Finite Element Modeling of Vibration of Bars - Dynamics of Aerospace Structures | ASEN 5022, Study notes of Aerospace Engineering

Download Finite Element Modeling of Vibration of Bars - Dynamics of Aerospace Structures | ASEN 5022 and more Study notes Aerospace Engineering in PDF only on Docsity! Dynamics of Mechanical and Aerospace Structures, Spring 2008 Finite Element Modeling of Vibration of Bars This lecture covers: FEM discretization of a bar with general boundary conditions. EA, m(x) L Bar with unknown boundary conditions M0 LM k01 L1 kk02 L2k Theoretical Basis: Same as the continuum bar model! Kinetic energy of the contimuun bar: Tbar = 12 ∫ L 0 m(x) ut(x, t) 2dx, ut(x, t) = ∂u(x, t) ∂t (1) Potential energy of the continuum bar: Vbar = 12 ∫ L 0 { E A(x) ux(x, t)2 } dx ux(x, t) = ∂u(x, t) ∂x (2) External energy due to distributed applied force f(x, t): δWbar = ∫ L 0 f(x, t) δu(x, t) dx (3) We now consider a completely free, isolated, small segment of a bar. EA, m(x) L Bar with unknown boundary conditions M0 LM k01 L1 kk02 L2k 1 2 1 2 x A completely free element of length taken out from the continuum bar an element isolated u2u1 l x Approximation of u(x, t) over the element segment, ` << L: u(x, t) = c0(t)+ c1(t)x + c2(t)x2 + . . .+ cn(t)xn (8) Class of elements: 1. Linear Element or Constant Strain Element : Retain the first two terms in (8), viz., (c0, c1) to result in: u(x, t) = c0(t)+ c1(t)x (9) 2. Quadratic Element or Linear Strain Element : Retain the first three terms in (8), viz., (c0, c1, c2) to result in: u(x, t) = c0(t)+ c1(t)x + c2(t)x2 (10) 3. Cubic Element or Quadratic Strain Element : Retain the first four terms in (8), viz., (c0, c1, c2, c3) to result in: u(x, t) = c0(t)+ c1(t)x + c2(t)x2 + c3(t)x3 (11) How does one determine the coefficients, (c0, c1, c2, c3, etc.)? u u1 2 l o x x=- /2l x= /2l Node 1 Node 2 Linear (2-Node) Element u u1 2 l o x x=- /2l x= /2l Node 1 Node 2 Quadratic (3-Node) Element u3 x=0 Bar Elements Construction of Linear Bar Element Elemental kinetic energy: T elbar = 12 ∫ 1 2` −12` m(x) u̇2(x, t)dx = 1 2 ∫ 1 −1 m(ξ) u̇2(ξ, t)(12`dξ), x = 12`ξ (16) Elemental potential energy: V elbar = 12 ∫ `/2 −`/2 E A(x) u2x(x, t) dx = 1 2 ∫ 1 −1 E A(ξ) ( 2 ` )2u2ξ (ξ, t) ( 1 2`dξ) (17) Construction of Linear Bar Element – Cont’d T elbar = 1 2 ∫ 1 −1 m(ξ) u̇2(ξ, t)(12`dξ) = 1 2 ∫ 1 −1 1 2ρA` q̇(t) T [NT · N] q̇(t)dξ = 1 2 q̇T [ ∫ 1 −1 1 2ρA` N T · N dξ ] q̇(t) = 12 q̇(t)T [mel] q̇(t) (18) where we utilized (15), viz.: u(ξ, t) = N(ξ) q(t) (19) Construction of Linear Bar Element – Cont’d V elbar = 1 2 ∫ 1 −1 E A(ξ) ( 2 ` )2u2ξ (ξ, t) ( 1 2`dξ) = 1 2 ∫ 1 −1 2E A(ξ) ` q(t)T [NTξ · Nξ ] q(t)dξ = 12q(t)T [ ∫ 1 −1 2E A ` NTξ · Nξ dξ ] q(t) = 12q(t)T [kel] q(t) (20) Summary of elemental energy expressions: T elbar = 12{q̇(t)(el)}T [mel] {q̇(t)(el)} V elbar = 12{q(t)(el)}T [kel] {q(t)(el)} δW elbar = {δq(t)(el)}T {fel} (24) Question: How do we model a long bar with bar elements? Answer: (1)Partition the long bar into small elements. This needs to be explained. (2)Generate the elemental energy expression. This is done (see (24)). (3)Sum up the elemental energy to form the total system energy. (4)Perform the variation of the total system energy to obtain the equations of motion! Let’s now work on items (1), (3) and (4). 1.A Partition a bar into many elements. element element 3element 2 global node 1 2 3 4 5 element 1 (2)(1) (3) (4) (. . . ) Partition a bar into finite elements elelental node global node 1 2 3 42 3 1 2 1 2 1 2 element 3element 2element 1 u1 u2 u3 u4u2 u3 q1 (1) q2 (1) q1 (2) q2 (2) q1 (3) q2 (3) Partitioning and establishing the Boolean relation between the global degrees of freedom, u, and elemental degrees of freedom, q 3.B Sum up the total system strain energy. V totalbar = n∑ el=1 V elbar b 1 2{q(1)}T [k(1)]{q(1)} + 12{q(1)}T [k(1)]{q(1)}+ ...+ 12{q(n)}T [k(n)]{q(n)} =  q(1) q(2) .. q(n)  T  k(1) 0 ... 0 0 k(2) ... 0 0 0 ... 0 0 0 .. k(n)   q(1) q(2) .. q(n)  T ⇓ V totalbar = 12{q`}T [k`] {q`} (27) 3.C Assemble the partitioned elements back into a bar. This corresponds to substituting the partitioned elemental degrees of freedom, q`, by the assembled (or global) degrees of freedom, uc, via (25): • Total assembled system kinetic energy: T totalbar = 12 q̇T` m` q̇` = 12 u̇Tc LTc m` Lcu̇c = 12 u̇Tc [LTc m` Lc] u̇c T totalbar = 12 u̇Tg Mc u̇c, Mc = LTc m` Lc (28) • Total assembled system potential energy: V totalbar = 12qT` k` q` = 12uTc LTc k` Lcuc = 12uTc [LTc k` Lc] uc V totalbar = 12uTc Kc uc, Kc = LTc k` Lc (29) • Total assembled system external work: δW total = n∑ el=1 {δq(t)(el)}T {fel} = δqT` f` = δuTc {LT f`} δW total = δuTc fc, fc = LTc f` (30) The Lagrangian for the two end-spring and mass models are given by L0 = 12 q̇T0 m0 q̇0 − 12qT0 k0 q0 LL = 12 q̇TL mL q̇L − 12qTL kL qL m0 = [ M0 0 0 0 ] , k0 = [ k01 + k02 −k02 −k02 k02 ] mL = [ 0 0 0 ML ] , kL = [ kL1 −kL2 −kL2 kL2 + kL1 ] q0 = { u0(t) u(0, t) } , qL = { u(L , t) uL(t) } (35) How do we append the two end-springs and masses? M0 LM k01 L1 kk02 L2 k m0 k0( ) mL kL( )Mc Kc( ) Unassembled Model (hardware) Unassembled Model (software) 1. Stack the unassembled matrices: Munassemb = [m0 0 0 0 Mc 0 0 0 mL ] Kunassemb = [k0 0 0 0 Kc 0 0 0 kL ] (36) How do we append the two end-springs and masses? – Cont’d 2. Construct the assembling Boolean matrix (or Assemble directly element-by-element into the assembled matrices): Lassemb =  1 0 | 0 . . . . . . 0 0 0 1 | 0 . . . . . . 0 0 −− −− −− −− −− −− −− 0 | 1 0 . . . . . . 0 | 0 0 | 0 1 . . . . . . 0 | 0 0 | 0 1 0 . . . 0 | 0 0 | 0 0 1 . . . . . . | 0 . . . | . . . . . . . . . . . . . . . | . . . 0 | 0 . . . . . . . . . . . . | . . . 0 | 0 . . . . . . 1 0 | 0 0 | 0 . . . . . . 0 1 | 0 −− −− −− −− −− −− −− 0 0 . . . . . . 0 | 1 0 0 0 . . . . . . 0 | 0 1  (37) An Example of Three Bar Elements with end masses and springs Kc = E A `  1 −1 0 0 −1 2 −1 0 0 −1 2 −1 0 0 −1 1  , ` = L/3 Mc = ρA` 6  2 1 0 0 1 4 1 0 0 1 4 1 0 0 1 1  , ` = L/3 m0 = [ M0 0 0 0 ] , k0 = [ k01 + k02 −k02 −k02 k02 ] mL = [ 0 0 0 ML ] , kL = [ kL1 −kL2 −kL2 kL2 + kL1 ] (42) Three Bar Elements with end masses and springs - cont’d Assembling Boolean matrix Lassemb: Lassemb =  left spring-mass system −− three element bar −− left spring-mass system  =  1 0 0 0 0 0 0 1 0 0 0 0 − − − − − − 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 − − − − − − 0 0 0 0 1 0 0 0 0 0 0 1  (43) Use now equation(38) to assemble the mass and stiffness matrices!