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MANE 7100: Mechanical Engineering Foundations II Chet VanGaasbeek Homework 2 9/29/2012 Problem 1 In two-dimensional elasticity theory, the stress function φ(x, y) defined by the relationships: σxx = ∂2φ ∂y 2 (1) σyy = ∂2φ ∂x2 (2) σxy = σyx = − ∂2φ ∂x∂y (3) Substitute the above expressions into the equilibrium equations and obtain the stress function equation. Take into account also the equations of compatibility. The equilibrium equations are as follows: ∂σxx ∂σxy ∂σxz + + + Xx = 0 ∂x ∂y ∂z (4) ∂σyx ∂σyy ∂σyz + + + Xy = 0 ∂x ∂y ∂z (5) ∂σzx ∂σzy ∂σzz + + + Xz = 0 ∂x ∂y ∂z (6) with body forces Xx , Xy , Xz . Applying the assumption of plane stress (given this is a two dimensional elastic problem, equations 4, 5, 6 become: ∂σxx ∂σxy + + Xx = 0 ∂x ∂y (7) 1 MANE 7100: Mechanical Engineering Foundations II Chet VanGaasbeek ∂σyx ∂σyy + + Xy = 0 ∂x ∂y (8) In the absence of body forces or assuming body forces are small with respect to the problem, the plane stress approximation becomes: ∂σxx ∂σxy + =0 ∂x ∂y (9) ∂σyx ∂σyy + =0 ∂x ∂y (10) Applying the condition of compatibility1 : 2 ∂2 ∂ (σxx + σyy ) = 0 + ∂x2 ∂y 2 (11) yields the equation: ∂4φ ∂4φ ∂4φ + 2 + =0 ∂x4 ∂x2 ∂y 2 ∂y 4 (12) which is the biharmonic equation∇4 φ = 0. Solutions for the stress field satisfy equation 12. A common approach is to assume a polynomial φ less than degree 4, which satisfies Equation 12. Problem 2 A simply supported beam has length L = 1, breadth b = 1, and height h = 0.1. The elastic modulus of the material is E = 1011 and its Poisson’s ratio is ν = 0.3. A downward load of P = 105 N is applied at the midpoint of the cross section along the upper surface of the beam. Use the finite element method to determine approximate solutions to this problem. Repeat with a distributed load across the entire cross section Q = 105 N/m. Compare both results to those obtained from elementary beam theory. Point Load, Midpoint of the Cross Section For a simply supported beam with a central load (fixed at x = 0, roller support at x = L), the loading condition is as follows: P x=0 x=L 1 see http://homepages.engineering.auckland.ac.nz/p̃kel015/SolidMechanicsBooks/Part II/03 ElasticityRectangular/ElasticityRectangular 02 StressFunction.pdf 2 MANE 7100: Mechanical Engineering Foundations II Chet VanGaasbeek x=L x = L/2 x=0 The maximum deflection for this plate is: P L3 48EI where E is the modulus of elasticity, L is 1, P is 10000 N, and I is: umax = I= b ∗ h3 (1m)(0.1m)3 = = 8.333 ∗ 10−5 m4 12 12 The maximum deflection is: 100000N(1m)3 P L3 = = 2.5 ∗ 10−4 m umax = 11 −5 4 48EI 48 ∗ (10 P a)(8.333 ∗ 10 m ) Using Simulia’s Abaqus 6.10-1 FEA software, a model is constructed as follows: Part & Materials The part is a one dimensional line with length = 1 meter. A steel material is constructed with E = 101 1 Pa, ν = 0.3 in accordance with the problem statement. A homogeneous beam section is created and assigned a rectangular cross section (b = 1 meter, h = 0.1 meter). The beam section is orientated along the beam to accurately represent the geometric conditions of the problem. Step Definition, Loads, and Boundary Conditions The analysis is performed in a linear static load step, taking place over one second of application time. No nonlinear effects are considered2 . A point partition divides the part into two separate sections to define the load. The load is applied over the course of the step using a concentrated point load at the node midway along the span of 100,000 N. Boundary Conditions for simply supported beam: • The x = 0 node is fixed such that there is no displacement or rotation with the exception of the z axis rotation, which is allowed to freely vary. • The x = L node is fixed such that there is no displacement or rotation with the exception of the x direction which is allowed to slide and the z axis rotation, which is allowed to freely vary. 2 Second order terms in the solver are neglected. 3 MANE 7100: Mechanical Engineering Foundations II Chet VanGaasbeek Meshing & Results Initially the beam is meshed using 10 B31 2-node linear shear-flexible beam elements with no bias. The resulting maximum deflection is 0.0002571 meters, which is within 3% of the expected analytical result. A second run of the analysis is run using 10,000 B32 2 node quadratic beam elements. The resulting deflection produced is 0.0002576 m, which is not appreciably different from the initial run. Figure 1: 10 element beam analysis. Point load, midspan of the beam. Y X Z Line (Distributed) Load, Mid-Line of the Cross Section. 4