3. Numerical integration

Transcription

1 3. Numerical integration One-dimensional quadratures Two- and three-dimensional quadratures Exact Integrals for Straight Sided Triangles Reduced and Selected Integration Summary Exercises Numerical integration 3. 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 = Ω F(r) J e n d q q= F(r q ) J e (r q ) w q (3.-) 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 n q, 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, x, x,, x n, we obtain a set of n conditions that allow us to determine the n q, the abscissae, x k, and their corresponding weights, w k. The n q Gaussian one-dimensional quadrature points are symmetrically placed with respect to the center of the interval, and will exactly integrate a polynomial of degree (n q ). Gauss obtained the results as numbers defined by square roots or combinations thereof. For example, the two-point rule in natural coordinates gave a k = ± 3, w k =. 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 n q 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 r, or for a natural coordinate range of a, which are tabulated in Table 3.-. The first three integration points are sketched in Fig The unit coordinate values are

2 tabulated in function qp_rule_unit_gauss.m, a portion of which is shown in Fig The natural coordinate values of the quadrature data are stored in function qp_rule_gauss.m as seen in Fig 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 That script can be easily generalized to any D Lagrangian elements by calling the element library and the quadrature library as shown in Fig At each stage in the integration loop the location, r, and the number of nodes per element, n_n n n, 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 is Figure 3.- Natural,, and 3 point quadrature points, ξ, on a line actually shorter that the hard coded script. By changing the third line data, any interpolated onedimensional 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 4 significant figures. If the physical node spacing had not been equal (very undesirable except for the special case of linear fracture mechanics) in Ex 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 x e input data were equally spaced the larger matrix products would yield exactly the same result.

3 Figure 3.- Portion of the unit coordinate line quadrature data Table 3.- Abscissas and Weights for Gaussian Quadrature in Unit Coordinates 3

4 Figure 3.-3 Portion of the natural coordinate tabulated data 4

5 Example 3.- Given: Using numerical integration determine the physical length of the cubic line element in Ex..- with straight line coordinates of x et = [ 4 8] cm. Solution: The length is L e x = dx = dx(r) dr J e (r) dr. x Recall that the interpolation function and its derivative are H(r) = [( r + 8r 9r 3 ) (8r 45r + 7r 3 ) ( 9r + 3r 7r 3 ) (r 9r + 9r 3 )] H(r) r = [( + 3r 7r ) (8 9r + 8r ) ( 9 + 7r 8r ) ( 8r + 7r )] dr Since the coordinate is interpolated by the above cubic polynomial, x(r) = H(r) x e, the degree of the integrand is 3 and the number of Gauss points is only n q = to get the exact answer. The numerical integration is L e = dx(r) dr = dr n q q= dx(r q) w dr q = n q q= 5 dh xe dr (r q ) w q = [ where the determinant of the Jacobian at each point in the summation is J e (r q ) = dx(r q) dr = dh dr (r q) x e n q q= dh dr (r q ) w q ] x e Here, that product will be evaluated at each quadrature point. The two tabulated quadrature locations in unit coordinates are r =.3, and r =.7888, and the two tabulated weights are the same w = w =.5. Set the sum total initially to zero, L=, and begin the summation loop: set q =, substituting r =.3 into the derivative the Jacobian is J(r ) e = [ ] { 4 } cm =. cm 8 Multiply by the tabulated weight and add to the sum: L e = + J(r ) e w = + (. cm).5 = 3. cm. At the second point the Jacobian is J(r ) e = [ ] { 4 } cm =. cm 8 Multiply this by the tabulated weight and add to the sum L e = 3. cm + J(r ) e w = 3. + (. cm).5 =. 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

6 linear interpolation x(r) = ( r)x + rx 4. For a linear polynomial or a constant the exact integration requires only one quadrature point. For a one-point rule the tabulated data are r =.5, w =.. Because the physical coordinates were equally spaced the element will have a constant Jacobian: J e (r) = dh(r) dr x e = ( )x + x 4 = (x 4 x ) = L. Continuing with the numerical evaluation of the length: L e = + [ ] {. } =. cm. 8. Example 3.- Given: Automate the integration in Ex. 3.- of the cubic solution with nodal values of u et = [ 3 ] and nodal coordinates of x et = [ 4 8]m by applying Gaussian quadrature in the unit coordinate space. Solution: The integral is I u = u(r)dx = L e u(r) J(r) dr = [ H(r) J(r) e dr ] u e. Previously, it was noted that the Jacobian in this example is constant ( L e ) 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 ( n q ), so two quadrature points are required (and more are okay). The integral becomes I u = [ H(r) J(r) e dr n q ] u e = [ H(r q ) J(r q ) w q ] u e 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(r q ) e = x(r q) = H(r q) x e r r Here, that product will be evaluated at each quadrature point. The analytic expressions for H(r) and H(r q ) r are in Ex..-. Here the arrays are evaluated numerically at each point. The two tabulated quadrature locations in unit coordinates are r =.3, and r =.7888, and the two tabulated weights are the same w = w =.5. Begin the summation loop: set q =, the Jacobian there is J(r ) e = [ ] { 4 } cm =. cm 8 The function at the point is u(r ) = [ ] { 3 } = 3.45 and the contribution from this point is q=

7 I q = u(r ) J(r ) e w q = (. cm)(3.45 ).5 =.45 cm. Set q =, the Jacobian there is J(r ) e = [ ] { 4 } cm =. cm. The function at the point is u(r ) = [ ] { 3 } so u(r ) =.95 and the contribution from this point is I q = u(r ) J(r ) e w q = (. cm)(.95 ).5 =.745 cm. Therefore, the total integral is I u =.45 cm.745 cm = 7.5 cm, which agrees with the analytically integrated value given later in Ex Figure 3.-3 Evaluating Ex. 3.- by explicit numerical integration 7

8 Figure 3.-4 Numerical line integration using tables and the element library Example 3.-3 Given: Using numerical integration determine the physical y-moment of inertia of the cubic line element in prior Ex. 3.- with nodal coordinates of x et = [ 4 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 I Y = x dx = x(r) dx(r) n L e dr dr = q x(r dx(r q) q ) w q= dr q The exact integral for a specific element is I e Y = L e x + L e x + L e3 3 = 8 cm 3. For equal spaced physical nodes on a straight line the geometry is linear: x(r) = ( r)x e + rx e 4 = [( r) r] { x x } e, and dx(r) = (x e 4 dr 4 x e ) L e. Since x(r) is degree one, and the Jacobian is constant (degree ) the total polynomial degree to be integrated is ( + + ) = so a two-point rule (n q = ) will exactly integrate the moment of inertia. Recalling that in matrix notation x(r) = x(r) T x(r) = [[( r) r] { x x 4 } e ] T [( r) r] { x x 4 } e x(r) = [x x 4] e ( { r) } [( r) r] { x r x } e = [x x 4] e [ ( r) ( r)r 4 ( r)r r ] { x x } e 4 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, 8

9 I e Y = x(r) Let S e be the square matrix: Lecture Notes, Finite Element Analysis, J.E. Akin, Rice University dx(r) dr dr = [ x x 4 ] e [ ( r) ( r)r ( r)r r ] S e = L e ( r) ( r)r [ ( r)r r ] dr = Le [ ] Then the matrix notation of the exact moment of inertia is I e Le Y = [x x 4 ] e [ ] {x x } e 4 ( ) = ( )( )( ) L e dr { x e x } 4 The given statement requires the integration (of S e )to be done numerically. n q S e = L e [ ( r q) ( r q )r q ] q= ( r q )r q r q Again, r =.3, and r =.7888, and w = w =.5. The numerical integral is S e = + L e (.3) [ (.3).3 (.3).3.3 ].5 + L e [ (.7888) (.7888).7888 symmetric.7888 ].5 = L e [ ] Finally, substituting the other numerical constants I e Y = [ 8]cm ( cm) [ ] { } cm = 8. cm3 8 Of course, if the interpolations are done using natural coordinate interpolations the exact same matrix integrals and results are obtained. w q Example 3.-4 Given: The generalized mass matrix for a line element is defined (later) as x 3 M e = H(r) T c(x) H(r) dx x 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 x 3 M e = H(r) T c(x) H(r) dx = H(r) T c(r) H(r) J(r) e x dr 9

10 First, it is necessary to establish the polynomial degree of the integrand. Here, H(r) is of degree (quadratic): H(r) = [( 3r + r ) (4r 4r ) ( r + r )] The source term c(r) is constant (degree ). Require equal physical node spacing so the determinant of element Jacobian is also constant (degree ), J(r) e = L e. For these restrictions the degree of the polynomial to be integrated is Therefore n q Degree = ( ) = 4 (n q ) 5 n q, n q =.5, rounding up n q = 3 n q =3 w q= q M e = H(r q ) T c(r q ) H(r q ) J(r q ) e w q= q = c e L e H(r q ) T H(r q ) (3 3) = [(3 )( )( 3)( )] where the locations and weights are: e e r q = { 5.e }, w q = { e } e e Starting with M e =, loop over each Gaussian quadrature point, numerically form H(r q ), numerically calculate the matrix product, H(r q ) T H(r q ), add that matrix to M e. After that loop, multiply (numerically) the final square matrix by the element constants for the final result which approximates the exact integral: M e = c e L e [ ] ce L e 4 3 [ ] 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. 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 twoand three-point rules (in each parametric direction) in a quadrilateral are illustrated in Fig The Matlab script qp_rule_nat_quad.m shown in Fig. 3.- 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

11 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 7 points, but there is a special rule that uses only 4 points for the same accuracy. Figure 3.- Tensor product of, and 3 quadrature point rules in a quadrilateral Figure 3.- Creating quadrilateral integration rule from the -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 to 8 are, 3, 4,, 7,, 3, and, respectively are available from the function qp_rule_unit_tri.m (Fig. 3.-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

12 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 The two simplest quadrature rules are the internal (versus internal and boundary points) one-point rule (n q = ) located at the centroid with r = 3 = s and w = for integrating constant and linear functions; and the midangle symmetric three-point rule (for degree ) with r =, s = and r = 4, s =, r 3 =, s 3 = 4, and w = w = w 3 =. Figure 3.-3 Selected triangular quadrature data values

13 Degree n q Figure 3.-4 Some symmetric quadrature locations for unit triangle Example 3.- Given: For a four node quadrilateral element in local natural coordinates evaluate the local moment-of inertia with respect to the a-axis: I aa = b 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 (n q = 4) like Fig. 3.3-: I aa = b d = b n q =4 da db = b q w q From the Introduction and Summary the points are at a q and b q are located a ± 3 and all have a weight of w q =. I aa = + ( 3 ) + ( 3 ) + ( 3 ) + ( 3 ) = 4 3 Checking a geometric properties table for common shapes gives the inertia of a right rectangle with respect to its centerline as I aa = base height 3 = () ()3 = = 4 3 q= Example 3.- 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. A e = da = dx dy = A e A e J e d = J e d = J e 3

14 and any constant can be integrated exactly by a one-point rule (n q = ). 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 n q = A e = J e d = J e w q q= = J e 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 J e = A e for straight triangles (only). In the following chapters it will be shown that the algebra reduces to J e = (x x ) e (y 3 y ) e (x 3 x ) e (y y ) e = A e when the corners are numbered counter-clockwise. Example 3.-3 Given: For a linear unit triangle element evaluate the local polar-moment-ofinertia I rs = 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 (n q = 3): I rs = r s d I rs = + r = r s ds dr n q = r q s qw q q= = 9 = 4 Checking a geometric properties table for common shapes gives the polar inertia of a right triangle with respect to the right angle corner as I rs = 4 base height = 4 = 4 Example 3.-4 Given: For a four node quadrilateral element in local natural coordinates evaluate the local moment-of inertia with respect to the a-axis: I aa = b d 4

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 (n q = 4) like Fig. 3.3-: I aa = b d = b n q =4 da db = b q w q q= From the Introduction and Summary the points are at a q and b q are located at ± 3 and all have a weight of w q =. I aa = + ( 3 ) + ( 3 ) + ( 3 ) + ( 3 ) = 4 3 Checking a geometric properties table for common shapes gives the inertia of a right rectangle with respect to its centerline as I aa = base height 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 = A e r m s n da = Ae m! n! (+m+n)! Ae K m n (3.3-) where the constants K m n are listed in Table 3.3- for terms up to degree eight polynomials. The constant K m 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 = r m s n t p dv V e = V e m! n! p! (3 + m + n + p)! Table 3.3- Exact integrals for polynomials in straight triangles 5

16 K(m, n) = factorial ( + m + n) / ( * factorial (m) * factorial (n)) Example 3.3- Given: The generalized mass matrix for a triangular element is defined (later) as M e = H(r, s) T c H(r, s) da A e 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 K mn values. Solution: In unit coordinate parametric space the interpolation functions are: H(r, s) = [( r s) (r) (s)]. The matrix to be integrated is H H H H H H 3 M e = c [ H H H H H H 3 ] da = M et A e H 3 H H 3 H H 3 H 3 Due to the symmetry of the matrix only six integrals need to be evaluated. The term M e = c H (r, s) H (r, s) da A e = c r da A e Since this triangle has straight sides Table 3.3- is valid to use. Using m =, n =, for this term the table gives K =, so exact area integral is M e = c A e. Likewise, the matrix term M e = c H (r, s) H (r, s) da A e and the table gives three terms M e = c r( r s) da A e = c ( Ae 3 Ae Ae = c (r r rs) da A e Ae ) = c. The final mass matrix result is

17 M e = cae [ ] = M e T 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 7 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: B T E B. Let n q denote the number of element quadrature points while n r represents the number of independent relations at each of the integration points. The rank of the element is R = n q n r. Generally, n r corresponds to the number of rows in the B matrix in the usual triple product integrand of the element square matrix: B T E B. For a typical element with n c constraint equations on the n i element degrees of freedom the number of quadrature points needs to be selected such that n q (n r n c ) n i. (3.4-) For a non-singular system square matrix, assembled from n e elements, a similar requirement is n e (n q n r n c ) (n d n EBC ) (3.4-) where n EBC denotes the number of essential boundary conditions. These relations can be used as guides in selecting a minimal value of integration points, n q, 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 δ et S e δ e = for δ e. Usually these zero energy modes, δ e, 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 (S + αs )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 S δ = c α where the solution approaches the trivial result, δ =. That behavior is called solution locking. To obtain a non-trivial solution in this limit it is necessary for the square matrix S to be singular (rank deficient) and the matrix S to be non-

18 singular. Therefore, the two contributing element parts, S and S, are selectively integrated. That is, S matrix is under integrated so as to be rank deficient (singular) while S is integrated with a rule which renders S non-singular. A typical application of selective integration is plate bending where the bending stiffness contributions are in S while the shear stiffness contributions are in S which have the constant α = /t where t is the plate thickness. 8

19 3.5 Summary Notation n e Number of elements n m Number of mesh nodes n p Dimension of parametric space n s Dimension of physical space n g Number of generalized DOF per node n n Number of nodes per element n q Number of total quadrature points Element interpolations utilized in this chapter Element Type Example Number Polynomial Degree n p n g n n n i = n g n n Continuity Level Lagrange L C Lagrange L4 3.-,, C Lagrange T3 3.-, C Lagrange Q4 3.-,4 4 4 C Definition of Jacobian: [J] [ Numerical integration: (x, y, z) (r, s, t) ], -D: [J] [ x(r)], -D: [J] [ x r y r r x s y s ] F(r) n d q q= F(r q ) w q F(r) dω = F(r) Ω Je d n q q= F(r q ) J e (r q ) w q Exact unit coordinate integral for straight triangle (only): I = A e r m s n da = A e m! n! ( + m + n)! Number of points for exact -D polynomial numerical integration: Degree (n q ) Gaussian -D one-point line rule data, in unit coordinates and natural coordinates: r =.5, w = ; a =, w = Gaussian -D two-point line rule data, in unit coordinates and natural coordinates: r = ( 3) =.3, r = ( + 3) =.7887, w = w =.5 9

20 a = 3 =.57735, a = 3 =.57735, w = w =. Number of points needed for integrating polynomials on triangles: Degree n q Quadrature unit triangle one-point rule (for degree ): r = 3, s = 3, w = Quadrature unit triangle three-point rule (for degree ) r =, s =, w = r = 4, s =, w = r 3 =, s 3 = 4, w 3 = Lagrange linear C line interpolations (n p = ), in unit coordinates:, local nodes at (r = ) and (r = ), n n =, n g =, δ et = [u u ], H(r) = [( r) r], H(r) r = [ ] Lagrange quadratic C line interpolations (n p = ), in unit coordinates: local nodes at (r = ), (r = /) and (r = ), n n = 3, n g =, δ et = [u u u 3 ] H(r) = [( 3r + r ) (4r 4r ) ( r + r )], H(r) r = [(3 + 4r) (4 8r) ( + 4r)] Lagrange cubic C line interpolations (n p =, n n = 4), in unit coordinates: with local nodes at (r = ), (r = /3),(r = /3) and (r = ). DOF, δ et = [u u u 3 u 4 ] H(r) = [( r + 8r 9r 3 ) (8r 45r + 7r 3 ) ( 9r + 3r 7r 3 ) (r 9r + 9r 3 )] H(r) r = [( + 3r 7r ) (8 9r + 8r ) ( 9 + 7r 8r ) ( 8r + 7r )]. Lagrange linear triangle C interpolation (n p =, n n = 3, n g = ), in unit coordinates: with local nodes at (r =, s = ), (r =, s = ), and (r =, s = ). H(r, s) = [( r s) (r) (s)] H r [ H s ] = [ ]. Lagrange linear line element consistent mass matrix: M e = ρe A e L e [ ]

21 4 4 Lagrange quadratic line consistent mass: M e = ρe A e L e [ ] = me [ ] Lagrange cubic line element consistent mass matrix: M e = ρe A e L e 8 [ ] Lagrange linear triangle consistent mass matrix: M e = ρe A e t e [ ] = me [ ]

22 3. Exercises. Repeat Ex using a one-point quadrature rule.. Use numerical integration to verify that the local inertia of a unit triangle, with respect to the r- axis is I rr = s d = 3. Use numerical integration to verify that the local inertia of a unit triangle, with respect to the s- axis is I ss = r d = 4. Use numerical integration to form the 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, 9 constant Jacobian, 3, 4 cubic interpolation, 5 element constraints, else, elseif, exact integrals, 5 factorial, 5 for, 8, fracture mechanics, Gaussian quadrature, generalized mass matrix, 5 hexahedra, incompressible materials, integration loop, Jacobian, 5,, L4_C, 5, 8, 9 Lagrange_D_libary.m, 8 locking, 7 mass matrix, 9, minimal order integration, moment of inertia, 8, 9 natural coordinates, 8 number of independent relations, number of quadrature points,, numerical integration,, 5, 7, 8, 9, 8 physical length, 5 plate bending, 7 polar-moment-of- inertia, 3 Q4_C,, 4, 8 qp_rule_gauss.m,, 4 qp_rule_nat_quad.m,, qp_rule_unit_gauss.m,, 3, 8 qp_rule_unit_tri.m, quadratic interpolation, 9 quadrature, quadrature abscissas, 3, 4 quadrature point locations,, quadrature weights, 4 quadratures, quadrilateral quadrature, rank deficient, reduced integration, rigid body modes, selective integration rules, straight edged tetrahedral, 4 straight side triangle, 3 straight sided triangle, 4, 8 straight triangles, 3 strain energy density per unit volume, Summary, 8 switch, 3 T3_C, 9 tensor product, three-point rule, 3, 9 triangle quadrature, 3 triangular quadrature rule selection, 9 two-point rule, unit coordinates, 4, 5,, 5 unit triangle, 3 zero energy mode,