University of Illinois at Urbana-Champaign College of Engineering

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1 University of Illinois at Urbana-Champaign College of Engineering CEE 570 Finite Element Methods (in Solid and Structural Mechanics) Spring Semester 2014 Quiz #2 April 14, 2014 Name: SOLUTION ID#: PS1.: ALL the pages must be stapled at all times. PS2: You are allowed to use pen, pencil, eraser, and a calculator. Mobile phones are NOT allowed during the examination. Please turn your cell phones off. PS3.: This is a closed book, closed notes, open minds exam. Show ALL work on the exam sheets SCORE: Problem 1: 20 / 20 points Problem 2: 20 / 20 points TOTAL : 100 Problem 3: 10 / 10 points Problem 4: 50 / 50 points Page 1 of 12

2 Problem 1 (20 points total) In each question below, define or describe the term given and explain its significance in relation to finite element analysis, theory, or practice. Use a few sentences, several phrases, or an outline form. You will be graded for content and logical presentation of the topics. Topic (1): Galerkin Method It is a general method of formulation of finite element (FE) equations, which is a particular version of the Method of Weighted Residuals (MWR). It starts with governing differential equations (DEs). and that is an approximate FE model, Nd. The residual is defined as 0, and the MWR seeks a solution such that 0, where {W} is a vector of weighting functions. In the Galerkin method, the weight is taken as the shape function. Another characteristic of the Galerkin FE method is that 0 is usually integrated by parts (using the divergence theorem) to obtain lower order derivatives in the integral and weak requirements on BCs and condition of continuity. This weak formulation is then equivalent to an energy (variational) formulation, if such exists. The advantage of the Galerkin method is that generality it is applicable to any problem which can be formulated in terms of differential equation regardless of whether a corresponding variational formulation exists. The Galerkin method yields a symmetric coefficient matrix if the system of DE+BCs is self-adjoint. Page 2 of 12

3 Problem 1 (continued) Topic (2): C 0 (i.e. continous) triangular elements with complete polynomial models (Hint: remember Mr. Pascal and the famous Pascal triangle) Complete 2D polynomials can be constructed from the Pascal triangle. The addition of each full row gives a successively higher-order complete polynomial while parts of rows represent incomplete polynomials. The only polynomial 2D C elements which embody complete polynomial and which have only exterior nodes are: (1) The 3-node linear-polynomial triangle & (2) The 6-node quadratic-polynomial triangle. All other shapes and orders have either interior nodes or incomplete polynomials or perhaps both. Nevertheless, the triangular shapes are best able to accommodate higher-than-quadratic-order complete polynomials. Examples are the 10-node complete cubic element with 9 exterior nodes and 1 interior node, and the 15-node complete quartic triangle with 12 exterior and 3 interior nodes. Page 3 of 12

4 Problem 2 (20 points) MW.R: Consider the differential equation: u, x + 2u 15x = 0 in the range 0<x<1, with the boundary condition u=0 at x=0. An admissible 2-parameter approximating polynomial is: u* = a 1 x + a 2 x 2 Determine a 1 and a 2 using the Galerkin method. [ Acknowledgement: this question is based on Homework #4, Problem of the textbook by Cook et al ] Residual methods: ;, 2, ; Galerkin method: d 0 (1) wrdx x x a x x a xdx x x a x x a x dx a1 a a1 a (2) wrdx x x a x x a xdx x x a x x a x dx a1 a a1 a (3) From Eq. (1) and Eq. (2) , Page 4 of 12

5 Problem 3 (10 points) - I-CLICKER QUESTIONS 1. ku 2, xx u, t a. Hyperbolic PDE b. Elliptic PDE c. Parabolic PDE d. Higher-order PDE e. None of the above 2. Positive Discriminant a. Hyperbolic PDE b. Parabolic PDE c. Elliptic PDE d. Higher-order PDE e. None of the above au 2, xx u, tt a. Laplace Equation b. Abel Equation c. Wave Equation d. Diffusion Equation e. None of the above au 2, xx u, tt a. Hyperbolic PDE b. Elliptic PDE c. Parabolic PDE d. Higher-order PDE e. None of the above 5. Which method produces a symmetric matrix a. Petrov-Galerkin b. Collocation c. Least Squares d. Sub-domain e. All the above 6. The most popular version of the boundary element method (BEM) is based on a. collocation b. sub-domain c. least squares d. Galerkin e. None of the above Page 5 of 12

6 7. Which method is associated to a Self-Adjoint operator a. collocation b. sub-domain c. Galerkin d. Petrov-Galerkin e. None of the above 8. The main advantage of an FEM topological data structure is a. Efficient determination of adjacency relationships b. Effective for remeshing / dynamic remeshing c. Effective for mesh modification / handling d. Especially suited to treat evolution problem (fragmentation) e. All of the above Reference: W. Celes, G.H. Paulino and R. Espinha, "A compact adjacency-based topological data structure for finite element mesh representation." IJNME, 64(11), , What is the advantage of a radially graded mesh? a. Provides smooth mesh transition b. Alleviates mesh ill-conditioning c. Reduces the problem size with reasonable solution accuracy d. Effective to connect coarse & refined portions of a mesh e. All of the above 10. Which statement is FALSE regarding the nature of FEM solutions: a. Compatibility prevails at nodes b. Compatibility is satisfied within elements c. Equilibrium of nodal forces and moments is satisfied d. Equilibrium is usually satisfied within elements e. Equilibrium is usually not satisfied across inter-element boundaries Page 6 of 12

Problem 4 (50 points total) [Patran/ABAQUS/CEE471/FEA] [Acknowledgement: This question is related to your project, and is intended to assess your knowledge of FEM to solve practical engineering

The pipe is made of AISI 4130 steel; 29 10 and 0.285, with yield strength of 72,000. The pipe has an internal pressure of 3,000. Assume the pipe is very long.

7 Problem 4 (50 points total) [Patran/ABAQUS/CEE471/FEA] [Acknowledgement: This question is related to your project, and is intended to assess your knowledge of FEM to solve practical engineering problems.]. A seamless pressure pipe with 1 diameter is modeled using finite element analysis. The wall thickness is 0.12 and the internal diameter is The pipe is made of AISI 4130 steel; and 0.285, with yield strength of 72,000. The pipe has an internal pressure of 3,000. Assume the pipe is very long. The pipe is modeled as a quarter ring, with dimensions and boundary conditions as in the figure below. Part 1 Review of CEE471 a) What is the expected value of at points a and b indicated in the Figure (assume zero shear) Points a and b are at the boundary, where we can use Cauchy s relations (from CEE471) to Φ Φ determine the stress state: Φ Φ Thus, the values for at points a and b are: 3,000 ; 0 These values are theoretical, and the FEM solution is not expected to have these exact values (FEM will converge to these values). Page 7 of 12

Resulting in: 0 ; 0, with oriented tangent to the pipe, thus: Sample sketch c) Calculate the expected horizontal force

8 b) Can the direction of the principal stresses and at point c be known a priori? If so, sketch the orientation at point c in the sketch given below. Point c is at the boundary, where 0, 0 and 0. Resulting in: 0 ; 0, with oriented tangent to the pipe, thus: Sample sketch c) Calculate the expected horizontal force in section a b, and compare with the previous value. The expected horizontal force can be calculated from equilibrium: 2 cos dθ 2 sin 2 3, ,140 Page 8 of 12

9 Part 2 Finite Element Analysis The domain is discretized with Q4 elements. The stresses for the nodes in section a b are given in Table 1. Table 1: Stresses for nodes in section a b y [in] σ xx [psi] σ yy [psi] τ xy [psi] ,975-2, ,358-2, ,621-1, , , d) What type of ABAQUS element is appropriate for this analysis? Why? The pipe is very long, and thus we can assume 0: Plane strain is appropriate. The element in ABAQUS should be CPE4. e) Calculate the resultant force in section a b in the X direction using from the FE analysis (Table 1). The Q4 elements are linear on their edges, thus we can integrate by trapezoids: 10,975 10,358 10,358 9,621 9,621 8, ,851 8, , , , , ,140.7 Page 9 of 12

10 f) Based on the following Von Mises stress plot, what is the stress concentration factor for the given pressure load? Use this value to calculate the internal pressure that initiates yielding of the pipe. The maximum Von mises stress is 12,000 The stress concentration factor is: 4 The yield strength for AISI 4130 steel is 72,000. The pressure that initiates yielding is: 72, ,000 g) Based on your knowledge of CEE471 and the values in Table 1: What is the value of (out of plane) at point a ( 0.38? Hooke s law for Isotropic material is given as xx xx yy yy zz zz yz E (1 ) 0 0 yz xz (1 ) 0 xz xy (1 ) xy Page 10 of 12

Assuming plane strain condition, and knowing that: 1 0, we can solve for : 0.28510,975 2771 0.2858,204 2,338.1 h) How can the stress solution be improved?

11 Assuming plane strain condition, and knowing that: 1 0, we can solve for : , ,204 2,338.1 h) How can the stress solution be improved? h-, p-, r-, s- refinements or a combination of these. i) What stress is plotted in the following Figure? Why? Note that the stress is negative everywhere. This is a plot of the shear stress. This can be inferred in a number of ways: The maximum stress is at 45 and has its minimum at 0 and 90, as is expected for the shear in this type of problems. In a pipe problem, only is expected to have symmetry with the and axis. Knowing that the principal stresses are oriented in the and directions, then is expected to have extreme values at 0 and 90, with a smooth gradation Page 11 of 12

12 between these points. The same applies to, only that the extreme values will be reversed. Therefore, it cannot be any of these two: it can only be a plot of the shear stress. Recalling Mohr s circle, the shear is responsible for rotating the principal stress directions. The principal stresses and need no rotation at 0 and 90. However, at 45, Mohr s circle will have maximum shear and the angle in the circle will be 2 90, in accordance with the Figure. j) For the sake of simplicity, we are adopting a simplified mesh (Q4 elements) for the same problem. Draw (neatly) the DUAL GRAPH (DG) and the COMMUNICATION GRAPH (CG) associated to the simplified FEM mesh below. Using the graph of your preference, provide a feasible element numbering to be used a frontal solution scheme solver. [Hint: Notice that both DG and CG are element graphs (not a nodal graph). ] Page 12 of 12

University of Illinois at Urbana-Champaign College of Engineering

University of Illinois at Urbana-Champaign College of Engineering CEE 570 Finite Element Methods (in Solid and Structural Mechanics) Spring Semester 03 Quiz # April 8, 03 Name: SOUTION ID#: PS.: A the