UNIVERSITY OF GLASGOW FINITE ELEMENT ANALYSIS WEEKLY PROBLEM SET 1 September 24, 2020 The format for the problem set is identical to that of the exam. The numbers in square brackets in the right-hand margin indicate the marks allotted to the part of the question against which the mark is shown. This problem set test your understanding of the process of deriving the governing equations (know as the strong form) for linear elasticity in 1D. Q1 Consider the 1D elastic bar with boundary conditions, loading, and material properties shown in Fig. 1. x t = ?n = 1N/m 2 E = 8N/m 2 b = 3N/m A = 2m 2 l = 4m Figure 1: 1D elastic bar. The strong form (S) of the relation governing the displacement u of the bar is given by d dx ? AE du dx ? + b = 0 Ω := 0 < x < l (1) σn = ? E du dx ? n = t on Γ t (2) u = u on Γ u . (3) (a) Why is it allowed to approximate this problem as a 1D problem? [1] Page 1 of 4 Continued overleaf Solution: The solution u(x), the material properties E(x), and the geometry A(x) only vary in one dimension (b) At which end of the bar is the natural boundary condition applied? [1] Solution: At x = 4 m (c) At which end of the bar is the essential boundary condition applied? [1] Solution: At x = 0 m (d) What is the total force acting at the natural boundary? Is the applied traction t causing a tensile or a compressive stress? Why is this the case? [2] Solution: Total force = tA = 2 N Tensile as t = σn > 0. For simple problems in 1D we can derive the exact solution. This is generally not possible for all but the simplest problems - hence the need to solve the equations approximately using the FEM. Let’s first consider the steps in deriving the exact solution to our simple problem. Equation (1) can be expressed by d 2 u dx 2 = − b EA | {z } g . (4) (i) From Eq. (4), we obtain du dx = gx + c 1 , (5) where c 1 is a constant of integration. Use the natural boundary condition to determine the value of c 1 . [2] Solution: du dx = gx + c 1 natural bc: E du dx = E [gx + c 1 ] = t = 1 at x = 4 c 1 = 7/8 Page 2 of 4 Continued overleaf (ii) Furthermore, from Eq. (5), we obtain u(x) = 1 2 gx 2 + c 1 x + c 2 , where c 2 is another constant of integration. Use the essential boundary con- dition to determine the value of c 2 . [2] Solution: u = 1 2 gx 2 + c 1 x + c 2 essential bc: u = u = 0 at x = 0 c 2 = 0 (iii) Hence show that the displacement and the stress along the bar are given by u(x) = − 3 32 x 2 + 7 8 x σ(x) = Eε = − 3 2 x + 7. [3] (iv) Show that the expression for the stress above could have been obtained from basic mechanics as σ(x) = t + b(l − x) A . What is the value of the stress at x = 0 and x = l? Does this make sense given the boundary conditions and loading? [5] Solution: The stress acting at a point x is due to the traction and the body force and can be determined from a force balance as: σ(x) = t + b(l − x) A The stress at x = 0 is σ = t + bl/A. This makes sense as it is the stress due to the traction and the body force. The stress at x = l is σ = t. This makes sense as it is the applied traction. (v) Plot the distribution of the displacement and the stress along the length of the bar. [4] Solution: Page 3 of 4 Continued overleaf Page 4 of 4 End of problem set