CIVL 7/87 Chapter - /39 Chapter Learning Objectives To introduce concepts of three-dimensional stress and strain. To develop the tetrahedral solid-element stiffness matri. To describe how bod and surface tractions are treated. To illustrate a numerical eample of the tetrahedral element stiffness matri. Chapter Learning Objectives To describe the isoparametric formulation of the stiffness matri for three dimensional heahedral (brick) elements, including the linear (eight-noded) brick, and the quadratic (20 noded) brick. To present some commercial computer program eamples of three-dimensional solid models and results for real-world applications. To present a comparison of the four-noded tetrahedral, the ten-noded tetrahedral, the eightnoded brick, and the twent-noded brick.
CIVL 7/87 Chapter - 2/39 Introduction In this chapter, we consider the three-dimensional, or solid, element. This element is useful for the stress analsis of general threedimensional bodies that require more precise analsis than is possible though two-dimensional and/or aismmetric analsis. Eamples of three dimensional problems are arch dams, thickwalled pressure vessels, and solid forging parts as used, for instance, in the heav equipment and automotive industries. Introduction The tetrahedron is the basic three-dimensional element, and it is used in the development of the shape functions, stiffness matri, and force matrices in terms of a global coordinate sstem. We follow this development with the isoparametric formulation of the stiffness matri for the heahedron, or brick element. Finall, we will provide some tpical three-dimensional applications.
CIVL 7/87 Chapter - 3/39 Introduction Finite elements can be use for analing linear and nonlinear stress characteristics of structures and mechanical components. Introduction Finite element discretiation, stresses, and deformations of a wheel rim in a structural analsis.
CIVL 7/87 Chapter - 4/39 Introduction Cooling fan blade vibration, which was predicted b structural mechanics mode shape analsis. Introduction The model of a wrench and bolt that has been imported, meshed, and solved for an applied load.
CIVL 7/87 Chapter - 5/39 Introduction Below is a cross brace, supported at the left-hand edge b two lugs and loaded at the right-hand edge through two additional lugs. The structure is sitting in the global XY plane. Introduction The figure shows stress contour plot of the maimum principal stress P. This is a useful wa of seeing the biggest tensile stress flow direction.
CIVL 7/87 Chapter - 6/39 Introduction The figure shows the P3 stress distribution through the structure. The plot shows onl compressive stresses. Introduction A mesh with smaller elements will allow the solver to more accuratel calculate stress distribution across an area. Smaller elements mean more nodes for calculation. More calculation helps to average the stresses in a more accurate manner through the material.
CIVL 7/87 Chapter - 7/39 Introduction Finite element model of human skull showing stress distribution in bite at 2 nd molar. Introduction We begin b considering the three-dimensional infinitesimal element in Cartesian coordinates with dimensions d, d, and d. Normal stresses are perpendicular to the faces of the elements given b,, and.
CIVL 7/87 Chapter - 8/39 Introduction We begin b considering the three-dimensional infinitesimal element in Cartesian coordinates with dimensions d, d, and d. Shear stresses act parallel to the faces and are represented b,, and. Introduction From moment equilibrium of the element, we show in Appendi C that: The element strain-displacement relationships are: u v w
CIVL 7/87 Chapter - 9/39 Introduction Where u, v, and w are the displacements associated with the,, and directions. The shears strains are now given b: u v v w w u Introduction The stresses and strains can be represented b column matrices as: For an isotropic material the stress/strain as: D
CIVL 7/87 Chapter - 0/39 Introduction The constitutive matri [D] is given b: D E 2 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 0 0 2 2 0 0 0 0 0 2 We now develop the tetrahedral stress element stiffness matri b again using the steps outlined in Chapter. The development is seen to be an etension of the plane element previousl described in Chapter 6.
CIVL 7/87 Chapter - /39 Step - Discretie and Select Element Tpes Consider the tetrahedral element with corner nodes -4. The nodes of the element must be numbered such that when viewed from the last node, the first three nodes are numbered in a counterclockwise manner. This ordering of nodes avoids calculation of negative volumes. Step - Discretie and Select Element Tpes Consider the tetrahedral element with corner nodes -4. If the last node is sa, node 4, the first three nodes are numbered in a counterclockwise manner, such as, 2, 3, 4 or 2, 3,, 4. Using an isoparametric formulation to evaluate the [k] matri for the tetrahedral element enables us to use the element node numbering in an order (we will discuss this later).
CIVL 7/87 Chapter - 2/39 Step - Discretie and Select Element Tpes The unknown nodal displacements are given b: u v w d u 4 v 4 w 4 Hence, there are 3 degrees of freedom per node, or 2 total degrees of freedom per element. Step 2 - Select Displacement Functions For a compatible displacement field, the element displacement functions u, v, and w must be linear along each edge. We select the linear displacement function as:,, 2 3 4,, 5 6 7 8,, 9 0 2 u a a a a v a a a a w a a a a In the same manner as in Chapter 6, we can epress the a's in terms of the known nodal coordinates (,,,..., 4 ) and the unknown nodal displacements (u, v, w,..., w 4 ) of the element.
CIVL 7/87 Chapter - 3/39 Step 2 - Select Displacement Functions Skipping the straightforward but tedious details, we obtain: u,, u 6V u 2 2 2 2 2 u 3 3 3 3 3 4 4 4 4 u4 where 6V is obtained b evaluating the determinant: 6V 2 2 2 3 3 3 4 3 4 Step 2 - Select Displacement Functions The coefficients i, i, i, and i (i =, 2, 3, 4) are given b; 2 2 2 3 3 3 4 4 4 2 2 3 3 4 4 2 2 3 3 4 4 2 2 3 3 4 4
CIVL 7/87 Chapter - 4/39 Step 2 - Select Displacement Functions The coefficients i, i, i, and i (i =, 2, 3, 4) are given b; 2 3 3 3 4 4 4 2 3 3 4 4 2 3 3 4 4 2 3 3 4 4 Step 2 - Select Displacement Functions The coefficients i, i, i, and i (i =, 2, 3, 4) are given b; 3 2 2 2 4 4 4 3 2 2 4 4 3 2 2 4 4 3 2 2 4 4
CIVL 7/87 Chapter - 5/39 Step 2 - Select Displacement Functions The coefficients i, i, i, and i (i =, 2, 3, 4) are given b; 4 2 2 2 3 3 3 4 2 2 3 3 4 2 2 3 3 4 2 2 3 3 Step 2 - Select Displacement Functions Epressions for v and w are obtained b simpl substituting v i s for all u i s and then w i s for all u i s. The displacement epression for u can be written in epanded form in terms of the shape functions and unknown nodal displacements as: u v u N 0 0 N2 0 0 N3 0 0 N4 0 0 w v 0 N 0 0 N2 0 0 N3 0 0 N4 0 w 0 0 N 0 0 N2 0 0 N3 0 0 N 4 u 4 v 4 w 4
CIVL 7/87 Chapter - 6/39 Step 2 - Select Displacement Functions The shape functions are given b: N 6V N 2 2 2 2 2 6V N 3 3 3 3 3 6V N 4 4 4 4 4 6V Step 3 - Define the Strain-Displacement and Stress-Strain Relationships The element strains for the three-dimensional stress state are given b: u v w u v v w w u { } [ B]{ d} { } d d B B B B d 4 2 2 3 4 d3
CIVL 7/87 Chapter - 7/39 Step 3 - Define the Strain-Displacement and Stress-Strain Relationships The submatri [B ] is defined b: B N, 0 0 0 N, 0 0 0 N N, N, 0 0 N N N 0 N,,,,, where the comma indicates differentiation with respect to that variable. Step 3 - Define the Strain-Displacement and Stress-Strain Relationships Submatrices [B 2 ], [B 3 ], and [B 4 ] are defined b simpl indeing the subscript from to 2, 3, and then 4, respectivel. Substituting in the shape functions, [B ] is epressed as: 0 0 4 0 0 0 0 0 4 0 0 0 0 0 4 B B4 6V 0 6V 4 4 0 0 0 4 4 0 4 0 4
CIVL 7/87 Chapter - 8/39 Step 4 - Derive the Element Stiffness Matri and Equations The element stresses are related to the element strains b: D DBd The stiffness matri can be defined as: T k B D B dv V Because both matrices [B] and [D] are constant for the simple tetrahedral element, the element stiffness matri can be simplified to: T k B D B V where V is the volume of the tetrahedral element. Step 4 - Derive the Element Stiffness Matri and Equations Treatment of Bod and Surface Forces The element bod force matri is given b: where Xb X Yb Z b T fb N X dv For constant bod forces, the nodal components of the total resultant bod forces can be shown to be distributed to the nodes in four equal parts, that is: V 4 f X Y Z X Y Z X Y Z X Y Z b b b b b b b b b b b b b V T
CIVL 7/87 Chapter - 9/39 Step 4 - Derive the Element Stiffness Matri and Equations Treatment of Bod and Surface Forces The surface force matri is given b: T s S f N T ds where [N S ] is the shape function matri evaluated on the surface where the surface traction occurs. For eample, consider the case of a uniform pressure p acting on a face with nodes -3 of the element. p f N pds S surface,2,3 p T s evaluatedon S Step 4 - Derive the Element Stiffness Matri and Equations Treatment of Bod and Surface Forces Simplifing and integrating we can show that: p p p p p S p 3 p p p 0 0 0 23 f s Where S 23 is the area of the surface associated with nodes -3. The use of volume coordinates facilitates the integration surface forces.
CIVL 7/87 Chapter - 20/39 Step 5 - Assemble the Element Equations and Introduce Boundar Conditions The final assembled or global equation written in matri form is: [ ] F K d where {F} is the equivalent global nodal loads obtained b lumping distributed edge loads and element bod forces at the nodes and [K] is the global structure stiffness matri. Step 6 - Solve for the Nodal Displacements Once the element equations are assembled and modified to account for the boundar conditions, a set of simultaneous algebraic equations that can be written in epanded matri form as: [ ] F K d Step 7 - Solve for the Element Forces (Stresses) For the structural stress-analsis problem, important secondar quantities of strain and stress (or moment and shear force) can be obtained in terms of the displacements determined in Step 6.
CIVL 7/87 Chapter - 2/39 Eample Evaluate the matrices necessar to determine the stiffness matri for the tetrahedral element shown below. Let E = 30 0 6 psi and = 0.30. The coordinates are in units of inches. Eample To evaluate the element stiffness matri, we first determine the element volume V and all s, s, s, and s we have: 6V Also, we obtain: 2 2 2 3 3 3 4 3 4 2 0 0 0 0 2 0 2 0 3 8in 0 0 0 0 0 0 2 0 0 2 0 0 2 0 0
CIVL 7/87 Chapter - 22/39 Eample To evaluate the element stiffness matri, we first determine the element volume V and all s, s, s, and s we have: 6V Also, we obtain: 2 2 2 3 3 3 4 3 4 2 0 0 0 0 2 0 2 0 3 8in 2 0 0 0 0 0 0 2 4 0 0 2 Eample The remaining values for the s, s, s, and s are: 8 2 4 2 2 2 2 0 2 4 3 3 3 3 0 4 0 2 4 4 4 4 Note that s tpicall have units of cubic inches or cubic meters, where s, s, and s have units of square inches or square meters.
CIVL 7/87 Chapter - 23/39 Eample Net, the shape functions are: 4 8 N N2 N 2 4 8 3 N4 82 4 8 4 2 8 Note that N + N 2 + N 3 + N 4 = is again satisfied. Eample The 6 3 submatrices of the matri [B] are now evaluated as: B 0 0 0 0 0 0 0 0 2 0 0 0 0 2 0 2 0 0 B 2 4 0 0 0 2 0 0 0 8 2 4 0 0 8 2 8 0 4
CIVL 7/87 Chapter - 24/39 Eample The 6 3 submatrices of the matri [B] are now evaluated as: B 3 4 0 0 0 2 0 0 0 8 2 4 0 0 8 2 8 0 4 B 4 2 0 0 0 0 0 0 0 4 0 2 0 0 4 0 4 0 2 Eample Net, the [D] matri is evaluated as: D 0.7 0.3 0.3 0 0 0 0.3 0.7 0.3 0 0 0 6 30 0 0.3 0.3 0.7 0 0 0 0.30.6 0 0 0 0.2 0 0 0 0 0 0 0.2 0 0 0 0 0 0 0.2
CIVL 7/87 Chapter - 25/39 Eample Finall, substituting the results for V, [B], and [D], we obtain the element stiffness matri. T k B D B V 6 k 0 Isoparametric Formulation and Heahedral Element The basic (linear) heahedral element has eight corner nodes with isoparametric natural coordinates given b s, t, and. The formulation of the stiffness matri follows step analogous to the isoparametric formulation of the stiffness matri for the plane element in Chapter 0.
CIVL 7/87 Chapter - 26/39 Isoparametric Formulation and Heahedral Element The function used to describe the element geometr for in terms of the generalied degrees of freedom a i s is: a a sa t a ' a st a t' a ' sa st' 2 3 4 5 6 7 8 a a sa t a ' a st a t' a ' sa st' 9 0 2 3 4 5 6 a a sa t a ' a st a t' a ' sa st' 7 8 9 20 2 22 23 24 First, epand the coordinate definition from Chapter 0 to include the coordinate as follows: N 0 0 8 i i 0 Ni 0 i i 0 0 N i i Isoparametric Formulation and Heahedral Element The shape functions are: Ni ssi tti ' ' i 8 where s i, t i, and i = and i =, 2,, 8. For N : N ssi tti ' ' i 8 where s = -, t = - and =, we obtain: N s t ' 8
CIVL 7/87 Chapter - 27/39 Isoparametric Formulation and Heahedral Element The shape functions are: N s t ' 8 Isoparametric Formulation and Heahedral Element Eplicit forms of the other shape functions follow similarl. The shape functions map the natural coordinates (s, t, ') of an point in the element to an point in the global coordinates (,, ). For instance, when we let i = 8 and substitute s 8 =, t 8 =, and 8 =, we obtain: N8 s t ' 8
CIVL 7/87 Chapter - 28/39 Isoparametric Formulation and Heahedral Element Similar epressions are obtained for the other shape functions. Then evaluating all shape functions at node 8, we obtain N 8 =, and all other shape functions equal ero at node 8. We see that N = 0 when s = or when t =. Therefore, we obtain: 8 8 8 We see that indeed our geometric shape functions map an point in the natural-coordinate sstem to one in the globalcoordinate sstem. Isoparametric Formulation and Heahedral Element The displacement functions in terms of the generalied degrees of freedom are of the same form as used to describe the element geometr given b: ua asata' astat' as ' ast' 2 3 4 5 6 7 8 v a a sa t a ' a st a t' a ' sa st' 9 0 2 3 4 5 6 w a a sa t a ' a st a t' a ' sa st' 7 8 9 20 2 22 23 24 There are now a total of 24 degrees of freedom in the linear heahedral element. Therefore, we use the same shape functions as used to describe the geometr.
CIVL 7/87 Chapter - 29/39 Isoparametric Formulation and Heahedral Element The displacement functions is given b: u N 0 0 u 8 i i v 0 Ni 0 v i i w 0 0 N i w i with the same shape functions, the sie of the shape function matri now 3 X 24. Isoparametric Formulation and Heahedral Element The Jacobian matri in three dimensions is: J s s s t t t ' ' '
CIVL 7/87 Chapter - 30/39 Isoparametric Formulation and Heahedral Element The strain-displacement relationships, in terms of global coordinates are: f s s s f t t t f f ' ' ' J f s s s f t t t f f ' ' ' J f s s s f t t t f f ' ' ' J Substituting u, v, and then w for f and using the definitions of the strains, we can epress the strains in terms of natural coordinates (s, t, ). Isoparametric Formulation and Heahedral Element The matri [B] is now a function of s, t, and ' and is of order 6 24. The 24 24 stiffness matri is now given b: T [ k] B D B J dsdtd' It is best to evaluate [k] b numerical integration; that is, we evaluate (integrate) the eight-node heahedral element stiffness matri using a 2 2 2 rule (or two-point rule). Actuall, eight points defined in Table -are used to evaluate [k] as
CIVL 7/87 Chapter - 3/39 Isoparametric Formulation and Heahedral Element Gauss points and weights for the linear heahedral are: Points, i s t ' w i i i i 3 3 3 2 3 3 3 3 3 3 3 4 3 3 3 5 3 3 3 6 3 3 3 7 3 3 3 8 3 3 3 Isoparametric Formulation and Heahedral Element Using the eight Gauss points to evaluate [k] give: 8 i T [ k] B si, ti, ' i D B si, ti, ' i J si, ti, ' i Wi As is true with the bilinear quadrilateral element, the eightnoded linear heahedral element cannot model beambending action well because the element sides remain straight during the element deformation. During the bending process, the elements will be stretched and can shear lock. The quadratic heahedral element described subsequentl remedies the shear locking problem.
CIVL 7/87 Chapter - 32/39 Quadratic Heahedral Element The quadratic heahedral element has a total of 20 nodes with the inclusion of a total of 2 mid-side nodes Quadratic Heahedral Element The function describing the element geometr for in terms of the 20 a s is: a asata' astat' as ' as at 2 2 2 3 4 5 6 7 8 9 a ' a s t a st a t ' a t' a ' s 2 2 2 2 2 2 0 2 3 4 5 a ' s a st ' a s t ' a st ' a st ' 2 2 2 2 6 7 8 9 20 Similar epressions describe the and coordinates.
CIVL 7/87 Chapter - 33/39 Quadratic Heahedral Element The function describing the element geometr for in terms of the 20 a s is: a asata' astat' as ' as at 2 2 2 3 4 5 6 7 8 9 a ' a s t a st a t ' a t' a ' s 2 2 2 2 2 2 0 2 3 4 5 a ' s a st ' a s t ' a st ' a st ' 2 2 2 2 6 7 8 9 20 The -displacement function u is described b the same polnomial used for the element geometr. Similar epressions are used for displacement functions v and w. Quadratic Heahedral Element The function describing the element geometr for in terms of the 20 a s is: a asata' astat' as ' as at 2 2 2 3 4 5 6 7 8 9 a ' a s t a st a t ' a t' a ' s 2 2 2 2 2 2 0 2 3 4 5 a ' s a st ' a s t ' a st ' a st ' 2 2 2 2 6 7 8 9 20 In order to satisf interelement compatibilit, the three cubic terms s 3, t 3, and ' 3 are not included. Instead the three quartic terms s 2 t, st 2 ', and st' 2 are used.
CIVL 7/87 Chapter - 34/39 Quadratic Heahedral Element The development of the stiffness matri follows the same steps we outlined before for the linear heahedral element, where the shape functions now take on new forms. Quadratic Heahedral Element Again, letting s i, t i, and i =, we have for the corner nodes (i =, 2,..., 8): ssitti ' ' i ' ' 2 Ni ssi tti i 8
CIVL 7/87 Chapter - 35/39 Quadratic Heahedral Element For the mid-side nodes at s i = 0, t i =, and i =, (i = 7, 8, 9, and 20), we get: N i 2 s tt ' ' i i 4 Quadratic Heahedral Element For the mid-side nodes at s i =, t i = 0, and i =, (i = 0, 2, 4, and 6), we get: N i 2 ss t ' ' i 4 i
CIVL 7/87 Chapter - 36/39 Quadratic Heahedral Element For the mid-side nodes at s i =, t i =, and i = 0, (i = 9,, 3, and 5), we get: N i 2 ssitti ' 4 Quadratic Heahedral Element The [B] matri is now a 60 60 matri. The stiffness matri of the quadratic heahedral element is of order 60 60. This is consistent with the fact that the element has 20 nodes and 3 degrees of freedom (u i, v i, and w i ) per node
CIVL 7/87 Chapter - 37/39 Quadratic Heahedral Element The stiffness matri for this 20-node quadratic solid element can be evaluated using a Gaussian quadrature with ther 3 3 3 rule (27 points). However, a special 4-point rule ma be a better choice. As with the eight-noded plane element of Chapter 0, the 20- node solid element is also called a serendipit element. Quadratic Heahedral Element The 20-node solid uses a different tpe of integration point scheme. This scheme places points close to each of the 8 corner nodes and close to the centers of the 6 faces for a total of 4 points.
CIVL 7/87 Chapter - 38/39 Quadratic Heahedral Element The 20-node solid uses a different tpe of integration point scheme. This scheme places points close to each of the 8 corner nodes and close to the centers of the 6 faces for a total of 4 points. Tpe Integration Point Location Weighting Factor 8 Corner Points s = ±0.75878 6906 39328 0.3358 00554 0662 t = ±0.75878 6906 39329 = ±0.75878 6906 39329 6 Center Points s = ±0.79582 24257 54222, t= =0.0 0.88642 65927 97784 t = ±0.79582 24257 54222, s= =0.0 = ±0.79582 24257 54222, s=t=0.0 Isoparametric Elements Problems 25. Work problems.,.3,.6a, and.9 in our tetbook. 26. Use the solid elements in SAP2000 to solve problem.4 in our tetbook.
CIVL 7/87 Chapter - 39/39 End of Chapter