## B Analysis Ii Thermal Shock

In this second analysis, we explore a different potential cracking scenario Thermal Shock during construction. That is we speculate that during construction, possibly in winter, the heat 62,640 N m2 (estimated weight of concrete slab ( (6)m(5.8)m(.3)m(2400)A'( m3 (10)N Kg of hydration caused by the 280 Kg m3 concrete caused a thermal gradient large enough in the downstream face to induce cracking. Hence, our first step was to create a boundary file which accounts for the different concrete...

## Bzw Vl trdt W

Where r denotes the gamma function which satisfies r(n + 1) n for integers n > 0. Thus f Uiti iA 2Ar(* + i)r(i + i)r(m + i) J A r(fc + l + m + 3) v ' Since k, l, and m are nonnegative integers, 56 It should be noted that the above relation is used only if we have straight sided elements and the element formulation can thus be analytically derived. Alternatively, if we have a curvilinear side, this formula will not be used, and the element will be numerically integrated within the context of...

## C

5 With reference to Fig. 6.1 we start with 6 Since we have 2 d.o.f's, we will assume a linear deformation state where u can be either u or 0, and the B.C.'s are given by u u at x 0, and u U2 at x L. Thus we have r Solving for ai and a,2 in terms of U and 112 we obtain 8 Substituting and rearranging those expressions into Eq. 6.4 we obtain 9 The previous derivation can be generalized by writing where p corresponds to the polynomial approximation, and a is the coefficient vector. io We next apply...

## D y [qy

Da -daxiw + dwda ox Oy where we define the specific heat c as the amount of heat required to raise a unit mass by one degree. Figure 3.6 Flux Through Sides of Differential Element Figure 3.7 *Flow through a surface r 93 From the first law of thermodynamics, energy produced I2 plus the net energy across the boundary I1 must be equal to the energy absorbed I3, thus - dxdy + 1-dydx + Qdxdy cp dxdy 0 (3.105-b) 94 The amount of flow per unit time into an element of volume Q and surface r is Ii J q(...

## F F

Pipe Network where the displacement correspond to the pressure p 1. Element stiffness matrices Spring 1 2. Assemble active degrees of freedom 4. Rearrange state variables and multiply by 30 DC Network where the displacement u correspond to the current intensity I 1. Element stiffness matrices Spring 1 2. Assemble active degrees of freedom 2.3.2.1 Nonlinear Elastic Spring 2.3.3 Propagation Problems 18 The main characteristic of a propagation dynamic problem is that the response of the system...

## F l

Where u is the internal energy per unit mass or specific internal energy. We note that U appears only as a differential in the first principle, hence if we really need to evaluate this quantity, we need to have a reference value for which U will be null. The dimension of U is one of energy dim U ML2T-2, and the SI unit is the Joule, similarly dim u L2T-2 with the SI unit of Joule Kg. 58 The Generalized Hooke's Law can be written as 59 The (fourth order) tensor of elastic constants Dijkl has 81...

## K Un F j

(e - - 6) ( - - a+o (e - e ) ,(), h , (6- - 6X6- - 6) (6- - 6-1)(6- - 6-+0 (6- - 1 For a three noded quadratic element 1, 2 +1, and 3 0. Substituting, we obtain the three shape functions - (i2_5l)(52_53) - (l+l)(l-0) - 2 N,tn - _ (g+D(g-D _ irf - (i3-il)(S3-i2) (0+l)(0-l) - 1 5 LNJ L x-2 x-i T and u(0 LNJ L JT (9.7) 11 The strain displacement, relation is given by Eq. 7.4, e Lu LNue Bue, and the differential operator L is equal to For this one dimensional case, this reduces to 12 We invoke the...

## R dJVi d No l

4 For the truss element, the constitutive matrix D reduces to the scalar E Hence, substituting into Eq. 1.15 ke B T D B dQ and with dQ Adx for element with constant cross J n 5 We observe that this stiffness matrix is identical to the one earlier derived in Eq. . 8.2.2 Beam Element 6 For a beam element, for which we have previously derived the shape functions in Eq. 6.41 and the B matrix in Eq. 1.4-d, substituting in Eq. 1.15 and noting that y dA Iz Eq. 1.15 reduces to J A 7 For this simple...

## Parasitic Shear Incompatible Elements Q The Problem

37 The shape functions for the bilinear element are given by Eq. 9.19 Ni Z,rj 1 1 - 0 1 - 7 1 1 0 1 - 7 mt,v 1 1 0 1 v 1 1- 0 1 v 38 Imposing the displacement field field of Fig. 13.5 the bilinear element and correct displacements and strains are given by Table 13.2. Table 13.2 Bilinsear and Exact Displacements Strains Table 13.2 Bilinsear and Exact Displacements Strains 39 From this table we note that parasitic shear contributes to the strain energy stored in the element which is equal to PT.d...

## Strong Form

3 Column buckling theory originated with Leonhard Euler in 1744. 4 An initially straight member is concentrically loaded, and all fibers remain elastic until buckling occur. For buckling to occur, it must be assumed that the column is slightly bent as shown in Fig. 14.2. Note, in reality no column is either perfectly straight, and in all cases a minor 14.1.1.1 Lower Order Differential Equation 5 At any location x along the column, the imperfection in the column compounded by the concentric load...

## Ujhk uj skuj

33 For k 0 and k 1, the values of sk are 0.0 and 1.0, respectively. Therefore, uJ 1'0 uJ and uj hl uj l. The orthogonality condition is quantified by a. scalar value gk representing the iterative change in energy, which is defined as n3 hk Txt -rV U' 16.31 are the residual loads at the end of solution iteration j and line search iteration k. 34 gk can be expressed as a function of sk see Figure 16.6 and the object of the line search is to find sk such that gk is zero. An estimate of sk l such...

## Indirect Displacement Control

56 Direct displacement control can be applied only on structures loaded only at one point, or when the load is transmitted by a stiff platen so that all points on the loaded surface exhibit the same displacements. 57 However, this is not always the case. As an example, consider a dam loaded by hydrostatic pressure due to reservoir overflow see Fig. 16.9. Here, the load is applied along a large portion of Figure 16.9 Hydrostatically Loaded Gravity Dam Figure 16.9 Hydrostatically Loaded Gravity...

## Computer Implementation Algorithm

107 The computer implementation of a numerically integrated isoparametric element is summarized as follows. But first, it is assumed that this operation is to be performed in a function called stiff and it takes as input arguments elcod, young, poiss, type, ndime, ndofn, ngaus. In turn it will compute the stiffness matrix KELEM of element ielem. 1. Retrieve element geometry and material properties for the current element 3. Call function dmat to set the constitutive matrix De of the element 4....

## Vector Differentiation

2 A field is a function defined over a continuous region. This includes, Scalar Field g x , Vector Field v x , or Tensor Field T x . 3 We first introduce the differential vector operator Nabla denoted by V 4 We also note that there are as many ways to differentiate a vector field as there are ways of multiplying vectors, the analogy being given by Table A.1. V-v divergence V X v curl Vv gradient. Table A.1 Similarities Between Multiplication and Differentiation Operators 5 The derivative of a...

## List of Tables

1.1 Summary of Variational Terms Associated with One Dimensional Elements . . . 1-4 3.1 Selected Examples of Diffusion Problems 3-19 3.2 Comparison of Scalar and Vector Field Problems 3-26 3.3 Classification of various Physical Problems, Kardestuncer 1987 3-27 5.1 Functionals in Linear 5.2 Comparison Between Total Potential Energy and Hu-Washizu Formulations . . . 5-12 6.1 Characteristics of Beam Element Shape Functions 6-9 6.2 Interpretation of Shape Functions in Terms of Polynomial Series 1D...

## Direct Displacement Control

Adapted from Jirasek and Bazant 2001 43 Independently of the choice of iterative algorithm, any solution strategy using load control fails if the prescribed external loads cannot be maintained in equilibrium by the internal forces. This would typically occur if the load is monotonically increased until the load-carrying capacity of the structure is exhausted, Fig. 16.8 Figure 16.8 Divergence of Load-Controled Algorithms Figure 16.8 Divergence of Load-Controled Algorithms 44 In most engineering...

## Introduction

1 Whereas the first course focused exclusively on one dimensional rod elements, this course will greatly expand our horizons by considering introducing a methodology to solve partial differential equations, with special emphasis on solid mechanics. 2 The field of mechanics, can itself be subdivided into four major disciplines Theoretical which deals with the fundamental laws and principles of mechanics. A Continuum Mechanics course is a must. Applied mechanics seeks to apply the theoretical...

## Elastic Instability Bifurcation Analysis

41 In elastic instability, the intensity of the axial load system to cause buckling is yet unknown, the incremental stiffness matrix must first be numerically evaluated using an arbitrary chosen load intensity since Kg is itself a function of P . 42 For buckling to occur, the intensity of the axial load system must be A times the initially arbitrarily chosen intensity of the force. Note that for a structure, the initial distribution of P must be obtained from a linear elastic analysis. Hence,...

## Tangent Stiffness Matrix

23 It should be noted that each iteration involves three computationally expensive steps 1. Evaluation of internal forces fmt or reactions 2. Evaluation of the global tangent stiffness matrix Kt 3. Solution of a system of linear equations 24 This method is essentially the same as the Newton-Raphson however in Eq. 16.23 K is replaced by Kt which is the tangent stiffness matrix of the first iteration of either 1 the first increment Kt Ky0, Fig. 16.4, or 2 current increment, Fig. 16.3 Kt Kyn Fig....

## [ n n n n

1 - 1 - n 1 - n n n 1 - n 15.56-a 15.56-b 15.56-c 15.56-d 43 In order to facilitate the integration of Ks, a one point Gauss integration used. Upon substitution where a and 5 a 3. 44 The final element stiffness matrix is and in terms of the global stiffness matrix we will have 45 Through inspection of the previous equation, and noting that we observe that for very thin plates is very large, hence as t gt 0, gt 00. Hence, unrealistic u 0 can be obtained independently of the load. This phenomena...

## [uTn UTm

It is these general forms of the constraint equations that are implemented in MERLIN. Figure 16.13 Flow chart for line search with IDC methods 1 1, k l j l,k l Compute D , CT 1 1, k l j l,k l Compute D , CT Figure 16.13 Flow chart for line search with IDC methods Abramowitz, M. and Stegun, I. 1970, Handbook of mathematical functions, Technical report, National Bureau of Standard. Applied Mathematics Series, No. 55. Babuska, I. 1971, Error-bounds for finite element methods, Numer. Math 20 3 ,...

## Element Evaluation Patch Test

6 The patch test is a check which ascertains whether a patch of infinitesimally small elements subjected to constant strain reproduces exacly the constitutive behavior of the material through correct stresses. 7 It has been argued that an element which passes the patch test satisfies the two essential conditions for convergence. 8 First we assemble a small number of elements into a patch in such a way that there is at least one internal node shared by two or more elements, and that one or more...

## Fundamental Relations Equilibrium

2 Considering an arbitrary plate, the stress are given by, Fig. 15.2, resultants per unit width 3 Note that in plate theory, we ignore the effect of the membrane forces, those in turn will be accounted for in shells. 4 The equation of equilibrium is derived by considering an infinitesimal element tdx dy subjected to an applied transverse load pz. We would have to consider three equations of equilibrium, Fig. 15.3 Summation of Forces in the z direction dxdy - z-dxdy pAxdy 0 dx dy Figure 15.3...

## List of Figures

1.1 Summary of Variational 1.2 Duality of Variational Principles 1-3 1.3 Frame 1.4 Example for ID Matrix 1.5 Simple Frame Analyzed with the MATLAB Code 1-11 1.7 Simple Frame Analyzed with the MATLAB Code 1-18 1.8 Stiffness Analysis of one Element Structure 1-21 2.1 Finite Element Process, Bathe 1996 2-2 2.2 Seepage 2.3 One Dimensional Heat 2.4 Rod subjected to Step 2.5 System of Rigid Carts Interconnected by Linear Springs, Bathe 1996 2-9 2.6 Slab Subjected to Temperature Boundary Conditions,...

## Eigenvalue Test

12 The stiffness matrix is, by definition, singular due to the fact that equilibrium relations are embedded in its formulation, or alternatively, the assumed displacement field on which it is based is supposed to provide for rigid body motions translations and or rotations . 13 The augmented stiffness matrix may be expressed as where B is the statics or equilibrium matrix, relating external nodal forces to internal forces d is a flexibility matrix, and d_1 is its inverse or reduced stiffness...

## Mesh Generation

Requires further editing 4.1 Introduction 1 Finite element mesh generation is now an integral part of a finite element analysis. With the increased computational capabilities, increasingly more complex structures are being analysed. Those structures must be discretized. 2 The task is one of developing a mathematical model discretization or tessalation of a continuum model. This is not only necessary in finite elment analysis, but in computer graphics rendering also. 3 In computer graphics, we...

## F tndtdn nm P t ndtdn E wmwf f ti Vj

And the total number of integration points will thus be m x n, Fig. 9.16. 9.3.4 Triangular Regions 85 For the numerical integration over a triangle, the Gauss points are shown in Fig. 9.17, and the corresponding triangular coordinates are given by Table 9.5. Figure 9.16 Gauss-Legendre Integration Over a Surface Figure 9.17 Numerical Integration Over a Triangle Table 9.5 Coordinates and Weights for Numerical Integration over a Triangle Table 9.5 Coordinates and Weights for Numerical Integration...

## Fundamental Relations

1 Whereas, ideally, a course in Continuum Mechanics should be taken prior to a finite element course, this is seldom the case. Most often, students have had a graduate course in Advanced Strength of Materials, which can only provide limited background to a solid finite element course. 2 Accordingly, this preliminary chapter mostly extracted from the author's lecture notes in Continuum Mechanics will partially remedy for occasional defficiencies and will be often referenced. 3 It should be noted...

## Derivation of the Weak Form

3 Our starting point will be the field equation, Eq. 3.113 -div DV lt gt -Q pM 12.1 4 Recalling Eq. 3.98 that q -DV0, and that for steady state problems the right hand side reduces to zero, and finally for the mere sake of clarity substituting 0 by T, this equation reduces to 5 The boundary conditions are given by Essential, Temperature rT T g Natural, Flux qn q n qf 12.3 Natural, Convection Flux rc qc h T - T 6 Note that in a fin, the convection heat loss can be considered as a negative heat...

## Finite Element Discretization and Requirements

This section is mostly extracted from Reich 1993 7.1.1 Discretization of the Variational Statement for the General TPE Variational Principle 1 The discretization of Equation 5.8 will be performed on an element domain Qe using the procedures described in Chapter 2 of Zienkiewicz and Taylor 1989 2 The surface of the element subjected to surface tractions r comprises one or more surfaces of the element boundary r. For the present time this discussion will be kept on a very general level with no...

## L

Show that the kinetic energy and the strain energy are given by 2. For Wnc 0 derive an expression for the first variation of the Hamilton's functional 3. If we are interested in determining the periodic motion, which has the form where w is the frequency of natural vibration, and uo x is the amplitude. Show that 4. Show that the Euler equations of the preceding functional are EA PAlo2u0 0 for 0 lt x lt L it - on n - m. - i l k2 , a j , EA , a E 5. Rewrite Eq. 5.24 in terms of x u, 6. Show...

## Order of Integration Full Integration

22 For numerically integrated elements, a question of paramount importance is the order of the numerical integration. If the order of integration is too High it would certainly result in an exact formulation, which however may be computationally expensive. Low Then we may have rank deficiency. Hence, we ought to properly select the order of numerical quadrature. 1. We seek to integrate F BTDBdetJ. 2. The Jacobian is constant only for parallelograms. 3. In Gaussian Quadrature, integration with n...

## FQ UdQ Jr uitidr

Figure 1.1 Summary of Variational Methods Cinematically Admissible Displacements Displacements satisfy the kinematic equations and the the kinematic boundary conditions Principle of Statio Complementary En iciple of Stationary Potential Energy Stresses satisfy the equilibrium conditions and the static boundary conditions Figure 1.2 Duality of Variational Principles Table 1.1 Summary of Variational Terms Associated with One Dimensional Elements Table 1.1 Summary of Variational Terms Associated...

## Variational and Rayleighritz Methods

5.1 Multifield Variational Principles 1 A Multifield variational principle is one that has more than one master field or state variable , that is more than one unknown field is subject to independent variations. 2 In linear elastostatics, we can have displacement, u, strains e, or stress a as potential candidates for master fields. Hence seven combinations are possible, Felippa 2000 , Table 5.1. 3 In this course, we shall focus on only the Total Potential Energy, and the Hu-Washizu varia-tional...

## [k e [reT [ke[re

Where c cos a X2 TXl s sin a Element lL 20',c 0.8, s 0.6, 3 ' 00 in3 Element 2 L 16' , c 1 , s 0 , 18,750 k ft. Element 4 L 16' , c 1 , s 0 , 18,750 k ft. 18, 750 0 18, 750 0 9, 600 -7, 200 -9, 600 _ 7, 200 -7, 200 -9, 600 5, 400 7, 200 7, 200 9, 600 -5, 400 -7, 200 7, 200 -9, 600 5, 400 -7, 200 -7, 200 9, 600 -5, 400 7, 200 4. Assemble the global stiffness matrix in k ft Note that we are not assembling the augmented stiffness matrix, but rather its submatrix Kti . -25, 000 7, 200 - 7, 200 25,...