformation:𝑥(𝑅, 𝜃) = 𝑅 cos 𝜃, 𝑦(𝑅, 𝜃) = 𝑅 sin 𝜃. Solution:

For these two parameters the Jacobian is

𝑱 = [𝜕𝑥 𝜕𝑅⁄ 𝜕𝑦 𝜕𝑅⁄

𝜕𝑥 𝜕𝜃⁄ 𝜕𝑦 𝜕𝜃⁄] = [

cos 𝜃 sin 𝜃−𝑅 sin 𝜃 𝑅 cos 𝜃

]

and its determinant is |𝐽| = 𝑅 𝑐𝑜𝑠2𝜃 + 𝑅 𝑠𝑖𝑛2𝜃 = 𝑅. Therefore, the physical differential area in

polar coordinates is 𝑑𝐴 = 𝑑𝑥 𝑑𝑦 = |𝐽(𝑅, 𝜃)| 𝑑𝑅 𝑑𝜃 = 𝑅 𝑑𝑅 𝑑𝜃.

Example 4.2-5 Given: An area of interest is a square with a side length of √2 𝑚 which has been

rotated CCW about the origin by 45 degrees. The location of all points in the region can be

defined parametrically as 𝑥(𝑟, 𝑠) = (𝑟 + 𝑠) and 𝑦(𝑟, 𝑠) = (𝑠 − 𝑟) where r and s are unit

coordinates, 0 ≤ (𝑟, 𝑠) ≤ 1. Determine the Jacobian and its determinant for that transformation

and the physical area of the region. Solution: By definition the Jacobian matrix is

𝑱(𝑟, 𝑠) = [𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄

𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄] = [

1 −11 1

]𝑚, |𝐽 (𝑟, 𝑠)| = (1)(1) − (−1)(1) = 2 𝑚2

The Jacobian is constant because the region in Fig. 4.2-2 has orthogonal straight sides. The area

of the physical region is

𝐴 = ∫ 𝑑𝐴

𝐴= ∫ 𝑑𝑥 𝑑𝑦 = ∫ ∫ |𝐽 (𝑟, 𝑠)|

1

0

1

0

𝐴 𝑑𝑟 𝑑𝑠 = 2 𝑚2 ∫ ∫ 𝑑𝑟 𝑑𝑠 = (

1

0

1

02 𝑚2)(1).

If the interpolations are done using natural coordinate interpolations the exact same matrix

integrals and results are obtained. Such integrals are usually evaluated by numerical integration.

Example 4.2-6 Given: A quadrilateral area in physical space is defined in ‘natural coordinate’

parametric space by the mapping 𝑥(𝑎, 𝑏) = (−3 + 27 𝑎 − 27 𝑏 + 3 𝑎 𝑏) and 𝑦 (𝑎, 𝑏) =(51 + 31 𝑎 + 19 𝑏 − 𝑎 𝑏) where −1 ≤ 𝑎, 𝑏 ≤ +1. Verify that the variable Jacobian matrix of

the mapping is

𝐽(𝑎, 𝑏) = [(27 + 3 𝑏) (31 − 𝑏)(−27 + 3 𝑎) (19 − 𝑎)

],

and determine its determinant. Solution: Calculating the partial derivatives with respect to a:

𝜕𝑥 𝜕𝑎⁄ = (0 + 27 − 0 + 3 𝑏) and 𝜕𝑦 𝜕𝑎⁄ = (0 + 31 + 0 − 𝑏). These are the terms in the first

row of the Jacobian matrix. Repeating the process foe derivatives with respect to b gives the

cited matrix:

𝑱(𝑎, 𝑏) = [𝜕𝑥 𝜕𝑎⁄ 𝜕𝑦 𝜕𝑎⁄

𝜕𝑥 𝜕𝑏⁄ 𝜕𝑦 𝜕𝑏⁄] = [

(27 + 3 𝑏) (31 − 𝑏)(−27 + 3 𝑎) (19 − 𝑎)

].


9

Thus, |𝐽(𝑎, 𝑏)| = (27 + 3 𝑏)(19 − 𝑎) − (31 − 𝑏)(−27 + 3 𝑎) = 1,350 − 120 𝑎 + 30 𝑏. The

determinant is not constant, but it is positive for the given range of (a, b) values. Since the

determinant is positive, |𝐽(𝑎, 𝑏)| > 0, this mapping is called invertible. Valid finite element

shapes should be invertible.

Example 4.2-7 Given: The natural coordinate square −1 ≤ 𝑎, 𝑏 ≤ +1 is mapped into a

physical quadrilateral by 𝑥(𝑎, 𝑏) = (−3 + 27 𝑎 − 27 𝑏 + 3 𝑎 𝑏) and 𝑦 (𝑎, 𝑏) = (51 + 31 𝑎 +19 𝑏 − 𝑎 𝑏), in meters. Evaluate the mapping at the four parametric corners to find the physical

coordinates of the corners of the quadrilateral. Use the mapping to determine the physical area of

the quadrilateral. Solution: Directly substituting a variable b at 𝑎 = ±1 gives two sides

connecting the physical coordinates of the vertices. Substituting a variable a at 𝑏 = ±1

completes the figure below:

The area is given by the integral 𝐴 = ∫ 𝑑𝐴 = ∫ ∫ |𝐽(𝑎, 𝑏)| 𝑑𝑎 𝑑𝑏1

−1

1

−1

𝐴. The determinant for this

area, with the units 𝑚2, was developed in Ex. 4.2-4. Substituting gives

𝐴 = ∫ 𝑑𝐴 = ∫ ∫ |𝐽(𝑎, 𝑏)| 𝑑𝑎 𝑑𝑏1

−1

1

−1

= 𝑚2∫ ∫ (1,350 − 120 𝑎 + 30 𝑏) 𝑑𝑎 𝑑𝑏1

−1

1

−1

𝐴

= 5,400 𝑚2

4.2.1 Linear triangle geometry: The three-node linear triangle is the simplest two-dimensional

element and is probably the most common element presented in the finite element literature. It

always has a constant Jacobian. As shown in Fig. 4.2-1 the two-dimensional Jacobian matrix

depends on the nodal coordinates of the triangle:

𝑱𝒆(𝑟, 𝑠) = [𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄

𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄]𝑒

The corner locations are counted counter-clockwise, starting from ant vertex. Since the two

coordinates of a physical location of any point are found by interpolation:

[𝑥(𝑟, 𝑠) 𝑦(𝑟, 𝑠)] = 𝑯(𝑟, 𝑠)𝑒[𝒙𝒆 ⋮ 𝒚𝒆]

or

a b x (a, b) y (a, b)

-1 -1 0 0

-1 +1 48 64

+1 +1 0 100

+1 -1 -60 40


10

[𝑥(𝑟, 𝑠) 𝑦(𝑟, 𝑠)] = [(1 − 𝑟 − 𝑠) 𝑟 𝑠] [

𝑥1 𝑦1𝑥2 𝑦2𝑥3 𝑦3

]

𝑒

and the first row of the Jacobian matrix is

[𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄ ] = [(−1) 1 0] [


]

𝑒

= [(𝑥2 − 𝑥1) (𝑦2 − 𝑦1)]𝑒

while the second row is

[𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄ ] = [(−1) 0 1] [


]

𝑒

= [(𝑥3 − 𝑥1) (𝑦3 − 𝑦1)]𝑒.

Of course, the numerical value of the Jacobian for a straight sided triangle can be obtained by

simple matrix multiplication of the two constant matrices:

𝑱𝒆 = [𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄

𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄]𝑒

= [−1 1 0−1 0 1

] [


]

𝑒

= [(𝑥2 − 𝑥1) (𝑦2 − 𝑦1)

(𝑥3 − 𝑥1) (𝑦3 − 𝑦1)]𝑒

. (4.2-6)

Its determinant is also constant with a value of twice the physical area of the triangle:

|𝑱𝒆| = (𝑥2 − 𝑥1)𝑒(𝑦3 − 𝑦1)

𝑒 − (𝑥3 − 𝑥1)𝑒(𝑦2 − 𝑦1)

𝑒 = 2𝐴𝑒. (4.2-7)

Since the straight side triangle has a constant 2 × 2 matrix its inverse matrix can be written in

closed form:

𝑱𝒆−1 =1

2𝐴𝑒[(𝑦3 − 𝑦1) (𝑦1 − 𝑦2)

(𝑥1 − 𝑥3) (𝑥2 − 𝑥1)]𝑒

. (4.2-8)

The differences in corner coordinates occur so often in the literature they have been assigned

shorthand notations using subscripts i, j, k that follow a cyclic permutation through 123: 𝑖𝑗𝑘 →123123 so the nine element shape constants are

𝑎𝑖𝑒 = 𝑥𝑗

𝑒𝑘 − 𝑥𝑘𝑒𝑦𝑗

𝑒 , 𝑏𝑖𝑒 = 𝑦𝑗

𝑒 − 𝑦𝑘𝑒 , 𝑐𝑖

𝑒 = 𝑥𝑘𝑒 − 𝑥𝑗

𝑒 (4.2-9)

which give an alternate form for the area as 2𝐴𝑒 = 𝑎1𝑒 + 𝑎2

𝑒 + 𝑎3𝑒.

4.2.2 Linear tetrahedron: The simplest solid element is the four node linear tetrahedron

(pyramid). That element has straight edges and flat faces and clearly has a constant Jacobian

matrix. The volume of any straight sided tetrahedron is one-third the base area times the height.

In unit coordinates, the unit triangle base area is 1 2⁄ and the parametric tetrahedron has a unit

height, therefore its parametric measure (the volume in parametric space) is: |⊡| =(1 3⁄ )(1 2⁄ )1 = 1 6⁄ , so it will yield the constant determinant of the Jacobian as |𝐉𝐞| =|Ω| |⊡|⁄ = 𝑉𝑒 (1 6⁄ ) = 6𝑉𝑒⁄ , where 𝑉𝑒 is the volume of the tetrahedron in physical space. A

process similar to that just applied to the straight triangle gives the constant Jacobian of any

straight edged tetrahedron:


11

𝑱𝒆 = [

(𝑥2 − 𝑥1) (𝑦2 − 𝑦1) (𝑧2 − 𝑧1)

(𝑥3 − 𝑥1) (𝑦3 − 𝑦1) (𝑧3 − 𝑧1)

(𝑥4 − 𝑥1) (𝑦4 − 𝑦1) (𝑧4 − 𝑧1)]

𝑒

,

and its determinant is six times the physical volume: |𝑱𝒆| = 6𝑉𝑒.

4.3 Inverse Jacobian: The inverse of the Jacobian matrix is needed when the physical

derivatives must be obtained from known parametric derivatives:

𝝏𝛀( ) = 𝑱(⊡)−𝟏 𝝏⊡( ),

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦

𝜕𝑢

𝜕𝑧}

=

[ 𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝑠

𝜕𝑦

𝜕𝑠

𝜕𝑧

𝜕𝑠𝜕𝑥

𝜕𝑡

𝜕𝑦

𝜕𝑡

𝜕𝑧

𝜕𝑡] −1

{

𝜕𝑢

𝜕𝑟𝜕𝑢

𝜕𝑠𝜕𝑢

𝜕𝑡}

. (4.3-1)

In other words, (4.3-1) is the inverse of (4.2-1). In the one-dimensional case it simplifies to

𝜕

𝜕𝑥=

1

𝜕𝑥(𝑟) 𝜕𝑟⁄ 𝜕

𝜕𝑟=

𝜕

𝜕𝑟 𝜕𝑟

𝜕𝑥.

For line elements with equally spaced physical nodes the Jacobian is a constant proportional

to the physical length of the element. A unit-coordinate element with equally spaced nodes has a

Jacobian of 𝜕𝑥 𝜕𝑟⁄ = 𝐿𝑒 1⁄ . In such an element the physical derivative is

𝜕

𝜕𝑥=

1

𝐿𝑒 1⁄ 𝜕

𝜕𝑟=

𝜕

𝜕𝑟 1

𝐿𝑒,

which should have been expected since every derivative with respect to x must introduce a term

having the units of length. Likewise, using the popular natural coordinates form, with equal node

spacing, 𝜕

𝜕𝑥=

1

𝐿𝑒 2⁄ 𝜕

𝜕𝑎=

2

𝐿𝑒 𝜕

𝜕𝑎 .

For future analytic examples, in two-dimensions the inverse Jacobian matrix and its determinant

are

𝑱−𝟏 =1

|𝑱|[ 𝜕𝑦

𝜕𝑠−𝜕𝑦

𝜕𝑟

−𝜕𝑥

𝜕𝑠

𝜕𝑥

𝜕𝑟 ] , |𝑱| = (

𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑠−𝜕𝑦

𝜕𝑟

𝜕𝑥

𝜕𝑠). (4.3-2)

The inverse of the three-dimensional Jacobian is also known analytically, but in practice it is

usually inverted numerically.

EXAMPLES

4.4 Parametric substitution in integrals: To be able to treat curvilinear regions with the

finite element method, it is necessary to define the physical space in terms of non-dimensional

parametric coordinates. Therefore, the original governing integrals must employ a change of


12

variable of integration and carry out the integrals in parametric space. Let x be the original

independent spatial variable and replace it with 𝒙 = 𝒈(𝒖) where u is the choice for a new

independent variable. Such a change is called a transformation or mapping of x, and 𝒇(𝒙) =𝒇{𝒈(𝒖)} is called a composite function. Then, it can be shown that

∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎= ∫ 𝑓{𝑔(𝑢)}

𝛽

𝛼

𝜕𝑔

𝜕𝑢𝑑𝑢 = ∫ 𝑓{𝑔(𝑢)}

𝛽

𝛼|𝐽(𝑢)| 𝑑𝑢 (4.4-1)

Here 𝛽 is the number giving the upper limit and 𝛼 is the lower limit of the parametric space. For

a linear transform x a general parametric form is 𝑥 = 𝑔(𝑢) = [𝛽𝑎 − 𝛼𝑏 + (𝑏 − 𝑎)𝑢] (𝛽 − 𝛼)⁄ .

Evaluating this mapping at the lower limit 𝑢 = 𝛼 gives

𝑥 = 𝑔(𝛼) = [𝛽𝑎 − 𝛼𝑏 + (𝑏 − 𝑎)𝛼]

(𝛽 − 𝛼)=[𝛽𝑎 − 𝛼𝑏 + (𝑏𝛼 − 𝑎𝛼)]

(𝛽 − 𝛼)=𝑎(𝛽 − 𝛼)

(𝛽 − 𝛼)= 𝑎

Likewise at the upper limit: 𝑔(𝑢 = 𝛽) = 𝑏. Unit parametric coordinates are defined by 𝛽 = 1

and 𝛼 = 0 while the natural parametric coordinates are defined by 𝛽 = 1 and 𝛼 = −1. The

Jacobian of this transformation is 𝐽(𝑢) = 𝜕𝑔 𝜕𝑢 =⁄ (𝑏 − 𝑎) (𝛽 − 𝛼) ⁄ = |𝐽(𝑢)|. In FEA, x has physical units and u is a non-dimensional parametric coordinate. Of course, the

change of variables does not change the units of the original integral. Let Ω be the original

physical domain and let ⊡ be a parametric space and define their generalized volumes as

|Ω| = ∫ 𝑑𝑥𝑏

𝑎= (𝑏 − 𝑎) ≡ 𝐿 𝑎𝑛𝑑 | ⊡ | = ∫ 𝑑𝑢

𝛽

𝛼= (𝛽 − 𝛼) ≡ 𝜆.

Here, x is typically a physical coordinate and (𝑏 − 𝑎) = 𝐿 would be a physical length while. The

parametric measure (𝛽 − 𝛼) = 𝜆 is just a non-dimensional number which is either 1 or 2 for unit

or natural coordinates, respectively.

In other words, when the one-dimensional geometry mapping is linear then the Jacobian will

be a constant. Two examples are the unit-coordinate, 0 ≤ 𝑟 ≤ 1, form: 𝑥(𝑟) = 𝑥1𝑒 + 𝑟 𝐿𝑒 so the

Jacobian matrix is 𝑑𝑥 𝑑𝑟 = 𝐿𝑒 1⁄⁄ , and the natural-coordinate form, −1 ≤ 𝑎 ≤ 1, 𝑥(𝑎) = 𝑥1𝑒 +

(𝑎 + 1)𝐿𝑒/2 , which gives the Jacobian as 𝑑𝑥 𝑑𝑎⁄ = 𝐿𝑒/2.

Likewise, a typical change of variable of integration in two-dimensions becomes

∬ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑑𝑦

Ω= ∬ 𝑓[𝑥(𝑟, 𝑠), 𝑦(𝑟, 𝑠)]| 𝐽(𝑟, 𝑠)|

⊡𝑑𝑟 𝑑𝑠. (4.4-2)

In FEA, each element usually has a different Jacobian matrix. In other words, x(r) depends

on the control point values 𝑥1…𝑥𝑛 on a particular element. Therefore, in FEA the Jacobian

matrix is always element dependent and it is denoted here as 𝑱𝒆. Likewise, its determinant and

inverse are denoted in a similar fashion as |𝑱𝒆| and 𝑱𝒆−𝟏. Only on regular uniform meshes will

the Jacobian be the same for every element.

Example 4.4-1 Given: Ex. 4.2-3 describes a square rotated 45 degrees. Evaluate the polar-

moment-of-inertia ∫ (𝑥2 + 𝑦2)

𝐴𝑑𝑥 𝑑𝑦 of the square using the parametric substitution 𝑥(𝑟, 𝑠) =

(𝑟 + 𝑠) 𝑚 and 𝑦(𝑟, 𝑠) = (𝑠 − 𝑟) 𝑚, for the unit coordinates 0 ≤ (𝑟, 𝑠) ≤ 1. Solution: In Ex. 4.2-

3 the determinant of the Jacobian of that mapping is shown to be a constant |𝐽| = 2𝑚2. The

transformation of the integral is


13

∫(𝑥2 + 𝑦2)

𝐴

𝑑𝑥 𝑑𝑦 = ∫ ∫ [𝑥(𝑟, 𝑠)2 + 𝑦(𝑟, 𝑠)2]1

0

1

0

|𝐽(𝑟, 𝑠)| 𝑑𝑟 𝑑𝑠

= ∫ ∫ [(𝑟 + 𝑠)2𝑚2 + (𝑠 − 𝑟)2𝑚2]1

0

1

0

2𝑚2 𝑑𝑟 𝑑𝑠

= 2𝑚4∫ ∫ [2𝑟2 + 2𝑠2]1

0

1

0

𝑑𝑟 𝑑𝑠 = 2𝑚4∫ [2

3𝑟3 + 2𝑠2𝑟]

1

0

1 0𝑑𝑠 = 2𝑚4∫ [

2

3+ 2𝑠2]

1

0

𝑑𝑠

∫(𝑥2 + 𝑦2)

𝐴

𝑑𝑥 𝑑𝑦 = 2𝑚4 [2

3𝑠 +

2

3𝑠3]

1 0= 2𝑚4 [

4

3] =

8

3𝑚4

4.5 Integration by parts: The use of the popular Galerkin’s method of finite element analysis

generally requires the use of integration by parts, or Green’s Theorem in two- and three-

dimensional space in order to reduce the continuity requirements of the weak solution and to

render even order equations to a symmetric integral form. In one-dimension the most commonly

used relations are

∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥𝑏

𝑎= 𝑓(𝑥)𝑔(𝑥)|𝑎

𝑏 − ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥𝑏

𝑎

∫ 𝑓(𝑥)𝑔′′(𝑥)𝑑𝑥𝑏

𝑎= 𝑓(𝑥)𝑔′(𝑥)|𝑎

𝑏 − ∫ 𝑓′(𝑥)𝑔′(𝑥)𝑑𝑥𝑏

𝑎. (4.5-1)

For a volumetric region, 𝑉, bounded by the surface, 𝑆, Green’s Theorem gives the following

commonly used relations:

∭ 𝐴 ∇2𝐵

𝑉 𝑑𝑉 = ∬ 𝐴 ∇⃗⃗ 𝐵 ∙ �⃗�

𝑆𝑑𝑆 −∭ ∇⃗⃗ 𝐴 ∙ ∇⃗⃗ 𝐵 𝑑𝑉

𝑉= ∬ 𝐴

𝜕𝐵

𝜕𝑛

𝑆𝑑𝑆 −∭ ∇⃗⃗ 𝐴 ∙ ∇⃗⃗ 𝐵 𝑑𝑉

𝑉 (4.5-2)

∭ (𝐴 ∇2𝐵 − 𝐵 ∇2𝐴 )𝑑𝑉

𝑉= ∬ (𝐴

𝜕𝐵

𝜕𝑛− 𝐵

𝜕𝐴

𝜕𝑛) 𝑑𝑆

𝑆.

Example 4.5-1 Given: Evaluate the following integral using integration by parts

𝐼 = ∫𝑢(𝑥) [𝑑2𝑢

𝑑𝑥2+ 𝑐𝑥]

𝐿

0

𝑑𝑥 = 0

Solution: The first term can be integrated by parts

∫𝑢(𝑥) [𝑑2𝑢

𝑑𝑥2]

𝐿

0

𝑑𝑥 = [𝑢 𝑑𝑢

𝑑𝑥]0

𝐿

−∫𝑑𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥𝑑𝑥

𝐿

0

so an alternate form of the integral formulation is

𝐼 = [𝑢 𝑑𝑢

𝑑𝑥]0

𝐿− ∫

𝑑𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥𝑑𝑥 + ∫ 𝑢(𝑥) 𝑐𝑥 𝑑𝑥

𝐿

0

𝐿

0= 0.


14

Note that the first term introduces the derivative of the solution normal to the boundary.

Often such derivatives appear as nonessential boundary conditions. It also lowered the highest

derivative in the integrand from the second to the first.

Example 4.5-2 Given: The Poisson equation, ∇2𝑢 (𝑥, 𝑦) + 𝑄(𝑥, 𝑦) = 0 occurs commonly in

engineering and physics applications. The Galerkin weighted residual method approximation

begins with the integral ∫ 𝑢(𝑥, 𝑦)[∇2𝑢 (𝑥, 𝑦) + 𝑄(𝑥, 𝑦)]𝑑Ω = 0

Ω. By using Green’s Theorem

reduce the second derivatives in the first integral. Solution: The first form of the theorem is

∫𝑢(𝑥, 𝑦)

Ω

∇2𝑢 (𝑥, 𝑦) 𝑑Ω = ∫𝑢(𝑥, 𝑦)

Γ

𝜕 𝑢(𝑥, 𝑦)

𝜕𝑛 𝑑Γ − ∫ ∇⃗⃗ 𝑢 ∙ ∇⃗⃗ 𝑢

Ω

𝑑Ω

= ∫𝑢(𝑥, 𝑦)

Γ

𝜕 𝑢(𝑥, 𝑦)

𝜕𝑛 𝑑Γ −∫ [

𝜕𝑢

𝜕𝑥

𝜕𝑢

𝜕𝑥+𝜕𝑢

𝜕𝑦

𝜕𝑢

𝜕𝑦] 𝑑Ω

Ω

Note that this process automatically brings in the derivative normal to the boundary. Those

terms are often specified in the secondary boundary conditions. It also lowered the highest

derivative in the integrand from the second to the first.

4.6 Integral of point sources*: Wherever there is a discontinuity in the coefficients

(material properties) in the differential equations calculus requires the placement of an element

interface at (along) that location in the mesh. It is possible to violate that guideline for point

sources added to the interior of an existing element. However, that renders the post-processing

data from such an element quite inaccurate.

To add a point source to an existing element requires starting with the definition of the

consistent resultant matrix (derived later) due to a distributed source per unit length, 𝑤(𝑥):

𝑪𝒆 = ∫ 𝑯(𝑟)𝑇𝑤(𝑥) 𝑑𝑥

𝐿𝑒

A point source (load, or sink) at point 𝑥𝑘 is mathematically defined using the Dirac Delta

Distribution, 𝛿(𝑥 − 𝑥𝑘), as 𝑤(𝑥) = 𝑃𝑘 𝛿(𝑥 − 𝑥𝑘). The distribution is defined as

𝛿(𝑥 − 𝑥𝑘) = {

∞, 𝑥 = 𝑥𝑘

0, 𝑥 ≠ 𝑥𝑘

and it has the integral property of ∫ 𝐹(𝑥) 𝛿(𝑥 − 𝑥𝑘) 𝑑𝑥 = 𝐹(𝑥𝑘)∞

−∞. Then the source vector

becomes

𝑪𝒆 = ∫ 𝑯(𝑟)𝑇𝑃𝑘 𝛿(𝑥 − 𝑥𝑘) 𝑑𝑥

𝐿𝑒= 𝑃𝑘 𝑯(𝑟𝑘)

𝑇 (4.6-1)

where 𝑟𝑘 is the parametric coordinate of the physical location of the point source. In other words,

an internal source would just fill the load vector with the non-dimensional interpolations

functions at the point times the magnitude of the source. It is best to avoid internal point sources

and have the mesh generator place a node at each point source. Then, the full value of its source

is assigned to that node.


15

Example 4.6-1 Given: A point load of P=12 N is applied interior to a two-node linear element at

location 0.6 𝐿𝑒. Compute the corresponding element load vector. Solution: Clearly 𝑟𝑘 = 0.6,

and for this element 𝑯(𝑟) = [(1 − 𝑟) 𝑟], therefore the resultant consistent element nodal forces

are

𝑪𝒆 = 𝑃 {1 − 0.60.6

} = 𝑃 {0.40.6} = {

4.87.2}𝑁

Example 4.6-2 Given: A point load of P=12 N is applied to the third (right) node of a quadratic

line element. Compute the corresponding element load vector. Solution: Clearly 𝑟𝑘 = 1, and for

this element 𝑯(𝑟) = [(1 − 3𝑟 + 2𝑟2) (4𝑟 − 4𝑟2) (−𝑟 + 2𝑟2)], therefore the resultant

consistent element nodal forces are

𝑪𝒆 = 𝑃𝑘 𝑯(𝑟𝑘)𝑇 = 𝑃 {

1 − 3 + 24 − 4−1 + 2

} = 𝑃 {001} = {

0012}𝑁

This reinforces the concept that external point sources at a node a just directly added to that

row in the system source vector, C.

Example 4.6-3 Given: A point mass of m = 12 kg is applied to the first (left) node a two-node

linear element. Compute the corresponding element consistent mass matrix. Solution: Clearly

𝑟𝑘 = 0, and for this element 𝑯(𝑟) = [(1 − 𝑟) 𝑟], therefore the resultant consistent element mass

matrix (derived later) is

𝑴𝒆 = ∫ 𝑯(𝑟)𝑇𝑚(𝑥) 𝑯(𝑟)𝑑𝑥

𝐿𝑒= 𝑚𝑘 ∫ 𝑯(𝑟)𝑇𝛿(𝑥 − 𝑥𝑘) 𝑯(𝑟)𝑑𝑥

𝐿𝑒

𝑴𝒆 ≡ 𝑚𝑘𝑯(𝑟𝑘)𝑇 𝑯(𝑟𝑘) = 𝑚𝑘 {

1 − 00

} [(1 − 0) 0] = 𝑚𝑘 [1 00 0

]

Therefore, the point mass at any node is simply added to the diagonal of the corresponding DOF

in the system mass matrix, M.

4.7 Axisymmetric integrals*: Many finite element applications involve volumes and surfaces

of revolution. Those problems are formulated in the radial and axial dimensions (that is the r-z

plane), as a curve or closed area to be revolved about the z-axis. They are treated by the Theorem

of Pappus. For an angle of revolution of 𝟎 < 𝜽 ≤ 𝟐𝝅 the volume and surface area (excluding

end faces when 𝜽 ≠ 𝟐𝝅) are

𝑆 = ∫ 𝑑𝐴

𝑆= Θ∫ 𝑅 𝑑𝐿 = Θ �̅�

𝐿𝐿, 𝑉 = ∫ 𝑑𝑉

𝑉= Θ∫ 𝑅 𝑑𝐴 = Θ �̅�

𝐴𝐴 (4.7-1)

where R is the radial coordinate and 𝑅 ̅is the centroid radius to the shape in the r-z plane.


16

Figure 4.7-1 Planar curves revolve to form partial or complete surfaces of revolution

Figure 4.7-2 Planar areas revolve to form partial or complete solids of revolution

For a complete revolution of an area the volume becomes 𝑉 = 2𝜋 𝑅 ̅𝐴. Of course, in finite

element analysis the radial coordinate is interpolated from the radial coordinates of the nodes of

the element (as is the axial coordinate): 𝑅(𝑟, 𝑠) = 𝑯(𝑟, 𝑠) 𝑹𝒆 and the volume is

𝑉 = 2𝜋 ∫ 𝑅(𝑟, 𝑠) 𝑑𝐴

𝐴= 2𝜋 ∫ 𝑅(𝑟, 𝑠) |𝐽𝑒(𝑟, 𝑠)| 𝑑 ⊡

⊡

𝑉 = 2𝜋 ∫ 𝑯(𝑟, 𝑠) |𝐽𝑒(𝑟, 𝑠)| 𝑑 ⊡

⊡𝑹𝒆 . (4.7-2)


17

4.8 Summary 𝑛𝑏 ≡ Number of boundary segments 𝑛𝑑 ≡ Number of system unknowns

𝑛𝑒 ≡ Number of elements 𝑛𝑔 ≡ Number of unknowns per node

𝑛𝑖 ≡ Number of unknowns per element 𝑛𝑚 ≡ Number of mesh nodes

𝑛𝑛 ≡ Number of nodes per element 𝑛𝑝 ≡ Dimension of parametric space

𝑛𝑞 ≡ Number of total quadrature points 𝑛𝑠 ≡ Dimension of physical space

b = boundary segment number e = element number

⊂ = subset of a larger set ∪ = union of two sets

Boundary, element, and system unknowns: 𝜹𝒃 ⊂𝑏 𝜹𝒆 ⊂𝑒 𝜹

Boolean extraction arrays: 𝜹𝒃 ≡ 𝜷𝒃 𝜹, 𝜹𝒆 ≡ 𝜷𝒆 𝜹

Geometry: Γ𝑏 ⊂ Ω𝑒 ≡ Boundary segment Ω𝑒 ≡ Element domain

Ω = ∪𝒆 Ω𝑒 ≡ Solution domain Γ = ∪𝒃 Γ

𝑏 ≡ Domain boundary

Element interpolations utilized in this chapter

Element

Type

Example

Number

Polynomial

Degree

𝑛𝑝

𝑛𝑔

𝑛𝑛

𝑛𝑖 =

𝑛𝑔 × 𝑛𝑛

Continuity

Level

Lagrange L2 4.2-5, 4.6-1,3 1 1 1 2 2 𝐶0

Lagrange L3 4.1-4,4.6-2 2 1 1 3 3 𝐶0

Lagrange L4 4.2-4 3 1 1 4 4 𝐶0

Lagrange T3 4.2-1 1 2 1 3 3 𝐶0

Lagrange Q4 4.2-6 1 2 1 4 4 𝐶0

Matrix multiplication: 𝑪 = 𝑨 𝑩 ≠ 𝑩 𝑨, 𝐶𝑖𝑗 ≡ ∑ 𝐴𝑖𝑘𝐵𝑘𝑗 𝑘

Transpose of a product: (𝑨 𝑩 𝑪)𝑇 = 𝑪𝑇 𝑩𝑇 𝑨𝑇

Common constants for straight side triangles:

𝑖𝑗𝑘 → 12312, cyclic permutations

𝑎𝑖𝑒 = 𝑥𝑗

𝑒𝑘 − 𝑥𝑘𝑒𝑦𝑗

𝑒 , 𝑏𝑖𝑒 = 𝑦𝑗

𝑒 − 𝑦𝑘𝑒 , 𝑐𝑖

𝑒 = 𝑥𝑘𝑒 − 𝑥𝑗

𝑒,

2𝐴𝑒 = 𝑎1𝑒 + 𝑎2

𝑒 + 𝑎3𝑒.

Vector cross product: 𝒄 = 𝒂 × 𝒃

𝑐𝑖 = 𝑎𝑗𝑏𝑘 − 𝑎𝑘𝑏𝑗 , 𝑖𝑗𝑘 → 12312, cyclic permutations

Jacobian matrix:

{

𝜕𝑢

𝜕𝑟𝜕𝑢

𝜕𝑠𝜕𝑢

𝜕𝑡}

=

[ 𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝑠

𝜕𝑦

𝜕𝑠

𝜕𝑧

𝜕𝑠𝜕𝑥

𝜕𝑡

𝜕𝑦

𝜕𝑡

𝜕𝑧

𝜕𝑡]

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦

𝜕𝑢

𝜕𝑧}

≡ [𝑱(𝑟, 𝑠, 𝑡)]

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦

𝜕𝑢

𝜕𝑧}


18

Calculation of element Jacobian: 𝑱𝒆(𝑟) =𝜕𝑯(𝒓 )

𝜕𝑟𝒙𝒆

Jacobian determinant: | 𝑱 | ≡ |𝜕(𝑥,𝑦,𝑧)

𝜕(𝑟,𝑠,𝑡)|

𝐽 = | 𝑱 | =|

|

𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝑠

𝜕𝑦

𝜕𝑠

𝜕𝑧

𝜕𝑠𝜕𝑥

𝜕𝑡

𝜕𝑦

𝜕𝑡

𝜕𝑧

𝜕𝑡

|

|

Differential volume Relations: 𝑑Ω = |𝐽(⊡)| 𝑑 ⊡

𝑑𝑥 = |𝐽(𝑟) |𝑑𝑟 = |𝜕𝑥(𝑟)

𝜕𝑟| 𝑑𝑟

𝑑𝐴 = 𝑑𝑥 𝑑𝑦 = |𝐽(𝑟, 𝑠)| 𝑑𝑟 𝑑𝑠 = |𝜕𝑥

𝜕𝑟 𝜕𝑦(𝑟,𝑠)

𝜕𝑠−𝜕𝑦

𝜕𝑟 𝜕𝑥(𝑟,𝑠)

𝜕𝑠| 𝑑𝑟 𝑑𝑠

𝑑𝑉 = 𝑑𝑥 𝑑𝑦 𝑑𝑧 = |𝐽(𝑟, 𝑠, 𝑡)| 𝑑𝑟 𝑑𝑠 𝑑𝑡

Tangents to a surface: 𝑻𝒓 =𝜕𝑹

𝜕𝑟=

𝜕𝑥

𝜕𝑟�̂� +

𝜕𝑦

𝜕𝑟𝒋̂ +

𝜕𝑧

𝜕𝑟 �̂� , 𝑻𝒔 =

𝜕𝑹

𝜕𝑠=

𝜕𝑥

𝜕𝑠�̂� +

𝜕𝑦

𝜕𝑠𝒋̂ +

𝜕𝑧

𝜕𝑠 �̂�

[𝑻𝑟𝑻𝑠] = [

𝜕𝑯𝑏 𝜕𝑟⁄

𝜕𝑯𝑏 𝜕𝑠⁄] [𝒙𝑏 | 𝒚𝑏 | 𝒛𝑏]

Surface area normal: 𝑑𝑆𝑏𝒏(𝑟, 𝑠) = 𝑻𝑟(𝑟, 𝑠) × 𝑻𝑠(𝑟, 𝑠) 𝑑𝑟 𝑑𝑠

Non-flat surface area: 𝑑𝑆𝑏 = ‖𝑻𝑟(𝑟, 𝑠) × 𝑻𝑠(𝑟, 𝑠)‖ 𝑑𝑟 𝑑𝑠

𝑑𝑆𝑏 = √(𝑥, 𝑟 𝑦, 𝑠 − 𝑦, 𝑟 𝑥, 𝑠)2 + (𝑦, 𝑟 𝑧, 𝑠 − 𝑧, 𝑟 𝑦, 𝑠)2 + (𝑧, 𝑟 𝑥, 𝑠 − 𝑥, 𝑟 𝑧, 𝑠)2 𝑑𝑟 𝑑𝑠

Inverse Jacobian matrix: 𝝏𝛀( ) = 𝑱(⊡)−𝟏 𝝏⊡( )

{

𝜕𝑢

𝜕𝑥𝜕𝑢

𝜕𝑦𝜕𝑢

𝜕𝑧}

=

[ 𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑟

𝜕𝑧

𝜕𝑟𝜕𝑥

𝜕𝑠

𝜕𝑦

𝜕𝑠

𝜕𝑧

𝜕𝑠𝜕𝑥

𝜕𝑡

𝜕𝑦

𝜕𝑡

𝜕𝑧

𝜕𝑡] −1

{

𝜕𝑢

𝜕𝑟𝜕𝑢

𝜕𝑠𝜕𝑢

𝜕𝑡}

𝑱−𝟏 =1

|𝑱|[ 𝜕𝑦

𝜕𝑠−𝜕𝑦

𝜕𝑟

−𝜕𝑥

𝜕𝑠 𝜕𝑥

𝜕𝑟

] , |𝑱| = (𝜕𝑥

𝜕𝑟

𝜕𝑦

𝜕𝑠−𝜕𝑦

𝜕𝑟

𝜕𝑥

𝜕𝑠)

Integral transformations (2D): ∬ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑑𝑦

Ω= ∬ 𝑓[𝑥(𝑟, 𝑠), 𝑦(𝑟, 𝑠)]| 𝐽(𝑟, 𝑠)|

⊡𝑑𝑟 𝑑𝑠


19

Green’s theorem:

∭𝐴 ∇2𝐵

𝑉

𝑑𝑉 = ∬𝐴 ∇⃗⃗ 𝐵 ∙ �⃗�

𝑆

𝑑𝑆 −∭∇⃗⃗ 𝐴 ∙ ∇⃗⃗ 𝐵 𝑑𝑉

𝑉

=∬ 𝐴𝜕𝐵

𝜕𝑛

𝑆

𝑑𝑆 −∭∇⃗⃗ 𝐴 ∙ ∇⃗⃗ 𝐵 𝑑𝑉

𝑉

Volume of revolution: 𝑉 = 2𝜋 ∫ 𝑅(𝑟, 𝑠) 𝑑𝐴

𝐴= 2𝜋 ∫ 𝑯(𝑟, 𝑠) |𝐽𝑒(𝑟, 𝑠)| 𝑑 ⊡

⊡𝑹𝒆

4.9 List of Examples Ex 4.1-1 Proof that global x-y interpolation fails for quadrilaterals Ex 4.1-2 Proof that local r-s interpolation works for quadrilaterals Ex 4.1-3 Shape of any edge of a four-node quadrilateral, Q4 Ex 4.1-4 Shape of any edge of an eight-node quadrilateral, Q8 or Q9 Ex 4.2-1 Variable Jacobian of a line element with unequal spacing, L3 Ex 4.2-2 Physical line integral of interpolated quantity, L4 Ex 4.2-3 Moment integral of interpolated line force, L2 Ex 4.2-4 Jacobian matrix in cylindrical coordinates Ex 4.2-5 Area of a rotated square by integration Ex 4.2-6 Jacobian matrix of a quadrilateral Ex 4.2-7 Area calculation for the same quadrilateral Ex 4.5-1 Integration by parts of an ODE over a line Ex 4.5-2 Integration by parts of a PDE over a volume (Green’s Theorem) Ex 4.6-1 Resultant of a point load interior to a line element, L2 Ex 4.6-2 Resultant of a point load at a line element node point, L3 Ex 4.6-3 Mass matrix of a point mass at a line element node point, L2


20

Table 4.1 Interpolation column and symmetric matrix integrals for constant Jacobian line elements:

Type ∫ 𝑯𝑻

Ω𝑑Ω ∫ 𝑯𝑻𝑯

Ω𝑑Ω ∫ �⃗⃗� 𝑯𝑻 �⃗⃗� 𝑯

Ω𝑑Ω

L2C0 𝐿𝑒

2{11}

𝐿𝑒

6[2 11 2

] 1

𝐿𝑒[ 1 −1−1 1

]

L3C0 𝐿𝑒

6{141}

𝐿𝑒

30[ 4 2 −1 2 16 2−1 2 4

] 1

3𝐿𝑒[ 7 −8 1−8 16 −8 1 −8 7

]

L4C0 𝐿𝑒

8{

1331

} 𝐿𝑒

1680[

128 99 99 648

−36 19−81 −36

−36 −81 19 −36

648 99 99 128

] 1

40𝐿𝑒[

148 −189−189 432

54 −13−297 54

54 −297 −13 54

432 −189 −189 148

]

Notes: Line elements have a constant Jacobian only if the physical nodes are equally spaced, like the parametric nodes, along a straight physical line. Curved line elements do not have a constant Jacobian and require numerical integration. If an integral includes constant properties, then the element matrices are simply obtained by multiplying the above integrals by that property. For the first two integrals the sum of all the terms in the brackets is one. In the last integral the sum of all the terms in brackets is zero.

Table 4.2 Unsymmetric constant Jacobian line element integrals: 𝑼 = ∫ 𝑯𝑻

Ω�⃗⃗� 𝑯𝑑Ω:

Type L2C0 L3C0 L4C0

Matrix 𝑼 =1

2[−1 1−1 1

] 𝑼 =1

6[−3 4 −1−4 0 4 1 −4 3

] 𝑼 =1

80[

−40−57 24 −7

57 0−81 24

−24 81 0−57

7−24 57 40

]

Notes: Line elements have a constant Jacobian only if the physical nodes are equally spaced, like the parametric nodes, along a straight physical line. Curved line elements do not have a constant Jacobian and require numerical integration. If an integral includes constant properties, then the element matrices are simply obtained by multiplying the above integrals by that property. The sum of the row terms for this matrix is zero.


21

4.10 Exercises

1. Repeat Ex. 4.2-1 using natural coordinates −1 ≤ 𝑎 ≤ 1 so the quadratic interpolation

functions are 𝑯(𝑎) = [𝑎(𝑎 − 1) 2⁄ (𝑎 + 1)(1 − 𝑎) 𝑎(𝑎 + 1) 2⁄ ].

4.11 List of examples Ex 4.1-1 Proof that global x-y interpolation fails for quadrilaterals Ex 4.1-2 Proof that local r-s interpolation works for quadrilaterals Ex 4.1-3 Shape of any edge of a four-node quadrilateral, Q4 Ex 4.1-4 Shape of any edge of an eight-node quadrilateral, Q8 or Q9 Ex 4.2-1 Variable Jacobian of a line element with unequal spacing, L3 Ex 4.2-2 Physical line integral of interpolated quantity, L4 Ex 4.2-3 Moment integral of interpolated line force, L2 Ex 4.2-4 Jacobian matrix in cylindrical coordinates Ex 4.2-5 Area of a rotated square by integration Ex 4.2-6 Jacobian matrix of a quadrilateral Ex 4.2-7 Area calculation for the same quadrilateral Ex 4.5-1 Integration by parts of an ODE over a line Ex 4.5-2 Integration by parts of a PDE over a volume (Green’s Theorem) Ex 4.6-1 Resultant of a point load interior to a line element, L2 Ex 4.6-2 Resultant of a point load at a line element node point, L3 Ex 4.6-3 Mass matrix of a point mass at a line element node point, L2

4.12 Subject Index calculus review, 1 circular arc, 2 cross product, 17 curve tangent, 5 cyclic permutation, 10, 17 differential area, 8 differential geometry, 5 differential volume, 5, 18 Dirac Delta distribution, 14 distorted elements, 6 edge interpolation, 1, 2 edge shape, 3 exact integral, 5 exercises, 21 gap, 2 Green’s Theorem, 13, 19 integration by parts, 13 inverse Jacobian, 5, 11 inverse Jacobian matrix, 11 invertible map, 9 Jacobian, 8, 12 Jacobian determinant, 5

Jacobian matrix, 4, 5, 9 L2_C0, 7, 15 L3_C0, 3, 6, 15 L4_C0, 6 Lame’ constant, 5 line element integrals, 11, 20 linear triangle, 9 mass matrix, 15 material properties, 14 matrix multiplication, 17 matrix transpose, 17 natural coordinates, 8, 9 P4_C0, 10 parabola, 2 parametric coordinates, 1 parametric derivatives, 4 parametric transformation, 8 physical area, 9, 10 physical derivatives, 4, 11 point source, 14 Poisson equation, 14 polar coordinates, 1, 8


22

polar-moment-of-inertia, 12 Q4_C0, 2, 3, 4, 8 Q8_C0, 3 solid of revolution, 16 Summary, 17 surface area, 18 surface normal, 18

surface of revolution, 16 surface tangent, 5, 18 symmetric integrals, 20 T3_C0, 9 unsymmetric integrals, 20 variable Jacobian, 6 zero Jacobian, 6

4. calculus review · edge at node 3 ( (1), (1))=( 3, 3). therefore, the edge goes exactly through...

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