EML4507 Finite Element Analysis and Design EXAM 1

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2-17-15 Name (underline last name): EML4507 Finite Element Analysis and Design EXAM 1 In this exam you may not use any materials except a pencil or a pen, an 8.5x11 formula sheet, and a calculator. Whenever you use a formula, you need to write the formula first, before substituting numbers into it. For full credit, you need to explain what you do. Please sign a pledge that you have not violated any of the university honesty provisions concerning this quiz.

1. (33%) The only non-zero stress components are σxx 60 MPa, σzz 30MPa. a. What is the normal stress in the direction of the vector i+2j+2k? We first normalize to unit length n( i+2j+2k)/3 60 1 1 σn [ σ] [ 1 2 2 ] 0 n n 2 /3 3 30 2 This is a quadratic form equal to 60(1/3) 2-30(2/3) 2-60/9MPa b. What are the three principal stresses? The principal stresses are the eigenvalues of the stress matrix. The eigenvalues of a diagonal matrix are the entries on the diagonal. So σ1 60, σ2 0, σ3 30 c. If the yield stress is 180MPa, what is the safety factor? If we use the Tresca criterion /2 /2 180 SF σ σ 2 τ σ σ / 2 60 ( 30) ( ) max 1 3 If we use the von Mises criterion 180 180 SF σ σ 2.27 σ 2 2 2 2 2 2 0.5 ( 1 2) ( 1 3) ( 3 2) VM σ σ σ σ σ σ ( 60 90 30 ) / 2 6300 + + + + We cannot use Rankine, because it is for brittle materials that do not yield.

2. (32%) A uniform cantilever beam of length L is supported at the loaded end, so that this end cannot rotate, as shown in the figure. oung s modulus is E, the moment of inertia is I, and the load is P. a. Calculate the displacement at the loaded end using one beam finite element. With one element, there is no assembly, so the FE equation is [ kq ] F 12 6L 12 6L v1 F1 2 2 6L 4L 6L 2L EI θ 1 C 1 3 L 12 6L 12 6L v 2 F2 P 2 2 6L 2L 6L 4L θ 2 C 2 The boundary conditions give us v1 θ1 θ2 0. So we strike out the first, second and fourth rows and 3 12EI PL columns to have v 3 2 P v2 L 12EI b. For some value of the load, the answer to part (a) is 0.05L. Calculate the displacement in the middle of the beam using the appropriate interpolation function(s). 2 3 ( ) vs () N() sv+ N() sθ + N() sv + N() sθ N() sv 3s 2s v 1 1 2 1 3 2 4 2 3 2 2 v 0.5 3 / 4 2 / 8 0.05L 0.025L Substituting s0.5 and the value of v2, we get ( ) ( ) c. Calculate the bending moment in the middle of the beam using the same interpolation function(s). 2 2 d v EI d v EI M ( s) EI 2 2 2 2 ( 6 12s) v2 dx L ds L M (0.5) 0 d. Is that solution exact? Why? Because it is a uniform beam with loads applied at the nodes, the solution is exact for the Euler-Bernoulli model. However, the Euler Bernoulli model is not exact because it neglects shear stresses.

3. (30%) Answer the following questions: a. Statically determinate trusses do not develop thermal stresses when heated. Explain what does it mean for a truss to be statically determinate, and why this definition leads to zero thermal stresses. Statically determinate means that the element forces can be calculated from equilibrium alone. There are enough equations to do that, which means that the number of elements is limited. Since the element forces are determined by the equations of equilibrium, and these are determined by the applied forces alone, temperature field cannot change them. So that means that there is enough freedom in the truss to allow free expansion of members. b. What are the steps involved in the Rayleigh-Ritz method, and how does that apply to the Finite Element method, making it a Rayleigh-Ritz method. 1. Assume a displacement shape that satisfies the displacement boundary conditions as linear combination of known functions time unknown coefficients. 2. Calculate the potential energy in terms of the unknown coefficients. 3. Determine the coefficients by minimizing the potential energy (differentiating it with respect to coefficients). For the FE method, the assumed functions are the interpolation functions, and the unknown coefficients are the nodal degrees of freedom, and we find them by minimizing the potential energy. c. What is the geometrical meaning of yz γ? It is the change in angle between the y-axis and z-axis due to the deformation.

4. (5%) Answer the following questions a. What can I do to make it easier for you to prepare for questions such as the ones in Problem 3 above (besides putting them in blue in the lecture slides)? b. The audio part of the Power Point requires a lot of work on my part. Do you use it? Should I use the time instead to add some other feature to the course? c. Any other thing that you would like to change or not to change in the way I run the course.