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SOLUTIONS TO CHAPTER 3 PROBLEMS Problem 3.1 According to Table 3.1 and given the example’s specification, it follows that: 33 13 mc = mb 140 35

(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: mb = 0.635mc

(P3.2)

lb wb = 0.635lc wc

(P3.3)

which results in:

where l and w are length and width. As shown in Appendix D, the mass moments of inertia of the two members are: mc ( wc2 + h 2 ) mb ( wb2 + h 2 ) ; Jb = Jc = 12 12

(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 mc ( wc + h ) 1 mb ( wb + h ) × = × 3 12 3 12

(P3.5)

Taking into account that the two masses are: = mc ρ= lc wc h; mb ρ lb wb h

(P3.6)

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

= wb 1.255 wc2 + 0.365h 2

(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).

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.2 The mass contribution from the four fixed-guided beams is calculated as: me = 2 ( me,long + me, short ) = 2 × 3 × me, short = 6 × me, short

(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: me, short =

13 π d 2 ρ l1 35 4

(P3.2)

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

39 ρπ d 2l1 70

(P3.3)

The total moving mass adds the beam contributions to the actual mass of the plate, which yields: mt= m plate + me= ρ l 2 (5d ) +

39 39π   ρπ d 2l1= ρ d  5l 2 + dl1  70 70  

(P3.4)

Taking into account that me = 0.1 mt leads to: 0.9mt mt = m plate + 0.1mt or m plate =

(P3.5)

Combining now Eqs. (P3.4) and (P3.5) yields: dl1 = 0.3174l 2

(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).

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.3 The equivalent stiffness of the original bridge is determined based on Table 3.2 as: 2GI p 4GI p k= 2× = 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: GI p 4GI p k ' 2= = l − lc l − lc 2

(P3.2)

The following stiffness ratio can be formulated by means of Eqs. (P3.1) and (P3.2): k' l 1 = = k l − lc 1 − lc / l

(P3.3)

Figure P3.1 shows the variation of the stiffness ratio in terms of the length ratio lc/l. 2 1.9 1.8

stiffness ratio

1.7 1.6 1.5 1.4 1.3 1.2 1.1 1

0

0.05

Figure P3.1

0.1

0.15

0.2

0.25 lc /l

0.3

0.35

0.4

0.45

0.5

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:

π d 4 πρ d 4l J 1 J e = = × ρl × = 3 3 32 96

(P3.4)

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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: 4 πρc d c4lc 1 πρ d ( l − lc ) / 2 = J + 2× × 32 3 32 ' e

(P3.5)

The ratio of the two equivalent mechanical moments of inertia is obtained from Eqs. (P3.4) and (P3.5): lc J e' lc 3ρc =1 − + × l ρ  d 4 Je l    dc 

(P3.6)

Mechanical moment of inertia ratio

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

1200 1000 800 600 400 200 0 0.5 0.5

0.4

0.4

0.3 d/dc

Figure P3.2

0.3 0.2

0.2

lc /l

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:

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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>> [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.

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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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. y or z

Figure P3.1

k2

k2

k1

k1

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

The stiffnesses about the two directions are calculated as: k1 y k2 y k k ; k z 2 1z 2 z = k y 2= k1 y + k2 y k1z + k2 z

(P3.1)

where the stiffnesses of the flexible segments of length l1 and l2 (which are actually fixedguided beams) are determined based on Table 3.2 as:   = k1 y     = k1z 

w3h w3h 12 E 12 E 3 12 EI z 12 EI z Ew h Ew3h 12 12 ; k2 y = = = = = l13 l13 l13 l23 l23 l23 wh3 wh3 12 EI y 12 E 12 Ewh3 12 EI y 12 E 12 Ewh3 = = = = = ; k 2z l13 l13 l13 l23 l23 l23

(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: 2 Ew3h 2 Ewh3 = ky = ; k z l13 + l23 l13 + l23

(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.

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Problem 3.5 The lumped-parameter model of the microspring of Fig. 3.30 is shown in Fig. P3.1.

Figure P3.1

k2

k2

k2

k2

k1

k1

x

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: = ke 2

k1k2 + 2k 2 k1 + k2

(P3.1)

where: 12 EI1 12 EI 2 = k1 = ; k2 3 l1 l23

(P3.2)

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

24 EI 2  I1l23  + 1   l23  I1l23 + I 2l13 

(P3.3)

The moments of inertia of the two beams are:

π d14 π d 24 = I1 = ; I2 64 64

(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.

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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.

k1

k1

k2

k1

k1

k1

k2

k1

(a) Figure P3.1

(b)

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: ka = 2k1

(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: 12 EI 24 EI 2 × 3 =3 ka = 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 fixedfixed beams of Fig. 3.31(b). The six beams are connected in parallel and therefore the equivalent stiffness is: = kb 4k1 + 2k2

(P3.3)

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

432 EI l3

(P3.4)

By comparing Eqs. (P3.2) and (P3.4) results in: kb 432 = = 18 ka 24

(P3.5)

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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

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Problem 3.7 The out-of-plane bending natural frequencies for a cantilever and for a bridge are: ke , c ωn , c = ; ωn ,b = me,c

k e ,b me,b

(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:

) = ω − (ω ) ∆ (ω ) ω − (ω ) ∆ (ω

2 n ,c

2 n ,c

* 2 n ,c

2 n ,b

2 n ,b

2 * n ,b

ke , c ke , c − me c me,c + m p ke,c me,b me,b + m p = , = × × k e ,b k e ,b ke,b me,c me,c + m p − me,b me,b + m p

(P3.2)

where mp is the mass of the particle. Taking into account that:  = ke,c  m =  e,c

3EI 192 EI = ; k e ,b ; 3 l l3 33 13 = m; me,b m 140 35

(P3.3)

where I is the beam cross-sectional moment of inertia and m is the beam mass, it follows that: 13 m + mp 13 me,b + m p 13 35 = × = × ∆ (ωn2,b ) 528 me,c + m p 528 33 m + m p 140 ∆ (ωn2,c )

(P3.4)

Equation (P3.4) can be written as: 13 ∆ (ωn2,c ) 13 35 + cm = × ∆ (ωn2,b ) 528 33 + c m 140

(P3.5)

with cm =

mp m

(P3.6)

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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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.

0.039 0.038 0.037

(∆ωn,c )2/(∆ωn,b)2

0.036 0.035 0.034 0.033 0.032 0.031 0.03 0.029

0

Figure P3.1

0.05

0.1

0.15

0.2

0.25 mp/m

0.3

0.35

0.4

0.45

0.5

Ratio of squared natural frequency ratio in terms of the mass ratio

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.8 The microbridge-mass system is sketched in Fig. P3.1 below. y

w/2

x

z

w

b

mp

l/2 l

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: 192 EI y 16 Ewh3 = k e ,b = l3 l3

(P3.1)

The equivalent mass at the bridge midpoint is given in Table 3.1 and for the particular rectangular cross-section is: me,b =

13 ρ whl 35

(P3.2)

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

ωn ,b =

k e ,b me,b

(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:

ωb* =

k e ,b me,b + m p

(P3.4)

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

∆ωb = ωn ,b − ωb*

(P3.5)

Eqs. (P3.3), (P3.4) and (P3.5) enable expressing the deposited mass as: Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

12

= mp

k e ,b  k e ,b  − ∆ωb    me,b   

2

− me,b

(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 lumpedparameter 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: 4GI t 4Gwh3 = k e ,t = l 3l

(P3.7)

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

ρ whl ( w2 + h 2 ) 1 J= = J e ,t 3 36

(P3.8)

The original torsional natural frequency is therefore:

ωn ,t =

k e ,t J e ,t

(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:

ωt* =

k e ,t J e ,t + m p b 2

(P3.10)

The change in the torsional frequency is:

∆ωt = ωn ,t − ωt*

(P3.11)

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

      k 1 e ,t × − J e ,t  b= 2 mp   k   e ,t − ∆ωt        J e,t  

(P3.12)

with a numerical value of b = 22.7 μm.

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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. θx

Je ke,t

Figure P3.1

ke,t

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: GI p π Gd 4 = ke,t 2= 16l1 l1

(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: Je =

ρ l1l2 h ( l22 + h 2 ) 12

2 2 2 1 1  d  ρ l1l2 h ( l2 + h ) 1 1 ρl π d 2 d 2 + 2× = × m  + 2× × × 1 × 3 2 2 12 3 2 4 4

ρ l1  πd4  2 2 = l h l + h + ( ) 2 2 12  4 

(P3.2)

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

ωn , x =

k e ,t Je

(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.

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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y

me ke,b

Figure P3.2

ke,b

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: 12 EI 3π Ed 4 k e ,b = 2 × 3 z =3 l1 8l1

(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:

= me ρ l1l2 h + 2 ×

 πd2 13 13π d 2  × ρ l1 × = ρ l1 l2 h + 35 4 70  

(P3.5)

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

ωn , y =

k e ,b me

(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.

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Problem 3.10 In terms of the z-axis plate translation (out-of-plane motion) the hinges bend as fixedguided beams (see Table 3.2) and therefore, act as four springs connected in parallel; the corresponding bending stiffness is: 12 EI y 3π Ed 4 12 EI x = ke,tr 4= 4 = l13 l13 4l13

(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:

m= ρ l1l2 h + 4 × e ,tr

 13 πd2 13π d 2  × ρ l1 × = ρ l1  l2 h +  35 4 35  

(P3.2)

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

ωe,tr =

ke,tr me,tr

(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: 12 EI z 3π Ed 4 = ke,rot 4= = ke,tr l13 4l13

(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: J p,z =

ρ l1l2 h ( l12 + l22 ) 12

(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: mb ,e =

13 π d 2 13ρπ d 2l1 × ρ l1 × = 35 4 140

(P3.6)

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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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y l1 l2 x

mb,e

l2

Figure P3.1

l1 mb,e mb,e mb,e

Lumped-parameter model showing the inertia contributions by the four beams

J b , z = 2mb ,e ×

13ρπ d 2l1 ( l12 + l22 ) l12 l2 m + 2mb ,e × 2 = b ,e × ( l12 + l22 )= 4 4 2 280

(P3.7)

where the equivalent mass of Eq. (P3.6) has been used. The equivalent (total) mass moment of inertia is therefore: J e, z = J p , z + J b, z

ρ l1 ( l12 + l22 )  l2 h 13π d 2  = × +  4 70   3

(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:

ωe,rot =

ke,rot J e, z

(P3.9)

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

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Problem 3.11 By transferring the lumped-parameter spring from the cantilever’s midpoint to the free end the following stiffness is obtained: 2

k l /2 = k ' =  k 4  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:  π d 2 33πρ ld 2 33 33 = = × × = ρ m m l  e 140 140 4 560  4 4 k =3EI =3E × π d =3π Ed  e l3 l3 64 64l 3

(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:

ωn =

k '+ ke me

(P3.3)

whose value is ωn = 1,389,300 rad/s.

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.12 The main coordinates of motion are highlighted in Fig. P3.1 below.

m R

φ

k

θ y

x

l

y 2l k

m

Figure P3.1

z

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: = y 2= lϕ ; x lϕ

(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 = Rθ

(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 1 1 1 1 4l 2 1 T =mz 2 + mR 2θ 2 =mz 2 + mR 2 2 ϕ 2 =mz 2 + ml 2ϕ 2 2 4 2 4 R 2

(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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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1 1 1 1 2 2 U = kx 2 + k ( z − y ) = kl 2ϕ 2 + k ( z − 2lϕ ) 2 2 2 2

(P3.4)

The mechanical system being conservative, it follows that d 0 (T + U ) = dt

(P3.5)

0 ( mz + kz − 2klϕ ) z + ( 2ml 2ϕ + 5kl 2ϕ − 2klz ) ϕ =

(P3.6)

which leads to

Equation (P3.6) is satisfied at all times when: 0 mz + kz − 2klϕ =  0 2mlϕ − 2kz + 5klϕ =

(P3.7)

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

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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: 1 1 m1 x12 + m2 x22 2 2

= T

(P3.1)

and the potential energy when only considering the elastic spring contributions (gravity effects are neglected) is: = U

1 2 1 k1 x1 + k2 x22 2 2

(P3.2)

The system being conservative, the time derivative of the total energy E = T + U is zero, which results in: x1 ( m1  x1 + k1 x1 ) + x2 ( m2  x2 + k2 x2 ) = 0

(P3.3)

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

x1 + k1 x1 = 0 m1   x2 + k2 x2 = 0 m2 

(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:

m1 gl1 = m2 gl2  l l1 + l2 =

(P3.5)

m2  l1 = m + m l  1 2  l = m1 l  2 m1 + m2

(P3.6)

which results in:

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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: x1 − x2   tan θ = l l  2 = x − x2  l x1 − x2

(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:  x1= x + l1θ   x2= x − l2θ

(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: T=

(

)

(

2 1 1 m1 x + l1θ + m2 x − l2θ 2 2

)

2

(P3.9)

and U=

1 1 2 2 k1 ( x + l1θ ) + k2 ( x − l2θ ) 2 2

(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: x ( m1 + m2 )  x + ( m1l1 − m2l2 ) θ + ( k1 + k2 ) x + ( k1l1 − k2l2 ) θ 

(P3.11) +θ ( m1l1 − m2l2 )  x + ( m1l12 + m2l22 ) θ + ( k1l1 − k2l2 ) x + ( k1l12 + k2l22 ) θ  = 0 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: ( m1 + m2 )  x + ( m1l1 − m2l2 ) θ + ( k1 + k2 ) x + ( k1l1 − k2l2 ) θ = 0 (P3.12)  x + ( m1l12 + m2l22 ) θ + ( k1l1 − k2l2 ) x + ( k1l12 + k2l22 ) θ = 0 ( m1l1 − m2l2 ) 

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).

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22

Problem 3.14 The mathematical model is: 0 mz + kz − 2lkϕ =  0 2mlϕ − 2kz + 5lkϕ =

(P3.1)

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

z = Z sin (ωt ) ; ϕ = Φ sin (ωt )

(P3.2)

which, after simplification by sin(ωt), reduces to: ( k − mω 2 ) Z − 2kl Φ =0   2 −2kZ + ( 5k − 2mω ) l Φ =0

(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

ωn1 =

1 2

(7 +

)

41 ×

k 1 ; ωn 2 = m 2

(7 −

)

41 ×

k m

(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: Y= 2l Φ

(P3.5)

and this is used in the first Eq. (P3.3) to get the following amplitude ratio Y k − mω 2 = Z k

(P3.6)

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

= −2.35

(P3.7)

ω =ωn1

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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

23

larger than Z. The resulting normalized eigenvector is found by solving the system formed of Eq. (P3.7) and the equation

(Y

) + (Z 2

= ω ω= ω ωn 1 n1

)

2

= 1

(P3.8)

The eigenvector is therefore: Y  0.9203  =     Z ω =ωn1 −0.3916 

(P3.9)

For ω = ωn2, Eq. (P3.6) changes to Y Z

= 0.8508

(P3.10)

ω =ωn 2

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

(Y

) + (Z 2

= ω ω= ω ωn 2 n2

)

2

= 1

(P3.11)

in conjunction with Eq. (P3.10) yields the eigenvector Y  0.648  =     Z ω =ωn 2 0.762 

(P3.12)

In order to conveniently use MATLAB® to determine the eigenvalues and eigenvectors, the two Eqs. (P3.1) are written as: 0 mz − ky + kz =   5 0 my + 2 ky − 2kz =

(P3.13)

where the connection y = 2lφ has been used. Equation (P3.13) can be written as:  −k y   0 m    + 5  m 0      z  k 2

k  0    y  =   −2k   z  0  

(P3.14)

Therefore, the mass matrix [M] and the stiffness matrix [K] are:  −k  0 m  = [ M ] =  ; [K ]  5 0 m k   2

k   −2k  

(P3.15)

which enables calculation of the dynamic matrix [D] = [M]-1[K] with MATLAB® as Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

24

>> 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].

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

25

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: ( m1 + m2 )  x + m2lθ + kx = 0  x + lθ + gθ = 0  

(P3.1)

The sinusoidal solution:  x = X sin (ωt )  θ = Θ sin (ωt )

(P3.2)

is substituted in Eqs. (P3.1), which transform into:   − ( m1 + m2 ) ω 2 + k  X − m2lω 2 Θ =0    2 2 −ω X + ( −lω + g ) Θ =0

(P3.3)

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

− ( m1 + m2 ) ω 2 + k −ω 2

−m2lω 2 =0 −lω 2 + g

(P3.4)

which is written as: m1lω 4 − ( m1 + m2 ) g + kl  ω 2 + kg = 0

(P3.5)

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

λ= ω= 1,2 2 1,2

( m1 + m2 ) g + kl ± ( m1 + m2 )

g 2 + k 2l 2 − 2k lg ( m1 − m2 )

2

2m1l

(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: m2ω 2 X = l Θ k − ( m1 + m2 ) ω 2

(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: X lΘ ω =

X lΘ ω ω= 1 = 0.0124;

= −0.9424

(P3.8)

ω2

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

26

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: X Θω =

X1 X = 0.0099; Θ1 Θω ω= 1 =

= ω2

X2 = −0.7539 Θ2

(P3.9)

To determine the unit-norm eigenvector corresponding to the natural frequency ω1, the following two equations system needs to be solved:  X1  Θ = 0.0099  1  X 2 + Θ 2 =1 1  1

(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: 0.0099  −0.602  = V1 =  ; V2    1   0.7985 

(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:  m1 + m2  1 

m2l    x  k   +   l  θ   0

0   x  0   =   g  θ  0 

(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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

27

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

212.5998

0

0

12.1002

d =

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.

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28

Problem 3.16 The lumped-parameter model of the mechanical microsystem is shown in Fig. P3.1. x1

x2

m

m

k1

Figure P3.1

k1

k2

Lumped-parameter model for the y-axis translation

The total energy of the system is T +U =

1 2 1 2 1 2 1 1 2 mx1 + mx2 + k1 x1 + k2 ( x1 − x2 ) + k1 x22 2 2 2 2 2

(P3.1)

Making zero the time derivative of the energy of Eq. (P3.1) because the system is conservative, results in: 0  mx1 + ( k1 + k2 ) x1 − k2 x2  x1 +  mx2 + ( k1 + k2 ) x2 − k2 x1  x2 =

(P3.2)

The equation above is zero at all times only when: mx1 + ( k1 + k2 ) x1 − k2 x2 = 0  0 mx2 − k2 x1 + ( k1 + k2 ) x2 =

(P3.3)

Harmonic solution of the following type is sought for x1 and x2 of Eqs. (P3.3): = x1 X= X 2 sin (ωt ) 1 sin ( ω t ) ; x2

(P3.4)

Substitution of Eqs. (P3.4) in Eqs. (P3.3) results in: ( k1 + k2 − mω 2 ) X 1 − k2 X 2 = 0   2 0 −k2 X 1 + ( k1 + k2 − mω ) X 2 =

(P3.5)

The nonzero solution of Eq. (P3.5) results from annulling the determinant of that equations determinant, which yields the following natural frequencies: = ωn1

k1 = ; ωn 2 m

k1 + 2k2 m

(P3.6)

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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

29

k1 =2 × k1* =2 ×

w3 h 3 12 =2 Ew h 3 3 l1 l1

12 E

(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

(k )

2× k2 =

* 2 2

2k2*

w3 h 12 E 12 EI Ew3 h = k2* =3 z = 3 12 =3 l2 l2 l2

(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: m ' = m + 2 ( m1 + m2 ) = m + 2 ×

13 ρ wh ( l1 + l2 ) 35

(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: X1 k2 = X 2 k1 + k2 − mω 2

(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:  X1  1 =     X 2 ω =ωn1 1

(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:  X1  −1 =     X 2 ω =ωn 2 1 

(P3.12)

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30

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:  2   X   X1   2  1 =   = ;    X 2 ω ω=  2   X 2 ω ωn 2 = n1  2 

 2 −   2     2   2 

(P3.13)

The mass and stiffness matrices are expressed from Eqs. (P3.3) as: k + k m 0  = ; [K ]  1 2 [ M ] =   0 m  − k2

− k2  k1 + k2 

(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.

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.17 (a) The free-body diagrams of the two rods are shown in Fig. P3.1.

θ1

θ2 fe

fe f

fd1

fd2 m1g

Figure P3.1

m2g

Free-body diagrams of two-pendulum mechanical system

Newton’s second law of motion is applied in the rotation variant, which yields the equations: m1 ( 2l )2 θ1 = − f d 1 ( 2l ) − m1 g ( 2l ) sin θ1 + f el cos θ1  2 − f d 2 ( 2l ) − m2 g ( 2l ) sin θ 2 − f el cos θ 2 + f ( 2l ) cos θ 2 m2 ( 2l ) θ2 =

(P3.1)

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

= f d 1 c ( 2l= l ) θ2 ; f e k ( l sin θ 2 − l sin θ1 ) )θ1; f d 2 c ( 2=

(P3.2)

By using the small-displacement approximations: cos θ ≈ 1; sin θ ≈ θ

(P3.3)

in conjunction with Eqs. (P3.2), Eqs. (P3.1) can be written in the form: 4m1lθ1 + 4lcθ1 + ( 2m1 g + kl ) θ1 − klθ 2 = 0  2f 4m2lθ2 + 4lcθ2 − klθ1 + ( 2m2 g + kl ) θ 2 =

(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:

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

32

θ1 = 0 −220θ1 − 101.12θ1 + 95θ 2 =   −91.67θ2 + 39.58θ1 − 45.71θ 2 + 78.13sin(10t ) θ 2 =

(P3.5)

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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

33

θ1

θ2

Figure P3.3

Simulink® time response of the two-pendulum mechanical system

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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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: x1 = R2θ x2

fe2

m2

fe2

m1 R1

x1

(P3.1)

fd m3

R2 fe1

Figure P3.1

f

θ

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: Jθ f e 2 R1 − f e1 R2 =  x2 =f − f e 2 − f d m3 

(P3.2)

where: 1  J m1 R12 + m2 R22 ) = (  2  = f k1 R2θ 1 x1  e1 k=  f e 2 k2 ( x2 − R1θ ) =   f d = cx2

(P3.3)

Substitution of Eqs. (P3.3) into Eqs. (P3.2) results in: 1 2 2 2 2 0  ( m1 R1 + m2 R2 ) θ + ( k1 R2 + k2 R1 ) θ − k2 R1 x2 = 2 m3  x2 + cx2 − k2 R1θ + k2 x2 = f

(P3.4)

Equations (P3.4) can be written with the numerical values of the problem as: θ = −258θ + 2963 x2 = 0  x2 = −64 x2 − 200 x2 + 3θ + 2 f  

(P3.5)

The forcing input can be specified by means of the Signal Builder block in the Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

35

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.

pr2_28/Signal Builder : Group 1 Signal 1

5

4

3

2

1

0 1

0

2

Figure P3.2

3

4

5 Time (sec)

6

7

8

9

10

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.

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

36

Figure P3.3

Simulink® diagram of the pulley-mass mechanical system

(a)

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

37

(b)

Figure P3.4

Simulink® time response: (a) pulley rotation angle θ(t); (b) linear-motion body displacement x2(t)

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.19 The total energy of the system shown in Fig. P3.1 is: 1 1 1 1 1 2 E =T + U = Jθ 2 + m3 x22 + k1 x12 + k2 ( x2 − R1θ ) + k3 x22 2 2 2 2 2 1 1 1 1 1 2 = m1 R12 + m2 R22 ) θ 2 + m3 x22 + k1 R22θ 2 + k2 ( x2 − R1θ ) + k3 x22 ( 4 2 2 2 2

(P3.1)

x2

k1

k3

k2 m2

m3

m1 R1

x1

R2 θ

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: 1  m1 R12 + m2 R22 ) θ + ( k1 R22 + k2 R12 ) θ − k2 R1 x2  ( 2  x2 − k2 R1θ + ( k2 + k3 ) x2  = 0 + x2  m3 

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: 1 2 2 2 2 0  ( m1 R1 + m2 R2 ) θ + ( k1 R2 + k2 R1 ) θ − k2 R1 x2 = 2 m3  0  x2 − k2 R1θ + ( k2 + k3 ) x2 =

(P3.3)

The solution to this mathematical model is of sinusoidal form: θ = Θ sin (ωt )   x2 = X 2 sin (ωt )

(P3.4)

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

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

39

 1 2 2 2 2 2   − 2 ( m1 R1 + m2 R2 ) ω + k1 R2 + k2 R1  Θ − k2 R1 X 2 = 0   −k R Θ + ( −m ω 2 + k + k ) X = 0 3 2 3 2  2 1

(P3.5)

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

1 m1 R12 + m2 R22 ) ω 2 + k1 R22 + k2 R12 ( 2 −k2 R1

−k2 R1

=0

(P3.6)

−m3ω + k2 + k3 2

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:

θ R1 X2

=

k2 + k3 − m3ω 2 k2

(P3.7)

which, for the two natural frequencies, becomes: θ R1 = 1.375;  X 2  ω =ωn1   θ R1 = −0.145  X 2 ω = ω n2 

(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: Θ Θ1 = = 91.6667   X 2 ω =ω1 X 21  2 2  Θ1 + X 21 =1

(P3.9)

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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: 0.9999  V1 =   0.0109 

(P3.10)

The unit-norm eigenvector corresponding to ω2 is found by solving the system: Θ Θ = 2 = −9.6667   X 2 ω =ω2 X 22  2 2  Θ 2 + X 22 =1

(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:  0.9947  V2 =   −0.1029 

(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: 1   2 2 2 2 0   2 ( m1 R1 + m2 R2 ) 0   θ  +  k1 R2 + k2 R1 −k2 R1   θ  = (P3.13)        x2   −k2 R1 k2 + k3   x2  0  m3  0 

where the mass (inertia) matrix [M] and the stiffness matrix [K] are: 1   k1 R22 + k2 R12 −k2 R1  m1 R12 + m2 R22 ) 0  (  ; [K ]  = 2 [ M ] =   −k2 R1 k 2 + k3   0 m3  

(P3.14)

The dynamic matrix [D] is obtained by using MATLAB®’s symbolic calculation capability as:  2 ( k1 R22 + k2 R12 )  2k2 R1   − −1 m1 R12 + m2 R22 m1 R12 + m2 R22   M ] [K ] = [ D] [=   k 2 + k3 kR   − 2 1 m3 m3  

(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:

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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v = -0.9999

0.9957

-0.0109

-0.0923

225.4067

0

0

532.3711

d =

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.

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 3

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Problem 3.20 (a) Figure P3.1 highlights the relevant angles of rotation.

c

N2 θ1

θ2 N3

J2 ma

c

k J3

J1 N1

J4

ml

N4

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:   N  2  N2   2  J1 + J 2  θ = ma − cθ1 − k (θ1 − θ 2 ) 1 N1  N1    2    N 3   − N 3 m − cθ − k (θ − θ )    J 4 + J3  θ2 = l 2 2 1 N4    N 4 

(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:

θ1 = −17,778θ1 − 177,780θ1 + 177,780θ 2 + 3.33 ×106   −14,693θ2 + 146,930θ1 − 146,930θ 2 − 2.5187 ×105 θ 2 =

(P3.2)

The Simulink® diagram and time response curves are shown in Figs. P3.2, P3.3, and P3.4.

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Figure P3.2

Simulink® diagram of the gear mechanical system

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(a)

Figure P3.3

(b)

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

(a) Figure P3.4

(b)

Simulink® time response of the gear mechanical system – angular velocities (in rad/s): (a) ω1; (b) ω2

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