




Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Some basic concept Vibration of Structures are Harmonic Waves, Influence of Axial Force, Initial Value Problem, Mathematical Modeling, Modal Analysis, Motion of Material Points, Orthogonality Relations, Projection Methods.Main points of this lecture are: String Vibrations, Variational Formulation, Lagrangian, Bar Vibrations, Equation of Motion, Boundary Condition, Transverse Dynamics of Taut String, Variational Principle, Dirac Delta Distribution, Kinetic Energy
Typology: Study notes
1 / 8
This page cannot be seen from the preview
Don't miss anything!





Contents:
Keywords: Variational formulation, Lagrangian, String vibrations, Bar vi- brations, Equation of motion, Boundary condition
Transverse dynamics of a taut string:
Lagrangian: L =^12
∫ (^) l 0 (ρAw
(^2) ,t − T w ,x (^2) ) dx
Equation of motion: Applying the variational principle δ ∫^ Ldt = 0 using the string Lagrangian leads to
1 2 δ
∫ (^) t 2 t 1
∫ (^) l 0 (ρAw
(^2) ,t − T w ,x (^2) ) dxdt = 0
⇒
∫ (^) t 2 t 1
∫ (^) l 0 (ρAw,tδw,t^ −^ T w,xδw,x) dxdt^ = 0. Integrating by parts gives ∫ (^) l 0 ρAw,tδw
∣∣t 2 t 1 dx^ −
∫ (^) t 2 t 1 T w,xδw
∣∣l 0 dt^ +
∫ (^) t 2 t 1
∫ (^) l 0 (−ρAw,tt^ +^ T w,xx)δw^ dxdt^ = 0. The integrand in the first integral is zero since δw(x, t 1 ) = δw(x, t 2 ) = 0. Following the arguments of the variational principle, the equation of motion is obtained as ρAw,tt − T w,xx = 0
2
Dirac delta distribution: δ(x−a) = 0 for x 6 = a, and ∫^0 l δ(x−a)f (x)dx = f (a) where f (x) is any sufficiently smooth function.
Potential energy:
V =^12 (kw^2 (a, t) +
∫ (^) l 0 T w ,x^2 dx^ =^1 2
∫ (^) l 0 (kw
(^2) δ(x − a) + T w ,x (^2) ) dx (2)
Lagrangian:
L =^12
∫ (^) l 0 [(mδ(x^ −^ a) +^ ρA)w ,t^2 −^ (kw^2 δ(x^ −^ a) +^ T w^2 ,x)]dx.^ (3)
Equation of motion: Using the variational procedure, the equation of motion is given by
(mδ(x − a) + ρA)w,tt − T w,xx + kwδ(x − a) = 0 (4)
and boundary conditions are obtained as w(0, t) = 0 and w(l, t) = 0.
Axial vibrations of a circular bar with lateral deformations: Assuming axisymmetric radial strain and negligible radial stress, the consti- tutive relations are given as
x = u,x = σ Ex r = w,r = −ν σ Ex (5)
where ν is the Poisson ration and w(r, x, t) is the radial displacement field. Using the two relations in (5), we can write w(r, x, t) = −νru,x (using w(x, 0 , t) = 0). 4
Kinetic energy:
T =^12
∫ (^) l 0 (ρAu
(^2) ,t +^ ∫ A^ ρw ,t^2 dA) dx^ =^1 2
∫ (^) l 0 (ρAu
(^2) ,t + ρν (^2) Ipu (^2) ,xt) dx. (6)
Potential energy:
V =^12
∫ (^) l 0 EAu
(^2) ,x dx (7)
Lagrangian:
L =^12
∫ (^) l 0 [ρAu
(^2) ,t + ρν (^2) Ipu (^2) ,xt − EAu (^2) ,x]dx. (8)
Equation of motion: From the variational statement δ ∫^ Ldt = 0, one obtains 1 2 δ
∫ (^) t 2 t 1
∫ (^) l 0 [ρAu
(^2) ,t + ρν (^2) Ipu (^2) ,xt − EAu (^2) ,x]dxdt = 0
⇒
∫ (^) t 2 t 1
∫ (^) l 0 [ρAu,tδu,t^ +^ ρν
(^2) Ipu,xtδu,xt − EAu,xδu,x]dxdt = 0
Integrating by parts, we have ∫ (^) l 0 [ρAu,tδu^ +^ ρν
(^2) Ipu,xtδu,x]t t (^21) dx +^ ∫^ t^2 t 1 [−ρν
(^2) Ipu,xtt − EAu,x]δu∣∣l 0 dxdt
∫ (^) t 2 t 1
∫ (^) l 0 [−ρAu,ttδu,t^ +^ ρν
(^2) Ipu,xxtt + EAu,xx]δudxdt = 0.
Following the arguments of the variational formulation, the integrand in the first integral is zero. The equation of motion is obtained as
ρAu,tt − EAu,xx − ρν^2 Ipu,xxtt = 0. (9)
The possible boundary conditions are given by
EAu,x(0, t) + ρν^2 Ipu,xtt(0, t) = 0 or u(0, t) = 0 and EAu,x(l, t) + ρν^2 Ipu,xtt(l, t) = 0 or u(l, t) = 0. 5
and the boundary conditions are obtained as w(0, t) = 0 and w(l, t) = 0.
l
h(t) x
z, w (^) ρ, A, T
Figure 3: A string with a specified boundary motion
String with prescribed boundary motion: (see Fig. 3) We may define the field variable as a superposition of a general motion (with homogeneous boundary conditions) over the “static” configurations of the string at different time instants as w(x, t) = h(t)(1 − x/l) + u(x, t), where u(0, t) = u(l, t) = 0. Kinetic energy:
T =^12
∫ (^) l 0 ρAw ,t^2 dx^ =^1 2
∫ (^) l 0 ρA
h(t)
1 − x l
dx
Potential energy:
V =^12
∫ (^) l 0 T w
(^2) ,x dx =^1 2
∫ (^) l 0 T
u,x − h( lt)
dx
Equation of motion:
u,tt − c^2 u,xx = −
1 − x l
h(t).
The boundary conditions on u(x, t) are homogeneous, as already mentioned 7
above.
The variational method has the following features:
8