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Force/Deflection Relationships

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Force/Deflection Relationships Introduction to Finite Element Dynamics ⎡ x x ⎤ 1. Element Stiffness Matrices u()x = ⎢1− ⎥ u = []C u (3) ⎣ L L ⎦ 1.1 Bar Element (iii) Derive Strain - Displacement Relationship Consider a bar element shown below. At the two by using Mechanics Theory2 ends of the bar, axially aligned forces [F1, F2] are The axial strain ε(x) is given by the following applied producing deflections [u ,u ]. What is the 1 2 relationship between applied force and deflection? du d ⎡ 1 1 ⎤ ε()x = = ()[]C u = []B u = ⎢− ⎥u (4) dx dx ⎣ L L ⎦ Deformed shape u2 u1 (iv) Derive Stress - Displacement Relationship by using Elasticity Theory The axial stresses σ (x) are given by the following assuming linear elasticity and homogeneity of F1 x F2 material throughout the bar element. Where E is Element Node the material modulus of elasticity. There are typically five key stages in the σ (x)(= Eε x)= E[]B u (5) 1 analysis . (v) Use principle of Virtual Work (i) Conjecture a displacement function There is equilibrium between WI , the internal The displacement function is usually an work done in deforming the bar, and WE , approximation, that is continuous and external work done by the movement of the differentiable, most usually a polynomial. u(x) is applied forces. The bar cross sectional area the axial deflections at any point x along the bar A = ∫∫dydz is assumed constant with respect to x. element. WI = ∫∫∫σ (x).ε (x)(dxdydz = ∫σ x).ε (x)dx∫∫dydz ⎡a1 ⎤ (6) u()x = a1 + a2 x = []1 x ⎢ ⎥ = [N] a (1) = A∫σ ()x .ε ()x dx ⎣a2 ⎦ ⎡F ⎤ W = []u u 1 = uT F (7) E 1 2 ⎢F ⎥ (ii) Express u()x in terms of nodal ⎣ 2 ⎦ displacements by using boundary conditions. Hence since WE = WI by substituting (4) and (5) T T T i.e. u()0 = u1 , u()L = u2 where L is the bar length. Note [B] u is a scalar thus []B u = ()[]B u = u [B] It is clear that with only two nodes only two unknown coefficients for the polynomial can be uT F = EA [B] u.[B] udx = EAuT []B T [B]dxu defined i.e. a linear relationship. Thus if a higher ∫ ∫ order polynomial is required for u()x more boundary conditions would be required and hence F = {EA∫[]B T []B dx} u = []K u (8) intermediate nodes along the length of the bar are necessary. Equation (8) represents the force-displacement relationship for the bar element. The Stiffness ⎡u1 ⎤ ⎡1 0⎤⎡a1 ⎤ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ so u = []A a (2) matrix [K] is derived by integration thus ⎣u2 ⎦ ⎣1 L⎦⎣a2 ⎦ L 1 ⎡− L ⎤ []K = EA ⎢ ⎥[]− 1 1 dx Substituting (2) into (1) ∫ 0 1 L L ⎣ L ⎦ −1 (9) −1 ⎡1 0⎤ 1 1 u()x = []N [A]u = []1 x ⎢ ⎥ u L ⎡ 2 − 2 ⎤ EA ⎡ 1 −1⎤ 1 L = EA ⎢ L L ⎥dx = ⎣ ⎦ ∫ 0 1 1 ⎢ ⎥ − 2 2 L −1 1 ⎣⎢ L L ⎦⎥ ⎣ ⎦ 1 1.2 Beam Element (iii) Derive “Strain”-Displacement Relationship A similar approach can be adopted to deriving the Beam curvature which turns out to be the proxy force-displacement relationship for a beam for “strain” in the virtual work equation. The element, i.e. one that is subject to flexural rather beam curvature c(x) is approximated by the than axial deformation. second derivative of the deflection. This assumes Consider a beam element shown below. At the that the flexural deflections are small, i.e the two ends of the bar, normally aligned forces and beam remains almost flat. T moments F = [F1 M1 F2 M 2 ]are applied to d 2u d 2 produce deflections and rotations given by c()x = = ()[]C u = []N′′ [A]−1u = []B u (14) T dx2 dx2 u = [v1 θ1 v2 θ 2 ] What is the relationship between applied Force/Moment and Deflection/Rotations? (iv) Derive “Stress”-Displacement Relationship Here, the relationship between moment m(x) and curvature c(x) is the proxy for “stress”-“strain”. For a beam element the Bernoulli-Euler simple θ2 beam theory result is used. This assumes no shear θ1 F2 deformations are present. F v v1 1 2 M (x)(= EI.c x)= EI[]B u (15) M1 y M2 x (v) Use principle of Virtual Work The internal virtual work is given by the same fundamental expression as (6). However it is (i) Conjecture a displacement function modified by application of Bernoulli-Euler The normal displacement u(x) is a cubic simple beam theory. The flexurally induced u′ x polynomial. The slope () of the beam is given normal stresses σ (x, y)(= M x)y Iz where the by its first derivative. origin of the y axis is the centroidal axis 2 3 u()x = a1 + a2x + a3x + a4x 2 and Iz = ∫∫ y dydz is the second moment of area ⎡a1 ⎤ about the z axis; assuming homogeneity. Shear ⎢a ⎥ (10) stresses are neglected. = [1 x x2 x3 ]⎢ 2 ⎥ = []N a ⎢a3 ⎥ ⎢ ⎥ Normal stresses ⎣a4 ⎦ due to M(x) Tension[+] y u′()x = a + 2a x + 3a x2 2 3 4 (11) = 0 1 2x 3x2 a = N′ a [][] z M(x) (ii) Express u(x) in terms of nodal x displacements and rotations by using boundary conditions. compression i.e. u()0 = v1 , u′()0 = θ1 , u()L = v2 , u′()L = θ 2 , M ()x y M (x)y v 1 0 0 0 a WI = σ .ε dxdydz = dxdydz ⎡ 1 ⎤ ⎡ ⎤⎡ 1 ⎤ ∫∫∫ ∫∫∫ I EI ⎢ ⎥ ⎢ ⎥⎢ ⎥ z z θ1 0 1 0 0 a2 2 2 ⎢ ⎥ = ⎢ ⎥⎢ ⎥ so u = []A a (12) L M ()x L M ()x ⎢v ⎥ ⎢1 L L2 L3 ⎥⎢a ⎥ = dx. y 2dydz = dx 2 3 ∫ 0 2 ∫∫ ∫ 0 ⎢ ⎥ ⎢ ⎥⎢ ⎥ EI EI z θ 0 1 2L 3L2 a z ⎣ 2 ⎦ ⎣ ⎦⎣ 4 ⎦ L = M ()x c()x dx ∫ 0 hence substituting (12) into (10) and (11) Since WE = WI and by substituting (14) and (15) −1 u()x = []N [A]u = []C u (13) −1 u′()x = []N′ [A]u uT F = ∫ M ()x c()x dx 2 T ⎡a1 ⎤ F = {}EI ∫[]B [B]dx u = []K u (16) ⎡u ()x ⎤ ⎡1 x x2 x3 0 0⎤ y = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ M ⎥ ⎣ux ()x ⎦ ⎣⎢0 0 0 0 1 x⎦⎥ (18) By comparing (16) and (8) it can be observed that ⎣⎢a6 ⎦⎥ the general form of the stiffness matrix is given U ()x = []N a by a similar matrix integral equation. Thus a vector of displacement functions is Hence stiffness matrix [K] is derived thus conjectured. −1 T T −1 []K = EI ()[A] ∫[]N′′ []N′′ dx[A] Fy1 Fy2 M1 F ⎡ 0 ⎤ y Fx1 M2 x2 ⎢ 0 ⎥ x []N′′ T [N′′]dx = L ⎢ ⎥[0 0 2 6x] dx ∫ ∫ 0 ⎢ 2 ⎥ ⎢ ⎥ ⎣6x⎦ u1 ⎡0 0 0 0 ⎤ θ2 ⎢0 0 0 0 ⎥ θ1 = ⎢ ⎥ u ⎢ 2 ⎥ 2 v2 0 0 4L 6L v1 ⎢ 2 3 ⎥ ⎣0 0 6L 12L ⎦ Using matrix algebra (ii) Express U(x) in terms of nodal displacements and rotations by using boundary conditions. ⎡ 1 0 0 0 ⎤ The boundary conditions are the same as those ⎢ 0 1 0 0 ⎥ −1 ⎢ ⎥ used in the previous sections []A = ⎢ 3 2 3 1 ⎥ − 2 − 2 − ⎢ L L L L ⎥ 2 1 2 1 u 0 0 0 0 1 0 a ⎢ 3 2 − 3 2 ⎥ ⎡ 1 ⎤ ⎡ ⎤⎡ 1 ⎤ ⎣ L L L L ⎦ ⎢ ⎥ ⎢ ⎥⎢ ⎥ Hence ⎢v1 ⎥ ⎢1 0 0 0 0 0⎥⎢a2 ⎥ ⎢θ1 ⎥ ⎢0 1 0 0 0 0⎥⎢a3 ⎥ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ so u = [A] a ⎡ 12 6L −12 6L ⎤ ⎢u2 ⎥ ⎢0 0 0 0 1 L⎥⎢a4 ⎥ ⎢ 2 2 ⎥ 2 3 EI 6L 4L − 6L 2L ⎢v ⎥ ⎢1 L L L 0 0⎥⎢a ⎥ K = ⎢ ⎥ ⎢ 2 ⎥ ⎢ ⎥⎢ 5 ⎥ [] 3 (17) 2 L ⎢−12 − 6L 12 − 6L⎥ ⎣⎢θ2 ⎦⎥ ⎣⎢0 1 2L 3L 0 0⎦⎥⎣⎢a6 ⎦⎥ ⎢ 2 2 ⎥ ⎣ 6L 2L − 6L 4L ⎦ hence U (x) = [N][A]−1u (19) 1.3 Combined Beam/Bar Element This element contains both axial and flexural (iii) Derive “Strain”-Displacement Relationship displacements. The stiffness matrix that relates action and deformation can be derived by ⎡d 2u ⎤ inspection from results (17) and (9). However, it ⎢ y ⎥ ⎡c()x ⎤ 2 ⎡0 0 2 6x 0 0⎤ is instructive to derive it using the five stages ⎢ ⎥ = ⎢ dx ⎥ = ⎢ ⎥a ⎣ε ()x ⎦ ⎢ dux ⎥ ⎣0 0 0 0 0 1⎦ (20) used previously. ⎢ ⎥ ⎣ dx ⎦ (i) Conjecture displacement functions ε = []N′′ [A]−1u = []B u In this problem, because there are two distinct and separate deformations i.e. axial and flexural, (iv) Derive “Stress”-Displacement Relationship there is a need for two separate displacement functions ⎡ M (x) ⎤ ⎡EI 0 ⎤⎡c()x ⎤ 2 3 ⎢ ⎥ = ⎢ ⎥⎢ ⎥ so σ = [D] ε (21) uy ()x = a1 + a2x + a3x + a4x Aσ x 0 EA ε x ⎣ ()⎦ ⎣ ⎦⎣ ()⎦ ux ()x = a5 + a6x 3 ⎡ EA EA ⎤ (v) Use principle of Virtual Work ⎢ 00− 00⎥ ⎢ ⎥ ⎢ L L ⎥ ⎢ ⎥ ⎢ EI EI EI EI ⎥ ⎢ z z z z ⎥ T ⎢ 012 6 0 −12 6 ⎥ T T T ⎢ 3 2 3 2 ⎥ u F = ε σdx = u [B][D][B] dx u ⎢ L L L L ⎥ ∫ ∫ ⎢ ⎥ ⎢ EI EI EI EI ⎥ ⎢ z z z z ⎥ ⎢ 06 4 0 −6 2 ⎥ ⎢ 2 2 ⎥ ⎢ L L L L ⎥ T K := ⎢ ⎥ F = B D B dx u = K u ⎢ ⎥ {∫[][][] } [] (22) ⎢ EA EA ⎥ ⎢− 00 00⎥ ⎢ L L ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ EI EI EI EI ⎥ T ⎢ z z z z ⎥ ⎢ 0 −12 3 −6 2 0123 −6 2 ⎥ Where F = [Fx1 Fy1 M1 Fx2 Fy2 M 2 ] ⎢ L L L L ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ EI EI EI EI ⎥ ⎢ 06z 2 z 0 −6 z 4 z ⎥ ⎢ 2 L 2 L ⎥ −1 T T −1 ⎣ L L ⎦ (24) []K = ()[A] ∫[]N′′ [D][]N′′ dx[A] (23) 1.4 General Element Equation (23) can be evaluated by using Maple3.
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