3. numerical integration 1 · 2018-02-06 · lecture notes, finite element analysis, j.e. akin,...

Lecture Notes, Finite Element Analysis, J.E. Akin, Rice University

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3. Numerical integration ........................................................................................................................................ 1

3.1 One-dimensional quadratures ............................................................................................................................. 1

3.2 Two- and three-dimensional quadratures ......................................................................................................... 10

3.3 Exact Integrals for Straight Sided Triangles....................................................................................................... 15

3.4 Reduced and Selected Integration .................................................................................................................... 17

3.5 Summary ........................................................................................................................................................... 19

3.6 Exercises ............................................................................................................................................................ 22

3. Numerical integration

3.1 One-dimensional quadratures: Since the finite element method is based on integral

relations it is logical to expect that one should strive to carry out the integrations as efficiently as

possible. In some cases we will employ exact integration. In others we may find that the integrals

can become too complicated to integrate exactly. In such cases the use of numerical integration

will prove useful or essential. The important topics of local coordinate integration and Gaussian

quadratures will be introduced here. They will prove useful when dealing with higher order

interpolation functions in complicated element integrals and allow the whole process to be

automated.

𝑰 = ∫ 𝑭(𝑟) 𝑑Ω = ∫ 𝑭(𝑟)

⊡

Ω|𝑱𝒆|𝑑 ⊡ ≈ ∑ 𝑭(𝑟𝑞)

𝑛𝑞𝑞=1 |𝑱𝒆(𝑟𝑞)| 𝑤𝑞 (3.1-1)

Numerical integration is simply a procedure that approximates an integral by a summation.

However, it can yield the exact integral and often does when the integrand is a polynomial. The

famous mathematician Gauss posed this question: What is the minimum number of points, say

nq, required to exactly integrate a polynomial, and what are the corresponding non-dimensional

abscissae (locations) and weights for those points? If we require the summation to be exact

when the integrand f(x) is any one of the power functions 1, x, x2, … , x2n, we obtain a set of 2n

conditions that allow us to determine the nq, the abscissae, xk , and their corresponding weights,

wk. The nq Gaussian one-dimensional quadrature points are symmetrically placed with respect

to the center of the interval, and will exactly integrate a polynomial of degree (2nq − 1).

Gauss obtained the results as numbers defined by square roots or combinations thereof. For

example, the two-point rule in natural coordinates gave 𝑎𝑘 = ±1 √3⁄ , 𝑤𝑘 = 1. For efficiency

and fast access, all quadrature data are tabulated as finite numbers with many significant figures.

The center point location is included in the abscissae list only when 𝑛𝑞 is odd, and the end points

are never utilized in a Gauss rule.

The Gauss rule data are usually tabulated for a non-dimensional unit coordinate range of

0 ≤ 𝑟 ≤ 1, or for a natural coordinate range of −1 ≤ 𝑎 ≤ 1, which are tabulated in Table 3.1-1.

The first three integration points are sketched in Fig. 3.1-1. The unit coordinate values are


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tabulated in function qp_rule_unit_Gauss.m, a portion of which is shown in Fig. 3.1-2. The

natural coordinate values of the quadrature data are stored in function qp_rule_Gauss.m as seen

in Fig. 3.1-3. The number of integration points is the input argument, and the arrays of positions

and weights are the returned items, ready to be used in a loop over those integration points.

A Matlab script for the same ‘hard coded’ calculation is given in Fig. 3.1-3. That script can be

easily generalized to any 1D Lagrangian elements by calling the element library and the

quadrature library as shown in Fig. 3.1-4. At each stage in the integration loop the location, r,

and the number of nodes per element, 𝑛_𝑛 ⟷ 𝑛𝑛, are supplied as arguments. The value of n_n

is used to select which interpolation member of the element library is desired. Then that

interpolation function, and its parametric derivative, is evaluated at the input local coordinate and

those two arrays of numbers are the returned items from the function. The script in Fig. 3.1-4 is

Figure 3.1-1 Natural 1, 2, and 3 point quadrature points, 𝝃, on a line

actually shorter that the hard coded script. By changing the third line data, any interpolated one-

dimensional cubic quantity can be integrated numerically.

Next a few more examples of numerical integration of interpolation functions are presented. Note that in the above examples only four significant digits were used in the integrations to keep

the examples simple. The actual data are tabulated to 24 significant figures.

If the physical node spacing had not been equal (very undesirable except for the special case

of linear fracture mechanics) in Ex. 3.1-3 the four term cubic interpolation would have been

needed and then the square matrix would have been 4 x 4. Even in that case, if the 𝒙𝒆 input data

were equally spaced the larger matrix products would yield exactly the same result.


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Figure 3.1-2 Portion of the unit coordinate line quadrature data

Table 3.1-1 Abscissas and Weights for Gaussian Quadrature in Unit Coordinates


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Figure 3.1-3 Portion of the natural coordinate tabulated data


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Example 3.1-1 Given: Using numerical integration determine the physical length of the cubic

line element in Ex. 2.2-1 with straight line coordinates of 𝒙𝒆𝑇 = [2 4 6 8] 𝑐𝑚. Solution:

The length is

𝐿𝑒 = ∫ 𝑑𝑥𝑥2

𝑥1= ∫

𝑑𝑥(𝑟)

𝑑𝑟

1

0𝑑𝑟 ≡ ∫ |𝐽𝑒(𝑟)|𝑑𝑟

1

0.

Recall that the interpolation function and its derivative are

𝑯(𝑟) = [(2 − 11𝑟 + 18𝑟2 − 9𝑟3) (18𝑟 − 45𝑟2 + 27𝑟3) …

(−9𝑟 + 36𝑟2 − 27𝑟3) (2𝑟 − 9𝑟2 + 9𝑟3)] 2⁄

𝜕𝑯(𝑟) 𝜕𝑟⁄ = [(−11 + 36𝑟 − 27𝑟2) (18 − 90𝑟 + 81𝑟2)…

(−9 + 72𝑟 − 81𝑟2) (2 − 18𝑟 + 27𝑟2)] 2⁄

Since the coordinate is interpolated by the above cubic polynomial, 𝑥(𝑟) = 𝑯(𝑟) 𝒙𝒆, the

degree of the integrand is 3 and the number of Gauss points is only 𝑛𝑞 = 2 to get the exact

answer. The numerical integration is

𝐿𝑒 = ∫𝑑𝑥(𝑟)

𝑑𝑟

1

0

𝑑𝑟 =∑𝑑𝑥(𝑟𝑞)

𝑑𝑟

𝑛𝑞

𝑞=1 𝑤𝑞 =∑

𝑑𝑯 𝒙𝒆

𝑑𝑟

𝑛𝑞

𝑞=1(𝑟𝑞 ) 𝑤𝑞 = [∑

𝑑𝑯

𝑑𝑟

𝑛𝑞

𝑞=1(𝑟𝑞 ) 𝑤𝑞] 𝒙

𝒆

where the determinant of the Jacobian at each point in the summation is

|𝐽𝑒(𝑟𝑞)| =𝑑𝑥(𝑟𝑞)

𝑑𝑟=𝑑𝑯

𝑑𝑟(𝑟𝑞) 𝒙

𝒆

Here, that product will be evaluated at each quadrature point. The two tabulated quadrature

locations in unit coordinates are 𝑟1 = 0.21132, and 𝑟2 = 0.78868, and the two tabulated

weights are the same 𝑤1 = 𝑤2 = 0.50000. Set the sum total initially to zero, L=0, and begin the

summation loop: set q =1, substituting 𝑟 = 0.21132 into the derivative the Jacobian is

𝑱(𝑟1)𝑒 = [−2.2990 1.2990 1.2990 − 0.2990] {

2468

} 𝑐𝑚 = 6.0000 𝑐𝑚

Multiply by the tabulated weight and add to the sum:

𝐿𝑒 = 0 + 𝑱(𝑟1)𝑒𝑤1 = 0 + (6.0000 𝑐𝑚)0.5000 = 3.0000 𝑐𝑚.

At the second point the Jacobian is

𝑱(𝑟2)𝑒 = [0.2990 − 1.2990 − 1.2990 2.2990]{

2468

} 𝑐𝑚 = 6.0000 𝑐𝑚

Multiply this by the tabulated weight and add to the sum

𝐿𝑒 = 3.0000 𝑐𝑚 + 𝑱(𝑟2)𝑒𝑤2 = 3.0000 + (6.0000 𝑐𝑚)0.5000 = 6.0000 𝑐𝑚

That yields the exact physical length of the cubic line element.

However, since the physical nodes are equally spaced on a straight line the physical location

only depends on the first and last node. In other words, the geometry mapping degenerates to a


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linear interpolation 𝑥(𝑟) = (1 − 𝑟)𝑥1 + 𝑟𝑥4. For a linear polynomial or a constant the exact

integration requires only one quadrature point. For a one-point rule the tabulated data are

𝑟1 = 0.5000, 𝑤1 = 1.0000. Because the physical coordinates were equally spaced the element

will have a constant Jacobian:

𝐽𝑒(𝑟) = 𝑑𝑯(𝑟) 𝑑𝑟⁄ 𝒙𝒆 = (−1)𝑥1 + 1𝑥4 = (𝑥4 − 𝑥1) = 𝐿.

Continuing with the numerical evaluation of the length: 𝐿𝑒 = 0 + [−1 1] {2.8.} = 6. 𝑐𝑚.

Example 3.1-2 Given: Automate the integration in Ex. 3.2-1 of the cubic solution with nodal

values of 𝒖𝒆𝑇= [2 3 0 −1]℃ and nodal coordinates of 𝒙𝒆

𝑇= [2 4 6 8]𝑚 by

applying Gaussian quadrature in the unit coordinate space. Solution: The integral is 𝐼𝑢 =

∫ 𝑢(𝑟)𝑑𝑥

𝐿𝑒= ∫ 𝑢(𝑟) |𝐽(𝑟)| 𝑑𝑟

1

0= [∫ 𝑯(𝑟) |𝐽(𝑟)𝑒| 𝑑𝑟

1

0] 𝒖𝒆. Previously, it was noted that the

Jacobian in this example is constant ( 𝐿𝑒) but this example will treat it as a variable. The degree

of the polynomial is 3. The number of necessary quadrature points is determined from

𝐷𝑒𝑔𝑟𝑒𝑒 ≤ (2 𝑛𝑞 − 1), so two quadrature points are required (and more are okay). The integral

becomes

𝐼𝑢 = [∫ 𝑯(𝑟) |𝐽(𝑟)𝑒| 𝑑𝑟 1

0

] 𝒖𝒆 = [∑𝑯(𝑟𝑞)

𝒏𝒒

𝒒=𝟏

|𝑱(𝑟𝑞 )|𝑤𝑞] 𝒖𝒆

where, in general, the Jacobian matrix at the quadrature point is calculated (numerically) as the

product of the local derivative of the interpolation functions times the input nodal coordinate

vector:

𝑱(𝑟𝑞)𝑒 =

𝜕𝑥(𝑟𝑞)

𝜕𝑟 =𝜕𝑯(𝑟𝑞)

𝜕𝑟 𝒙𝒆

Here, that product will be evaluated at each quadrature point. The analytic expressions for

𝐇(r) and ∂𝐇(rq) ∂r⁄ are in Ex. 2.2-1. Here the arrays are evaluated numerically at each point.

The two tabulated quadrature locations in unit coordinates are 𝑟1 = 0.21132, and 𝑟2 = 0.78868,

and the two tabulated weights are the same 𝑤1 = 𝑤2 = 0.50000. Begin the summation loop: set

q = 1, the Jacobian there is

𝑱(𝑟1)𝑒 = [−2.2990 1.2990 1.2990 − 0.2990] {

2468

} 𝑐𝑚 = 6.0000 𝑐𝑚

The function at the point is

𝑢(𝑟1) = [ 0.1972 1.0245 − 0.2745 0.0528] {

2 3 0−1

}℃ = 3.4151 ℃

and the contribution from this point is


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𝐼𝑞 = 𝑢(𝑟1) 𝑱(𝑟1)𝑒 𝑤𝑞 = (6.0000 𝑐𝑚)(3.4151 ℃) 0.5000 = 10.2452 ℃ 𝑐𝑚.

Set q = 2, the Jacobian there is

𝑱(𝑟2)𝑒 = [0.2990 − 1.2990 − 1.2990 2.2990] {

2468

} 𝑐𝑚 = 6.0000 𝑐𝑚.

The function at the point is 𝑢(𝑟1) = [0.0528 − 0.2745 1.0245 0.1972] {

2 3 0−1

}℃

so 𝑢(𝑟1) = −0.9151 ℃ and the contribution from this point is

𝐼𝑞 = 𝑢(𝑟1) 𝑱(𝑟1)𝑒 𝑤𝑞 = (6.0000 𝑐𝑚)(−0.9151 ℃)0.5000 = −2.7452 ℃ 𝑐𝑚.

Therefore, the total integral is

𝐼𝑢 = 10.2452 ℃ 𝑐𝑚 − 2.7452 ℃ 𝑐𝑚 = 7.5000 ℃ 𝑐𝑚,

which agrees with the analytically integrated value given later in Ex. 4.2-2.

Figure 3.1-3 Evaluating Ex. 3.1-1 by explicit numerical integration


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Figure 3.1-4 Numerical line integration using tables and the element library

Example 3.1-3 Given: Using numerical integration determine the physical y-moment of inertia

of the cubic line element in prior Ex. 3.1-2 with nodal coordinates of 𝒙𝒆𝑇 = [2 4 6 8] 𝑐𝑚

with respect to the first node. Solution: The definition of the moment of inertia with respect to

the y-axis (x-origin) is

𝐼𝑌 = ∫ 𝑥2

𝐿𝑒 𝑑𝑥 = ∫ 𝑥(𝑟)2

1

0

𝑑𝑥(𝑟)

𝑑𝑟𝑑𝑟 =∑ 𝑥(𝑟𝑞)

2𝑛𝑞

𝑞=1

𝑑𝑥(𝑟𝑞)

𝑑𝑟𝑤𝑞

The exact integral for a specific element is 𝐼𝑌𝑒 = 𝐿𝑒𝑥1

2 + 𝐿𝑒2𝑥1 + 𝐿𝑒3 3 = 168 𝑐𝑚3⁄ . For equal

spaced physical nodes on a straight line the geometry is linear:

𝑥(𝑟) = (1 − 𝑟)𝑥1𝑒 + 𝑟𝑥4

𝑒 = [(1 − 𝑟) 𝑟] {𝑥1𝑥4}𝑒

, and 𝑑𝑥(𝑟)

𝑑𝑟= (𝑥4

𝑒 − 𝑥1𝑒) ≡ 𝐿𝑒.

Since x(r) is degree one, and the Jacobian is constant (degree 0) the total polynomial degree

to be integrated is (1 + 1 + 0) = 2 so a two-point rule (𝑛𝑞 = 2) will exactly integrate the moment

of inertia. Recalling that in matrix notation

𝑥(𝑟)2 = 𝑥(𝑟)𝑇𝑥(𝑟) = [[(1 − 𝑟) 𝑟] {𝑥1𝑥4}𝑒

]

𝑇

[(1 − 𝑟) 𝑟] {𝑥1𝑥4}𝑒

𝑥(𝑟)2 = [𝑥1 𝑥4]𝑒 {(1 − 𝑟)𝑟

} [(1 − 𝑟) 𝑟] {𝑥1𝑥4}𝑒

= [𝑥1 𝑥4]𝑒 [(1 − 𝑟)2 (1 − 𝑟)𝑟

(1 − 𝑟)𝑟 𝑟2] {𝑥1𝑥4}𝑒

Now, for the first time the integrand is a square matrix and the resultant scalar will come

from a row matrix times the square matrix times a column matrix. Namely,


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𝐼𝑌𝑒 = ∫ 𝑥(𝑟)2

1

0

𝑑𝑥(𝑟)

𝑑𝑟𝑑𝑟 = [𝑥1 𝑥4]𝑒∫ [

(1 − 𝑟)2 (1 − 𝑟)𝑟

(1 − 𝑟)𝑟 𝑟2]

1

0

𝐿𝑒𝑑𝑟 {𝑥1𝑥4}𝑒

Let 𝑺𝒆 be the square matrix:

𝑺𝒆 = 𝐿𝑒∫ [(1 − 𝑟)2 (1 − 𝑟)𝑟

(1 − 𝑟)𝑟 𝑟2]

1

0

𝑑𝑟 =𝐿𝑒

6[2 11 2

]

Then the matrix notation of the exact moment of inertia is

𝐼𝑌𝑒 = [𝑥1 𝑥4]𝑒

𝐿𝑒

6[2 11 2

] {𝑥1𝑥4}𝑒

(1 × 1) = (1 × 2)(2 × 2)(2 × 1)

The given statement requires the integration (of 𝑺𝒆)to be done numerically.

𝑺𝒆 = 𝐿𝑒∑ [(1 − 𝑟𝑞)

2(1 − 𝑟𝑞)𝑟𝑞

(1 − 𝑟𝑞)𝑟𝑞 𝑟𝑞2

]𝑛𝑞

𝑞=1𝑤𝑞

Again, 𝑟1 = 0.21132, and 𝑟2 = 0.78868, and 𝑤1 = 𝑤2 = 0.50000. The numerical integral is

𝑺𝒆 = 𝟎 + 𝐿𝑒 [(1 − 0.21132)2 (1 − 0.21132)0.21132

(1 − 0.21132)0.21132 0.211322] 0.5000

+ 𝐿𝑒 [(1 − 0.78868)2 (1 − 0.78868)0.78868

𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 0.788682] 0.5000 = 𝐿𝑒 [

0.3333 0.16670.1667 0.3333

]

Finally, substituting the other numerical constants

𝐼𝑌𝑒 = [2 8]𝑐𝑚 (6 𝑐𝑚) [

0.3333 0.16670.1667 0.3333

] {28} 𝑐𝑚 = 168.0 𝑐𝑚3

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

same matrix integrals and results are obtained.

Example 3.1-4 Given: The generalized mass matrix for a line element is defined (later) as

𝑴𝒆 = ∫ 𝑯(𝑟)𝑻 𝑐(𝑥) 𝑯(𝑟) 𝑑𝑥𝑥3

𝑥1

where c(x) is a known property (like the mass per unit length). Evaluate this integral for a

quadratic line element, with constant property c, by using exact numerical integration. Solution:

In parametric coordinate space the definition of the matrix is

𝑴𝒆 = ∫ 𝑯(𝑟)𝑻 𝑐(𝑥) 𝑯(𝑟) 𝑑𝑥𝑥3

𝑥1= ∫ 𝑯(𝑟)𝑻 𝑐(𝑟) 𝑯(𝑟) |𝐽(𝑟)𝑒|⏟ 𝑑𝑟

1

0


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First, it is necessary to establish the polynomial degree of the integrand. Here, 𝑯(𝑟) is of degree

2 (quadratic):

𝑯(𝑟) = [(1 − 3𝑟 + 2𝑟2) (4𝑟 − 4𝑟2) (−𝑟 + 2𝑟2)]

The source term c(r) is constant (degree 0). Require equal physical node spacing so the

determinant of element Jacobian is also constant (degree 0), |𝐽(𝑟)𝑒| = 𝐿𝑒. For these restrictions

the degree of the polynomial to be integrated is

𝐷𝑒𝑔𝑟𝑒𝑒 = (2 + 0 + 2 + 0) = 4 ≤ (2𝑛𝑞 − 1)

5 ≤ 2𝑛𝑞 , 𝑛𝑞 = 2.5, rounding up 𝑛𝑞 = 3

Therefore

𝑴𝒆 = ∑ 𝑯(𝑟𝑞)𝑻 𝑐(𝑟𝑞) 𝑯(𝑟𝑞) |𝐽(𝑟𝑞)

𝑒|𝑛𝑞𝒒=𝟏

𝑤𝑞 = 𝑐𝑒 𝐿𝑒 ∑ 𝑯(𝑟𝑞)

𝑻 𝑯(𝑟𝑞) 𝑛𝑞=3

𝒒=𝟏 𝑤𝑞

(3 × 3) = ∑ [(3 × 1)(1 × 1)(1 × 3)(1 × 1)]

where the locations and weights are:

𝒓𝒒 = {1.1270166537925831148208𝑒 − 01 5.0000000000000000000000𝑒 − 018.8729833462074168851792𝑒 − 01

}, 𝒘𝒒 = {2.7777777777777777777778𝑒 − 014.4444444444444444444445𝑒 − 012.7777777777777777777778𝑒 − 01

}

Starting with 𝑴𝒆 = 𝟎, loop over each Gaussian quadrature point, numerically form 𝑯(𝑟𝑞),

numerically calculate the matrix product, 𝑯(𝑟𝑞)𝑻 𝑯(𝑟𝑞), add that matrix to 𝑴𝒆. After that loop,

multiply (numerically) the final square matrix by the element constants for the final result which

approximates the exact integral:

𝑴𝒆 = 𝑐𝑒 𝐿𝑒 [ 0.13333 0.06667 −0.03333 0.06667 0.53333 0.06667−0.03333 0.06667 0.13333

] ≈𝑐𝑒𝐿𝑒

30[ 4 2 −1 2 16 2−1 2 4

]

Note: In higher dimensions the Jacobian is usually not a constant. However, in practical

problems where the mesh has many small elements the Jacobian will be very close to a constant.

Thus, the number of quadrature points is often increased by one to account for a not exactly

constant Jacobian.

3.2 Two- and three-dimensional quadratures: The quadrature rules for quadrilaterals

and hexahedra are simply tensor products of the one-dimensional rule appropriate for the degree

of polynomial in each direction (which are almost always the same degree). The products of two-

and three-point rules (in each parametric direction) in a quadrilateral are illustrated in Fig. 3.3-1.

The Matlab script qp_rule_nat_quad.m shown in Fig. 3.2-2 creates the numerical integration

data for a quadrilateral in natural coordinates. The only input argument is the total number of

two-dimensional integration points. That value is the square of the number of one-dimensional

integration points to be use in each parametric direction. The return items are the two parametric

coordinates and the combined weight at each quadrature point. Only one line needs to be

changed to create equivalent unit coordinate quadrature data. For a hexahedron a similar script


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would just use three loops. Of course, its input argument would have to be the cube of the

number of one-dimensional points used in each parametric direction.

It is not unusual for a finite element analysis to include over a million elements. Then, the

numerical integration loop involves a significant amount of computations. Therefore, some

specialized integration rules have been developed that give the same accuracy with fewer points.

For example, in a hexahedra element the Gauss process often requires 27 points, but there is a

special rule that uses only 14 points for the same accuracy.

Figure 3.2-1 Tensor product of 1, 2 and 3 quadrature point rules in a quadrilateral

Figure 3.2-2 Creating quadrilateral integration rule from the 1-D rule

The quadrature rules for triangles and tetrahedra do not have a simple equation relating the

number of quadrature points to the polynomial degree. The number of quadrature points in a

triangle required to integrate polynomials from degree 1 to 8 are 1, 3, 4, 6, 7, 12, 13, and 16,

respectively are available from the function qp_rule_unit_tri.m (Fig. 3.2-3), and data for higher

degrees are in function qp_rule_high_unit_tri.m. In each case, the single argument to the script is


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the total number of required quadrature points in the triangle (and not the degree of the

polynomial being integrated). The quadrature points in triangles fall along the lines that bisect

the corner angles of the element, as shown in Fig. 3.2-4. The two simplest quadrature rules are

the internal (versus internal and boundary points) one-point rule (𝑛𝑞 = 1) located at the centroid

with 𝑟1 = 1 3⁄ = 𝑠1 and 𝑤1 = 1 2⁄ for integrating constant and linear functions; and the mid-

angle symmetric three-point rule (for degree ≤ 2) with 𝑟1 = 1 6⁄ , 𝑠1 = 1 6⁄ and 𝑟2 = 4 6⁄ ,𝑠2 = 1 6⁄ , 𝑟3 = 1 6⁄ , 𝑠3 = 4 6⁄ , and 𝑤1 = 𝑤2 = 𝑤3 = 1 6⁄ .

Figure 3.2-3 Selected triangular quadrature data values


13

Degree 0 1 2 3 4 5 6 7 8

𝑛𝑞 1 1 3 4 6 7 12 13 16

Figure 3.2-4 Some symmetric quadrature locations for unit triangle

Example 3.2-1 Given: For a four node quadrilateral element in local natural coordinates

evaluate the local moment-of inertia with respect to the a-axis:

𝐼𝑎𝑎 = ∫ 𝑏2 𝑑 ⊡

⊡

by numerical integration and check it with handbook values for right angled rectangles.

Solution: The integral is quadratic in b, so a two-point rule in each local direction will give the

exact results. That is, use the four-point rule (𝑛𝑞 = 4) like Fig. 3.3-1:

𝐼𝑎𝑎 = ∫ 𝑏2 𝑑 ⊡

⊡

= ∫ ∫ 𝑏2 𝑑𝑎 𝑑𝑏1

−1

=1

−1

∑ 𝑏𝑞2 𝑤𝑞

𝑛𝑞=4

𝑞=1

From the Introduction and Summary the points are at 𝑎𝑞and 𝑏𝑞are located a ±1 √3⁄ and all have

a weight of 𝑤𝑞 = 1.

𝐼𝑎𝑎 = 0 + (−1

√3)2

× 1 + (−1

√3)2

× 1 + (1

√3)2

× 1 + (1

√3)2

× 1 =4

3

Checking a geometric properties table for common shapes gives the inertia of a right

rectangle with respect to its centerline as

𝐼𝑎𝑎 =1

12𝑏𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡3 =

1

12(2) (2)3 =

16

12=4

3

Example 3.2-2 Given: All straight side triangles have a constant Jacobian matrix and

determinant. Use this fact to calculate the area of the triangle. Solution: The calculation of the

area is simple.

𝐴𝑒 = ∫ 𝑑𝐴 =

𝐴𝑒∫ 𝑑𝑥 𝑑𝑦 =

𝐴𝑒∫ |𝑱𝒆| 𝑑 ⊡

⊡

= |𝑱𝒆| ∫ 𝑑 ⊡

⊡

= |𝑱𝒆| 1

2


14

and any constant can be integrated exactly by a one-point rule (𝑛𝑞 = 1). Here, the chapter

summary shows that the one-point weight is one-half (and for linear functions is evaluated at the

centroid of the parametric triangle) so the triangle physical area integral reduces to

𝐴𝑒 = |𝑱𝒆|∫ 𝑑 ⊡

⊡

= |𝑱𝒆| ∑ 𝑤𝑞

𝑛𝑞=1

𝑞=1

= |𝑱𝒆|1

2

Another way of saying this is that the determinant of the Jacobian of all straight sided

triangles is just twice the physical area of the triangle |𝑱𝒆| = 2 𝐴𝑒 for straight triangles (only). In

the following chapters it will be shown that the algebra reduces to

|𝑱𝒆| = (𝑥2 − 𝑥1)𝑒(𝑦3 − 𝑦1)

𝑒 − (𝑥3 − 𝑥1)𝑒(𝑦2 − 𝑦1)

𝑒 = 2 𝐴𝑒

when the corners are numbered counter-clockwise.

Example 3.2-3 Given: For a linear unit triangle element evaluate the local polar-moment-of-

inertia

𝐼𝑟𝑠 = ∫ 𝑟 𝑠 𝑑 ⊡

⊡

and check it with handbook values for right triangles. Solution: The integral is only linear in r

and s, so both the one-point and three-point rules will give the exact results. Use the three-point

rule (𝑛𝑞 = 3):

𝐼𝑟𝑠 = ∫ 𝑟 𝑠 𝑑 ⊡

⊡

= ∫ ∫ 𝑟 𝑠 𝑑𝑠 𝑑𝑟1−𝑟

0

1

0

=∑ 𝑟𝑞 𝑠 𝑞𝑤𝑞𝑛𝑞

𝑞=1

𝐼𝑟𝑠 = 0 +1

6 1

6 1

6 +4

6 1

6 1

6+1

6 4

6 1

6=

9

216=1

24

Checking a geometric properties table for common shapes gives the polar inertia of a right

triangle with respect to the right angle corner as

𝐼𝑟𝑠 =1

24𝑏𝑎𝑠𝑒2 ℎ𝑒𝑖𝑔ℎ𝑡2 =

1

2412 12 =

1

24

Example 3.2-4 Given: For a four node quadrilateral element in local natural coordinates

evaluate the local moment-of inertia with respect to the a-axis:

𝐼𝑎𝑎 = ∫ 𝑏2 𝑑 ⊡

⊡


15

by numerical integration and check it with handbook values for right angled rectangles.

Solution: The integral is quadratic in b, so a two-point rule in each local direction will give the

exact results. That is, use the four-point rule (𝑛𝑞 = 4) like Fig. 3.3-1:

𝐼𝑎𝑎 = ∫ 𝑏2 𝑑 ⊡

⊡

= ∫ ∫ 𝑏2 𝑑𝑎 𝑑𝑏1

−1

=1

−1

∑ 𝑏𝑞2 𝑤𝑞

𝑛𝑞=4

𝑞=1

From the Introduction and Summary the points are at 𝑎𝑞and 𝑏𝑞are located at ±1 √3⁄ and all have

a weight of 𝑤𝑞 = 1.

𝐼𝑎𝑎 = 0 + (−1

√3)2

× 1 + (−1

√3)2

× 1 + (1

√3)2

× 1 + (1

√3)2

× 1 =4

3

Checking a geometric properties table for common shapes gives the inertia of a right

rectangle with respect to its centerline as

𝐼𝑎𝑎 =1

12𝑏𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡3 =

1

12(2) (2)3 =

16

12=4

3

3.3 Exact Integrals for Straight Sided Triangles:* Since straight sided triangles have a

constant Jacobian integrals of parametric polynomial terms can be integrated in closed form. A

typical polynomial integral can be expressed in terms of factorials of its exponents:

𝐼 = ∫ 𝑟𝑚 𝑠𝑛 𝑑𝐴

𝐴𝑒=

2𝐴𝑒 𝑚! 𝑛!

(2+𝑚+𝑛)! ≡

𝐴𝑒

𝐾𝑚 𝑛 (3.3-1)

where the constants 𝐾𝑚 𝑛 are listed in Table 3.3-1 for terms up to degree eight polynomials.

The constant 𝐾𝑚 𝑛 which divides into the physical area can also be expressed as a simple Matlab

script appended at the bottom of the table. A similar analytic volume integral is available for a

straight edged tetrahedral element:

𝐼 = ∫ 𝑟𝑚 𝑠𝑛 𝑡𝑝 𝑑𝑉

𝑉𝑒= 6𝑉𝑒

𝑚! 𝑛! 𝑝!

(3 + 𝑚 + 𝑛 + 𝑝)!

Table 3.3-1 Exact integrals for polynomials in straight triangles


16

K(m, n) = factorial (2 + m + n) / (2 * factorial (m) * factorial (n))

Example 3.3-1 Given: The generalized mass matrix for a triangular element is defined (later) as

𝑴𝒆 = ∫ 𝑯(𝑟, 𝑠)𝑻 𝑐 𝑯(𝑟, 𝑠) 𝑑𝐴

𝐴𝑒

where c is a known constant property (like the mass density). Evaluate this integral for a linear

triangular element by exact integration using the above table of 𝐾𝑚𝑛 values. Solution: In unit

coordinate parametric space the interpolation functions are: 𝑯(𝑟, 𝑠) = [(1 − 𝑟 − 𝑠) (𝑟) (𝑠)]. The matrix to be integrated is

𝑴𝒆 = 𝑐 ∫ [𝐻1𝐻1 𝐻1𝐻2 𝐻1𝐻3𝐻2𝐻1 𝐻2𝐻2 𝐻2𝐻3𝐻3𝐻1 𝐻3𝐻2 𝐻3𝐻3

] 𝑑𝐴

𝐴𝑒

= 𝑴𝒆𝑇

Due to the symmetry of the matrix only six integrals need to be evaluated. The term

𝑀22𝑒 = 𝑐∫ 𝐻2(𝑟, 𝑠) 𝐻2(𝑟, 𝑠) 𝑑𝐴

𝐴𝑒= 𝑐∫ 𝑟2 𝑑𝐴

𝐴𝑒

Since this triangle has straight sides Table 3.3-1 is valid to use. Using m = 2, n = 0, for this term

the table gives K = 6, so exact area integral is 𝑀22𝑒 = 𝑐 𝐴𝑒 6⁄ . Likewise, the matrix term

𝑀12𝑒 = 𝑐∫ 𝐻1(𝑟, 𝑠) 𝐻2(𝑟, 𝑠) 𝑑𝐴

𝐴𝑒= 𝑐∫ 𝑟(1 − 𝑟 − 𝑠) 𝑑𝐴

𝐴𝑒= 𝑐 ∫ (𝑟 − 𝑟2 − 𝑟𝑠) 𝑑𝐴

𝐴𝑒

and the table gives three terms 𝑀12𝑒 = 𝑐 (

𝐴𝑒

3−𝐴𝑒

6−𝐴𝑒

12) = 𝑐

𝐴𝑒

12. The final mass matrix result is


17

𝑴𝒆 =𝑐𝐴𝑒

12[2 1 11 2 11 1 2

] = 𝑴𝒆𝑇

3.4 Reduced and selective integration*: Since the numerical integration of the element

square matrix can represent a large part of the total cost it is desirable to use the lowest order

integration rules. Care must be taken when selecting the minimal order of integration. Usually

the integrand will contain global derivatives so that in the limit, as the element size h approaches

zero, the integrand can be assumed to be constant, and then only the volume integral

𝑉 = ∫ 𝑑𝑣

𝑉= ∫ |𝑱|

⊡ 𝑑𝑟 𝑑𝑠 𝑑𝑡

remains to be integrated exactly. Such a rule could be considered the minimal order. However,

that order is often too low to be practical since it may lead to a rank deficient element (and

system) square matrix, if the rule does not exactly integrate the equations. Typical integrands

involve terms such as the strain energy density per unit volume: 𝑩𝑻 𝑬 𝑩 2⁄ .

Let 𝑛𝑞 denote the number of element quadrature points while 𝑛𝑟 represents the number of

independent relations at each of the integration points. The rank of the element is 𝑅 = 𝑛𝑞 ∗ 𝑛𝑟.

Generally, 𝑛𝑟 corresponds to the number of rows in the 𝑩 matrix in the usual triple product

integrand of the element square matrix: 𝑩𝑻𝑬 𝑩. For a typical element with 𝑛𝑐 ≥ 0 constraint

equations on the 𝑛𝑖 element degrees of freedom the number of quadrature points needs to be

selected such that

𝑛𝑞(𝑛𝑟 − 𝑛𝑐) ≥ 𝑛𝑖. (3.4-1)

For a non-singular system square matrix, assembled from 𝑛𝑒 elements, a similar requirement is

𝑛𝑒(𝑛𝑞𝑛𝑟 − 𝑛𝑐) ≥ (𝑛𝑑 − 𝑛𝐸𝐵𝐶) (3.4-2)

where 𝑛𝐸𝐵𝐶 denotes the number of essential boundary conditions. These relations can be used

as guides in selecting a minimal value of integration points, 𝑛𝑞, in the element.

It has long been known that a finite element model gives a stiffness which is too high. Using

reduced integration so as to underestimate the element stiffness has been accepted as one way to

improve the results. A danger of low order integration rules is that zero energy modes may arise

in an element. That is, the element energy is 𝜹𝒆𝑻𝑺𝒆𝜹𝒆 = 0 for 𝜹𝒆 ≠ 𝟎. Usually these zero

energy modes, 𝜹𝒆, are incompatible with the same modes in an adjacent element. Thus, the

assembly of elements may have no zero energy modes (except for the standard rigid body

modes). An eigen-analysis of the element can be used as a check since eigenvalues of zero

correspond to zero energy modes.

For element formulations involving element constraints, or penalties, it is considered best to

employ selective integration rules. For penalty formulations, such as incompressible materials,

it is common to have equations of the form (𝑺𝟏 + 𝛼𝑺𝟐)𝜹 = 𝒄 where the global constant 𝛼 → ∞

in the limit where the penalty constraint is exactly satisfied. In the limit as 𝛼 → ∞ the system

degenerates to 𝑺𝟐 𝜹 = 𝒄 𝛼⁄ ≈ 𝟎 where the solution approaches the trivial result, 𝜹 = 𝟎. That

behavior is called solution locking. To obtain a non-trivial solution in this limit it is

necessary for the square matrix 𝑺𝟐 to be singular (rank deficient) and the matrix 𝑺𝟏 to be non-


18

singular. Therefore, the two contributing element parts, 𝑺𝟏 and 𝑺𝟐, are selectively integrated.

That is, 𝑺𝟐 matrix is under integrated so as to be rank deficient (singular) while 𝑺𝟏 is integrated

with a rule which renders 𝑺𝟏 non-singular. A typical application of selective integration is plate

bending where the bending stiffness contributions are in 𝑺𝟏 while the shear stiffness

contributions are in 𝑺𝟐 which have the constant 𝛼 = 1/𝑡 where t is the plate thickness.


19

3.5 Summary Notation

𝑛𝑒 ≡ Number of elements 𝑛𝑔 ≡ Number of generalized DOF per node

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

Element interpolations utilized in this chapter

Element

Type

Example

Number

Polynomial

Degree

𝑛𝑝

𝑛𝑔

𝑛𝑛

𝑛𝑖 =

𝑛𝑔 × 𝑛𝑛

Continuity

Level

Lagrange L3 3.1-4 2 1 1 3 3 𝐶0

Lagrange L4 3.1-1,2,3 3 1 1 4 4 𝐶0

Lagrange T3 3.2-2,3 3.3-1 1 2 1 3 3 𝐶0

Lagrange Q4 3.2-1,4 1 2 1 4 4 𝐶0

Definition of Jacobian:

[𝑱] ≡ [𝜕(𝑥, 𝑦, 𝑧)

𝜕(𝑟, 𝑠, 𝑡)], 1-D: [𝑱] ≡ [

𝜕𝑥(𝑟)

𝜕𝑟], 2-D: [𝑱] ≡ [

𝜕𝑥 𝜕𝑟⁄ 𝜕𝑦 𝜕𝑟⁄

𝜕𝑥 𝜕𝑠⁄ 𝜕𝑦 𝜕𝑠⁄]

Numerical integration:

∫ 𝑭(𝑟)

⊡ 𝑑 ⊡ ≈ ∑ 𝑭(𝑟𝑞)

𝑛𝑞𝑞=1 𝑤𝑞

∫ 𝑭(𝑟) 𝑑Ω = ∫ 𝑭(𝑟)

⊡

Ω|𝑱𝒆| 𝑑 ⊡ ≈ ∑ 𝑭(𝑟𝑞)

𝑛𝑞𝑞=1 |𝑱𝒆(𝑟𝑞)|𝑤𝑞

Exact unit coordinate integral for straight triangle (only):

𝐼 = ∫ 𝑟𝑚 𝑠𝑛 𝑑𝐴

𝐴𝑒= 2𝐴𝑒𝑚! 𝑛! (2 + 𝑚 + 𝑛)!⁄

Number of points for exact 1-D polynomial numerical integration:

𝐷𝑒𝑔𝑟𝑒𝑒 ≤ (2𝑛𝑞 − 1)

Gaussian 1-D one-point line rule data, in unit coordinates and natural coordinates:

𝑟1 = 0.5, 𝑤1 = 1 ; 𝑎1 = 0, 𝑤1 = 2

Gaussian 1-D two-point line rule data, in unit coordinates and natural coordinates:

𝑟1 = (1 − 1 √3⁄ ) 2⁄ = 0.21132, 𝑟2 = (1 + 1 √3⁄ ) 2⁄ = 0.78867, 𝑤1 = 𝑤2 = 0.50000


20

𝑎1 = − 1 √3⁄ = −0.57735, 𝑎2 = 1 √3⁄ = 0.57735, 𝑤1 = 𝑤2 = 1.00000

Number of points needed for integrating polynomials on triangles:

Degree 0 1 2 3 4 5 6 7 8

𝑛𝑞 1 1 3 4 6 7 12 13 16

Quadrature unit triangle one-point rule (for degree ≤ 1):

𝑟1 = 1 3⁄ , 𝑠1 = 1 3⁄ , 𝑤1 = 1 2⁄

Quadrature unit triangle three-point rule (for degree ≤ 2)

𝑟1 = 1 6⁄ , 𝑠1 = 1 6⁄ , 𝑤1 = 1 6⁄

𝑟2 = 4 6⁄ , 𝑠2 = 1 6⁄ ,𝑤2 = 1 6⁄

𝑟3 = 1 6⁄ , 𝑠3 = 4 6⁄ ,𝑤3 = 1 6⁄

Lagrange linear 𝐶0 line interpolations (𝑛𝑝 = 1), in unit coordinates: ,

local nodes at (r = 0) and (r = 1), 𝑛𝑛 = 2, 𝑛𝑔 = 1, 𝜹𝒆𝑻 = [𝑢1 𝑢2],

𝑯(𝑟) = [(1 − 𝑟) 𝑟], 𝜕𝑯(𝑟) 𝜕𝑟⁄ = [−1 1]

Lagrange quadratic 𝐶0 line interpolations (𝑛𝑝 = 1), in unit coordinates:

local nodes at (r = 0), (r = 1/2) and (r = 1) , 𝑛𝑛 = 3, 𝑛𝑔 = 1, 𝜹𝒆𝑻 = [𝑢1 𝑢2 𝑢3]

𝑯(𝑟) = [(1 − 3𝑟 + 2𝑟2) (4𝑟 − 4𝑟2) (−𝑟 + 2𝑟2)], 𝜕𝑯(𝑟) 𝜕𝑟⁄ = [(3 + 4𝑟) (4 − 8𝑟) (−1 + 4𝑟)]

Lagrange cubic 𝐶0 line interpolations (𝑛𝑝 = 1, 𝑛𝑛 = 4), in unit coordinates:

with local nodes at (r = 0), (r = 1/3),(r = 2/3) and (r = 1). DOF, 𝜹𝒆𝑻 = [𝑢1 𝑢2 𝑢3 𝑢4]

𝑯(𝑟) =1

2[(2 − 11𝑟 + 18𝑟2 − 9𝑟3) (18𝑟 − 45𝑟2 + 27𝑟3)…

(−9𝑟 + 36𝑟2 − 27𝑟3) (2𝑟 − 9𝑟2 + 9𝑟3)]

𝜕𝑯(𝑟) 𝜕𝑟⁄ = 1

2[(−11 + 36𝑟 − 27𝑟2) (18 − 90𝑟 + 81𝑟2)…

(−9 + 72𝑟 − 81𝑟2) (2 − 18𝑟 + 27𝑟2)].

Lagrange linear triangle 𝐶0 interpolation (𝑛𝑝 = 2, 𝑛𝑛 = 3, 𝑛𝑔 = 1 ), in unit coordinates:

with local nodes at (r = 0, s = 0), (r = 1, s = 0), and (r = 0, s = 1).

𝑯(𝑟, 𝑠) = [(1 − 𝑟 − 𝑠) (𝑟) (𝑠)]

[𝜕𝑯 𝜕 𝑟⁄

𝜕𝑯 𝜕⁄ 𝑠] = [

−1 1 0−1 0 1

].

Lagrange linear line element consistent mass matrix: 𝑴𝒆 =𝜌𝑒𝐴𝑒𝐿𝑒

6[2 11 2

]


21

Lagrange quadratic line consistent mass: 𝑴𝒆 =𝜌𝑒𝐴𝑒𝐿𝑒

30[4 2 −1 2 16 2−1 2 4

] =𝑚𝑒

30[4 2 −1 2 16 2−1 2 4

]

Lagrange cubic line element consistent mass matrix: 𝑴𝒆 =𝜌𝑒𝐴𝑒𝐿𝑒

1680[

128 99 99 648

−36 19−81 −36

−36 −81 19 −36

648 99 99 128

]

Lagrange linear triangle consistent mass matrix: 𝑴𝒆 =𝜌𝑒𝐴𝑒𝑡𝑒

12[2 1 11 2 11 1 2

] =𝑚𝑒

12[2 1 11 2 11 1 2

]


22

3.6 Exercises 1. Repeat Ex. 3.3-1 using a one-point quadrature rule.

2. Use numerical integration to verify that the local inertia of a unit triangle, with respect to the r-

axis is

𝐼𝑟𝑟 = ∫ 𝑠2 𝑑 ⊡

⊡

=1

12

3. Use numerical integration to verify that the local inertia of a unit triangle, with respect to the s-

axis is

𝐼𝑠𝑠 = ∫ 𝑟2 𝑑 ⊡

⊡

=1

12

4. Use numerical integration to form the 2 × 2 Lagrange linear line element consistent mass

matrix given in the summary.

5. Use numerical integration to form the 4 × 4 Lagrange cubic line element consistent mass

matrix given in the summary.

Index case, 3 consistent mass matrix, 19 constant Jacobian, 13, 14 cubic interpolation, 5 element constraints, 16 else, 11 elseif, 11 exact integrals, 15 factorial, 15 for, 8, 11 fracture mechanics, 2 Gaussian quadrature, 1 generalized mass matrix, 15 hexahedra, 10 incompressible materials, 16 integration loop, 11 Jacobian, 5, 6, 10 L4_C0, 5, 8, 19 Lagrange_1D_libary.m, 8 locking, 17 mass matrix, 9, 16 minimal order integration, 16 moment of inertia, 8, 9 natural coordinates, 18 number of independent relations, 16 number of quadrature points, 10, 11 numerical integration, 1, 5, 7, 8, 9, 18 physical length, 5 plate bending, 17 polar-moment-of- inertia, 13 Q4_C0, 12, 14, 18

qp_rule_Gauss.m, 2, 4 qp_rule_nat_quad.m, 10, 11 qp_rule_unit_Gauss.m, 2, 3, 8 qp_rule_unit_tri.m, 12 quadratic interpolation, 9 quadrature, 2 quadrature abscissas, 3, 4 quadrature point locations, 11, 12 quadrature weights, 4 quadratures, 1 quadrilateral quadrature, 10 rank deficient, 16 reduced integration, 16 rigid body modes, 16 selective integration rules, 16 straight edged tetrahedral, 14 straight side triangle, 13 straight sided triangle, 14, 18 straight triangles, 13 strain energy density per unit volume, 16 Summary, 18 switch, 3 T3_C0, 19 tensor product, 11 three-point rule, 13, 19 triangle quadrature, 13 triangular quadrature rule selection, 19 two-point rule, 1 unit coordinates, 4, 5, 12, 15 unit triangle, 13 zero energy mode, 16

3. numerical integration 1 · 2018-02-06 · lecture notes, finite element analysis, j.e. akin,...

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