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 ⎦⎥ ⎣ ⎦

1.2 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 ⎡a ⎤ F = EI []B [B]dx u = []K u (16) 2 3 1 {}∫ ⎡uy ()x ⎤ ⎡1 x x x 0 0⎤⎢ ⎥ ⎢ ⎥ = ⎢ ⎥⎢ 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 ⎤ u ()x = a + a x + a x2 + a x3 ⎢ ⎥ = ⎢ ⎥⎢ ⎥ so σ = [D] ε (21) y 1 2 3 4 ⎣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. Consider an abstract, general, element with n The following code will evaluate the stiffness nodes and m nodal displacements/rotations, matrix. known as degrees of freedom. The process of

> with(linalg): deriving the stiffness matrix for this element is > A:=matrix(6,6,[[0,0,0,0,1,0], identical to that adopted in the previous sections. [1,0,0,0,0,0], [0,1,0,0,0,0], Nodes do not have to be at edges or corners, [0,0,0,0,1,L], [1,L,L^2,L^3,0,0], though normally they are so that elements can be [0,1,2*L,3*L^2,0,0]]); easily matched together in some mesh that will describe a structure. Also Nodes need not have ⎡00 0 0 10⎤ ⎢ ⎥ ⎢ ⎥ the same number of degrees of freedom though ⎢10 0 0 00⎥ ⎢ ⎥ ⎢01 0 0 00⎥ normally they do for a similar reason. ⎢ ⎥ A := ⎢ ⎥ ⎢00 0 0 1L⎥ ⎢ ⎥ ⎢ 2 3 ⎥ ⎢1 LL L 00⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎣012L 3 L 00⎦

> N2:=matrix(2,6,[[0,0,2,6*x,0,0], [0,0,0,0,0,1]]);

0026x 00 ⎡ ⎤ N2 := ⎢ ⎥ ⎣000 0 01⎦

> DD:=matrix(2,2,[[EI[z],0],[0,EA]]); (i) Conjecture displacement functions

⎡EI 0 ⎤ DD := ⎢ z ⎥ U (x, y, z) = []N a (a) ⎢ 0 EA⎥ ⎣ ⎦ (ii) Express U(x,y,z) in terms of nodal > NDN:=map(int,evalm( transpose(N2) &* displacements and rotations by using boundary DD &* N2 ), x=0..L); conditions ⎡00 0 0 0 0⎤ u = []A a (b) ⎢ ⎥ ⎢ ⎥ −1 ⎢00 0 0 0 0⎥ hence U x, y, z = N A u = C u (c) ⎢ ⎥ ( ) [][] [] ⎢00 4EI L 6 EI L2 00⎥ ⎢ z z ⎥ ⎢ ⎥ NDN := ⎢ ⎥ ⎢006EI L2 12 EI L3 00⎥ (iii) Derive “Strain”-Displacement Relationship ⎢ z z ⎥ ⎢ ⎥ ⎢ ⎥ ⎢00 0 0 0 0⎥ ⎢ ⎥ ε = N′′ a = N′′ A −1u = B u (d) ⎣00 0 0 0EA L⎦ [ ] [][] []

> K:=evalm(transpose(inverse(A)) &* NDN (iv) Derive “Stress”-Displacement Relationship &* inverse(A)); σ = [D] ε = []D [B] u (e)

4 (v) Use principle of Virtual Work > F:=evalm( transpose(inverse(A)) &* map(int, evalm( w*x * transpose(N)), T T T T x=0..L)); u F = ε σ dxdydz = u []B [D][]B dxdydz u ⎡ 3 2 ⎤ ∫∫∫ ∫∫∫ ⎢ wL ⎥ ⎢ ⎥ ⎢ 20 ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢ 3 ⎥ ⎢ wL ⎥ T ⎢ 30 ⎥ F = {} []B [D][]B dxdydz u = []K u (f) F := ⎢ ⎥ ∫∫∫ ⎢ ⎥ ⎢ 7 2 ⎥ ⎢ wL ⎥ ⎢ 20 ⎥ ⎢ ⎥ T ⎢ ⎥ ⎢ 1 3⎥ []K = ∫∫∫[B][D][B] dxdydz (g) ⎢− wL ⎥ ⎣ 20 ⎦ 2. Generalised Actions Using Maple it is not difficult to derive nodal action F in terms of any P(x). In the previous sections, F , the applied actions on the element are always applied at nodes. 2.1.2 Example 2, Point load P at position L 2 However in a more general case what will F be if If P(x) is discrete and discontinuous, i.e. the case actions are applied at any point on the element. of an applied point load, the integrals drop out of 2.1 Distributed Load for Beam Element (25) thus

The aim is to express the distributed load P(x) in W = u(L 2)P = []N()L 2 [A]−1u P terms of F the nodal actions. E T (26) = uT ()[]A −1 []N()L 2 T P = uT F P(x) Continue the Maple file from the previous section by adding the following line, thus

> F:=evalm( P* transpose(inverse(A)) &* y x transpose(map2(subs,x=L/2,N)));

⎡ 1 ⎤ ⎢ P ⎥ ⎢ ⎥ Let us reformulate the equation for the external ⎢ 2 ⎥ ⎢ ⎥ ⎢ 1 ⎥ virtual work, thus ⎢ ⎥ ⎢ PL ⎥ ⎢ 8 ⎥ F := ⎢ ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢ P ⎥ L L ⎢ 2 ⎥ −1 ⎢ ⎥ W = u()x P()x dx = N A u P()x dx ⎢ ⎥ E [][] ⎢ 1 ⎥ ∫ 0 ∫ 0 ⎢− PL⎥ (25) ⎣ 8 ⎦ T L = uT ()[]A −1 []N T P()x dx = uT F ∫ 0 2.2 General loads for a general 2.1.1 Example 1, linearly increasing element distributed load In a completely general case P(x, y, z)will be a Consider as an example, P(x) = w x i.e. a linear vector of loads, in different directions that are distributed load then (25) becomes distributed across the element. T L F = ()[]A −1 []N T w x dx which can be evaluated ∫ 0 T using Maple thus. WE = ∫∫∫U(x, y, z) P()x, y, z dxdydz

T −1 T T T > with(linalg): = u ()[]A ∫∫∫[]N P()x, y, z dxdydz = u F > A:=matrix(4,4,[[1,0,0,0], [0,1,0,0], [1,L,L^2,L^3], [0,1,2*L,3*L^2]]); −1 T T F = ([]A ) ∫∫∫[]N P()x, y, z dxdydz (h) ⎡10 0 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢01 0 0⎥ A := ⎢ ⎥ ⎢ 2 3 ⎥ ⎢1 LL L ⎥ ⎢ 2⎥ ⎣012L 3 L ⎦ > N:=matrix(1,4,[1,x,x^2,x^3]);

N := []1 xx2 x3

5 > with(linalg): 3. Element Mass Matrices > A:=matrix(4,4,[[1,0,0,0], [0,1,0,0], [1,L,L^2,L^3], [0,1,2*L,3*L^2]]);

3.1 Bar Element ⎡10 0 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢01 0 0⎥ In order to compute the element mass matrices it A := ⎢ ⎥ ⎢ 2 3 ⎥ is necessary to consider the work done by the ⎢1 LL L ⎥ ⎢ 2⎥ inertial accelerations. ⎣012L 3 L ⎦ > N:=matrix(1,4,[1,x,x^2,x^3]); WA = ∫∫∫ρu&&()x u()x dxdydz = ∫u&&(x)(u x)dx∫∫ ρdydz (27) N := []1 xx2 x3 = ρA∫u&&()x u(x)dx > NTN:=map(int,evalm( transpose(N) &* N),x=0..L); Where ρu&& is the inertial force per unit volume ⎡ 1 2 1 3 1 4 ⎤ and ρ is the density; which is assumed constant, ⎢ L L L L ⎥ ⎢ ⎥ ⎢ 2 3 4 ⎥ ⎢ ⎥ throughout the bar, in this example. ⎢1 1 1 1 ⎥ ⎢ 2 3 4 5 ⎥ ⎢ L L L L ⎥ ⎢2 3 4 5 ⎥ NTN := ⎢ ⎥ ⎢ ⎥ 2 ⎢1 3 1 4 1 5 1 6 ⎥ d u ⎢ L L L L ⎥ ⎢3 4 5 6 ⎥ u()x = = []C u ⎢ ⎥ && 2 && ⎢ ⎥ dt ⎢1 4 1 5 1 6 1 7 ⎥ ⎢ L L L L ⎥ T T ⎣4 5 6 7 ⎦ WA = ρA ()[]C u ()[]C u&& dx = u []C []C dxu&& (28) ∫ ∫

In the case of free vibrations > M:=evalm(rho*a* transpose(inverse(A)) &* NTN &* inverse(A) );

W +W = 0 ⎡ 156 22 L 54 −13 L⎤ A I ⎢ ⎥ ⎢ 2 2⎥ T T 1 ⎢ 22 L 4 L 13 L −3 L ⎥ (31) C C dx u + EA B B dx u = 0 (29) M := ρ aL⎢ ⎥ {}∫[][] && {∫[][] } ⎢ ⎥ 420 ⎢ 54 13 L 156 −22 L⎥ M u + K u = 0 ⎢ 2 2 ⎥ []&& [] ⎣−13 L −3 L −22 L 4 L ⎦ L x ⎡−1 L ⎤ ρAL ⎡2 1⎤ []M = ρA ⎢ ⎥[]1− x x dx = (30) ∫ 0 x L L ⎢ ⎥ 3.3 General Element ⎣ L ⎦ 6 ⎣1 2⎦ For a general element with n nodes and m degrees The mass matrix (30) is often termed a consistent of freedom, the work done by inertial mass matrix. The mass is distributed across the accelerations (27) is characterised thus element and is not lumped at any particular places T along the element. WA = ∫∫∫ρu&&(x, y, z) u()x, y, z dxdydz

Hence the mass matrix is given by 3.2 Beam Element If rotational accelerations of beam are neglected T [M][= ∫∫∫ρ C][C] dxdydz (i) then the formula (29) can be applied directly to the beam element with an alternate [C] matrix. hence the equation of free vibration is given by

the following W +W = 0 T −1 T T −1 A I [M] = ρA∫[]C []C dx = ρA()[A] ∫[]N []N dx[A] [M] u&& + []K u = 0 (j) Using Maple this can be evaluated

6 φ ≡ 0 1 0 −1 T and φ ≡ 0 1 0 1 T 1 [ ] 2 [] 4 Free Vibration Examples Thus from equation (13) the modal shape is given 4.1 Simply supported beam by the following where the nodal displacement Determine periods of vibration of a simply- vector u is replaced by φ . Remember that as the supported beam with constant flexural rigidity EI magnitude of the eigenvector is unknown and is and weight w per unit length normalised. Thus, in general, mode shapes have 1 2 unknown magnitude. Only with knowledge of some initial conditions, velocity and displacement at time t, can the relative magnitude of each mode L shape be assessed for an actual free vibration problem.

(17) and (31) are substituted into equation (j) to u(x) = []N [A]−1φ gives the following. By applying support x restraints i.e. v1 = v2 = 0 −1 ⎛ ⎞ u1()x = []N [A]φ = x⎜1− ⎟ (33) 1 ⎝ L ⎠

⎡ 156 22L 54 −13L⎤⎡v&&1 ⎤ −1 ⎛ x ⎞⎛ 2x ⎞ u2 ()x = []N [A]φ = x⎜1− ⎟⎜1− ⎟ (34) ⎢ 2 2 ⎥⎢ ⎥ 2 ⎝ L ⎠⎝ L ⎠ wL ⎢ 54 4L 13L − 3L ⎥⎢θ&&1 ⎥

420g ⎢ −12 13L 156 − 22L⎥⎢v&&2 ⎥ ⎢ 2 2 ⎥⎢ ⎥ ⎣−13L − 3L − 22L 4L ⎦⎣θ&&2 ⎦ Mode 1 ⎡ 12 6L −12 6L ⎤⎡v1 ⎤ ⎢ 2 2 ⎥⎢ ⎥ EI 6L 4L − 6L 2L θ + ⎢ ⎥⎢ 1 ⎥ = 0 3 L ⎢−12 − 6L 12 − 6L⎥⎢v2 ⎥ ⎢ 2 2 ⎥⎢ ⎥ Mode 2 ⎣ 6L 2L − 6L 4L ⎦⎣θ2 ⎦

3 wL ⎡ 4 − 3⎤⎡θ&&1 ⎤ EI ⎡4 2⎤⎡θ1 ⎤ ⎢ ⎥⎢ ⎥ + ⎢ ⎥⎢ ⎥ = 0 4.1.1 Accuracy of results 420g ⎣− 3 4 ⎦⎣θ&&2 ⎦ L ⎣2 4⎦⎣θ2 ⎦ The accuracy of these results is dependant chiefly

on the how near the cubic displacement function Hence dynamic matrix [D] can be found. The (10) is to the actual dynamic displacement resulting eigenvalue problem on [D] is solved. function.

420EIg ⎡ 22 20 ⎤ D = M −1 K = 7 7 The exact shape for the first mode is sin(πx L) [] [ ][] 4 ⎢ 20 22 ⎥ wL ⎣⎢ 7 7 ⎦⎥ and for the second mode sin(2πx L); exact values 420EIg ⎡ 2 0⎤ ⎡ 1 1⎤ for first and second frequencies are Λ = 7 Φ = [] 4 ⎢ ⎥ [] ⎢ ⎥ wL ⎣0 6⎦ ⎣−1 1⎦ EIg EIg

f1 = 1.571 4 , f 2 = 6.283 4 [Hz] The natural frequencies for the first two modes of wL wL vibration are approximated from the eigenvalues in the standard way f = λ 2π . This means (32) overestimates the first mode by 11% and underestimates the second mode by

26.6%. The first mode is more accurate than the EIg EIg second mode. f1 = 1.74 , f 2 = 4.61 [Hz] (32) wL4 wL4 The discrepancy of (33) and (34) from the exact solutions are displayed by the following graphs. The mode shapes can be extracted from the eigenvectors. In this case the first mode is given by φ = 1 −1 T and if the degrees of freedom that 1 [] have been removed are now reinstated we get

7 wl ⎡312 0 ⎤⎡&y&2 ⎤ EI ⎡24 0 ⎤⎡y2 ⎤ φ1 + = 0 ⎢ 2 ⎥⎢ ⎥ 3 ⎢ 2 ⎥⎢ ⎥ 420g ⎣ 0 8l ⎦ θ&&2 l ⎣ 0 8l ⎦ θ2 ⎣ ⎦ ⎣ ⎦ 420EIg ⎡1 13 0⎤ D = M −1 K = [] [ ][] 4 ⎢ ⎥ exact wl ⎣ 0 1⎦

The resulting eigenvalue problem on [D] is solved and the approximate natural frequencies of φ2 vibration are determined; note that span L is exact reintroduced.

420EIg ⎡ 1 0⎤ ⎡1 0⎤ Λ = 13 Φ = [] 4 ⎢ ⎥ [] ⎢ ⎥ It is clear that the accuracy of the FE approach is wl ⎣ 0 1⎦ ⎣0 1⎦ dependant the displacement shape function. This is true if only one element is employed. To EIg EIg f1 = 3.618 , f2 = 13.05 [Hz] (35) improve accuracy there are two alternate wL4 wL4 strategies you could adopt. (a) increase order of polynomial displacement shape function; or (b) A more exact answer for these frequencies are use more that one element. (b) is the most given below from a 10 element model. commonly adopted strategy and is the basis of the finite element method4 EIg EIg f = 3.56 f = 9.81 1 4 2 4 [Hz] 4.2 Encastré beam wL wL Determine periods of vibration of a fixed-end beam with constant flexural rigidity EI and weight per unit length w. The span L=2l

1 2 3

l l This means (35) overestimates the first mode by 16.3% and overestimates the second mode by 33%. The first mode is more accurate than the ⎡ 156 22l 54 −13l 0 0 ⎤⎡ &y&1 ⎤ second mode. ⎢ 2 2 ⎥⎢ ⎥ ⎢ 4l 13 − 3l 0 0 ⎥⎢θ&&1 ⎥ wl ⎢ 312 0 54 −13l ⎥⎢y ⎥ ⎢ ⎥⎢&&2 ⎥ 2 2 Dr Nick A. Alexander (2004) 420g ⎢ 8l 13l − 3l ⎥⎢θ&&2 ⎥ ⎢ ⎥⎢ ⎥ Department of Civil Engineering 156 − 22l &y&3 ⎢ ⎥⎢ ⎥ University of Bristol ⎢ 2 ⎥⎢θ&& ⎥ ⎣ 4l ⎦⎣ 3 ⎦ ⎡ 12 6l −12 6l 0 0 ⎤⎡ y1 ⎤ ⎢ 2 2 ⎥⎢ ⎥ References ⎢ 4l − 6l 2l 0 0 ⎥⎢θ1 ⎥ ⎢ ⎥ EI 24 0 −12 6l ⎢y2 ⎥ + = 0 1 3 ⎢ 2 ⎥⎢ ⎥ K.C. Rockey, H.R.Evans, D.W.Griffiths and l ⎢ 8l − 6l 2 ⎥ θ2 ⎢ ⎥ D.A.Nethercot. (1975) “The Finite Element Method, A ⎢ 12 − 6l⎥⎢ y ⎥ ⎢ ⎥⎢ 3 ⎥ Basic introduction”, ISBN 0258968214, Granada ⎣⎢ 4 ⎦⎥⎣⎢θ3 ⎦⎥ Publishing Ltd. 2 J.M. Gere & S. P. Timoshenko, (2000) “Mechanics of Materials”, ISBN 0534371337, Brooks Cole. (17) and (31) are substituted into equation (j) to 3 MAPLE 7, (2001)Waterloo Maple Inc gives the above. By applying support restraints 4 O.C. Zienkiewicz, (1971) “The Finite Element Method in Engineering Science”. ISBN 070941386, McGraw-Hill i.e. v1 = v3 = θ1 = θ3 = 0 these degrees of freedom are removed from the problem.