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MECHANICAL VIBRATIONS
EXERCISES
\
tf#s
Autumn2014
Division of MechanicsLund University
Faculty of Engineering, LTH
GETTO
2014
MECHANICAL VIBRATIONSIntroductory exercises
The student may consult the basic course in mechanics for information on vibrations of one-dimensional systems, for instance the textbook Meriam & Kraige; Dynamics, Chapter 8 orNyberg C.; Mekanik, Grundkurs, Chapter 12.
Theory
1. Show that the initial value problem for a one-dimensional system performing freevibrations
mi+kx =0x(0)=xo, *(0)=vs
(1)
where m andft are positive constants and xst v6 are constants prescribing the initialconditions, has the solution
x(t) = xo cosaot+J9-sin aror
2, Show that the initial value problem for a one-dimensional system performing forcedvibrations
mi+kx: Psinatx(0) = "ro, *(0) : vo
(3)
where P and o are constants characteÅzing the 'external driJing force', has thesolution
a)o
.Ewhere ,r: l; is called the naturalfrequency of the system.
x(t) :xo cos arot + a (uo - s(9) + r) sin aror + 9191! tit r," a)o - -@r' k - 'or' k
.81where r, = l; and g(0) = ft is the so-called magnificationfactor.
(2)
(4)
3. Using the solution (4), in exercise 2 above, and the following data:
20r4 2
m = L0kg, k : t000Nmm-1 , xo = l.7mm, vo = l0mms-1 ,
f':100N, a=20rads-l
Plot (using a math software such as Matlab or Maple) the motion x = x(t)a) in the time interval 0 <t < 2 .
b) in the time interval 0 <t < l0 .
4. If viscous damping is introduced into the model then the initial value problem for thesystem performing forced vibrations may be written
mi+c*+kx= Psinafi
x(0)=xo, *(0)=vo(s)
where c is a positive constant ('the viscous damping coefficient'). Calculate (using amath software such as Matlab or Maple) the solution to the initial value problem. Usethe data in exercise 3 above along with the viscous damping coefficient c:3Nsmm-l .
Plot the motion x = x(t)
in the time interval 0 3t < 2 .
in the time interval 0 < t < I 0
Explain the performance of the vibration. What happens in the limit/ ) oo. Comparewith the un-damped vibration in exercise 3!
5. The so-call ed steady state solution to (5) may be written
a)
b)
x,(t) = | s(9, dsin(ar - ä)K a)o
(6)
where ( =;Lis the so-called relative damping. The magnification foctor g iszmoo
given by
g(0,6) =I
(I-0t)t +46202,0<e<a,0 <g'<oo (7)
andthe phase lag angle ä is given by
6(o,q):arctanffi) if 0 <e < I
arctanffi)+rif0>1(8)
a) Show that the steady state solution (6) is a solution to the differential equation in(5). Does it tulfil the initial conditions?
2014 3
b) Plot the solution to the initial value problem (5) x = x(t), as obtained in exercise 4,
and the steady state solution.rr -x"(r) in the same diagram for the time interval0 < t < 10 . How do the solutions differ? What happens in the long run?
c) Plot the magnification factors g=9(0,0.05),0<e<4 andg = g(0,0.5), 0 <0 < 4 inthe same diagram.
d) Plot the phase lag angles 6 =6(0,0.0D, 0<0 <4 and 6 =6(0,0.1), 0<0 <4 inthe same diagram.
Applications
6. Calculate the natural frequency of vertical oscillations of the 25-kgblock when it is set
in motion. Each spring has a stiffness of 1200 Nm-1 . Neglect the mass of the pulleys.(Meriam & Kraige, Dynamics (5m ed.), Problem 8123).
7. The system shown is released from rest from an initial positionxo. Determine the
'overshoot' displacementx, . Assume translational motion in the x-direction. (Meriam& Kraige, Dynamics (5ft ed.), Problem 8/38).
x.
t0å = 108 N/m c = t8 N.slm
I
8. The motion of the outer cart B is givenby
xr=xu(t)=bsin4x9t (e)
20t4 4
For what range of the driving frequency ar is the amplitude of the motion of the massm relative to the cart less than 2b. The damping of the system is assumed to benegligible. (Meriam & Kraige, Dynamics (5ft ed.), problem S/5S).
*l ", = å sin arf
9. Two identical uniform bars are welded together atight angle and are pivoted about ahorizontal axis through the point shown. Determine the critical driving frequency ofthe block B which will result in excessively large oscillations of the assembly. Themass of the welded assembly is z. (Meriam & Kraige, Dynamics(5tr ed.), problem8te2).
t/2 l/2
o
B
_l xn= b sin ot
10. (Rotating unbalance). A conterrotating eccentric weight exciter is used to produceforced oscillations of a spring-damper supported mass, as shown in the figure below.By varying the speed of rotation a resonant amplitude of 0.6 cm was recorded. Whenthe speed of rotation was increased considerably beyond the resonant frequency, theamplitude appears to approach a fixed value of 0.08cm. Determine the relativedamping of the system.
hh
a t
Irl
it Å.t
2014 1
MECHANICAL VIBRATIONSExercise l: Equations of motion
Theory
1. A mechanical system (body) rB is subjected to the extemal force system {af"}and the
internal force system{UU'} . Show that
dF'+ dF' =adm dF'+dFt =adm
dF'=0 adm (l)
rxdFt = 0 lrxadm
I
I
=JdF"<> Ja
dF"rX@
2. Consider a mechanical system (3 whose configuration space is l-dimensional, i.e. theposition vector of a material point P eE is given by r = r(q;P) where q is one scalarvaiable, q=q(t). We assume that the system has a potential energy U =(l(q) and
that Q*' :Q'nt =0.
a) Show that the kinetic enorgy of the system may be written
r(q,q)=Io@)d' where a(q)=t# #* (2)
b) With the Lagrangian L = L(q,q) =T -(J calculate the equation of motion forthe system.
c) Show that the mechanical energy E = E(q,Q) :T+U is a constant (an integralof motion). Use this to show that a solution q = q(t), / > 0 to the equation ofmotion with the initial condition q(0)= q", q(0) =ro ) 0 must (at least in thefirst phase of the motion) satisfiz
[-sG)
Ja(x)
(3)E -U(x))
Evaluate this relation for the case q(q)=m, (l(q)=XU'where m and k are
positive constants.
d) Show that q = 4o, Q = 0 is an equilibrium state if and only if
2014 2
(s)
3. A mechanical system has the Lagrangsan L=L(t,q,q)and the generalized forces
Q, =Q,(t,q,f) . Show that
e) Show that an equilibrium position is stable if
du(q) _ndq
d'u(g) >o
dq'
o,.=0. oL =o=L=const' oqo ö40
(4)
(6)
(7)
(8)
In this case we say that * tt an integral of motion.oQr
4. Given a mechanical system with Lagrangian function L=L(t,q1,...,g,,qr,...,qo) and
with the generalizedmechanical energyE=E(t,q,Q):t*A,-2. Show that in ai=t uQi
change of coordinates 4 -) q-, defined by q = q(t,q.), we have the relation
E*-r- a öL 0q,
k aq. a,
5. A mechanical system has the Lagrangian L = L(t,q,q) and the gyroscopic inertia force
K! =f frA,. The equations of motion are lineari zed, at an equilibrium statej=l
Q : eo, e : 0 . Show that the gyroscopic matrixG = ["u] may be written
.azL. ,ig,i = (;
- ), -(^-oQiosi i*'"-(E)"
20r4 -t
Applications
6, A physical pendulum in the form of a rigid body with mass m may rotate around afixed horizontal axis, with respect to which its moment of inertia is ,/. The centre ofmass c has the perpendicular distance h> 0 to the axis, see figure below. UseLagrange's method to formulate the equation of motion for the system:
a) if there is no friction in the revolute joint between body and axis.b) if there is a constant friction moment M, in the revolute joint between body
and axis.
Calculate the possible equilibrium configurations of the system. Are they Liapunovstable (asymptotically stable) or unstable?
J
gm
7, The pendulum system, moving in a vertical plane, consists of a small ball with mass mfixed with a string, of length l, to a carriage which may move, without friction, alonga horizontal line as shown in the figure below. The carriage, with mass rz", isconnected to a fixed point with a linear elastic spring, with spring constant k, andsubjected to an external force F = F(t) along its line of motion. Use Lagrange'smethod to formulate the equations of motion for the system, (-r=0corresponds tounstressed spring).
l.__-| mc
F:F(t)
I
X
k
a
-k\\ c
d
m
2014 4
8. A mechanical system consists of a thin hoop of radius Ä and mass mhthat may rotatewithout friction around a vertical axis. A bead of mass m can slide freely (no friction)around the hoop. The hoop is acted upon by the torque M : M(t) about the verticalaxis.
a) Use Lagrange's method to formulate the equations of motion for the system.b) Determine whether there are any integrals of motion.c) Assume that the hoop is rotating with the constant angular velocity .f)
Determine the possible equilibrium states of the system.
9. A double pendulum consists of two particles E,and,Qwith masses m.,andm,respectively connected by two strings of lengths l, and l, according to the figurebelow.
a) Use Lagrange's method to formulate the equations of motion for the system.b) Linearize the equations of motions at the equilibrium state:
0=ö:0,0=ö:0.
I uo)
X
v
0('b
[1Qr = $
9z=QPI lll 1
L2
s lil2P2
2014 5
10. A mechanical system consists of a particle pendulum with mass m, and cordJength /connected to a cariage, with mass mz, which may move along a vertical guide, seefigure below. The carriage is supported by a non-linear spring element with aretracting spring force equal to S=S(x) =krx+kzx3, kt,k2>0 where x is theelongation of the spring (measured from the unstressed state). The contact between thecarÅage and the guide walls gives rise to a constant frictional forceF, >0 opposing
the motion. Using the generalized coordinates x and 0
a) Formulate the equations of motion for the system.b) Calculate the dissipation function for the system.c) Find the equilibrium states and decide on their stability
S:S(x)
{rrll9
0I
I
I
gm1
2014 I
MECHANICAL VIBRATIONSExercise 2: Un-damped Vibrations
Theory
1. The free motion of an n-dimensional, lineanzed, mechanical system with mass-matrixM, damping/gyroscopic-matrix B and stiffness-matrix K is given by
Mci+Bq+Kq=g (1)
Assume thatq :q(l) =x€'t, s eC and x e IRo a constant vector, is a solution to thisdifferential equation. Show that
Z(s)x = 0 (2)
where Z(s) =Ms2 +Bs +K is the so-called mechanical impedance matrix.
2. Show that if detK=O then there is a constant vector u+0 such that the motion("rigid-body mode") q = q(/) =u.(at+ F), a andB constants, is a solution to
Mti+Kq=O (3)
3. Show that the so-called Rayleigh quotient (.(x) = # for an z-dimensional
mechanical system with mass-matrix M and stiffness matrix K satisfies
tt <q!)<r,t, VxelR' (4)
where c,f is the smallest eigenfrequency and c,fi the largest, (tt < r,t < ...< rlt l.
4. An n-dimensional lineaized, undamped, non-gyroscopic mechanical system withmass-matrix M and stiffness matrix K has the characteristic equation
det(-afM+K)= c,(o')' +c,-r(af)'-t+...+co=0 (5)
a) Show that c, = (-1)'detM, co = detK
b) Show that if detK = 0 then oz =}is a root to the characteristic equation.
c) Show that if tt,tir,...,r?, are the roots to the characteristic equation and
detM + 0 then for the product of all roots we have
ttrdetKo:@:...a'tz'detM (6)
2014 2
Applications
5. See Exercise 1:7. The pendulum system, moving in a vertical plane, consists of a smallball with mass m fixed with a string, of length l, to a carriage which may move,without friction, along a horizorrtal line as shown in the figure below. The carriage,with mass m", is connected to a fixed point with a linear elastic spring, with springconstant k (x:0 corresponds to the unstressed spring).
a) Formulate the linearized equations of free vibrations for the system around theequilibrium state:.g= x = 0, ci = * = 0 .
b) Calculate the system eigenfrequencies and the corresponding modes shapes forthe case ffi = ffi"= 1.Okg, k = 100N/m, I :1.0m, g = 9.81ms-2.
c) Show that the mode shapes are orthogonal with respect to the mass-matrix.d) Introduce the extemal force as in Exercise 1:7; F:Fosin(at)and calculate a
forced response (particulate solution).
t--| mc
m
X
k
(p
6. See Exercise 1:9. A double pendulum consists of two particles P, ndPrwithmassesffit = ffiz = m respectively, connected by two strings of lengths l, = l, =/ according tothe figure below.
a) Formulate the linearized equations of free vibrations for the system around theequilibrium state: 0 = ö = 0, 0 = ö = 0.
b) calculate the system eigenfrequencies and the corresponding modes shapes.
v
0 X(}6
tiQr = s
Qz=ÖPi 1tn
L2
+
P2
til2
2014 aJ
7. Consider a mechanical system consisting of two identical particle pendulums withmass m=0.5k9 and cord-length l=2.0m (meter) connected with a spring whoseunstressed length is equal to the distance between the points of suspension of thependulums and with a spring-constant = k. Denote by 0, and 0, the angles ofinclination of the pendulums, see figure below.
a) Formulate the lineanzed equations of free vibrations for the system around theequilibrium state:0, - 0, = 0, 0t = 0z = 0 .
b) Calculate the system eigenfrequencies and the corresponding modes shapes.c) Suppose that the pendulums arc atrest at the initial moment (l = 0), and one of
them is given the initial angular velocity 4(0)= at=lradls. Calculate thesubsequent motion in the cases fr=10N/m andk =0.1N/mrespectively. Usesome math code (Matlab, Mathcad, Maple, ...) to plot the motion in a diagram.
oöll2
t/2 U2
8. Consider the mechanical system in the figure below. The masses move along a fixedhorizontal line and the left mass is subjected to an external harmonic force withamplitude .{ and angular frequency ar.
a) Use the coordinatesr, and xrand formulate the equations of motion for thesystem.
b) Calculate the natural frequencies and mode shapes for the free vibrations(4 = o).
c) Calculate the forced vibration (4 >0). Is there any frequency a for which
the left mass is at rest (x,(t) = 0 X
xr
I el
kkkm m
F0
sin (ort)
X2
20t4 4
9. Consider a mechanical system consisting of three wagons with masses 2m,3m and mrespectively. The wagons are connected to each other and to the walls with springs ofstiffness 2k, 3k, 2k and ä respectively, see figure below.
a) Use the coordinates x, x, ardx, and formulate the equations of motion for the
system.b) Calculate the natural frequencies and mode shapes for the free vibrations.c) Show that the mode shapes are orthogonal with respect to the mass- and
stiffness-matrices.d) Calculate the component ar, = arr(at)of the admittance matrix. Plot this
function in a diagram. Find frequencies a)o satisfuing: arr(at")=g (anti-resonances).
e) Introduce the constraint x, = 0 (i.e. we fix the wagon on the left). Calculate thenatural frequencies for this constrained system (with two degrees of freedom).
kUse 10as-2 and m = lkg in the calculations.
m
]* "t l* x2 l* t,
k2k3k2k
2014
MECHANICAL VIBRATIONSExercise 3: Damped Vibrations
Theory
1. A mechanical system is defined by its mass-matriX M, its stiffness-matrix K and itsdamping matrix:
C = al|d+ BK, a, B (positive) real constants (1)
Show that
CM-tK: KM-rC
2. Let E denote the relative modal damping and coo the un-damped modal natural
frequency of the system in exercise 3:1 above with a =l.0and F=0.00I. Plot the
function: C = C(a), 10S aro <100.
3. The free vibrations of the system in exercise 3:1 above are given by the solution to theinitial value problem:
Mii+Cti*Kq=[q(o): Qo, Q(o) = 4o
where {0,40 are given constant vectors.
(3)
Show that the solution to this problem is given by:
(2)
q(r) = lin''^" r-*''lM-*, (a4,t).r,#+sin(ar,r))]qo +
.[i e-€,,o,txixlD[, 'lfr Fi@ai rin(ar''r)lqo
(4)
where x, denotes the classical mode shapes, pi,ooi,6i and roo,are the modal mass, un-
damped natural frequency, relative damping and damped natural frequencyrespectively, corresponding to the ith mode. We assume that 0 < 6 < I (weak
damping).
4. Let p(s)=det(Ms2+Cs+K) be the characteristic polynomial for a mechanical
system.
a) Show that if p(s) : 0 thenp(s*) = 0 (where s * is the complex conjugate of s)
b) Show that if p has a double root at s : s, (p(s, ) = 0) then
2014 2
p'(t,)=0, p'(s,)+o (5)
Applications
5. See Exercise l:7. The pendulum system, moving in a vertical plane, consists of a smallball with mass m fixed with a string, of length /, to a carriage which may move,without friction, along a horizontal line as shown in the figure below. The carriage,with mass m", is connected to a fixed point with a linear elastic spring, with spring
constant k (x:0 corresponds to the unstressed spring). Introduce a Rayleigh dampingto the system according to
C = 0.7M (6)
a. Formulate the equations of free vibrations for the system around theequilibrium state:g = )c = 0t Q = * = 0 .
b. Calculate the modal relative dampings, the un-damped natural frequencies, thedamped natural frequencies and the corresponding modes shapes for the case
ffi : ffi"= 1.0kg, fr = 100N/m, / = 1.0m, g = 9.81ms-2.
c. Show that the mode shapes are the same for the damped and the un-damped(C = 0) cases.
d. Assume initial conditions x(0) = 0, p(0):0, i(0) = 0, p(0) =lrad I s .
Calculate the subsequent motion in the cases C = 0 and C = 0.7Mrespectively. Plot the motions in a diagram.
Use some math code (Matlab, Mathcad, Maple, ...) to solve the problem.
l--|
6. Consider the mechanical system in Exercise 2:7 consisting of two identical particlependulums with mass m=0.5k9 and cord-len# I =2.0m (meter) connected with a
spring whose unstressed length is equal to the distance between the points ofsuspension of the pendulums and with a spring-constant:k=l0N/m. Now assumethat the spring has an internal viscous damping with damping constant c = 0.5Ns/m .
Denote by 0, and 0, the angles of inclination of the pendulums, see figure below.
a. Formulate the equations of free vibrations for the system around the
equilibrium state: Q - 0, = 0, 0, = 0z = 0.
X
k
(p
m
20t4 J
b. Is it possible to diagonalize this system using, for instance, the classical normalmodes?
(,öll2
U2 ll2
7. Consider the mechanical system in the figure below. The masses move along a fixedhorizontal line and the left mass is subjected to an external harmonic force withamplitude Fo and angular frequency ar. Introduce Rayleigh damping according to
C = 0.006K (7)
where C and K are damping- and stiffrress-matrices respectively
a. Use the coordinatesx, and xrand formulate the equations of motion for thesystem.
b. Calculate the modal relative dampings, the un-damped natural frequencies, thedamped natural frequencies and the corresponding modes shapes for the freevibrations (4 = 0).
c. Calculate the forced vibration (4 =100N) for the frequencies:
a=30radlsandrco =40radlsrespectively. Assume that we are starting fromrest, i.e x,(0) = xr(0) = 0, xr(Q) = ir(0) = 0. Plot the position of the two masses
as a function of time.d. Calculate the forced vibration (Fo :100N ) for the frequency at = 3}radls ,
with the initial conditions:x,(0) = -0.1m,xr(0):0.|m, ir(Q) = *z(0)= 0. Plotthe position of the two masses as a function of time.
Use the following data in the calculations:m=lkg, ft=1000N/m. Use some mathcode (Matlab, Mathcad, Maple, ...) to solve the problem.
Xr
I 0l
kk k
ll2 e2
k,c
m m
FO sin (r,rt)
X2
2014 4
8. Consider a mechanical system consisting of three wagons with masses 2m,3m and mrespectively. The wagons are connected to each other and to the walls with springs ofstiffness 2k, 3k, 2k and k respectively, see figure below. Introduce Rayleigh dampingaccording to
C = 0.00lK (8)
where C and K are damping- and stiffrress-matrices respectively
a. Use the coordinates x, x, and xrand formulate the equations of motion for the
system.b. Calculate the modal relative dampings, the un-damped natural frequencies, the
damped natural frequencies and the corresponding mode shapes for the freevibrations.
c. Calculate the absolute value of the component ar, = arr(ia) of the admittance
matrix. Plot this function in a diagram. Are there any frequencies a)o
satisfuing: lar,(af 1l = g 7
kl0as-2 and m= lkg in the calculations.Use -=m
f* tt l* x2 l- t,2k 3k k2k
2014
MECHANICAL VIBRATIONSExercise 4: Damped Vibrations
I
Theory
1. An n-dimensional mechanical system, with mass-matrix Mand frequency responsefunction F: F(ar) is subjected to an extemal force
P = Posin(ot) (1)
wherePois a constant amplitude vector and (D the angular velocity. The initialconditions are
q(0):e0,4(0)=do
Show that if the system is "diagonizable" with natural modes:
(2)
(4)
X1 r..'r X,
51 r. ' '., Sn
then the motion of the system is given by:
^-(it4,t XrXrtM
e-
Fi
si =-(i@0, +iao., @,t, = @o,\lr- (:, 0 <(, <I (3)
(cos(aru t) + (t9!isin(ato,t))(qo-f(0)) +Ij=l
q(r)
fi=l "-(iti",t
XrX,tMsin(aru r)(qo - f(0)) + fQ)
Hioa,
where
f(/): Im(F(a)Poei'') (s)
Applications
2. A particle with mass la is connected to a circular ring with four linear elastic springs.The ring is rotating in a plane, around its syrnmetry axis perpendicular to the plane andwith a constant angular velocity o. When the particle is positioned at the centre of thering the springs form a perpendicular cross according to the figure below. When theparticle is displaced from this position the potential energy in the spring system isgiven by
v =v(x,y)=vo*)*ofi1*'* y'1 (6)
2014 2
where Vo and co,are constants and x andydenote the displacement of the particle intwo perpendicular directions. It is presumed that the particle may only move in theplane of the circle.
a. Formulate the equations of motion for the system using the coordinates .r andyintroduced in a coordinate system rotating with the ring.
b. Show that if A'*rd thenx=!=0 is an (relative) equilibrium state for theparticle.
c. Show that the fficient potential energ/ (V.:V-4) has a minimum at the
equilibrium position if C)2 < r'fi and a maximum if A'z > c'Å .
d. Show that the equilibrium position is Dirichlet stable if A'z * r'$.e. Assume that the particle is subjected to a viscous frictional force with the
Rayleigh dissipation function
a(|n=ma4C(*2 + i,2) (7)
where the damping constant ( << I .Show that the equilibrium position is nowDirichlet stable if and only if Q' .rå.
20r4
MECHANICAL VIBRATIONSExercise 5: Continuous systems
I
Theory
1. Show that the system of functions
is orthonormal, i.e.
g,(x)rp^(x)dx =
g,(x) : ffrin{!!!), o < x < L, n =0,r,. (1)
L
J0
0 if n*mI if n=m
(2)
(4)
2. The vibrational motion of a bar in extension is governed by the differential equation:
(3)
where Z is the length of the bar, E is the modulus of elasticity,A:l(x) is the
sectional area, k=k(x)is the external force density, m=m(x)the mass density, and
by the boundary-initial conditions:
*ruon Y)+k(x)=*<iffi, o( x<L, t)o
aru(o,t)* BrY=o, aru(L,t)* BrY=o
u(x,t):A@)fi(t)
a) show the following conditions are necessary:
- -r*rEA@)4#):Aa@)a,fi(o)* P,#=0, arfi(L)* Pr#=o
0)u(x,O) =uo(x), = io(x)
where a1,B,dz,F, are given constants ((a,F) +(0,0) and (a,F)*(0,0)) and
uo = ur(x) and io = åo(x) are given functions.
Assume that k = 0 (oofree vibrations") and consider a solution on the form ("separationof variables")
x)
0t
öu
(s)
(6)
2014 2
where )" is a constant and
d2QQ)+ ),QQ)=0
dt2
b) Show that the differential operator ("The Sturm-Liouville operator")
(7)
(8)
(e)
(10)
is symmetric, i.e. that:
tt=- | dtrot*ldfr(*)lm(x) dx dx
JI0
(rfi(x))$(x)m(x)dx = (rö(x))fi(x)m(x)dx0
for all fi =fi(x) and f = i(x) satisffing the boundary conditions in (6).
c) Show that if A, and fi, are solutions to (6) then
4* 4+ !d,1*1tt,(x)m(x)dx =0L
0
Applications(All beams below may be modelled by the Euler-Bernoulli theory)
3. A piano string is fixed at both ends. It has a length of 1.4m and a mass of 1109.
Calculate the required pre-tension 4 so that the first natural frequency will be 424
Hz.
4. A string of steel with a circular cross section is fixed at one end and attached to aspring at the other end with a frictionless slide. See figure below! The spring constant
fr = 8000Nm-t. The pre-tension of the string is { = 10.0.104N . Calculate the mode
shapes and the natural frequencies for the string-spring system. Use the followingdata: stringdensity = p=8000kgm-3, radius r = 1.0mm and length Z = 1.0m.
L' To
k>:i1
!
2014 -t
5. A homogeneous bar with total mass ftt, sectional area Aoand length d is clamped at
both its ends. Calculate the mode shapes and natural frequencies for the system inlongitudinal vibrations.
m,4, Lo
6. (Revised ) A homogeneous cantilevered ("clamped-free") bar of length Z = 5.0m and
sectional area A=4.10-amz is made of a material with density p=8000kgrn-3 and
modulus of elasticity E=200GPa. The bar is initially (at time l=0) at rest and isthen subjected to a step loading at the oofree end"
P(t1=0 t<0Po t>0
where Po =10000N. Calculate the longitudinal response of the bar. Plot the response
at the mid-point of the bar (using some math code).
P,E, A, LP
7. Calculate the natural frequencies and mode shapes for the first three modes of thehansverse (bending) vibrations of a beam that is clamped (fixed) at one end andsimply supported at the other. The beam length Z : 1.0m, sectional area
A=2.6.10-3m2, sectional area moment-of-inertia I =4.7.10-6ma. The beam is made
of a material with density p = 8000kgmi and modulus of elasticity E =200GPa .
P,E,A, L, I
8. Calculate the three lowest natural frequencies for the system in the figure below. Themass of the weight fixed to the end of the beam is equal to 10 kg. The beam is thesame as the one specified in the previous problem 5:7. Assume that there is noinfluence from gravitational forces.
20t4 4
9. The simply supported beam, withdensity = p, modulus of elasticity = E,length= L,sectional area = A and sectional area moment-of-inertia = 1 (assumed to be constant),is subjected to an external transverse force
p = p(x,t) = p(x)sincot (11)
Calculate the forced "steady-state" (only consider the particulate solution, assume thatthe homogeneous solution is damped out in the long run) response of the midpoint
* = L of the beam.2
a) t@) = po = const.
b) b@) = Pr6(x-*1. t... a point force .S applied at the midpoint of the beam.z
---->
Po
2014 I
MECHANICAL VIBRATIONSExercise 6: Continuous systems
and Approximate methods
Theory
1. Let J be a surface in iR3and let f =f(r) andA=A(r)be a scalar- and tensor-field(2nd order) respectively defined onJ . Show the calculation rule:
div,(f A)= f div,A.+AV./ (1)
and use this to show that for a membrane
div r(Tol r) = To2rc ^n+Y rTo
where d is the pre-tension of the membrane, 1,, the unit tensor s1 J, r- is the mean-
curvature and n = n(r) is the unit normal vector field on J .
2. Let
4 = e, @e,{ +e" 8er,S, (3)
denote the traction tensor for a thin plate. Show that
(2)
3. Let
(4)
Mo = -M*",@e,-Mre" @u, +M,erE e,+Moer@", (5)
denote the moment tensor for a thin plate. Show that
div,^To:.,(**91"oxoy
div,^M o == ".7-öY
o -Y.l + " "(% * %->"oxoyoydx (6)
20r4 2
Applications
4. We consider a uniform rectangular membrane with mass per unit area m extending, inits un-deformed configuration, over a domain Jo in the x-y-plane defined by
s, ={(",-v) elR2 :0<x< a, 0 1 y.b} (7)
The boundary of J, is the "curve" f0 consisting of the straight
linesx =0, x = a andy:0, y =b .The membrane is supported along fo and the pre-
tension of the membrane is a constant equal to {
a) Using membrane theory, formulate the equations of motion for the free transversalvibrations of the membrane.
b) Formulate the boundary conditions corresponding to the membrane fixed at theboundary fo.
c) Look for standing wave solutions and derive the necessary eigenvalue problems.d) Calculate the eigenmodes and the corresponding eigenfrequencies.
5. We consider a uniform rectangular plate with mass per unit area m and thickness å
extending, in its un-deformed configuration, over a domain J, in the x-y-plane
defined by
so = l@,y)e R.2 : 0 < x < a, 0 I y <bl (8)
The boundary of J, is the "curve" f0 consisting of the straight
linesx=O, x=aandy:0, y=b . The plate is simply supported along fo. The
modulus of elasticity and the Poisson's ratio of the plate material are denoted E and vrespectively.
a) Using thin plate theory, formulate the equations of motion for the free transversalvibrations of the plate.
b) Formulate the boundary conditions corresponding to the simply supported plate.c) Look for standing wave solutions and derive the necessary eigenvalue problems.d) Calculate the eigenmodes (mode shapes) and the corresponding eigenfrequencies.
6. We consider a uniform circular plate with mass per unit area m and thickness åextending, in its un-deformed configuration, over a domain Jo in the x-y-plane
defined by
J, = {(x,.v) e lR2 : x' + y' < a'} (e)
20t4 -t
The boundary of Jo is the circle
The plate is clamped along fo. The modulus of elasticity and the Poisson's ratio ofthe plate material are denoted E and v respectively.
a) Using thin plate theory, formulate the equations of motion for the freetransversal vibrations of the plate.
b) Formulate the boundary conditions corresponding to the simply supportedplate.
c) Look for standing wave solutions and derive the necessary eigenvalueproblems.
d) Calculate the eigenmodes and the corresponding eigenfrequencies.
7. A homogeneous clamped-free bar of length Z=5.0m is made of a material withdensity p = 8000kgm-3 and modulus of elasticity E =200GPa . Calculate the first twoeigenmodes in longitudinal vibrations using the Rayleigh-Ritz method and the two-dimensional approximation of the displacement field:
fo = {(",y) e lR2 : x2 + y2 = a2l
tt(x) = |,n,
*(;)' u,, o < x < L
tu(x):(;)' r,*(;)' q,, o<x<L
(10)
(r2)
(1 1)
m,A,L
8. The homogeneous cantilevered beam in the previous exercise has a bending stiffnessEI . Calculate the first two eigenfrequencies in transversal vibrations using theRayleigh-Ritz method and the two-dimensional approximate displacement field
where e, ffid q2are generalized coordinates. Compare these with the exact modes!
where 8, ffid q2are generalized coordinates. Compare these with the exact
frequencies!
9. A homogeneous bar with total mass m, sectional area A and length Z is clamped atboth its ends. The modulus of elasticity of the bar material is denoted E. Calculate thenatural frequencies for the system in longitudinal vibrations using the FEA with a
two-element approximation and both a consistent-mass and a lumped-mass matrix.
201,4 4
Compare the nafural frequencies of these two approximations and the exactfrequencies obtained in Exercise 4:5.
mrArL
10. The homogeneous beam in the previous exercise has the bending stiffrress^Ul.Calculate the natural frequencies for the system in transversal vibrations using theFEA with first a two-element approximation and then a three-element approximation.Compare the natural frequencies of the two approximations with the exact frequencies.
Usethe data: EI =9.4.105Nm2, ffi=l0kg, L:2.0m
MECHANICAL VIBRATIONS
SOLUTIONSTO EXERCISES
e
E Gf,T]O
Autumn2014
Division of MechanicsLund University
Faculty of Engineering, LTH
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Exercise 3:5
lnput:
mc:= I m:= 1
l:= 1 g:= 9.81
Galculations:
k:= 100
)"(;l)
Coefficientveclor:
Ia;= lul r+ = r
13:= 1.4
I2:= 12O.11
11 :; E3.734
b,= lrl Io = srr
Mass-, stiffness- and damping-matrices:
I=
981
M:=m"+m m.l
m l ^-12
I1
12
b14
1 _1ilut, t
83.734
x2=-1.097
t20.1
7.4
I
(t o\K:=l I
\0 m.g.l/ K.=100 -0
0 9.81(
(.t
C:= 0.7.M(t.+ o.t\
" =
[0., o.r]
Undamped eigenfrequencies:
6l;= genvals(K,M)
Roots to the charactedstic equation:
| ;= polyroots(I)
c) Mode shapes
Claeslcal mode shapee:
lJ;= genve(K,M)
f-o.rsoooooorooet 7 I + 2.9ss3724s420042i\
. _ | -o.rrooooooo z8z7o4 - to.slssglrorrnrr, I
| 4349999999717299 + l0-51s59180959s5i I
\ -0.34999999E993 82 - 2s5537z4s4zoolli )
Rayleigh damping
(i
o2=110.763
8.857 V., =10.524
2.976
'6, ,= ('/äz), .6r,= (./.2), 2.976
10.524oo=
a) Damped eigenfrequencies:
lmpedance mairix:
Z(s):- M.s2 + C.s + K
ChaEcteristicpolynomial:
p(s):= lz(tl
p(s) + s4 + 1.4.s3 + 120.11's2 + 83.?34.s + 981.00
/ o.oz+ -o.roz\u=t I
[+.zrs -o.ela./
t (o)
-.u-ut,z .'=[;,)xl :=corespondin g to o' =2.976
coresponding to o =10.524x2:=
3
Modal matrix: *(il ,= *, *6 ,= *, "= [n.]r, _r.t rr)
I7
Dampad mode shapes:
Firsl eioenvalue:
Ir -- -O35 + 2.95ii
161,= n{f1) odl =2.955 dampednaturaltrrequency
,Or=2.926
,,,= l(rilltot(l =0.ll8 relalivedamplng
Second eioenvalue:
lz=-O35-10.5t9i
.r '= lL(r,r)l odz = to.jte
tor= to'52q
e, ,= J{tt[ (r = 0.033r0,
damped natuEl frequenc!/
undamp€d natuEl frcquency
rehtirre damping
s:= z(r1)
The adjoint matrk e:
s'Q =
Comparison with the classical normal mode:
.,=(,;,)
"_(rrt, -r.r89ix r0-8 -r.852-s.vzix r0-el
(.-r.rr, - ,.rnr, * t,-9 o.ssz - s.g4z, , ,o- n J
s,= fi2) -121.527 -110J63-1t0.763 -100.953
s=
Qr,r:= (-1)I+r's2, , e",r:= 1-r;2+r.sr,,
Q,,r:=1-I;t+2.sr,t ez,z,=(-t)t*2.sr,r
o_(o.nt - r.rorrx r0- e
8.857 + s.rlzi x ro- e l
(r.osz * s.eezi * tf 9 tz.zel - r.lesi x to- 8J
a=
The adjoint matrk:
et,r = (-l)l+1.s2,2
Qt,2:= (-l)l+2.s2,1
Qr,r:= {-t12+1.s,,,
Q2,2F (-t)2+2.sl.r
qa-(r)o,,, - (-r.orJ
-t00.953 n0.763
110.763 -121.527
(-t.a, rc-6 -s.o6rix ro-7 o )[ . -3.64x 10-6-6.06 ,r, ro-r)
,.o= [-t.*. Io-6 - c.o6ti * to-7
Io
q(il=rr )Qr,r [-l.orz/
"=(-,.'rr)
0
,]-3.c4x to-6-6.oetix to-
o('=r r )Qr, r [e:et + e.+zi * to 8J
d1Ql,z -7.5a2ix to-9
(n,n
Compadson wittl the classioal normal modes:
3
7/o
d) F.ee vibrations:
* '= (l) "'=
(r) lnitial data
The motion for the undamped system (C = O):
x(ilT
q(t):= 'M
,"Hilr.M.x(D
* *(d (*('f *""{'or')
(*(rr.M.*(,
*(il
o0 'q0 -.The motion for the damped system (C=0.7M):
q(t):=
+
-0.2
q0 ....M
+
-0.2
-0.4
.r.r(r).0,.) ..
'.0,
(
0.4
0.2
t:= 0r0.001..20 q(t)r
q(t)z0
I]I
q(t)r
q(t)z 15
i
0.2
0.1
0 10 20
-0.1
10
t
3
V\/
7//L
/1t(
V\/
A/1
\/V
/r /\ t
VV
'V:-
u)e,7/1
A
VV
V
-0.3l5 20
//
Exercise 3:6
lnput:m1 := 0.5
l:= 2.0
k:= t0
c := 0.5
g:= 9.81
Calculations:
a) Mass- and stiffness- and damping-matrices
xl :=
/e
Classical normal mode shapes
U:= genves(K,M)
(o.toz o.to't \u=t I
\0.707 4.707)
r (r)_.u*U,,,
I .u(l)ur,
r
Modatmatrix: *(t) ,= *, *() ,= *, )(= (i l)
Damped eigenfrequencies:
lmpsdanco matrix: ,
z(s) := M.s2 + c.s + K
Charact€ristic polynomial:
p1.1:= lzlsyl
p(s) '+ 4.000000's4 + 2'0000000000000000000.s3 + 79.240000000000000000.s2 + 29.il000i000.s + zsz a36;000i
Coeflicienlvector
I4:= lMl r4=t
13:= 2-0
l2:= 79.24
11 := 29.81
h ,= I rl ro = 2ez.tz6
l'zgz.qtsLn.r'
t=l n.z+
t:
' = (1,)
., = (i)x2i=
{.'^r't2
-1.e.t + 1.t.r2 i.n.t'
m1.B.l4
M:= K:=1.t r' * 1.t t'
4
(r r
"=(; :) '.=(':;' ,;:,) ",=1":; *,;
l't.- c.-\44
b) ls th9 system diagonalizable?
..r- r.* _ f z.+sr -z.as:) x.v- l.c _( z.tst -z.nsz\
\-2.4s3 2.4s3 / \_z.ast z.+st )
o m1.12
Yesl
Undamped eigenfrequencies:
rr;=genwls(K,M) *=(t1;rT)
*,=(""\ M-(4s05.) -
*'- l..,r,J " = [ro.rorJ '6,= ./.2- .. = (X.l]i)
Ilt2
I.J
3
/S4
Roota to the chatacterlsflc equafion:
l, = polyrcots(D
Damped mode shapes:
FlEt elqenvälue:
ll=-0.25+2.2ori
.a1 ,= fn(11) odt = 22OI damp€d natural hequsncy
.0, = 2rt5
lnJr,ll(1:= J---'!-lll- qI =0.113 relativedamping
'0,
er'=.l$^')l Ez=t3txto-s rctarivedampinst0,
s:= z(1r1 s = f-to - l.e3i lo + l.e3i ')
\ l0 + 1.93i _tO _ t.et)
Socond oloonvaluo:
\l = -5.t72 x to- 8 - f .foti
oor,= lm(13)l o&=3.86r
to, = g'86t
Tho adjolnt maHx:
Ql, I := (-l)l+1.s2,2
Q,,r:= {-t)l+2.sr,,
o=|.-to - r.eai -ro- r.l:i)
\-t0 - 1.93i _ro _ t.gri,/
damped naiural fraquenci
undamp€d natural frequency
s'= z(r1)
The adrolnt matrlr e:
Qt,l:= (-t)l+1.s2,2
Qr,2:= (-r)l+2.s2, r
s _ fro.l25 - l.li lo.r25 - l.ri)\ro.l25 - r.ti ro.lzs _ r.riJ
Q2,r:= (-t)2+l.st,2
Qr,r:= (-t;2+2.Sr,,
x to- 5 - z.esai r lo- 5
0 -5.t86x t0-5 - 2.65ti x l0-
Qr,, := {-t)2+1.s,,,
Q2,2:= (-1)2+2.sl,t
['
a_( to.t2s - r.ri -ro.r25 + t.ti)\-10.125 + t.ti I0.t25 - r.ri ,/
s'Q =
a(t)
[''* ,]
s'Q =-t.429x l0- 13
+ 2.,Kaix !0- 14
(l)Qr,z
Comparleon wfth tha claaslcal nonnal mode:
-1.429 x to- 13 + 2.a6gi x to- la
Qr I + 2.5tti r to- 8a('Qt,z - 2.5t8i x 1O-
ta(il _Qr,
r
e.= tl)Comparlson wlth tha clålslcel normel mod6!
" =(i)xl=
-t
J
.4i/6
(
Exercise 3:7
lnput:
ml:= I
k;= 1000
g:= 9.81
Calculations:
Mass- ånd stiffness- and damping-matrices:
b) Undamped eiEenfniquencies:
02:= genvals(K,M)
a2=x 103
x 103
Classical normal mode shapes:g3= gavecs(K,M)
Modat matrix: x(l) ,= *r x() ,= *, * = f l t )' [l -1./
Damped eigenfrequencies:
lmpedance matrix:
Z(s):= M.s2 + C.s + K
Characteristicpolynomial:
pG):= lzG)l
p(s) -+ s4 + 24.000.s3 + 410E.000000,s2 + 36000.000.s + 3000000
Coefficient vector:
Ia:= lvl
Ig:= 24.O
12 ;= 4108.0
.I1 := 36000,0
16:= lrl
In=t
I=
3x106
3.6 x 104
4.108x 103
'24I
(,
t,
') -
/zt.e:J *,= r ,,=11',;.1:,i,)
((-t o\
"=[.; ;,j "=(; l)
K,= f2'k -k') *-( ,, rot -r * ro3)*'= (+ z.r,J ^={._, . ro3 z, # )
c:=o.oo6.K "=ftt
*')\-6 12,
Io=3*to6
"'[;:::]l-, =[:;;)
I1
12
I.(
(Rools to the characteristic equalion:
], ;= polyroots(I)
Damped mode shapes:
First eioEnvalue:
13 =:3 * 3r.Ot'
o61 := m(r) @d1 = 3r.48
f-9.0ooo00295oo4ls - sq.ozzzz3o8+regri\
^ _ | -e.oooooors, Bg*ss + 54.o2i.t.too3z.37 ti
I
| + +3t.4s0t524773s44r I
\-2.9999994130973t - lt.taotee+zaoelsi )
(
( o.tot -.o.zoz\u=t I
\4.707 4.707)
r .u(r)Ut,z
r .u(r)ur.,
xl := .,=(l)
" = (1,)x2:=
damped natural fEquency
3
/7 /g
to, = 3l'623
q,,= l*{t')l.0,
s:= z(r,3)
Th6 adjoint matrix Q:
(-r)l*Lsz,:
Q,,, := {-t)l+2.Sr, ,
(r = 0.095 relative damping
C:=
s'Q =
e= 0J6a, -, f*(^r))'
IR{rr)J
( Qr,r'=
*=($r* t88.88ri -982 - rre.eeli)
\-982 - t88.8Sli eu + rte.ttri J
Qr,, := {-t)2+l.sr,,
Qr,r:= {-t)2+2'sr,,
The adjoint matrix:
er,r r= (-r)1+r.s2 ,, Qr,,,= 1-t)2*l'sr,,
el,2:= (-l)l+2.s2, r ez,z,= (-r)2*2.sr, r
a =f*u - 324.16ii xo + rzn.rezi')
\946 + 324.167i _ett _ lz+.rc2i)
o.orrr)
(
q = lntt * l88.t8ri e82 + rrs.roli)
\982 + l68.E6Ii 982 + tg8.88ti/
.O _ fo.rrr " to- ro + t.+rzi " ro- lo
\oq(D -r')'or,, -(l/
-0.144 + 0.0ui 0
0 -{.t44 +
4.358x lo-'o l,.orr,',t ioJ
# r;l
q(il_[ t ) q(t (a,,, -l-t + s.ri* ro-eJ %
=l
Comparison with th6 classical normal modes:
"=[],J
-1 * 5.li x l0- 8
(
Comparison wilh th€ classicål normal mode:
/r\-'=[.tJ
Second eloenvalue:
I'2=-g + 54.028i
^a, ,= r"t(r2) 6dz= 54.oaa
damped natural frequency
ooz= 54'7n undamp€d naturalfrequency
q, '= l*{^t)l cz=o.t64 retative dampins,0,
s:= 4rz) "=(-xe
-noten -ue -tztxtt\\_946 _ 324.167i 446 _ 324.167i)
(
3
/72/
c) Forc6d vibrations:
*,= (;) '. '= (:)
100Load amplitude
0(
lnitial data
.ar.r) * (rto,
*(t) .M
q(t,30)r
q(t,40)r
q(t,30)z
q(t,a0)r
0.4
0.2
-0.2
-0.4
0.4
0.2
0
-0.2
-0.4
(
o := 30 Load frequency
The motion for th6 damped system:
q(t,.),=(1 o6r't
- (2 agr.tx<)
(*of."l,(r.o6,.t
X(il ,M
.M.x(il
x@ .M
(
odt x(l)r.M.x(il
0.5.M 1.5 2
e ."io(
.t
Y
j
I/\r\r r
^. AiV
0.5 1.5
T.M.x(,
( I+
+
{
"-(t.'0,.(*')..,n[.61.(t_sl.sin(o.s)a,drrTh
- (2'ogr'(Fs).sirfoa2.(r - sl.sint..O *.-{('glg-L -- r
(*(t)t.*.*(r)..0,(.
0
3
AA^
f\rVV
\l
AA
,,/L/l
/"\I rUVV
L il l\'\I
\J'
A ,,."ArJ
t := 0,0.01 .. 3
8/
d) Forced vibrations:
/-o-r \co'=
[ o.r J
"'= (T')
. l=
,o Load frequancY
The motion tor ths damped system:
q(r,'),=(t'oor x(t .M
'.'= (:) lnitial dala
t:= 0,0.01..3
0.4
0.2
q(l3o)r
q(t40)r
'0.2
-0.4
0.4
0.2
q(t,30)z
q(1,40)2
-0.2
-0.4
Load amplilude
e
+e
.ar.t) * (')F.M.x(r)
- (2 o6r't
l.oor.r x(t)T
.M
.M.x(r) od1
e2'agr't x<, .M
x(r.(x(rf,M , \
@ffi'"o1'o'ol (
e
+e
+I+ €
r 0,5
0.5
1.5
i.5
st(o62.t)
M.x('
-(l'o0 (Fs) o't10" *(ilHil)tto
"'€ , .sin[o61.(t _ s)].sin(, 11",0y.r.*(il)..0,
- (2'ogr.(t-s).sir{ro62.1t - sl.sinl''.1 *.---d1(ltf :"-L sz ' 'r
((*(r)'.r.*(r).0,(
(
J
dc
AAt,
,i\.. If\rVV
I
\l
AA
JI,/I,r\t IUVV
r/j l\,\I \J ',
/\ ,,'Att
J
23 2y'
Exercise 3:8
lnput
m:= I k:= 10000
Calculations
a) Mass- and stiffness- and damping+natrices:
/zm o o\
"'=l; ',- :.j/ s.r< -g.r o \
r:= | -3.p ,.n -r.u I
I o -r.n ,.o )
1xr:=-- ut,t ufi)
*,,= t .u(d' ur,l
"-h*]. []l
/zoo\
"=[;;:J
b) Damped eigenfrequencies:
lmpedance matrix:
Z(s):= M.s2 + C.s + K
Charactsristicpol omial:
ptr),= lzt.ll
p(s) + 6's6 + +30.000.s5 + 43E300.000000.s4 + 16628000.000000000.s3 + 8384000000.000000.s2 +
Coefiicient vectoc
Iu:= lul re = a
15 := 430
14:= 438300
13:= 16628000
12 ;= 8384000000.000000
\ := 84000000000.000
h= lrl ro=2.8' lol3
[email protected] + ZSOOOOO0OOO0OO
I=
2.s x tol3
E.4 x lol0
8.384x 109
1.663 x 107
4.3E3 x 105
430
6
.I
t,i
5xlo4 -3xlo4 o
-3 x lo4 5 x lo4 -2, to1
o exto4 3*!04
C:= 0.001.K-30 0
50 -20
-20 30
c
b) Undamped eigenfrequencies:
(r.tq * ,oo\Ctz:= senwls(K,M) * =
l o.r*. r0, I
[ .',0' J
- ffiil
4.256x 103
2.741x 104
4x 104
a2= <og:= JA oo =
65.24
16s.561
200
Glasslcal normal mode shapes Ilt2
13
t4
15
k
{J;= genvc(K,M) f-o.ez -o.+ee o.or )u=l o.r -o.ese -{.a08 I
I o.rr, *.r' o.r,u J
., = f, l,rlIr.oz+J
3
t (r)xr :=
-,IJ*' ur.r.
dc 2b
Roots lo the characteristic equation:
tr := polyroots(I) -20.0000008676844 +
-19,999996s190s87 _ 198.997493404432i
-13.7051889461608 - 164.99253166617i
-13.?051836?03884 + 164.g925304s6s27i
4.12814833772925 + 65.205 57997 6802|
-2.t2Et 4E32564526 - 65.205 579791 87 5i
damped natural frequency
relative damping
fr.rrr* rot* .1.268ix.107
2.65?x 108+ r.005ix t08 5.949x 108+2.808ix to7)a=l z.esz' 108+ l.005ix 108 r.059" r09+ r.39ix lo| B.zztxros+ t.oaix tos I
[s.l+s" to8*7.808ix l0? 8.227x108+ l.0Eix 108 6.391 x t08+ E.3asix ro7,l
a(t a('t
1.383
t.074
I1.383
1.074Qr.
r r, i,,l['.ro)
Qt,, bG) _Qr,,
b) Damped mode shapes:
First elosnvaluo:
Ä := 15 ly = 1.12t + 65.206i
Compadson wilh the classical normal mode:
'[*]
Second eioenvalue:
.ar '=
l*(r.)
^Or= 65-Zl
6,,=,Jx{l)ltot
s:= z(,t)
.d, = 6S.206
(l = 0.033
s=
4.14 x 104 + 2.705i x 103 -2.994x 104 - 1.956i x 103 0
-2.994x 104 - 1.956ix 103 3.715x t04 + 2.428ix t03 -1.996x I04 _ 1.304ix 103
0 -1,996 x t04 - 1.304i x 103 2.569 x 104 + l:679i x 103
Ä:= l:
ro, ,= ln(l)
o02 = 165.561
qr,= Jd4)lro,
s:= z(4.)
Å =-13.705 - 164.993i
,dr= -l6q.ggZ damped natural frequency
(2 = 0.083 relativg damping
The adioint matrix e:
Ql, | := (-1)l+1.(s2,u.s:,: - sz,:.s:,2)
Qr,2 := (-l)l+2.(E,,.sr,, - sr,r,sr, r)
Ql, 3 := (-1)l+1.(S2,
r.sr,, - sr,r,sr, r)
Q3, r := (-1)3+r'(sl,z'sr,l - sr,i.sz,z)
e3,2 r= (-l)3r2.(sr,r.sz,r - sz,r.sr,r)
Q3,3 := (-l)3+3.(sl, ,.sr,, - s,, r.sr, ,)
g, r ;= (-l)2+t'(sr,r's:,, - sr,r'si,r)
or;, := t-rt?+2.1sr, r.s:,r - sr,:.s:, r)
Q2,3 ;= (-l)2+1.(sr,,.sr,, - s,,r.sa,,)
s=
--4.?55 x 103 + 795.389i -2.9s9 x 104 + 4.95i x 103 0
i.959 x 104 +4.95i x.103 -3,179 x 104 +5.318i x 103 -,.nr, * roo * 3,3i x 103
o -1.973 x 104 + 3.3i x 103 2.554x rc3 _ 427,268i
The adjoint matrix e:
Qr,, := {-ryl+1.(sr,r.sr,i * sz,:.sr,z)
el,z := (-l)l+2.(s2, r.sr,r - sr,r.sr, J
el, 3 := (-l)1+3.(s2,
r.sr,z - sr,z.sr, r)
Q2, l := (-t)2r1.(sl,z.sr,r - sr,r.sr,z)
Qr,, := t-ry2+2.1sr, I.si,, - sr,r.sr, l)
Qr,, := {-t;2+3.1s,,,.sr,2 - s,,z.sl,,)
3
27 2B
Qr,, ;= {-r)3+1,(sr,z.s2,t - sr,r.sr,r)
e3,2 := (-l)3+2.(s,, r.sr,, - sr, r.sr,r)
e3,3 := (-1)3+3.(s,,,.sr,, - sr,r.sr, r)
a=
dlQr,r
=l*u, -nlr,* ,t'l[-,.r0, * r.os:i x ro- 7J
s,= z(n)
s=
-2.g4x to4 - s.glix to3 -z.g4x lo4 - s.9zi * t03 0
-z.g4x1o4 -s.giix 103 -6.86x to4-1.393i, to4 -1,96* 104-3.98i, to3
0 -1.96 x 104 - 3.98i x to3 -9.E x to3 - t.ggi " to3
-4.571 x 108+ l.574ix 108 7.346x107 - z.52nix 107 s.6?3x t08-1.9s3i" to8
't.346x107 -z.s2Eix 107 -t.tEx 107+e.063ix to6 -9,112* 107+3.138ix to7
5.6?3x 108- 1.953ix 108 -9.117x107 +3.138ix t07 -z.04lx 108+2.423ix lo8
The adjoint matrix Q:
er, I := (-1)l+1.(s2,:.sr,, - sr,r.sr,r)
Ql,2 := (-1)l+2.(S2, r.Sr,l - Sr,r.Sr, r)
Qr,, := {-r )l+3'(sr, r,sl,, - sr,r.sr, r)
Qr,, ;- {-t)3+1.(s,,2'sz,r - sr,:.sz,r)
e3,2 := (-1)3+2.(sl, r'sz, r - sz, r.sr,:)
Qr,, := {-t)3+3.(s,,,.%,, - r,,r.%,,)
e2, I := (-l)2+1.(sl,r.s:,, - s,,r.s:,r)
e2,2 := (-1)2+2.(sl, r.s:,: - sr,r.s:, r)
Qr,u := {-r)2+3'(sr, l.sr,, - sl,z.sr, t)(r)
a- =(-,r,*,1,,,' 't'l[-r.ro, * t.urr, ',0- u))Ql
d1=Qr,r
1
-0.16l + l.t?i x l0- ?
-1.241 -2.21ix lf 7
Comparison with the classical normal modes:
a(t)
I
-l - 5,054i x !0- 8
z + t.4q4ix to-7
I_e
-l + l.4,l4i x l0 '
z - t.abgi x ro- 8
Qr
a(,)Q1
Comparison wilh the classical normal modes:
"[]
?.762 x 108 + l.lzi x 108 -2.162x 108 - t.t7i x 108 5.525 x 108 + z.l4i x to8
-2,762x108 - l.l7ix 108 2.162x108 +t.ä* to8 -5.525x !06-2.34i x 108
5.525x 108+ 2.34i x 108 -5.s25xlor -2.34i x 108 t.losx 109+4.6Ei x 108
x2=I
-0.161
-1.241
t
t
/t\ot'=l-,+r.arai"ro-slar,, I -, I
\2+l..t44ixl0'/Third eigenvaluo:
Å := lt Ä = -20 + 198.997i
.6r ,= t-(,t ) .d, = 198.997 damped nalural frequency
o6, = 200
In{,r)lt0,
,3
3
Ca:= (: - o.t relative damping
2730
pt:= xl?.M.xl
1t2,= *.].u.*2
p3,= *aT.M.*3
-.( r)
.. [l:l]
/t.aqz\-=l':"J
Frequency response function:
r(.),= ro-4.;
hoJ'-r.
1e)X'' i= xz
^ :=x3
Pl = 8.892
P2 = 3.618
P3=9
x=ttl
1.383 -0.161 -11.074 -1.241 2
I ,=t,ffi]J I o' J
,oo
" loo
k=1
'[i]o := 0,0.00001 ..2,5
0.5
0.45
0.4
0.35
0.3
ro1 lr1,t')l o.zs
3.14
2.51
1.88
1.26
0.63
0-e(E,rrol)
*(d
-0.63
-t.26
-1.88
-2.51
-3.t40 0.5 t.5pk
f
2 2.5
rrr(.),= rr.r(r).,
0.2
0.r5
0.1
0.05
0
(
I
0
3
0.5 1.5 2.5
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Exercise 5:4 (Mathcad document)
INPUT:
T6:= 10.t04
k= 8000
p := 8000
L:= 1.0
r= 1.0.10-3
CALCULATIONS:
_2A:= r.r
mg:= p.A m0 = 0.025
z:= 0,0.001.. l0
kL
&)n : -(2n
r'ya
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t
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22a
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50
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2
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I
I
..i-.
I
l
I
0
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.//
4
zi= 2
Given
, To'"tan(z)+-=0
z := Find(z)
z= 1.62
i:= 5
Given
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z:= Find(z)
z= 4.729
z'.= 8
Civen
To'zt8n(z)+-=0
z:= Find(z)
n = 1.571
2.L
3' r = 4.7122.L
FJo:= l-J-o
o = 3.232x 103
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4
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To=1"105
Tor =12-sk-L
FJo:= l-J'o
Wo:= l-J'o
o = 9.434x 103
I = 1,569x 104
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Exercise 5:6 (Mathcad document)
INPUT:
E:= 200.109
p := 8000
L:= 5.0
A:= 2'10
P6 := 100000
CALGULATIONS;
m:= p.A m= 1.6
.on ((e/ -, g/ (at ant:*
ca/at/a-/tbn a/ r'Z* ,'rt /e7ra/ a/e
lEac:= l-Im_ ,-3
c = J x lu
'4 if-lnrno-,)g,z-'t4 b,ei-/)z "- /'d-,0,=
no .-- tno'".+E.A2Ll.E.A i=r
n:= 10
car(/ai-4 # */
hz4 cre ( :
"2o-ed
ll/n
t := 0,0.00001..0.05
0.0r5
0.01
€ 0.005
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dactz-z*z/,
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Exercise 5:7 (Mathcad document)
INPUT:
p := 8000
E:= 200.109
L;= 1.0
A:= 2.6.10- 3
I '=
4.?.10- 6
CALGULATIONS:
m:= p'A m = 20.8
Eigenfrequencies
z:= 2
Givm
tan(z)-tsnh(z)=0
z:= Find(z)
z=3.92i (4'l + 1)'n -g.n,
4
r,1 '=; \=3.s27
zt= 5
Giren
l,an(z)-tanh(z)a0
z:= Find(z)
z=7.069 Q'2 + t)'r -;.06s
4
z;= 8
Given
ta(z)-tanh(z)=g
z := Find(z)
z= 10.27(4.3 + ll.r' ' = 10.21
4
z4:= L P3 = 10.21
Modeshapes
x:= 010.001'L..L
o(*,ut) o.o
6-r),(*'*r)-o.o
-1.2
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2
1.2
x
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Exercise 5:8 (Mathcad document)
INPUT:
p:= 8000
E:= 200.I09
L:= 1.0
_2A:= 2.6-10 -
I:= 4.7.10 "
Q:= 10
CALGULATIONS:
z
3/
/arr * (f) n)
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t'= Q
m.L
m=2O.8
k = 0.481
Ds P 7/.a. /ds s-/Eigenfrequencles
z:= 2
Giva
1 .+ o{z).msh(z) + k'2.(cos{z).sinh(z) - sin(z).cosh(z)) = 0
z:= Fnd(z)
("\2 lwz= 1.42e ' '= EJ 'J; a = 434'313
z:= 4
Given
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z:= Find(z)
z = 4.1t7 . ,= (:)t. lE o = 3.603 x ro3\L/ .rl m
5
3f 2/JO
z;= 7
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6
25
Exercise 6.10
lnput:
m;= l0
EI ;= 9.4.105
L:= 2,0
CALCULATIONS
mmo:=;
Element stiffness matrix:
Structure mass matrix assembly:
N( *r,= I rl."r*,
f,.rzr* tot 5.64x 106 -l.l28x ,ot s.el, rou')
I s.ol' ro6 3.16x106 -5.6+x 106 r.8sx ro6 |
| -r.,rr-,ot -5.64x 106 r.l28x r0? -r.uo,,ou I
I s.el " ,ou r.88 x 106 -5.64 x to6 l.te * rc6 )
1.85't .0.262
0.262 0.048
0.643 0.155
-0.155 -0.036
0000
0.643 -.0.155 0 0
0.155 -0.036 0 o
3.114 0 0.643 -{.155
0 0.095 0.t55 _0.036
0.643 0.155 1.857 4.262
-0.155 -0.036 -0.262 0.048
Mass density
1.857 0.262 0.643 {.1550.262 0.048 0.155 -{.036
0.643 0.155 1.857 4.262
-0.155 -0.036 4-262 0.048
It:=
K=
12 6.le
6.le 44e2
-t2 -6.Ie
6.1" 2.r"2
l"'
-rz e.r.'1
-u.,. ,.r.' I
rz -e.r" lK"
=
4.t" 4.t"2 )
M=i=l
Two element model:
N:= 2 t-:= !
Lo€liätion operators:
[rooooo'1 [oorooo)",'=l:;i:;:l ",,=l:::;::l(oootoo.J looooo,J
StluotuEl mass matrix, (lnvoking boundary conditions, i.e, deleting apprcpriate rore andcolumns):
M,=(31t4 o )\ 0 o.ogs)
S{ructure stifness matrix assembty: r,= } n.T.r".n
t]]"[] n '|]]Element mass mtrix:
54 -t:.t")
,r'," -r.t"t l
156 -zztel ". =
-2z.le 4.te' )
l.l28x lo7 5.64x lo6 -l.l28x Io?
5.64x lo6 3.i6xlo6 -i.64x lo6
-l.l28x l0? -5.64x 106 2-256x1{
5.64 x lo6' t.BE x lo6 o
0 O -l.l28x 107
0 0 5.64xt06
00
00
-l.l28x 107 5.64x 106
-5.64 x 106 1.88 x to6
l.t28x 107 -5.64x lo6
-5.54 x lo6 3.?6 x lo6
5.64 x 106
l-88 x 106
0
7.52x 106
-5.61x 106
1.88 x 106mo'1"
156 22,1e
22,1" 4.1"2
54 t3.leM":=
42I
6
4r: StructuEl mass matdx, (invoklng boundary @nditions, l.€ dsleting app.oprialo rows andcolumns):
2B
Flnile €lement elg€nvaluas:
Cl2 := genv.ls(K,M)
'Exacl' åiganvalues:
Threo element modal:
O 7.52x to6
L
( 12 6.te -12 6.le
. ,= it.l 6 le 4'le2 -6'te 2 tez
" l"r | -tz -6le t2 -6.1e
1r.,, ,.,"t -6.t" 4.1.2
Element mass matrix:
M":=to le
420
Elem€nt stiffness matrix:
1.238 0.n6 0.429 _0.069
0.116 0.014 0.069 -{.01t0-429 0.069 2.476 0
-0.069 -0.0t1 0 0.028
0 0 0.429 0.069
0 0 -0.069 -O.0ll00000000
2.256x lO7 0K:=
t56 22.1e 54 _l3.le)
zzte 4.1e2 l3.re -3.1e2 I fl238 0116 0.42e -0.06e)
s4 I3'l- 156 -22t I "-=l 0116 0014 0'06e -0011
|| . | 0.429 0.069 r.zls _{.rre I
-l3.le -3.1e2 .'zz.te 4tezJ (-0.*, *.0,, *.,,. o.o,oJ*=[u.oto. toul
u-_[ro*.,, ,o'][z.qe, roTJ la.rrroz" ror.J
(
Structuremässmatrixassembly: l'tt= ! rl,T'Utq
3.807x.107 1.269x t07 -3.807x 107 t-26gx t{t.269x107 5.64x 106 -1.269x t07 2.82,106
-3.807x 107 -1.269x107 3.807x l0? -t-269x107
t.269x td 2-B2x to6 -t.269x 107 5.64 x 106
N
i=l
000000
-0.069 0 0
-{.0r I 0 0
0 0.429 4.0690.028 0.069 -0.0110.069 1.238 4.116
-0.011 -0.116 0.014
fsoo.ss.J!- =2.425x to3 l:sor.r.-I!-=u.urr,,otJ -0..' J .o.ro
( K"=
N;= 3 leN
Localization opsEtors: 0
0
0.429
0.069
2.476
0
0.429
-0.069
flooooooo\ /oorooooo\
",'=l: I I ; : ; : :l ",,=l: : : ; l: : :l. loooroooo.l looooo,oo]
foooorooo)rl_,=lo o o.o o t o ol5 lo o o o o o r ol
looooooo,J(
M=
( 2.476 o 0.429 -o.o6e\
,n,= I o 0.028 o.oun -0.0,, I
| 0.42e 0.069 2.4i6 o I
f-o.oen +.0' o o.or, )
Structural mass matdx, (invoking boundary @nditions, i.e. del€ting appropriato rows and@lumns):
4
5
6
'ttil
I
2
3
4
5
6
7
li*l:
t:
til
r:)
[;J NK:= \i 1T
/r I
i=l
6
Structurs mass matrix assembly: .K.,IL.Ll
2?
3-80ix 107 L269x l0? -3.807x 107 t.26gx td 0 0 0 0l.26gx td 5.64x 106 -l26gx t07 2.a2x 106 0 0 o o
-3.807x t07 -1.269xt01 7-6l4xt{ 0 _3.807x tol l.zeg*td 0 01269x101 2.82x 106 0 t.l28x l0? -t26.,xtol. 2.82xr06 o 0
0 0 -3.807x 107 _l,26gx t& 7.6t4x to7 0 _3.802x tf t.zdg, tolo 0 1.26gxld 2.82x t06 o l.t28x 107 -.t.26gx td 2.E2x 106
0 0 0 0 _3.80?x 107 _1269x1{ 3.80?x l0z _1.269xt07
0 o 0 o l.z6gx 107 2-g2xto6 _l.z6gx to? s.6l * to6
.. StrucfuBl mass mat.ix, (invoking boundary condidons, i.e dsl8ting appropriats ro\^rs and( olumns):
K=
(
( ,.n0,,o'
*,=l o
| -r.aoz' rot
[,r.r*,0'
o -3.802x tf t.zes*
t.t28x to7 -1.269x to? 2.82x to6
-1.269x 107 l.6tlx lo7 0
2-82x t06 0 t.l28x t07
Finita olsment sigsbvalues:
Q2:= gmvrlsK,M) c)2 =
'Exact' 6ig6nvalu€s:
l-037 x
2-503 x
4.651x
x
;$l
i;J
.3.22m1
1.58217
6.82019
x roa)
"r041
:;51
'l-c,.
=2.166x lO4
ffi=,..r,,,0, ffi=u.*,,,0,
li]lE=r,,,",0nJ 'o'to ho L'
(
(:
#
6
