Vibration of Continuum Beams - Dynamics Mechanical/Aerospace Structures | ASEN 5022, Study notes of Aerospace Engineering

Download Vibration of Continuum Beams - Dynamics Mechanical/Aerospace Structures | ASEN 5022 and more Study notes Aerospace Engineering in PDF only on Docsity! Dynamics of Mechanical and Aerospace Structures, Spring 2008 Vibration of Continuum Beams This lecture covers: • Formulate the vibration of a beam in bending. In doing so, we cover: 1. Starting with the strain-displacement relation and the constitu- tive relation, we derive the potential energy for beam bending including the effect of axial force. 2. We then formulate Hamilton’s principle for the dynamics of a beam in bending. 3. By carrying out the variation of Hamilton’s principle, we derive four boundary conditions. 4. We obtain the characteristic equation for free vibration (as op- posed to forced) and demonstrate for a free-free beam, that is, the beam is floating in the space. Summary of Elementary Beam Theory Strain-displacement: !xx = "u(x, t) "x ! z" 2w(x, t) "x2 + 1 2 ( "w(x, t) "x )2 Constitutive relation: #xx = E!xx Bending moment: My = ! ! A z#xxd A = E Iz "2w(x, t) "x2 (1) Kinetic energy: T = 1 2 ! L 0 m(x) (wt) 2dx, wt = "w(x, t) "t (2) Variation of the Potential energy: $V = { E I w(x, t)xx$w(x, t)x }|L0 {![E I w(x, t)xx ]x + [P w(x, t)x$w(x, t) }|L0 + ! L 0 { %[E I w(x, t)xx ]xx ! [P w(x, t)x ]x }$w(x, t) dx (7) Substituting (5)-(7) into Hamilton’s principle (4), we obtain the desired variational equation. ! t2 t1 # ! L 0 { m(x) w(x, t)t t + [E I w(x, t)xx ]xx ! [P w(x, t)x ]x ! f (x, t)}$w(x, t) dx + {E I w(x, t)xx$w(x, t)x }|L0 {!([E I w(x, t)xx ]x + P w(x, t)x)$w(x, t) } |L0 $ dt = 0 (8) As in the case of string equations, the first term provides the govern- ing continuum partial differential equation for the beam, the last two expressions yield the boundary conditions. Boundary-value problem for beam bending with P = 0 Governing equation of motion: m(x) w(x, t)t t + [E I w(x, t)xx ]xx = f (x, t) Moment or Slope Boundary Conditions: {M(x, t)$w(x, t)x }|L0 = 0, M(x, t) = E I w(x, t)xx Shear or Displacement Boundary Condition: {V (x, t)$w(x, t) } |L0 = 0, V (x, t) = ![E I w(x, t)xx ]x (9) Since the left-hand term is a function of time whereas the right-hand term is a function of only x, both terms must be independent of time, t , and the coordinate, x . Hence, both terms must be a constant if the equality is to hold for all time and for the entire spatial domain range:, viz., F(t)t t F(t) = !E I (x) m(x) W (x)xxxx W (x) = !&2, & being real (why?) (16) which yields the following two sets of ordinary differential equations: F(t)t t + &2F(t) = 0 (17) W (x)xxxx ! '4 W (x) = 0, '4 = m(x)&2 E I (x) (18) Solution of (17) can be expressed as F(t) = F̄est % s2 + &2 = 0 % " s1 s2 # = " + j ! j # & (19) Therefore, F(t) has the form of F(t) = A sin&t + B cos&t (20) which implies that w(x, t) oscillates with time with the corresponding period ( = 2)/&. Hence, & is its characteristic frequency. Solution of (18) assumes the following form: W (x) = C1 sin'x + C2 cos'x + C3 sinh'x + C4 cosh'x (21) Vibration of Beam Bending 1. Free-free beam: For this the boundary condition (10) gives " M(x, t) = E I (x)w(x, t)xx = 0 V (x, t) = !E I (x)w(x, t)xxx = 0 # & at x = 0 " {E I (x)W (x)xx}|x=0 = 0 {E I (x)W (x)xxx}|x=0 = 0 # at x = L " {E I (x)W (x)xx}|x=L = 0 {E I (x)W (x)xxx}|x=L = 0 # (22) Substituting W (x) (21) into the above four boundary conditions, we obtain with '̄ = 'L: 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Beam span M od e sh ap e Mode shape of free-free beam 2-th mode beta*L = 4.73 Second mode of a free-free beam 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Beam span M od e sh ap e Mode shape of free-free beam 3-th mode beta*L = 7.8532 Third mode of a free-free beam What are the mode shapes of 'L = 0 ? When 'L = 0, one can check the rank of the characteristic matrix A reduces from three to two. This is a degenerative case. Hence, the solution form assumed by (21) is not appropriate. Therefore, one must invoke the governing equation (18), which becomes with ' = 0: W (x)xxxx = 0 (25) that is subject to the boundary conditions (22). Integration of the above equation yields W (x) = c1 = c2x + c3x2 + c4x3 (26) which upon satisfying the boundary conditions (22), yields W (x)x = c2 % W (x) = c1 + c2x (27)