Chapter 03

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3 1

SOLUTIONS TO CHAPTER 3 PROBLEMS Problem 3.1

According to Table 3.1 and given the example’s specification, it follows that:

33 13140 35c bm m= (P3.1)

where the subscripts c and b refer to cantilever and bridge, respectively. The mass of the

bridge is therefore related to the mass of the cantilever as:

0.635b cm m= (P3.2)

which results in:

0.635b b c cl w l w= (P3.3)

where l and w are length and width. As shown in Appendix D, the mass moments of

inertia of the two members are:

( ) ( )2 2 2 2

;12 12

c c b bc b

m w h m w hJ J

+ += = (P3.4)

Table 3.1 shows that the equivalent mass moments of inertia for both a cantilever and a

bridge about their relevant points are equal to 1/3 of the corresponding mass moments of

inertia. On the basis of Eq. (P3.3), equality of the two equivalent mass moments of inertia

requires:

( ) ( )2 2 2 2

1 13 12 3 12

c c b bm w h m w h+ +× = × (P3.5)

Taking into account that the two masses are:

;c c c b b bm l w h m l w hρ ρ= = (P3.6)

in combination with Eqs. (P3.3) and (P3.5) results in:

2 21.255 0.365b cw w h= + (P3.7)

Provided Eq. (P3.6) is satisfied, it is possible that the cantilever and the bridge have

identical equivalent mass moments of inertia with respect to their relevant points (free

end point for the cantilever and midpoint for the bridge).


Problem 3.2

The mass contribution from the four fixed-guided beams is calculated as:

( ), , , ,2 2 3 6e e long e short e short e shortm m m m m= + = × × = × (P3.1)

where me,long, the equivalent mass of a long beam, is twice the equivalent mass of a short

beam, me,short, because the lengths of the two beams are in the same relationship, all other

mass parameters being identical. According to Table 3.1, the equivalent mass of a short

fixed-guided beam is:

2

, 11335 4e short

dm l=πρ (P3.2)

Combination of Eqs. (P3.1) and (P3.2) leads to:

21

3970em d l= ρπ (P3.3)

The total moving mass adds the beam contributions to the actual mass of the plate, which

yields:

2 2 21 1

39 39(5 ) 570 70t plate em m m l d d l d l dl = + = + = +

πρ ρπ ρ (P3.4)

Taking into account that me = 0.1 mt leads to:

0.1 or 0.9t plate t plate tm m m m m= + = (P3.5)

Combining now Eqs. (P3.4) and (P3.5) yields:

21 0.3174dl l= (P3.6)

which indicates that the beams’ diameter d and length l1 are related to the plate

dimension l. As a consequence, once l is specified, either the beam diameter or the beam

length can be selected; the other dimension has to be determined by using Eq. (P3.6).


Problem 3.3

The equivalent stiffness of the original bridge is determined based on Table 3.2 as:

2 4

2 p pGI GIk

l l= × = (P3.1)

because the two equivalent end springs act in parallel with respect to the bridge center.

The equivalent stiffness of the cylinder-bridge system results from the two end torsional

springs, each of length (l – lc)/2, which are connected in parallel, namely:

4

' 2

2

p p

c c

GI GIk l l l l= =

− − (P3.2)

The following stiffness ratio can be formulated by means of Eqs. (P3.1) and (P3.2):

' 11 /c c

k lk l l l l= =

− − (P3.3)

Figure P3.1 shows the variation of the stiffness ratio in terms of the length ratio lc/l.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.51

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

lc/l

stiff

ness

rat

io

Figure P3.1 Stiffness ratio as a function of the length ratio

The original equivalent mass moment of inertia with respect to the bridge midpoint is

calculated by means of Table 3.1:

4 41

3 3 32 96eJ d d lJ l= = × × =

π πρρ (P3.4)


After the addition of the cylinder, the equivalent mass moment of inertia is the sum of the

actual cylinder mass moment of inertia and the equivalent mass moments of inertia

corresponding to the two end elastic segments, namely:

( )44' / 212

32 3 32cc c c

e

d l ld lJ−

= + × ×πρπρ (P3.5)

The ratio of the two equivalent mechanical moments of inertia is obtained from Eqs.

(P3.4) and (P3.5):

'

4

31c

e c c

e

c

lJ l lJ l d

d

= − + ×

ρρ

(P3.6)

and the 3D plot of Fig. P3.2 illustrates the variation of this ratio.

0.2

0.3

0.4

0.5

0.2

0.3

0.4

0.50

200

400

600

800

1000

1200

lc/ld/dc

Mec

hani

cal m

omen

t of

iner

tia r

atio

Figure P3.2 Mechanical moment of inertia ratio as a function of the length and diameter ratios

For lc/l = 0.25 and d/dc = 0.2, which are the particular values of the problem, the stiffness

ratio of Eq. (P3.3) becomes 1.33 and the inertia ratio of Eq. (P3.6) becomes 581.1. The

3D plot has been obtained by means of the following MATLAB® code (used after

specifying the numerical values of the mass densities) where X = lc/l and Y = dc/d:


>> [X,Y]=meshgrid(0.2:0.01:0.5);

>> Z=1-X+3*rhoc/rho*X./Y.^4;

>> mesh(X,Y,Z)

Similarly, the MATLAB® command:

>> ezmesh('1-X+3*7800/6300*X/Y^4', [0.2, 0.5], [0.2, 0.5])

produces a similar plot.


Problem 3.4

For both the motion about the y direction and the motion about the z direction, the

lumped-parameter stiffness model is the one of Fig. P3.1.

Figure P3.1 Lumped-parameter stiffness model for the microspring of Fig. 3.29

The stiffnesses about the two directions are calculated as:

1 2 1 2

1 2 1 2

2 ; 2y y z zy z

y y z z

k k k kk kk k k k

= =+ +

(P3.1)

where the stiffnesses of the flexible segments of length l1 and l2 (which are actually fixed-

guided beams) are determined based on Table 3.2 as:

3 3

3 3

1 23 3 3 3 3 31 1 1 2 2 2

3 3

3 3

1 23 3 3 3 3 31 1 1 2 2 2

12 1212 1212 12;

12 1212 1212 12;

z zy y

y yz z

w h w hE EEI EIEw h Ew hk kl l l l l l

wh whE EEI EIEwh Ewhk kl l l l l l

= = = = = = = = = = = =

(P3.2)

In Eqs. (P3.2) w is the beams’ width (dimension in the plane of Fig. 3.29) and h is the

beams’ thickness. Substitution of Eqs. (P3.2) into Eqs. (P3.1) yields:

3 3

3 3 3 31 2 1 2

2 2;y zEw h Ewhk k

l l l l= =

+ + (P3.3)

By using Eqs. (P3.3) in conjunction with the problem condition, which is ky = 4kz, results

in w = 2h; this indicates the width of the beams needs to be twice the beams’ thickness.

Numerically, the width is w = 2 x 200 nm = 400 nm. The lengths l1 and l2 can be chosen

arbitrarily.

y or z

k1

k2

k1

k2


Problem 3.5

The lumped-parameter model of the microspring of Fig. 3.30 is shown in Fig. P3.1.

Figure P3.1 Lumped-parameter stiffness model for the microspring of Fig. 3.30

The stiffness of this device is calculated by considering that the two springs of stiffness k2

situated above the rigid connector (identified by the coordinate x) in Fig. P3.1 are coupled

in parallel; their resultant is also in parallel with each of the two spring series groups

formed of k1 and k2 that are situated underneath the rigid connector in Fig. P3.1. As a

consequence, the equivalent stiffness of the microspring is:

1 22

1 2

2 2ek kk k

k k= +

+ (P3.1)

where:

1 21 23 3

1 2

12 12;EI EIk kl l

= = (P3.2)

Substitution of Eq. (P3.2) in Eq. (P3.1), transforms the former equation in:

3

2 1 23 3 32 1 2 2 1

24 1eEI I lkl I l I l

= + +

(P3.3)

The moments of inertia of the two beams are:

4 4

1 21 2;

64 64d dI I= =

π π (P3.4)

Using the numerical values of this problem, yields k1 = 1.4608 N/m, k2 = 2.6952 N/m,

and ke = 7.285 N/m.

k2

k2

k2

k2

k1 k1

x


Problem 3.6

The lumped-parameter spring model of the device of Fig. 3.31(a) is shown in Fig.

P3.1(a), where the two identical springs act in parallel.

Figure P3.1 Lumped-parameter stiffness models for the microdevice shown in: (a) Fig. 3.31(a); (b) Fig.

3.31(b)

The stiffness of this device is therefore:

12ak k= (P3.1)

The two beams in Fig. 3.31(a) are fixed-guided and therefore their stiffness is the one

given in Table 3.2; as a consequence, the stiffness of Eq. (P3.1) is:

3 3

12 242aEI EIk

l l= × = (P3.2)

The lumped-parameter spring model of the device of Fig. 3.31(b) is sketched in Fig.

P3.1(b), where two more identical springs are added to the original springs of Fig.

P3.1(a), as well as two other identical springs resulting from the two horizontal fixed-

fixed beams of Fig. 3.31(b). The six beams are connected in parallel and therefore the

equivalent stiffness is:

1 24 2bk k k= + (P3.3)

where k2 is the stiffness of a fixed-fixed beam and is provided in Table 3.2. It follows

that:

3

432b

EIkl

= (P3.4)

By comparing Eqs. (P3.2) and (P3.4) results in:

432 1824

b

a

kk

= = (P3.5)

k1

k1 k2

k1

k1

k1

k1

k2

(a) (b)


which indicates that the microdevice of Fig. 3.31(b) is 18 times stiffer than the

microdevice of Fig. 3.31(a).


Problem 3.7

The out-of-plane bending natural frequencies for a cantilever and for a bridge are:

, ,, ,

, ,

;e c e bn c n b

e c e b

k km m

= =ω ω (P3.1)

where the c subscript denotes cantilever and the b subscript refers to bridge.

The better detection precision is offered by the mechanical member producing a

larger variation in the natural frequency after mass deposition, which also means a larger

variation in the square of the natural frequency; as such, the following ratio can be

formulated:

( )( )

( )( )

, ,22 2 *

, , , , , ,, ,22 2 * , , , , ,, , ,

, ,

e c e c

n c n c n c e c e c p e b pe c e b

e b e b e b e c e c pn b n b n b

e b e b p

k km m m m mk mk k k m m mm m m

−∆ − + +

= = = × ×+∆ − −

+

ω ω ω

ω ω ω (P3.2)

where mp is the mass of the particle. Taking into account that:

, ,3 3

, ,

3 192; ;

33 13;140 35

e c e b

e c e b

EI EIk kl l

m m m m

= = = =

(P3.3)

where I is the beam cross-sectional moment of inertia and m is the beam mass, it follows

that:

( )( )

2, ,

2,,

1313 13 35

33528 528140

pn c e b p

e c pn bp

m mm mm m m m

+∆ += × = ×

+∆ +

ω

ω (P3.4)

Equation (P3.4) can be written as:

( )( )

2,

2,

1313 35

33528140

mn c

n bm

c

c

+∆= ×

∆ +

ω

ω (P3.5)

with

pm

mc

m= (P3.6)


being the mass ratio. The ratio of Eq. (P3.5) is plotted in Fig. P3.1, which indicates that

Δωn,b > Δωn,c, and therefore the bridge offers a better detection precision compared to the

cantilever.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50.029

0.03

0.031

0.032

0.033

0.034

0.035

0.036

0.037

0.038

0.039

mp/m

( ∆ω

n,c)2 /(

∆ω

n,b)2

Figure P3.1 Ratio of squared natural frequency ratio in terms of the mass ratio


Problem 3.8

The microbridge-mass system is sketched in Fig. P3.1 below.

Figure P3.1 Microbridge with midpoint deposited mass

In bending, the equivalent stiffness connected to the bridge midpoint is given in Table 3.2,

and for a rectangular cross-section whose moment of inertia is Iy = wh3/12, the stiffness

is:

3

, 3 3

192 16ye b

EI Ewhkl l

= = (P3.1)

The equivalent mass at the bridge midpoint is given in Table 3.1 and for the particular

rectangular cross-section is:

,1335e bm whl= ρ (P3.2)

The original out-of-plane bending natural frequency is therefore:

,,

,

e bn b

e b

km

=ω (P3.3)

For the numerical values of this problem, Eqs. (P3.1), (P3.2) and (P3.3) yield: ke,b = 608

N/m, me,b = 2.85 x 10-9 kg, ωn,b = 0.46167 x 106 rad/s. The altered bending frequency is

expressed as:

,*

,

e bb

e b p

km m

=+

ω (P3.4)

Taking into account that the variation in the bending frequency is:

*,b n b b∆ = −ω ω ω (P3.5)

Eqs. (P3.3), (P3.4) and (P3.5) enable expressing the deposited mass as:

x

l/2

l

w z

b

w/2

y

mp


,,2

,

,

e bp e b

e bb

e b

km m

km

= −

− ∆

ω

(P3.6)

with a numerical value of mp = 1.94 x 10-12 kg.

A similar approach is applied to mass detection by torsion where the lumped-

parameter stiffness is expressed in Table 3.2 with the torsional moment of inertia given in

Appendix D for very thin cross-sections as It = wh3/3:

3

,4 4

3t

e tGI Gwhkl l

= = (P3.7)

The lumped mechanical moment of inertia is determined as in Table 3.1:

( )2 2

,13 36e t

whl w hJ J

+= =

ρ (P3.8)

The original torsional natural frequency is therefore:

,,

,

e tn t

e t

kJ

=ω (P3.9)

Equations (P3.7), (P3.8) and (P3.9) give the numerical values: ke,t = 2.22 x 10-5 N-m, Je,t

= 8.55 x 10-18 kg-m2, ωn,t = 1.6112 x 106 rad/s. The modified torsional frequency is:

,*2

,

e tt

e t p

kJ m b

=+

ω (P3.10)

The change in the torsional frequency is:

*,t n t t∆ = −ω ω ω (P3.11)

The parameter b can now be found through combining Eqs. (P3.9), (P3.10) and (P3.11)

as:

,,2

,

,

1 e te t

p e tt

e t

kb J

m kJ

ω

= × −

− ∆

(P3.12)

with a numerical value of b = 22.7 μm.


Problem 3.9

The rotation about the x axis can be lumped-parameter modeled as shown in Fig. P3.1

where the two end flexible bars behave as two torsional springs connected in parallel.

Figure P3.1 Lumped-parameter model for the x-axis rotation

The total stiffness results from the torsional springs and is based on Table 3.2, being

calculated as:

4

,1 1

216

pe t

GI Gdkl l

= =π (P3.1)

where Ip = πd4/32 is the cross-sectional polar moment of inertia. The equivalent mass

moment of inertia contains contributions from the middle plate as well as from the two

end flexible bars/cylinders (as indicated in Table 3.1 and in Appendix D) and is

calculated as:

( ) ( )

( )

2 2 2 22 2 21 2 2 1 2 2 1

42 21

2 2

1 1 1 12 212 3 2 2 12 3 2 4 4

12 4

e

l l h l h l l h l h l dd dJ m

l dl h l h

+ + = + × × = + × × × ×

= + +

ρ ρ ρ π

ρ π (P3.2)

The natural frequency corresponding to the system’s rotation about the x axis is

calculated as:

,,

e tn x

e

kJ

=ω (P3.3)

With the numerical values of the problem, it is obtained that ωn,x = 7,758.6 rad/s.

Figure P3.2 shows the lumped-parameter model corresponding to the plate translation

about the y axis.

θx

ke,t ke,t

Je


Figure P3.2 Lumped-parameter model for the y-axis translation

The two springs acting in parallel produce the total stiffness that is calculated based on

Table 3.1 as:

4

, 3 31 1

12 328

ze b

EI Edkl l

= × =π (P3.4)

with the cross-sectional moment of inertia being Iz = πd4/64. The equivalent mass is the

sum between the central plate mass and the contributions by the two end beams

(calculated as in Table 3.2 for a clamped-guided beam), namely:

2 2

1 2 1 1 213 13235 4 70e

d dm l l h l l l hπ πρ ρ ρ

= + × × × = +

(P3.5)

The natural frequency corresponding to the system’s translation about the y axis is:

,,

e bn y

e

km

=ω (P3.6)

and has a numerical value of ωn,y = 5,596.2 rad/s. The torsion-to-bending natural

frequency ratio is therefore ωn,x/ωn,y = 1.386.

y

ke,b ke,b

me


Problem 3.10

In terms of the z-axis plate translation (out-of-plane motion) the hinges bend as fixed-

guided beams (see Table 3.2) and therefore, act as four springs connected in parallel; the

corresponding bending stiffness is:

4

, 3 3 31 1 1

1212 34 44

yxe tr

EIEI Edkl l l

= = =π (P3.1)

and its value is ke,tr = 0.386 N/m. The equivalent mass associated with the out-of-plane

motion results from the actual plate mass and the equivalent mass of the four identical

bending beams (according to Table 3.1), namely:

2 2

, 1 2 1 1 213 13435 4 35e tr

d dm l l h l l l h

= + × × × = +

π πρ ρ ρ (P3.2)

and its value is me,tr = 6.1665 x 10-9 kg. The natural frequency corresponding to the z-xis

translation is:

,,

,

e tre tr

e tr

km

=ω (P3.3)

with a numerical value of ωe,tr = 7,912.2 rad/s.

When the central plate rotates about the z axis, the four hinges bend in the xy plane

(as fixed-guided beams for small plate rotations) and the equivalent stiffness is:

4

, ,3 31 1

12 344

ze rot e tr

EI Edk kl l

= = =π (P3.4)

The equivalent mass moment of inertia combines contributions from the plate and the

beams (in equivalent form). The plate’s moment of inertia is:

( )2 2

1 2 1 2, 12p z

l l h l lJ

+=ρ

(P3.5)

Figure P3.1 shows the four equivalent lumped (point) masses resulting from the fours

beams. For small motions, as mentioned before, the beams are fixed-guided and therefore,

their equivalent mass is determined according to Table 3.1:

22

1, 1

131335 4 140b e

d ldm l= × × =ρππρ (P3.6)

The mass moment of inertia produced by these four equivalent masses about the z axis is:


Figure P3.1 Lumped-parameter model showing the inertia contributions by the four beams

( ) ( )2 2 22 21 1 2, 2 21 2

, , , 1 2

132 2

4 4 2 280b e

b z b e b e

d l l lml lJ m m l l+

= × + × = × + =ρπ

(P3.7)

where the equivalent mass of Eq. (P3.6) has been used. The equivalent (total) mass

moment of inertia is therefore:

( )2 2 2

1 1 2 2, , ,

134 3 70e z p z b z

l l l l h dJ J J+

= + = × +

ρ π (P3.8)

with a numerical value of Je,z = 5.7019 x 10-17 kg-m2. The natural frequency

corresponding to the plate rotation in the xy plane is calculated as:

,,

,

e rote rot

e z

kJ

=ω (P3.9)

and has a numerical value of ωe,rot = 8.2282 x 107 rad/s.

y

x

l1 l1

l2

l2

mb,e

mb,e

mb,e

mb,e


Problem 3.11

By transferring the lumped-parameter spring from the cantilever’s midpoint to the

free end the following stiffness is obtained:

2/ 2'

4l kk k

l = =

(P3.1)

and its value is k’ = 2.5 N/m. At the same time, the distributed-parameter cantilever is

replaced by a lumped-parameter mass me and stiffness ke, both located at the free end.

These parameters are calculated based on Tables 3.1 and 3.2 as:

2 2

4 4

3 3 3

33 33 33140 140 4 5603 3 3

64 64

e

e

d ldm m l

EI E d Edkl l l

= = × × =

= = × =

π πρρ

π π (P3.2)

Their numerical values are me = 1.194 x 10-10 kg and ke = 227.898 N/m.

The mass and two parallel springs are collocated at the cantilever’s free end; as a

consequence, the natural frequency of the system is calculated as:

' en

e

k km

ω += (P3.3)

whose value is ωn = 1,389,300 rad/s.


Problem 3.12

The main coordinates of motion are highlighted in Fig. P3.1 below.

Figure P3.1 Pulley-lever-rod mechanical system with motion coordinates

Apparently, this system is defined by five coordinates: the angles φ and θ, and the linear

displacements x, y, and z. However, under the assumption of small motions, the vertical

motion of the rod’s end point and the vertical displacement of the rod’s midpoint are:

2 ;y l x l= =ϕ ϕ (P3.1)

The point identified by coordinate y also belongs to the vertical rod, which attaches to the

pulley whose rotary motion is defined by the angle θ relating to y as:

y R= θ (P3.2)

It follows that the actual number of DOF is equal to 2, which is the number of apparent

DOF (5) minus the number of rigid constraints (3).

The kinetic energy of the system is produced by the rotating pulley and translating

mass, namely

2

2 2 2 2 2 2 2 2 22

1 1 1 1 4 12 4 2 4 2

lT mz mR mz mR mz mlR

= + = + = +

θ ϕ ϕ (P3.3)

Equation (P3.3) used the relationship between the coordinates θ and φ from Eqs.

(P3.1) and (P3.2). The potential energy is elastic in the absence of gravity effects and is

stored by the two springs as

R k

k 2l

l y

y

θ φ

z m

x

m


( ) ( )2 22 2 21 1 1 1 22 2 2 2

U kx k z y kl k z l= + − = + −ϕ ϕ (P3.4)

The mechanical system being conservative, it follows that

( ) 0d T Udt

+ = (P3.5)

which leads to

( ) ( )2 22 2 5 2 0mz kz kl z ml kl klz+ − + + − = ϕ ϕ ϕ ϕ (P3.6)

Equation (P3.6) is satisfied at all times when:

2 0

2 2 5 0mz kz kl

ml kz kl+ − =

− + =

ϕϕ ϕ

(P3.7)

Equations (P3.7) form the mathematical model of the given mechanical system.


Problem 3.13

Let us use the energy method to derive the mathematical model of the mechanical

system for the two sets of coordinates.

(a)

For small motions, the displacements of the two point masses are vertical. The

kinetic energy is:

2 21 1 2 2

1 12 2

T m x m x= + (P3.1)

and the potential energy when only considering the elastic spring contributions (gravity

effects are neglected) is:

2 21 1 2 2

1 12 2

U k x k x= + (P3.2)

The system being conservative, the time derivative of the total energy E = T + U is zero,

which results in:

( ) ( )1 1 1 1 1 2 2 2 2 2 0x m x k x x m x k x+ + + = (P3.3)

As the velocities cannot be zero at all times, it follows that

1 1 1 1

2 2 2 2

00

m x k xm x k x

+ = + =

(P3.4)

The two differential Eqs. (P3.4) are the mathematical model for the case where the

motion coordinates are x1 and x2.

(b)

The position of the center of gravity CG can be found by using the conditions:

1 1 2 2

1 2

m gl m gll l l

= + =

(P3.5)

which results in:

21

1 2

12

1 2

ml lm m

ml lm m

= + = +

(P3.6)


When the coordinates of motion are x (the CG vertical displacement) and θ (the rod’s

tilt angle), it is necessary to express the coordinates x1 and x2 in terms of x and θ. The

geometry of Fig. 3.36, which indicates both the original (horizontal) position and the

displaced position, shows that:

1 2

2 2

1 2

tan x xl

l x xl x x

− = − =

−

θ (P3.7)

When small motions are only considered, tanθ is approximated to θ, and therefore x1 and

x2 are found from Eqs. (P3.7) as:

1 1

2 2

x x lx x l= +

= −

θθ

(P3.8)

Equations (P3.8) are substituted into Eqs. (P3.1) and (P3.2), which give the following

expressions for the kinetic and potential energies:

( ) ( )2 2

1 1 2 21 12 2

T m x l m x l= + + −

θ θ (P3.9)

and

( ) ( )2 21 1 2 2

1 12 2

U k x l k x l= + + −θ θ (P3.10)

The system being conservative, the time derivative of the total energy (the sum of kinetic

and potential energies) is zero, which leads to the equation:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 1 2 2

2 2 2 21 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 0

x m m x m l m l k k x k l k l

m l m l x m l m l k l k l x k l k l

+ + − + + + − + − + + + − + + =

θ θ

θ θ θ (P3.11)

Because the velocities pertaining to the coordinates x and θ cannot be zero at all times,

the only alternative for the left-hand side of Eq. (P3.11) to be zero is when:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 2 1 1 2 2 1 2 1 1 2 2

2 2 2 21 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2

0

0

m m x m l m l k k x k l k l

m l m l x m l m l k l k l x k l k l

+ + − + + + − =

− + + + − + + =

θ θ

θ θ (P3.12)

Equations (P3.12) are the mathematical model of the mechanical system of Fig. 3.36

when the coordinates of motion are x and θ and with l1, l2 of Eqs. (P3.6).


Problem 3.14

The mathematical model is:

2 0

2 2 5 0mz kz lk

ml kz lk+ − =

− + =

ϕϕ ϕ

(P3.1)

The following substitutions are used in the mathematical model of Eqs. (P3.1):

( ) ( )sin ; sinz Z t t= = Φω ϕ ω (P3.2)

which, after simplification by sin(ωt), reduces to:

( )

( )

2

2

2 0

2 5 2 0

k m Z kl

kZ k m l

− − Φ =− + − Φ =

ω

ω (P3.3)

The homogeneous Eqs. (P3.3) have nontrivial solutions (Z, Φ) when the determinant of

the system is zero (which is the characteristic equation). The following MATLAB® code

>> syms k m l om1

>> solve(det([k-m*om1,-2*k*l;-2*k,(5*k-2*m*om1)*l]),om1)

where om1 is the square of the natural frequency, generates the squares of the following

natural frequencies

( ) ( )1 21 17 41 ; 7 412 2n n

k km m

= + × = − ×ω ω (P3.4)

and the numerical values are ωn1 = 5.23 rad/s and ωn2 = 24.785 rad/s.

The following relationship applies between the amplitudes Y and Φ:

2Y l= Φ (P3.5)

and this is used in the first Eq. (P3.3) to get the following amplitude ratio

2Y k m

Z k−

=ω (P3.6)

For ω = ωn1, Eq. (P3.6) transforms into

1

2.35n

YZ =

= −ω ω

(P3.7)

which indicates that during the modal motion corresponding to ωn1, the translating body

and the point at the top of the vertical spring move in opposite directions, and that Y is


larger than Z. The resulting normalized eigenvector is found by solving the system

formed of Eq. (P3.7) and the equation

( ) ( )1 1

2 21

n nY Z= =+ =ω ω ω ω (P3.8)

The eigenvector is therefore:

1

0.92030.3916

n

YZ

=

= − ω ω

(P3.9)

For ω = ωn2, Eq. (P3.6) changes to

2

0.8508n

YZ =

=ω ω

(P3.10)

which indicates that the points denoted by the coordinates y and z move along the same

directions during this modal motion and that the amplitude Y is smaller than the

amplitude Z. Using the normalizing condition

( ) ( )2 2

2 21

n nY Z= =+ =ω ω ω ω (P3.11)

in conjunction with Eq. (P3.10) yields the eigenvector

2

0.6480.762

n

YZ

=

=

ω ω

(P3.12)

In order to conveniently use MATLAB® to determine the eigenvalues and

eigenvectors, the two Eqs. (P3.1) are written as:

0

5 2 02

mz ky kz

my ky kz

− + =

+ − =

(P3.13)

where the connection y = 2lφ has been used. Equation (P3.13) can be written as:

0 0

50 022

k km y ym z zk k

− + = −

(P3.14)

Therefore, the mass matrix [M] and the stiffness matrix [K] are:

[ ] [ ]0; 50 2

2

k kmM K

m k k

− = = −

(P3.15)

which enables calculation of the dynamic matrix [D] = [M]-1[K] with MATLAB® as


>> d = inv([0,m;m,0])*[-k,k;5/2*k,-2*k];

whose eigenvalues and eigenvectors are obtained by using

>> [V,D] = eig(d)

as

V =

0.9202 0.6480

-0.3914 0.7616

D =

614.3099 0

0 27.3568

It can be seen that the eigenvectors of [V] are also the ones obtained analytically, while

the squares of the analytical natural frequencies are the eigenvalues of [D].


Problem 3.15

In Example 3.7, Eqs. (3.41) have been determined to be the mathematical model of

the cart-pendulum mechanical system, namely:

( )1 2 2 0

0

m m x m l kx

x l g

+ + + =

+ + =

θ

θ θ (P3.1)

The sinusoidal solution:

( )( )

sin

sin

x X t

t

=

= Θ

ω

θ ω (P3.2)

is substituted in Eqs. (P3.1), which transform into:

( )

( )

2 21 2 2

2 2

0

0

m m k X m l

X l g

− + + − Θ = − + − + Θ =

ω ω

ω ω (P3.3)

The characteristic equation corresponding to Eqs. (P3.3) is:

( ) 2 21 2 2

2 20

m m k m ll g

− + + −=

− − +ω ω

ω ω (P3.4)

which is written as:

( )4 21 1 2 0m l m m g kl kg− + + + = ω ω (P3.5)

The eigenvalues (the roots) resulting from Eq. (P3.5) are:

( ) ( ) ( )2 2 2 2

1 2 1 2 1 221,2 1,2

1

2 lg2

m m g kl m m g k l k m mm l

+ + ± + + − −= =λ ω (P3.6)

The numerical values of the natural frequencies are ω1 = 3.48 rad/s and ω2 = 14.58 rad/s.

From the first Eq. (P3.3), the following ratio can be formed:

( )

22

21 2

mXl k m m

=Θ − +

ωω

(P3.7)

The specific ratio of Eq. (P3.7) has been chosen in order to compare two similar amounts

(linear displacements), X and lΘ; this equation results in:

1 2

0.0124; 0.9424X Xl l= =

= = −Θ Θω ω ω ω

(P3.8)

The first ratio of Eq. (P3.8) indicates that during the first modal motion, the rotation of

the pendulum is in the direction shown in the problem’s figure and that the lΘ > X.


During the second modal motion, the pendulum rotates in a direction opposite to the one

selected as being the positive rotation direction and the magnitude of lΘ is larger than X.

In order to determine the unit-norm eigenvectors (which are also the ones that are

provided by the eig MATLAB® command), the X/Θ ratio is needed; this can simply be

found by multiplying the X/(l Θ) ratio values, which have been determined analytically,

by the value of l; this results in:

1 2

1 2

1 2

0.0099; 0.7539X XX X= =

= = = = −Θ Θ Θ Θω ω ω ω

(P3.9)

To determine the unit-norm eigenvector corresponding to the natural frequency ω1, the

following two equations system needs to be solved:

1

1

2 21 1

0.0099

1

X

X

=Θ +Θ =

(P3.10)

which results in X1 = 0.0099, Θ1 = 1; similar calculations yield X2 = −0.602, Θ2 = 0.7985.

As a consequence, the two unit-norm eigenvectors are:

1 2

0.0099 0.602;

1 0.7985V V

− = =

(P3.11)

In order to use MATLAB® for eigenvalue and eigenvectors calculation, the

mathematical model – Eq. (P3.1) – is written in vector-matrix form as:

1 2 2 0 01 0 0

m m m l x k xl g

+ + =

θ θ (P3.12)

with the first matrix in the left hand side of Eq. (P3.12) being the mass (or inertia) matrix

[M] and the second one being the stiffness matrix [K]. The dynamic matrix [D] is

calculated as [D] = [M]-1[K]. The following MATLAB® code is used to determine the

dynamic matrix symbolically together with its numerical eigenvalues and eigenvectors:

>> syms m1 m2 k l g;

>> in=[m1+m2,m2*l;1,l];

>> stiff=[k,0;0,g];

>> d=inv(in)*stiff


This code returns

d =

[ 1/m1*k, -1/m1*m2*g]

[ -1/l/m1*k, (m1+m2)/l/m1*g]

The following code

>> m1=1; m2=0.2; k=210; l=0.8; g=9.8; dn=d;

>>[v,d]=eig(dn)

returns

v =

0.6020 0.0099

-0.7985 1.0000

d =

212.5998 0

0 12.1002

The elements on the diagonal matrix denoted by d are the eigenvalues λ1, λ2; it can be

checked that their square roots are the natural frequencies that have. The two columns of

the matrix denoted by v are the eigenvectors corresponding to the two natural

frequencies, that are, again, identical to the ones that have been obtained by analytic

calculation.


Problem 3.16

The lumped-parameter model of the mechanical microsystem is shown in Fig. P3.1.

Figure P3.1 Lumped-parameter model for the y-axis translation

The total energy of the system is

( )22 2 2 21 2 1 1 2 1 2 1 2

1 1 1 1 12 2 2 2 2

T U mx mx k x k x x k x+ = + + + − + (P3.1)

Making zero the time derivative of the energy of Eq. (P3.1) because the system is

conservative, results in:

( ) ( )1 1 2 1 2 2 1 2 1 2 2 2 1 2 0mx k k x k x x mx k k x k x x + + − + + + − = (P3.2)

The equation above is zero at all times only when:

( )

( )1 1 2 1 2 2

2 2 1 1 2 2

0

0

mx k k x k x

mx k x k k x

+ + − =

− + + =

(P3.3)

Harmonic solution of the following type is sought for x1 and x2 of Eqs. (P3.3):

( ) ( )1 1 2 2sin ; sinx X t x X t= =ω ω (P3.4)

Substitution of Eqs. (P3.4) in Eqs. (P3.3) results in:

( )

( )

21 2 1 2 2

22 1 1 2 2

0

0

k k m X k X

k X k k m X

+ − − =− + + − =

ω

ω (P3.5)

The nonzero solution of Eq. (P3.5) results from annulling the determinant of that

equations determinant, which yields the following natural frequencies:

1 1 21 2

2;n nk k km m

+= =ω ω (P3.6)

The spring k1 is formed of two fixed-guided beams and therefore its stiffness is

calculated as:

x1

k1

m m

k2 k1

x2


3

3*

1 1 3 31 1

12 2122 2

w hE Ew hk kl l

= × = × = (P3.7)

whose numerical value is k1 = 37.5 N/m. The middle spring is formed of two parallel

groups that are connected in parallel, each group being formed of two fixed-guided

beams that are in series; the middle spring stiffness is therefore

( )

32* 3

2 *2 2* 3 3 3

2 2 2 2

1212 1222

z

w hEk EI Ew hk kk l l l

= × = = = = (P3.8)

whose numerical value is k2 = 347.222 N/m. As a consequence, the natural frequencies of

Eqs. (P3.6) are: ωn1 = 559,020 rad/s and ωn2 = 1,326,400 rad/s.

When considering the inertia of the beams, each of two original masses transform

into:

( ) ( )1 2 1 213' 2 235

m m m m m wh l l= + + = + × +ρ (P3.9)

Because two long beams and two short beams add their contributions to each of the

masses. The beams being fixed-guided, Table 3.1 gives the mass fraction of Eq. (P3.9).

The following numerical value is obtained: m’ = 1.331 x 10-10 kg. With this mass instead

of m, Eqs. (P3.6) yield the following natural frequencies: ω’n1 = 530,790 rad/s and ω’n2 =

1,259,400 rad/s.

From the first Eq. (P3.5), the following ratio results:

1 22

2 1 2

X kX k k m

=+ − ω

(P3.10)

Fo r ω = ωn1, Eq. (P3.10) results in a value of 1. This indicates the two masses are

moving in a synchronized fashion during the first modal motion and therefore the

corresponding eigenvector is:

1

1

2

11

n

XX

=

= ω ω

(P3.11)

For ω = ωn2, Eq. (P3.7) results in a value of -1. This indicates the two masses are moving

in opposition during the second modal motion; the corresponding eigenvector is:

2

1

2

11

n

XX

=

− = ω ω

(P3.12)


If unit-norm vectors are sought instead of the straightforward ones of Eqs. (P3.11)

and (P3.12), whose sum of components squares is one, these vectors are:

1 2

1 1

2 2

2 22 2;2 2

2 2n n

X XX X

= =

− = =

ω ω ω ω

(P3.13)

The mass and stiffness matrices are expressed from Eqs. (P3.3) as:

[ ] [ ] 1 2 2

2 1 2

0;

0k k km

M Kk k km+ −

= = − + (P3.14)

And they enable calculating the dynamic matrix [D] = [M]-1 [K]. Using MATLAB® and

its eig command, the following eigenvectors and eigenvalues are obtained for the case

where the inertia of the flexures is neglected: V =

-0.7071 -0.7071

-0.7071 0.7071

D =

1.0e+012 *

0.3125 0

0 1.7593

which are the eigenvectors of Eq. (P3.13) and the squares of the natural frequencies ωn1

and ωn2. The values of ω’n1 and ω’n2 are obtained if m’ is used instead of m in the

MATLAB® matrix formulation.


Problem 3.17

(a)

The free-body diagrams of the two rods are shown in Fig. P3.1.

Figure P3.1 Free-body diagrams of two-pendulum mechanical system

Newton’s second law of motion is applied in the rotation variant, which yields the

equations:

( ) ( ) ( )( ) ( ) ( ) ( )

21 1 1 1 1 1

22 2 2 2 2 2 2

2 2 2 sin cos

2 2 2 sin cos 2 cosd e

d e

m l f l m g l f l

m l f l m g l f l f l

θ θ θ

θ θ θ θ

= − − +

= − − − +

(P3.1)

The damping forces and the elastic (spring) forces are:

( ) ( ) ( )1 1 2 2 2 12 ; 2 ; sin sind d ef c l f c l f k l lθ θ θ θ= = = − (P3.2)

By using the small-displacement approximations:

cos 1; sin≈ ≈θ θ θ (P3.3)

in conjunction with Eqs. (P3.2), Eqs. (P3.1) can be written in the form:

( )

( )1 1 1 1 1 2

2 2 2 1 2 2

4 4 2 0

4 4 2 2

m l lc m g kl kl

m l lc kl m g kl f

+ + + − =

+ − + + =

θ θ θ θ

θ θ θ θ (P3.4)

Equations (P3.4) are the mathematical model of the two-pendulum mechanical system.

(b)

With the numerical parameters of the problem, Eqs. (P3.4) are rewritten as:

fe

θ1

θ2

fe

fd1 fd2

f

m1g m2g


1 1 1 2

2 2 1 2

220 101.12 95 0

91.67 39.58 45.71 78.13sin(10 )t

θ θ θ θ

θ θ θ θ

= − − + =

= − + − +

(P3.5)

The Simulink® diagram that is based on the equations above and simulate the two-

pendulum mechanical system is shown in Fig. P3.2, and the time response of this system

to the sinusoidal input is given in Fig. P3.3.

Figure P3.2 Simulink® diagram of the two-pendulum mechanical system

θ1

θ2


Figure P3.3 Simulink® time response of the two-pendulum mechanical system

θ1

θ2


Problem 3.18

Figure P3.1 shows the free-body diagrams of the mechanical system. This system has

two DOF, which are the pulley rotation angle θ and the mass m3 translation x2. For small

motions, the coordinate x1 is:

1 2x R= θ (P3.1)

Figure P3.1 Free-body diagrams of pulley and translatory body

Newton’s second law of motion for the pulley rotation and the body translation is

expressed as:

2 1 1 2

3 2 2

e e

e d

J f R f Rm x f f f = −

= − −

θ (P3.2)

where:

( )

( )

2 21 1 2 2

1 1 1 1 2

2 2 2 1

2

12

e

e

d

J m R m R

f k x k Rf k x Rf cx

= + = = = −

=

θθ

(P3.3)

Substitution of Eqs. (P3.3) into Eqs. (P3.2) results in:

( ) ( )2 2 2 21 1 2 2 1 2 2 1 2 1 2

3 2 2 2 1 2 2

1 02

m R m R k R k R k R x

m x cx k R k x f

+ + + − = + − + =

θ θ

θ (P3.4)

Equations (P3.4) can be written with the numerical values of the problem as:

2

2 2 2

258 2963 064 200 3 2

xx x x f

= − + =

= − − + +

θ θθ (P3.5)

The forcing input can be specified by means of the Signal Builder block in the

x2

f

m3

fe2 fd x1

R2

m2

R1 m1

fe2

fe1 θ


Sources library of Simulink®. Horizontal and vertical plot segments can be displaced

and values of the function and variable can be changed interactively. The input force in

this problem has been defined as shown in Fig. P3.2.

0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5Signal 1

Time (sec)

pr2_28/Signal Builder : Group 1

Figure P3.2 Input created by means of a Signal Builder Simulink® source

The Simulink® diagram and time response curves θ(t) and x2(t) are shown in Figs. P3.3

and P3.4.


Figure P3.3 Simulink® diagram of the pulley-mass mechanical system

(a)


(b)

Figure P3.4 Simulink® time response: (a) pulley rotation angle θ(t); (b) linear-motion body

displacement x2(t)


Problem 3.19

The total energy of the system shown in Fig. P3.1 is:

( )

( ) ( )

22 2 2 23 2 1 1 2 2 1 3 2

22 2 2 2 2 2 21 1 2 2 3 2 1 2 2 2 1 3 2

1 1 1 1 12 2 2 2 2

1 1 1 1 14 2 2 2 2

E T U J m x k x k x R k x

m R m R m x k R k x R k x

= + = + + + − +

= + + + + − +

θ θ

θ θ θ (P3.1)

Figure P3.1 Two-DOF mechanical system

The time derivative of the total energy needs to be zero as the system is conservative,

which leads to:

( ) ( )

( )

2 2 2 21 1 2 2 1 2 2 1 2 1 2

2 3 2 2 1 2 3 2

12

0


x m x k R k k x

+ + + − + − + + =

θ θ θ

θ (P3.2)

The velocities cannot be zero at all times and therefore the brackets have to be zero,

which results in the following equations:

( ) ( )( )

2 2 2 21 1 2 2 1 2 2 1 2 1 2

3 2 2 1 2 3 2

1 02

0


m x k R k k x

+ + + − = − + + =

θ θ

θ (P3.3)

The solution to this mathematical model is of sinusoidal form:

( )( )2 2

sin

sin

t

x X t

= Θ

=

θ ω

ω (P3.4)

which, substituted into Eqs. (P3.3), result in:

k1

k2

x2

x1

m3

R2

m2

k3

R1 m1

θ


( )

( )

2 2 2 2 21 1 2 2 1 2 2 1 2 1 2

22 1 3 2 3 2

1 02

0

m R m R k R k R k R X

k R m k k X

− + + + Θ− = − Θ+ − + + =

ω

ω (P3.5)

The characteristic equation corresponding to the differential Eq. (P2.167) is:

( )2 2 2 2 21 1 2 2 1 2 2 1 2 1

22 1 3 2 3

102

m R m R k R k R k R

k R m k k

− + + + −=

− − + +

ω

ω (P3.6)

With the numerical values of this problem, Eq. (P3.6) yields the natural frequencies ω1 =

15 rad/s and ω2 = 23 rad/s.

The second Eq. (P3.5) can be written as:

2

2 3 31

2 2

k k mRX k

+ −=

ωθ (P3.7)

which, for the two natural frequencies, becomes:

1

2

1

2

1

2

1.375;

0.145

n

n

RX

RX

=

=

=

= −

ω ω

ω ω

θ

θ (P3.8)

Because X2 represents length in the numerator of Eq. (P3.8), the amount θR1, which is

also a length, has been used in the numerator of the same equation in order to enable

comparison of similar amounts. The first Eq. (P3.8) shows that the pulley rotation and

mass translation directions are in concert and that the tangential displacement at the

point where the spring k2 is applied (and which is θR1) is larger than the displacement of

the translating body X2. During the second modal motion, the two motions mentioned

above are in opposition; the tangential displacement at the point where the spring k2 is

applied is smaller than the displacement of the translating body, X2.

The unit-norm eigenvector corresponding to ω1 is found by solving the system:

1

1

2 21

2 21 21

91.6667

1

X X

X

=

ΘΘ= =

Θ + =

ω ω (P3.9)


The first ratio of Eqs. (P3.9) has been obtained through division by R1 of the first ratio of

Eq. (P3.8). The solution to Eqs. (P3.9) is: Θ1 = 0.9999 and X21 = 0.0109, such that the

eigenvector corresponding to the first natural frequency is:

1

0.99990.0109

V =

(P3.10)

The unit-norm eigenvector corresponding to ω2 is found by solving the system:

2

2

2 22

2 22 22

9.6667

1

X X

X

=

ΘΘ= = −

Θ + =

ω ω (P3.11)

and the solution to Eqs. (P2.173) is: Θ2 = 0.9947 and X22 = − 0.1029, such that the

eigenvector corresponding to the first natural frequency is:

2

0.99470.1029

V = −

(P3.12)

In order to use specialized MATLAB® commands to determine the eigenfrequencies

and eigenvectors, Eqs. (P3.3) are written in vector-matrix form as:

( )2 2 2 21 1 2 2 1 2 2 1 2 1

22 1 2 323

1 002

00

m R m R k R k R k Rxk R k kxm

+ + − + = − +

θθ (P3.13)

where the mass (inertia) matrix [M] and the stiffness matrix [K] are:

[ ] ( ) [ ]2 2 2 2

1 1 2 2 1 2 2 1 2 1

2 1 2 33

1 0;2

0

m R m R k R k R k RM K

k R k km

+ + − = = − +

(P3.14)

The dynamic matrix [D] is obtained by using MATLAB®’s symbolic calculation

capability as:

[ ] [ ] [ ]

( )2 21 2 2 1 2 1

2 2 2 21 1 1 2 2 1 1 2 2

2 32 1

3 3

2 2k R k R k Rm R m R m R m RD M K

k kk Rm m

−

+ −

+ + = = + −

(P3.15)

The command [v,D] = eig(d) with d being the matrix [D] defined in Eq. (P3.15)

produces the following result for the numerical parameters of this problem:


v =

-0.9999 0.9957

-0.0109 -0.0923

d =

225.4067 0

0 532.3711

The eigenvalues are the elements on the diagonal of the d matrix above; their square

roots are the natural frequencies ω1 = 15.0136 rad/s and ω1 = 23.0732 rad/s, which are

the ones that have been obtained analytically. The two columns of v are the eigenvectors,

and it can be seen that they have components that are very close to the ones of the

analytical unit-norm eigenvectors calculated previously. The minus sign in both

components of the first eigenvector (column) above is irrelevant as it can be changed to

plus without losing the physical significance that Θ and X2 produce displacements that

are in concert during the first modal motion.


Problem 3.20

(a)

Figure P3.1 highlights the relevant angles of rotation.

Figure P3.1 Gear system with inertia, damping, stiffness, loading and relevant rotation angles

Newton’s second law of motion is applied to the cylinders J2 and J3, while transferring

the load and physical properties from the actuation and load shafts; the following

equations are obtained:

( )

( )

2

2 21 2 1 1 1 2

1 1

2

3 34 3 2 2 2 1

4 4

a

l

N NJ J m c kN N

N NJ J m c kN N

+ = − − − + = − − − −

θ θ θ θ

θ θ θ θ

(P3.1)

Equations (P3.1) form the mathematical model of the mechanical system studied in this

problem.

(b)

In order to enable integrations to be performed by means of Simulink®, Eqs. (P3.1)

are expressed in terms of the numerical parameters of this problem as:

6

1 1 1 2

52 2 1 2

17,778 177,780 177,780 3.33 10

14,693 146,930 146,930 2.5187 10

= − − + + ×

= − + − − ×

θ θ θ θ

θ θ θ θ (P3.2)

The Simulink® diagram and time response curves are shown in Figs. P3.2, P3.3, and

P3.4.

θ2 θ1 k

N3

N4

ml

J3

c c N2

J2

J1 ma

J4 N1


Figure P3.2 Simulink® diagram of the gear mechanical system


(a) (b)

Figure P3.3 Simulink® time response of the gear mechanical system – angles (in radians): (a) θ1; (b) θ2

(a) (b)

Figure P3.4 Simulink® time response of the gear mechanical system – angular velocities (in rad/s): (a)

ω1; (b) ω2

Documents

Chapter 03