BALANCING OF ROTATING MASSES

Transcription

1 BALANCING OF ROTATING MASSES Course Cotets. Itroductio. Static Balacig.3 Types of Balacig.4 Balacig of Several Masses Rotatig i the Same Plae.5 Dyamic Balacig.6 Balacig of Several Masses Rotatig i the differet Plaes.7 Balacig Machies Darsha Istitute of Egieerig & Techology, Rajkot Page.

2 . Balacig of Rotatig Masses Dyamics of Machiery (690). Itroductio Ofte a ubalace of forces is produced i rotary or reciprocatig machiery due to the iertia forces associated with the movig masses. Balacig is the process of desigig or modifyig machiery so that the ubalace is reduced to a acceptable level ad if possible is elimiated etirely. Fig.. A particle or mass movig i a circular path experieces a cetripetal acceleratio ad a force is required to produce it. A equal ad opposite force actig radially outwards acts o the axis of rotatio ad is kow as cetrifugal force [Fig..(a)]. This is a disturbig force o the axis of rotatio, the magitude of which is costat but the directio chages with the rotatio of the mass. I a revolvig rotor, the cetrifugal force remais balaced as log as the cetre of the mass of the rotor lies o the axis of the shaft. Whe the cetre of mass does ot lie o the axis or there is a eccetricity, a ubalaced force is produced [Fig..(b)]. This type of ubalace is very commo. For example, i steam turbie rotors, egie crakshafts, rotary compressors ad cetrifugal pumps. Most of the serious problems ecoutered i high-speed machiery are the direct result of ubalaced forces. These forces exerted o the frame by the movig machie members are time varyig, impart vibratory motio to the frame ad produce oise. Also, there are huma discomfort ad detrimetal effects o the machie performace ad the structural itegrity of the machie foudatio. The most commo approach to balacig is by redistributig the mass which may be accomplished by additio or removal of mass from various machie members. There are two basic types of ubalace-rotatig ubalace ad reciprocatig ubalace which may occur separately or i combiatio.. Static Balacig: A system of rotatig masses is said to be i static balace if the combied mass cetre of the system lies o the axis of rotatio. Page. Darsha Istitute of Egieerig & Techology, Rajkot

3 Dyamics of Machiery (690). Balacig of Rotatig Masses.3 Types of Balacig: There are mai two types of balacig coditios (i) Balacig of rotatig masses (ii) Balacig of reciprocatig masses (i) Balacig of Rotatig Masses Wheever a certai mass is attached to a rotatig shaft, it exerts some cetrifugal force, whose effect is to bed the shaft ad to produce vibratios i it. I order to prevet the effect of cetrifugal force, aother mass is attached to the opposite side of the shaft, at such a positio so as to balace the effect of the cetrifugal force of the first mass. This is doe i such a way that the cetrifugal forces of both the masses are made to be equal ad opposite. The process of providig the secod mass i order to couteract the effect of the cetrifugal force of the first mass is called balacig of rotatig masses. The followig cases are importat from the subject poit of view:. Balacig of a sigle rotatig mass by a sigle mass rotatig i the same plae.. Balacig of differet masses rotatig i the same plae. 3. Balacig of differet masses rotatig i differet plaes..4 Balacig of Several Masses Rotatig i the Same Plae Cosider ay umber of masses (say four) of magitude m, m, m 3 ad m 4 at distaces ofr, r, r 3 ad r 4 from the axis of the rotatig shaft. Let,, 3 ad 4 be the agles of these masses with the horizotal lie OX, as show i Fig.. (a). Let these masses rotate about a axis through O ad perpedicular to the plae of paper, with a costat agular velocity of rad/s. (a) Space diagram. (b) Vector diagram. Fig.. Balacig of several masses rotatig i the same plae. The magitude ad positio of the balacig mass may be foud out aalytically or graphically as discussed below: Darsha Istitute of Egieerig & Techology, Rajkot Page.3

4 . Balacig of Rotatig Masses Dyamics of Machiery (690). Aalytical method Each mass produces a cetrifugal force actig radially outwards from the axis of rotatio. Let F be the vector sum of these forces. F m r + m r + m 3 r 3 + m 4 r 4 The rotor is said to be statically balaced if the vector sum F is zero. If F is ot zero, i.e., the rotor is ubalaced, the produce a couterweight (balace weight) of mass m c, at radius r c to balace the rotor so that m r + m r + m 3 r 3 + m 4 r 4 + m c r c 0 m r + m r + m 3 r 3 + m 4 r 4 + m c r c 0 The magitude of either m c or r c may be selected ad of other ca be calculated. I geeral, if mr is the vector sum of m.r, m.r, m 3.r 3, m 4.r 4, etc., the mr + m c r c 0 To solve these equatio by mathematically, divide each force ito its x ad z compoets, mrcos + m c r c cos c 0 ad mrsi + m c r c si c 0 m c r c cos c mrcos (i) ad m c r c si c mrsi...(ii) Squarig ad addig (i) ad (ii), Dividig (ii) by (i), m c r c mr cos ² + mr si ² tac mr si mr cos The sigs of the umerator ad deomiator of this fuctio idetify the quadrat of the agle.. Graphical method First of all, draw the space diagram with the positios of the several masses, as show i Fig.. (a). Fid out the cetrifugal force (or product of the mass ad radius of rotatio) exerted by each mass o the rotatig shaft. Now draw the vector diagram with the obtaied cetrifugal forces (or the product of the masses ad their radii of rotatio), such that ab represets the cetrifugal force exerted by the mass m (or m.r ) i magitude ad directio to some suitable scale. Similarly, draw bc, cd ad de to represet cetrifugal forces of other masses m, m 3 ad m 4 (or m.r,m 3.r 3 ad m 4.r 4 ). Now, as per polygo law of forces, the closig side ae represets the resultat force i magitude ad directio, as show i Fig.. (b). Page.4 Darsha Istitute of Egieerig & Techology, Rajkot

5 Dyamics of Machiery (690). Balacig of Rotatig Masses The balacig force is, the, equal to resultat force, but i opposite directio. Now fid out the magitude of the balacig mass (m) at a give radius of rotatio (r), such that m.r. Resultat cetrifugal force or m.r Resultat of m.r, m.r, m 3.r 3 ad m 4.r 4 (I geeral for graphical solutio, vectors m.r, m.r, m 3.r 3, m 4.r 4, etc., are added. If they close i a loop, the system is balaced. Otherwise, the closig vector will be givig m c.r c. Its directio idetifies the agular positio of the coutermass relative to the other mass.) Example. :A circular disc mouted o a shaft carries three attached masses of 4 kg, 3 kg ad.5 kg at radial distaces of 75 mm, 85 mm ad 50 mm ad at the agular positios of 45, 35 ad 40 respectively. The agular positios are measured couterclockwise from the referece lie alog the x-axis. Determie the amout of the coutermass at a radial distace of 75 mm required for the static balace. m 4 kg r 75 mm 45 m 3 kg r 85 mm 35 m 3.5 kg r 3 50 mm 3 40 m r 4 x kg.mm m r 3 x kg.mm m 3 r 3.5 x 50 5 kg.mm Aalytical Method: mr + m c r c cos cos cos40 + m c r c cos c 0 ad 300 si si si 40 + m c r c si c 0 Squarig, addig ad the solvig, mr C C (300 cos45 55 cos 35 5 cos40 ) (300 si45 55 si 35 5 si40 ) m 75 ( 30.68) (84.) C 85.8 kg.mm m c 3.8 kg mr si 84. tac 9.6 mr cos ( 30.68) c c lies i the fourth quadrat (umerator is egative ad deomiator is positive). c c 76 9 Darsha Istitute of Egieerig & Techology, Rajkot Page.5

6 . Balacig of Rotatig Masses Dyamics of Machiery (690) Graphical Method: The magitude ad the positio of the balacig mass may also be foud graphically as discussed below : Now draw the vector diagram with the above values, to some suitable scale, as show i Fig..3. The closig side of the polygo co represets the resultat force. By measuremet, we fid that co kg-mm. Fig..3 Vector Diagram The balacig force is equal to the resultat force. Sice the balacig force is proportioal to m.r, therefore m C 75 vector co kg-mm or m C 85.84/75 m C 3.8 kg. By measuremet we also fid that the agle of icliatio of the balacig mass (m) from the horizotal or positive X-axis, θ C 76. Example. :Four masses m, m, m 3 ad m 4 are 00 kg, 300 kg, 40 kg ad 60 kg respectively. The correspodig radii of rotatio are 0. m, 0.5 m, 0.5 m ad 0.3 m respectively ad the agles betwee successive masses are 45, 75 ad 35. Fid the positio ad magitude of the balace mass required, if its radius of rotatio is 0. m. m 00 kg r 0. m 0 m 300 kg r 0.5 m 45 m 3 40 kg r m m 4 60 kg r m m r 00 x r C 0. m m r 300 x m 3 r 3 40 x m 4 r 4 60 x mr + m c r c 0 40 cos0 + 45cos cos0 + 78cos55 + m c r c cos c 0 ad 40 si si si si 55 + m c r c si c 0 Page.6 Darsha Istitute of Egieerig & Techology, Rajkot

7 Dyamics of Machiery (690). Balacig of Rotatig Masses Squarig, addig ad the solvig, mr C C (40cos0 45 cos45 60 cos 0 78 cos55 ) (40si0 45 si45 60 si 0 78 si55 ) m 0. (.6) (8.5) C 3. kg.mm m c 6 kg mr si 8.5 tac mr cos.6 c 8 c lies i the third quadrat (umerator is egative ad deomiator is egative). c c 0 8 Graphical Method: For graphical method draw the vector diagram with the above values, to some suitable scale, as show i Fig..4. The closig side of the polygo ae represets the resultat force. By measuremet, we fid that ae 3 kg-m. Fig..4 Vector Diagram The balacig force is equal to the resultat force.sice the balacig force is proportioal to m.r, therefore m 0. vector ea 3 kg-m or m C 3/0. m C 5 kg. By measuremet we also fid that the agle of icliatio of the balacig mass (m) from the horizotal or positive X-axis, θ C 0. Darsha Istitute of Egieerig & Techology, Rajkot Page.7

8 . Balacig of Rotatig Masses Dyamics of Machiery (690).5 Dyamic Balacig Whe several masses rotate i differet plaes, the cetrifugal forces, i additio to beig out of balace, also form couples. A system of rotatig masses is i dyamic balace whe there does ot exist ay resultat cetrifugal force as well as resultat couple. I the work that follows, the products of mr ad mrl (istead of mr ad mrl ), usually, have bee referred as force ad couple respectively as it is more coveiet to draw force ad couple polygos with these quatities. Fig..5 If m, ad m are two masses (Fig..5) revolvig diametrically opposite to each other i differet plaes such that m r m r, the cetrifugal forces are balaced, but a ubalaced couple of magitude m r l ( m r l) is itroduced. The couple acts i a plae that cotais the axis of rotatio ad the two masses. Thus, the couple is of costat magitude but variable directio..6 Balacig of Several Masses Rotatig i the differet Plaes Let there be a rotor revolvig with a uiform agular velocity. m, m ad m 3 are the masses attached to the rotor at radii r, r ad r 3 respectively.the masses m, m ad m 3 rotate i plaes, ad 3 respectively. Choose a referece plae at O so that the distaces of the plaes, ad 3 from O are l, l ad l 3 respectively. Trasferece of each ubalaced force to the referece plae itroduces the like umber of forces ad couples. The ubalaced forces i the referece plae are m r, m r ad m 3 r 3 actig radially outwards. The ubalaced couples i the referece plae are m r l, m r l ad m 3 r 3 l 3 which may be represeted by vectors parallel to the respective force vectors, i.e., parallel to the respective radii of m, m ad m 3. For complete balacig of the rotor, the resultat force ad resultat couple both should be zero, i.e., m r + m r + m 3 r 3 0 (a) ad m r l + m r l + m 3 r 3 l (b) If the Eqs (a) ad (b) are ot satisfied, the there are ubalaced forces ad couples. A mass placed i the referece plae may satisfy the force equatio but Page.8 Darsha Istitute of Egieerig & Techology, Rajkot

9 Dyamics of Machiery (690). Balacig of Rotatig Masses the couple equatio is satisfied oly by two equal forces i differet trasverse plaes. Thus i geeral, two plaes are eeded to balace a system of rotatig masses. Therefore, i order to satisfy Eqs (a) ad (b), itroduce two couter-masses m C ad m C at radii r C ad r C respectively. The Eq. (a) may be writte as m r + m r + m 3 r 3 + m C r C + m C r C 0 m r + m r + m 3 r 3 + m C r C + m C r C 0 mr + m C r C + m C r C 0.(c) Let the two coutermasses be placed i trasverse plaes at axial locatios O ad Q, i.e., the coutermassm C be placed i the referece plae ad the distace of the plae of m C be l C from the referece plae. Equatio (b) modifies to (takig momets about O) m r l + m r l + m 3 r 3 l 3 + m C r C l C 0 m r l + m r l + m 3 r 3 l 3 + m C r C l C 0 mrl + m C r C l C 0 (d) Thus, Eqs (c) ad (d) are the ecessary coditios for dyamic balacig of rotor. Agai the equatios ca be solved mathematically or graphically. Dividig Eq. (d) ito compoet form Squarig ad addig (i) ad (ii) Dividig (ii) by (i), mrlcos + m C r C l C cos C 0 mrl si + m C r C l C si C 0 m C r C l C cos C mrlcos m C r C l C si C mrl si m C r C l C mrl cos ² + mrl si ² ta C mrl si mrl cos (i)..(ii) After obtaiig the values of m C ad C from the above equatios, solve Eq. (c) by takig its compoets, mrcos +m C r C cos C + m C r C cos C 0 mrsi +m C r C si C + m C r C si C 0 m C r C cos C ( mrcos + m C r C cos C ) m C r C si C ( mrsi + m C r C si C ) m C r C mr cos + m C r C cos C ² + mr si + m C r C si C ² ta C mr si + m Cr C si C mr cos + m C r C cos C Darsha Istitute of Egieerig & Techology, Rajkot Page.9

10 . Balacig of Rotatig Masses Dyamics of Machiery (690) Example.3 : A shaft carries four masses A, B, C ad D of magitude 00 kg, 300 kg, 400 kg ad 00 kg respectively ad revolvig at radii 80 mm, 70 mm, 60 mm ad 80 mm i plaes measured from A at 300 mm, 400 mm ad 700 mm. The agles betwee the craks measured aticlockwise are A to B 45, B to C 70 ad C to D 0. The balacig masses are to be placed i plaes X ad Y. The distace betwee the plaes A ad X is 00 mm, betwee X ad Y is 400 mm ad betwee Y ad D is 00 mm. If the balacig masses revolve at a radius of 00 mm, fid their magitudes ad agular positios. m A 00 kg r A 80 mm A 0 l A -00 mm m B 300 kg r B 70 mm B 45 l B 00 mm m C 400 kg r C 60 mm C l C 300 mm m D 00 kg r D 80 mm D l D 600 mm r X r Y 00 mm l Y 400 mm Let m X Balacig mass placed i plae X, ad m Y Balacig mass placed i plae Y. The positio of plaes ad agular positio of the masses (assumig the mass A as horizotal) are show i Fig..5 (a) ad (b) respectively. Assume the plae X as the referece plae (R.P.). The distaces of the plaes to the right of plae X are take as + ve while the distaces of the plaes to the left of plae X are take as ve. (a) Positio of plaes. Fig..6 (b) Agular positio of masses. m A r A l A 00 x 0.08 x (-0.) -.6 kg.m m B r B l B 300 x 0.07 x kg.m m C r C l C 400 x 0.06 x kg.m m D r D l D 00 x 0.08 x kg.m m A r A 00 x kg.m m B r B 300 x 0.07 kg.m m C r C 400 x kg.m m D r D 00 x kg.m Page.0 Darsha Istitute of Egieerig & Techology, Rajkot

11 Dyamics of Machiery (690). Balacig of Rotatig Masses Aalytical Method: For ubalaced couple mrl + m Y r Y l Y 0 m r l ( mrl cos ) ( mrl si ) Y Y Y my ry ly (.6cos0 4.cos45 7.cos5 9.6cos35 ) (.6si0 4.si45 7.si5 9.6si35 ) m r l ( 7.79) (.63) Y Y Y m Y m Y 84 kg. mrl si.63 tay 0.7 mrl cos ( 7.79) Y 47 Y lies i the fourth quadrat (umerator is egative ad deomiator is positive). Y Y 347 For ubalaced cetrifugal force mr +m X r X + m Y r Y 0 m r ( mr cos m r cos ) ( mr si m r si ) X X Y Y Y Y Y Y mr X X (6cos0 cos45 4cos5 6cos35 8.4cos347 ') (6si0 si45 4si5 6si35 8.4si347') mr (9.47) (9.4) X X m X m X 353 kg. mr si 9.4 ta X mr cos 9.47 X 33 X lies i the third quadrat (umerator is egative ad deomiator is egative). X X 3 Graphical Method: The balacig masses ad their agular positios may be determied graphically as discussed below : Darsha Istitute of Egieerig & Techology, Rajkot Page.

12 . Balacig of Rotatig Masses Dyamics of Machiery (690) Table. Plae Agle Mass (m) Radius Cet.force Distace from Couple kg (r)m (mr) kg-m Ref. Plae (l) m (mrl) kg-m A X (R.P.) X m X m X 0 0 B C Y Y m Y m Y m Y D First of all, draw the couple polygo from the data give i Table. (colum 7) as show i Fig..7 (a) to some suitable scale. The vector d o represets the balaced couple. Sice the balaced couple is proportioal to 0.04 m Y, therefore by measuremet, 0.04m Y vector d o 73 kg-m or m Y 8.5 kg The agular positio of the mass m Y is obtaied by drawig Om Y i Fig..6 (b), parallel to vector d o. By measuremet, the agular positio of m Y is θ Y i the clockwise directio from mass m A (i.e. 00 kg), so θ Y (a) Couple Polygo Fig..7 (b) Force Polygo Now draw the force polygo from the data give i Table. (colum 5) as show i Fig..7 (b). The vector eo represets the balaced force. Sice the balaced force is proportioal to 0. m X, therefore by measuremet, 0.m X vector eo 35.5 kg-m or m X 355 kg. The agular positio of the mass m X is obtaied by drawig Om X i Fig..6 (b), parallel to vector eo. By measuremet, the agular positio of m X is θ X 45 i the clockwise directio from mass m A (i.e. 00 kg), so θ X Page. Darsha Istitute of Egieerig & Techology, Rajkot

13 Dyamics of Machiery (690). Balacig of Rotatig Masses Example.4: Four masses A, B, C ad D carried by a rotatig shaft are at radii 00, 40, 0 ad 60 mm respectively. The plaes i which the masses revolve are spaced 600 mm apart ad the masses of B, C ad D are 6 kg, 0 kg ad 8 kg respectively. Fid the required mass A ad the relative agular positios of the four masses so that shaft is i complete balace. m A? r A 00mm m B 6 kg r B 40mm l B 600 mm m C 0 kg r C 0mm l C 00 mm m D 8 kg r D 60 mm l D 800 mm Table. Plae Agle Mass (m) Radius Cet.force Distace from Couple kg (r) m (mr) kg-m Ref. Plae (l) m (mrl) kg-m A (R.P.) A m A 0. 0.m A 0 0 B C C D D First of all, draw the couple polygo from the data give i Table. (colum 7) as show i Fig..8 (a) to some suitable scale. By measuremet, the agular positio of m C is θ C 5 i the aticlockwise directio from mass m B ad the agular positio of m D is θ D 63 i the aticlockwise directio from mass m B. (a) Couple Polygo Fig..8 (b) Force Polygo Now draw the force polygo from the data give i Table. (colum 5) as show i Fig..8 (b). The vector co represets the balaced force. Sice the balaced force is proportioal to 0. m A, therefore by measuremet, 0.m A vectorco.36 kg-m Or m A 3.6 kg. By measuremet, the agular positio of m A is θ A 08 i the aticlockwise directio from mass m B (i.e. 6 kg). Darsha Istitute of Egieerig & Techology, Rajkot Page.3

14 . Balacig of Rotatig Masses Dyamics of Machiery (690) Example.5 :Four masses 50 kg, 00 kg, 00 kg ad 50 kg are attached to a shaft revolvig at radii 50 mm, 00 mm, 00 mm ad 50 mm; i plaes A, B, C ad D respectively. The plaes B, C ad D are at distaces 350 mm, 500 mm ad 800 mm from plae A. The masses i plaes B, C ad D are at a agle 05, 00 ad 300 measured aticlockwise from mass i plae A. It is required to balace the system by placig the balacig masses i the plaes P ad Q which are midway betwee the plaes A ad B, ad betwee C ad D respectively. If the balacig masses revolve at radius 80 mm, fid the magitude ad agular positios of the balace masses. m A 50 kg r A 50mm A 0 m B 00 kg r B 00mm B 05 m C 00 kg r C 00mm C 00 m D 50 kg r D 50 mm D 300 r X r Y 80 mm Fig..9 Table.3 Plae Agle Mass (m) Radius Cet.force Distace from Couple kg (r) m (mr) kg-m Ref. Plae (l) m (mrl) kg-m A (R.P.) P P m P m P 0 0 B C Q Q m Q m Q m Q D Aalytical Method: Table.4 mrlcos ( H C ) mrl si ( V C ) mrcos ( H F ) mr si (V F ) m P cos P 0.8 m P si P m Q cos Q m Q si Q 0.8 m Q cos Q 0.8 m Q si Q Page.4 Darsha Istitute of Egieerig & Techology, Rajkot

15 Dyamics of Machiery (690). Balacig of Rotatig Masses H C m Q cos Q m Q cos Q 0.73 m Q cos Q (i) V C m Q si Q m Q si Q 8.8 m Q si Q (ii) m ( 5.497) (39.59) Q m Q kg. mqsiq m cos Q Q ta Q.66 Q 69.5 Q Q 0.84 H F m P cos P m Q cos Q m P cos P (35.67) cos m P cos P.46 m P cos P 63.4 V F m P si P m Q si Q m P si P (35.67) si m P si P m P si P 4.54 m ( 63.4) ( 4.54) P m P 33.3 kg. mpsip 4.54 m cos 63.4 P P ta P 3.54 P 74.3 P P 54.3 Darsha Istitute of Egieerig & Techology, Rajkot Page.5

16 . Balacig of Rotatig Masses Dyamics of Machiery (690) Graphical Method : (a) Couple Polygo Fig..0 (b) Force Polygo First of all, draw the couple polygo from the data give i Table.4 (colum 7) as show i Fig..0 (a) to some suitable scale. The vector do represets the balaced couple. Sice the balaced couple is proportioal to m Q, therefore by measuremet, m Q vector do 30.5 kg-m or m Q kg. By measuremet, the agular positio of m Q is Q i the aticlockwise directio from mass m A (i.e. 50 kg). Now draw the force polygo from the data give i Table.4 (colum 5) as show i Fig..0 (b). The vector eo represets the balaced force. Sice the balaced force is proportioal to 0.8 m P, therefore by measuremet, 0.8 m P vector eo 4.5 kg-m Or m P 30.5 kg. By measuremet, the agular positio of m P is θ P 56 i the aticlockwise directio from mass m A (i.e. 50kg). Page.6 Darsha Istitute of Egieerig & Techology, Rajkot

17 Dyamics of Machiery (690). Balacig of Rotatig Masses Example.6 : A shaft carries four masses i parallel plaes A, B, C ad D i this order alog its legth. The masses at B ad C are 8 kg ad.5 kg respectively, ad each has a eccetricity of 60 mm. The masses at A ad D have a eccetricity of 80 mm. The agle betwee the masses at B ad C is 00 ad that betwee the masses at B ad A is 90, both beig measured i the same directio. The axial distace betwee the plaes A ad B is 00 mm ad that betwee B ad C is 00 mm. If the shaft is i complete dyamic balace, determie:. The magitude of the masses at A ad D;. The distace betwee plaes A ad D; ad 3. The agular positio of the mass at D. m A? r A 80 mm A 90 m B 8 kg r B 60 mm B 0 m C.5 kg r C 60 mm C 00 m D? r D 80 mm D? X Distace betwee plaes A ad D. (a) Positio of plaes. (b) Agular positio of masses. Fig.. Plae The positio of the plaes ad agular positio of the masses is show i Fig.. (a) ad (b) respectively. The positio of mass B is assumed i the horizotal directio, i.e. alog OB. Takig the plae of mass A as the referece plae, the data may be tabulated as below: Agle Mass (m) kg Radius (r) m Table.5 Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m A (R.P.) 90 m A m A 0 0 B C D D m D m D X 0.08 m D X Darsha Istitute of Egieerig & Techology, Rajkot Page.7

18 . Balacig of Rotatig Masses Dyamics of Machiery (690) First of all, draw the couple polygo from the data give i Table.5 (colum 7) as show i Fig.. (a) to some suitable scale. The closig side of the polygo (vector c o ) is proportioal to 0.08 m D.X, therefore by measuremet, 0.08 m D X vector c o 0.35 kg-m.(i) By measuremet, the agular positio of m D is D 5 i the aticlockwise directio from mass m B (i.e. 8 kg). (a) Couple Polygo Fig.. (b) Force Polygo Now draw the force polygo, to some suitable scale, as show i Fig.. (b), from the data give i Table.5 (colum 5), as discussed below : i. Draw vector ob parallel to OB ad equal to.08 kg-m. ii. From poit b, draw vector bc parallel to OC ad equal to 0.75 kg-m. iii. For the shaft to be i complete dyamic balace, the force polygo must be a closed. Therefore from poit c, draw vector cd parallel to OA ad from poit o draw vector od parallel to OD. The vectors cd ad od itersect at d. Sice the vector cd is proportioal to 0.08 m A, therefore by measuremet 0.08 m A vector cd 0.77 kg-m or m A 9.65 kg. ad vector do is proportioal to 0.08 m D, therefore by measuremet, 0.08 m D vector do 0.65 kg-m or m D 8.5 kg. Distace betwee plaes A ad D From equatio (i), 0.08 m D.X 0.35 kg-m X 0.35 kg-m X m 36.5 mm Page.8 Darsha Istitute of Egieerig & Techology, Rajkot

19 Dyamics of Machiery (690). Balacig of Rotatig Masses Example.7 : A rotatig shaft carries four masses A, B, C ad D which are radially attached to it. The mass ceters are 30 mm, 40 mm, 35 mm ad 38 mm respectively from the axis of rotatio. The masses A, C ad D are 7.5 kg, 5 kg ad 4 kg respectively. The axial distaces betwee the plaes of rotatio of A ad B is 400 mm ad betwee B ad C is 500 mm. The masses A ad C are at right agles to each other. Fid for a complete balace, (i) the agles betwee the masses B ad D from mass A, (ii) the axial distace betwee the plaes of rotatio of C ad D, ad (iii) the magitude of mass B. Fig..3 Positio of plaes Table.6 Plae Agle Mass (m) Radius Cet.force Distace from Couple kg (r) m (mr) kg-m Ref. Plae (l) m (mrl) kg-m A B(R.P.) B m B m B 0 0 C D D X 0.5X (a) Couple Polygo Fig..4 (b) Force Polygo First of all, draw the couple polygo from the data give i Table.6 (colum 7) as show i Fig..4 (a) to some suitable scale. The vector bo represets the balaced couple. Sice the balaced couple is proportioal to 0.5X, therefore by measuremet, Darsha Istitute of Egieerig & Techology, Rajkot Page.9

20 . Balacig of Rotatig Masses Dyamics of Machiery (690) 0.5X vector bo 0.3 kg-m or X m. The axial distace betwee the plaes of rotatio of C ad D mm By measuremet, the agular positio of m D is D i the aticlockwise directio from mass m A (i.e. 7.5 kg). Now draw the force polygo from the data give i Table.6 (colum 5) as show i Fig..4 (b). The vector co represets the balaced force. Sice the balaced force is proportioal to 0.04 m B, therefore by measuremet, 0.04 m B vector co 0.34 kg-m or m B 8.5 kg. By measuremet, the agular positio of m B is θ B i the aticlockwise directio from mass m A (i.e. 7.5 kg). Example.8: The four masses A, B, C ad D revolve at equal radii are equally spaces alog the shaft. The mass B is 7 kg ad radii of C ad D makes a agle of 90 ad 40 respectively (couterclockwise) with radius of B, which is horizotal. Fid the magitude of A, C ad D ad agular positio of A so that the system may be completely balace. Solve problem by aalytically. Table.7 Plae Agle Mass (m) kg Radius (r) m Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m A (R.P.) A m A X m A 0 0 B 0 7 X 7 Y 7Y C 90 m C X m C Y m C Y D 40 m D X m D 3Y 3m D Y mrlcos ( H C ) mrl si ( V C ) mrcos ( H F ) mr si (V F ) 0 0 m A cos A m A si A 7Y m C Y 0 m C.5m D Y.59m D Y 0.5m D 0.866m D H C Y + 0.5m D Y 0 m D 7/.5 m D 4.67 kg Page.0 Darsha Istitute of Egieerig & Techology, Rajkot

21 Dyamics of Machiery (690). Balacig of Rotatig Masses V C m C Y.59m D Y 0 m C kg H F 0 m A cos A m D 0 m A cos A V F 0 m A si A m C 0.866m D 0 m A si A.0078 m ( 4.665) (.0078) A m A kg masi A.0078 ta A 0.43 m cos θ A 3.3 θ A θ A Balacig Machies A A A balacig machie is able to idicate whether a part is i balace or ot ad if it is ot, the it measures the ubalace by idicatig its magitude ad locatio..7.. Static Balacig Machies Static balacig machies are helpful for parts of small axial dimesios such as fas, gears ad impellers, etc., i which the mass lies practically i a sigle plae. There are two machie which are used as static balacig machie: Pedulum type balacig machie ad Cradle type balacig machie. (i) Pedulum type balacig machie Pedulum type balacig machie as show i Figure.5 is a simple kid of static balacig machie. The machie is of the form of a weighig machie. Oe arm of the machie has a madrel to support the part to be balaced ad the other arm supports a suspeded deadweight to make the beam approximately horizotal. The madrel is the rotated slowly either by had or by a motor. As the madrel is rotated, the beam will oscillate depedig upo the ubalace of the part. If the ubalace is represeted by a mass m at radius r, the apparet weight is greatest whe m is at the positio I ad least whe it is at B as the legths of the arms i the two cases will be maximum ad miimum. Darsha Istitute of Egieerig & Techology, Rajkot Page.

22 . Balacig of Rotatig Masses Dyamics of Machiery (690) A calibrated scale alog with the poiter ca also be used to measure the amout of ubalace. Obviously, the poiter remais statioary i case the body is statically balaced. (ii) Cradle type balacig machie Fig..5 Cradle type balacig machie as show i fig..6 is more sesitive machie tha the pedulum type balacig machie. It cosists of a cradle supported o two pivots P-P parallel to the axis of rotatio of the part ad held i positio by two sprigs S-S. The part to be tested is mouted o the cradle ad is flexibly coupled to a electric motor. The motor is started ad the speed of rotatio is adjusted so that it coicides with the atural frequecy of the system. Thus, the coditio of resoace is obtaied uder which eve a small amout of ubalace geerates large amplitude of the cradle. The momet due to ubalace (mr cos θ).l where is the agular velocity of rotatio. Its maximum value is mr l. If the part is i static balace but dyamic ubalace, o oscillatio of the cradle will be there as the pivots are parallel to the axis of rotatio. Fig..6 Page. Darsha Istitute of Egieerig & Techology, Rajkot

23 Dyamics of Machiery (690). Balacig of Rotatig Masses.7.. Dyamic Balacig Machies For dyamic balacig of a rotor, two balacig or coutermasses are required to be used i ay two coveiet plaes. This implies that the complete ubalace of ay rotor system ca be represeted by two ubalaces i those two plaes. Balacig is achieved by additio or removal of masses i these two plaes, whichever is coveiet. The followig is a commo type of dyamic balacig machie. Pivoted-cradle Balacig Machie Fig.7 shows a pivot cradle type dyamic balacig machie. Here, part which is required to be balaced is to be mouted o cradle supported by supported rollers ad it is coected to drive motor through uiversal couplig. Two plaes are selected for dyamic balacig as show i fig..7 where pivots are provided about which the cradle is allowed to oscillate. As show i fig.7, right pivot is released coditio ad left pivot is i locked positio so as to allow the cradle ad part to oscillate about the pivot. At the both eds of the cradle, the sprig ad dampers are attached such that the atural frequecy ca be adjusted ad made equal to the motor speed. Two amplitude idicators are attached at each ed of the cradle. The permaet maget is mouted o the cradle which moves relative to statioary coil ad geerates a voltage which is directly proportioal to the ubalaced couple. This voltage is amplified ad read from the calibrated voltmeter ad gives output i terms of kg-m. Whe left pivot is locked, the ubalaced i the right correctio plae will cause vibratio whose amplitude is measured by the right amplitude idicator. After that right pivot is locked ad aother set of measuremet is made for left had correctio plae usig the amplitude idicator of the left had side. Fig..7 Darsha Istitute of Egieerig & Techology, Rajkot Page.3

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25 DYNANICS OF RECIPROCATING ENGINES Course Cotets. Slider Crak Kiematics (Aalytical). Gas Force ad Gas Torque.3 Iertia ad Shakig Forces.4 Iertia ad Shakig Torques.5 Dyamically Equivalet Systems.6 Pi Forces i the Sigle Cylider Egie.7 Balacig of ubalaced forces i reciprocatig masses.8 Balacig of Locomotives.9 Balacig of Multi Cylider Egie.0 Balacig of V Egie. Balacig of Radial Egie Darsha Istitute of Egieerig & Techology, Rajkot Page.

26 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690). Slider Crak Kiematics (Aalytical) There are may applicatios where movig parts are havig reciprocatig motio. For example, IC egie, shaper machie, air compressors ad may more where parts are i reciprocatig motio where they are subjected to cotiuous acceleratio ad retardatio. Due to this motio, iertia force acts i the opposite directio of acceleratio of the movig parts. This opposite directio force which is termed as iertia force is the ubalaced dyamic force which is actig o the reciprocatig parts. Cosider a horizotal reciprocatig egie mechaism as show i fig... Let the crak radius be r ad the coectig rod legth be I. The agle of the crak is θ, ad the agle that the coectig rod makes with the X axis is φ. For ay costat crak agular velocity, the crak agle θ t. The istataeous pisto positio is x. Two right triagles rqs ad Iqu are costructed. The from geometry: q r si l si t r si sit l s r cost u lcos x s u r cost lcos (.) (.) cos si sit l r r x r cost l sit l (.3) (.4) Equ..4 is a exact expressio for the pisto positio x as a fuctio of r, I ad t. This ca be differetiated versus lime to obtai exact expressios for the velocity ad acceleratio of the pisto. For a steady-state aalysis we will assume to be costat. r sit x r sit l (.5) r sit l 4 r l cos t r si t x r cost 3 (.6) l r sit Page. Darsha Istitute of Egieerig & Techology, Rajkot

27 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies (a) The likage geometry (b) Free body diagrams Fig.. Applyig biomial theorem ad Fourier series rules for simplificatio, we get r r x l r cost cost (.7) 4l 4l Differetiate for velocity of the pisto (with costat ): r x rsit sit (.8) l Differetiate agai for acceleratio (with costat ): r x r cost cost (.9) l Darsha Istitute of Egieerig & Techology, Rajkot Page.3

28 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690). Gas Force ad Gas Torque The gas force is due to the gas pressure from the explodig fuel-air mixture impigig o the top of the pisto surface as show i Fig..(a). Let Fg gas force. Pg gas pressure, Ap area of pisto ad B bore of cylider, which is also equal to the pisto diameter. The: Fg Pg Ap where, Ap B 4 Fg PgB 4 The egative sig is due to the choice of egie orietatio i the coordiate system. The gas pressure P g i this expressio is a fuctio of crak agle t ad is defied by the thermodyamics of the egie. The gas torque is due to the gas force actig at a momet arm about the crak ceter i Fig.. This momet arm varies from zero to a maximum as the crak rotates..3 Iertia ad Shakig Forces The simplified lumped mass model of Fig.. ca be used to develop expressios for the forces ad torques due to the acceleratios of the masses i the system. The acceleratio for poit B is give i equatio r x r cost cost l The acceleratio of poit A i pure rotatio is obtaied by differetiatig the positio vector R A twice, assumig a costat crakshaft. which gives: R A r cost + r sit (.0) a A r cost r sit (.) The total iertia force F i is the sum of the cetrifugal (iertia) force at poit A ad the iertia force at poit B. F i m A a A m B a B (.) Breakig it ito x ad y compoets: F m r cost m x (.3) i x A B F m r sit (.4) i y A Note that oly the x compoet is affected by the acceleratio of the pisto. r Fi x ma r cost mb r cost cost l Fi y ma r sit The shakig force is defied as the sum of all forces actig o the groud plae. From the free-body diagram for lik i Fig.. r FS x ma r cost mb r cost cost l Page.4 Darsha Istitute of Egieerig & Techology, Rajkot

29 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies F m r sit i y A The shakig force F S is equal to the egative of the iertia force. F S F i (a)dyamic Model (b)free body diagrams Fig.. Darsha Istitute of Egieerig & Techology, Rajkot Page.5

30 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690).4 Iertia ad Shakig Torques The iertia torque results from the actio of the iertia forces at a momet arm. The iertia force at poit A i Fig.. has two compoets, radial ad tagetial. The radial compoet has o momet arm. The tagetial compoet has a momet arm of crak radius r. If the crak is costat, the mass at A will ot cotribute to iertia torque. The iertia force at B has a ozero compoet perpedicular to the cylider wall except whe the pisto is at TDC or BDC. As we did for the gas torque, we ca express the iertia torque i terms of the couple F i4, F i4 whose forces act always perpedicular to the motio of the slider (eglectig frictio), ad the distace x, which is their istataeous momet arm. The iertia torque is: T i F i4.x F i4.x Substitutig value of F i4 ad x, we get r r Ti mbx ta l r cos t cos t 4l 4l The shakig torque is equal to the iertia torque. T s T i..5 Dyamically Equivalet System The expressio for the turig momet of the crakshaft has bee obtaied for the et force F o the pisto. This force F may be the gas force with or without the cosideratio of iertia force actig o the pisto. As the mass of the coectig rod is also sigificat, the iertia due to the same should also be take ito accout. As either the mass of the coectig rod is uiformly distributed or the motio is liear, its iertia caot be foud as such. Usually, the iertia of the coectig rod is take ito accout by cosiderig a dyamicallyequivalet system. A dyamically equivalet system meas that the rigid lik is replaced by a lik with two poit masses i such, a way that it has the same motio as the rigid lik whe subjected to the same force, i.e., the cetre of mass of the equivalet lik has the same liear acceleratio ad the lik has the same agular acceleratio. Fig..3 Page.6 Darsha Istitute of Egieerig & Techology, Rajkot

31 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Fig..3(a) shows a rigid body of mass m with the cetre of mass at G. Let it be acted upo by a force F which produces liear acceleratio a of the cetre of mass as well as agular acceleratio of the body as the force F does ot pass through G. As we kow, F m.a ad F.e I. α Acceleratio of G, F a m Agular acceleratio of the body, Fe. I where e perpedicular distace of F from G ad I momet of iertia of the body about perpedicular axis through G Now to have the dyamically equivalet system, let the replaced massless lik [Fig..3(b)] has two poit masses m (at B ad m at D) at distaces b ad d respectively from the cetre of mass G as show i fig..3(b).. To satisfy the first coditio, as the force F is to be same, the sum of the equivalet masses m ad m has to be equal to m to have the same acceleratio. Thus, m m + m.. To satisfy the secod coditio, the umerator F.e ad the deomiator I must remai the same. F is already take same, Thus, e has to be same which meas that the perpedicular distace of F from G should remai same or the combied cetre of mass of the equivalet system remais at G. This is Possible if m b m d To have the same momet of iertia of the equivalet system about perpedicular axis through their combied cetre of mass G, we must have I m b + m d Thus, ay distributed mass ca be replaced by two poit masses to have the same dyamical properties if the followig coditios are fulfilled: (i) The sum of the two masses is equal to the total mass. (ii) The combied cetre of mass coicides with that of the rod. (iii) The momet of iertia of two poit masses about the perpedicular axis through their combied cetre of mass is equal to that of the rod..6 Pi Forces i the Sigle Cylider Egie I additio to calculatig the overall effects o the groud plae of the dyamic forces preset i the egie, we also eed to kow the magitudes of the forces at the pi joits. These forces will dictate the desig of the pis ad the bearigs at the joits. Though we were able to lump the mass due to both coectig rod ad pisto, or coectig rod ad crak at poits A ad B for a overall aalysis of the likage's effects o the groud plae, we caot do so for the pi force calculatios. Darsha Istitute of Egieerig & Techology, Rajkot Page.7

32 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) This is because the pis feel the effect of the coectig rod pullig o oe "side" ad the pisto (or crak) pullig o the other "side" of the pi. Thus we must separate the effects of the masses of the liks joied by the pis. We will calculate the effect of each compoet due to the various masses, ad the gas force ad the superpose them to obtai the complete pi force at each joit. The resultat bearig loads have the followig compoets: i. The gas force compoets. ii. The iertia force due to the pisto mass. iii. The iertia force due to the mass of the coectig rod at the wrist pi. iv. The iertia force due to the mass of the coectig rod at the crak pi. v. The iertia force due to the mass of the crak at the crak pi..7 Balacig of ubalaced forces i reciprocatig masses Acceleratio of reciprocatig mass of a slider-crak mechaism is give by cos ar cos Therefore, the force required to accelerate mass m is cos F mr cos cos F mr cos mr mr cosθ is called the primary acceleratig force ad mr cosθ/ is called the secodary acceleratig force. Maximum value of the primary force mr Maximum value of the secodary force mr / As is, usually, much greater tha uity, the secodary force is small, compared with the primary force ad ca be safely eglected for slow-speed egies. The iertia force due to primary acceleratig force is show i Fig..4(a). I Fig..4(b), the forces actig o the egie frame due to this iertia force are show. The force exerted by the crakshaft o the mai bearigs has two compoets, F h ad F v. The horizotal force F h is a ubalaced shakig force. The vertical forces F v ad F4 v balace each other, but form a ubalaced shakig couple. The magitude ad directio of this force ad couple go o chagig with the rotatio of the crak agle θ. The shakig force produces liear vibratio of the frame i the horizotal directio whereas the shakig couple produces a oscillatig vibratio. Thus, it is see that the shakig force F h is the oly ubalaced force. It may hamper the smooth ruig of the egie ad Thus, effort is made to balace the same. However, it is ot at all possible to balace it completely ad oly some modificatio ca be made. Page.8 Darsha Istitute of Egieerig & Techology, Rajkot

33 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Fig..4 The usual approach of balacig the shakig force is by additio of a rotatig coutermass at radius r directly opposite the crak which however, provides oly a partial balace. This coutermass is i additio to the mass used to balace the rotatig ubalace due to the mass at the crak pi. Fig..4(c) shows the reciprocatig mechaism with a coutermass m at the radial distace r. The horizotal compoet of the cetrifugal force due to the balacig mass is mr cosθ i the lie of stroke. This eutralizes the ubalaced reciprocatig force. But the rotatig mass also has a compoet mr siθ perpedicular to the lie of stroke which remais ubalaced. The ubalaced force is zero at the eds of the stroke whe 0 0 or 80 ad maximum at the middle whe θ 90. The magitude or the maximum value of the ubalaced force remais the same, i.e., equal to mr. Thus, istead of slidig to ad fro o its moutig, the mechaism teds to jump up ad dow. Darsha Istitute of Egieerig & Techology, Rajkot Page.9

34 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) To miimize the effect of the ubalaced force, a compromise is, usually, made, i.e., /3 of the reciprocatig mass is balaced (or a value betwee oe-half ad three-quarters). If c is the fractio of the reciprocatig mass Thus, balaced the primary force balaced by the mass cmr cosθ primary force ubalaced by the mass ( -c)mr cosθ Vertical compoet of cetrifugal force which remais ubalaced cmr siθ I fact, i reciprocatig egies, ubalaced forces i the directio of the lie of stroke are more dagerous tha the forces perpedicular to the lie of stroke. Resultat ubalaced force at ay istat ( c) mr cos cmr si The resultat ubalaced force is miimum whe c /. The method just discussed above to balace the disturbig effect of a reciprocatig mass is just equivalet to as if a revolvig mass at the crakpi is completely balaced by providig a coutermass at the same radius diametrically opposite the crak. Thus, if mp is the mass at the crakpi ad c is the fractio of the reciprocatig mass m to be balaced, the mass at the crakpi may be cosidered as (cm + mp) which is to be completely balaced. Example.: The followig data relate to a sigle - cylider reciprocatig egie: Mass of reciprocatig parts 40 kg Mass of revolvig parts 30 kg at crak radius Speed 50 rpm Stroke 350 mm If 60% of the reciprocatig parts ad all the revolvig parts are to be balaced, determie (i) balace mass required at a radius of 30 mm (ii) ubalaced force whe the crak has tured 45 from top dead cetre. m 40 kg m P 30 kg N 50 rpm r l/ 75 mm N rad / s (i) Mass to be balaced at crak pi cm + m P 0.6 x kg Page.0 Darsha Istitute of Egieerig & Techology, Rajkot

35 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies m C r C mr m C x x 75 m C 9.53 kg. (ii) Ubalaced force (at θ 45 ) ( c) mr cos cmr si ( 0.6) (5.7) cos (5.7) si N. Example.: A sigle cylider reciprocatig egie has speed 40 rpm, stroke 300 mm, mass of reciprocatig parts 50 kg, mass of revolvig parts at 50 mm radius 30 kg. If all the mass of revolvig parts ad two-third of the mass of reciprocatig parts are to be balaced, fid the balace mass required at radius of 400 mm ad the residual ubalaced force whe the crak has rotated 60 from IDC. N 40 rpm l 300 mm m 50 kg m P 30 kg r l/ 50 mm N rad / s (i) Mass to be balaced at crak pi cm + m P /3 x kg m C r C mr m C x x 50 m C 3.75 kg. (ii) Ubalaced force (at θ 45 ) ( c) mr cos cmr si ( ) (5.3) cos (5.3) si (789.36) (734.55) 846. N Darsha Istitute of Egieerig & Techology, Rajkot Page.

36 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690).8 Balacig of Locomotives Locomotives are of two types, coupled ad ucoupled. If two or more pairs of wheels are coupled together to icrease the adhesive force betwee the wheels ad the track, it is called a coupled locomotive. Otherwise, it is a ucoupled locomotive. Locomotives usually have two cyliders. If the cyliders are mouted betwee the wheels, it is called a iside cylider locomotive ad if the cyliders are outside the wheels, it is a outside cylider locomotive. The craks of the two cyliders are set at 90 to each other so that the egie ca be started easily after stoppig i ay positio. Balace masses are placed o the wheels i both types. I coupled locomotives, wheels are coupled by coectig their crakpis with couplig rods. As the couplig rod revolves with the crakpi, its proportioate mass ca be cosidered as a revolvig mass which ca be completely balaced. Thus, whereas i ucoupled locomotives, there are four plaes for cosideratio, two of the cyliders ad two of the drivig wheels, i coupled locomotives there are six plaes, two of cyliders, two of couplig rods ad two of the wheels. The plaes which cotai the couplig rod masses lie outside the plaes that cotai the balace (couter) masses. Also, i case of coupled locomotives, the mass required to balace the reciprocatig parts is distributed amog all the wheels which are coupled. This results i a reduced hammer blow. Locomotives have become obsolete owadays. (a) Iside cylider locomotives (b) Outside cylider locomotives Fig Effects of Partial Balacig i Locomotives Reciprocatig parts are oly partially balaced. Due to this partial balacig of the reciprocatig parts, there is a ubalaced primary force alog the lie of stroke ad also a ubalaced primary force perpedicular to the lie of stroke. I. Hammer-blow Hammer-blow is the maximum vertical ubalaced force caused by the mass provided to balace the reciprocatig masses. Its value is mr. Thus, it varies as a square of the speed. At high speeds, the force of the hammer-blow could exceed the static load o the Page. Darsha Istitute of Egieerig & Techology, Rajkot

37 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies wheels ad the wheels ca be lifted off the rail whe the directio of the hammer-blow will be vertically upwards. Hammer blow mr II. Variatio of Tractive Force Fig..6 A variatio i the tractive force (effort) of a egie is caused by the ubalaced portio of primary force which acts alog the lie of stroke of a locomotive egie. If c is the fractio of the reciprocatig mass that is balaced the ubalaced primary force for cylider ( c) mr cosθ ubalaced primary force for cylider ( c) mr cos (90 + θ) ( c) mr si θ Total ubalaced primary force or the variatio i the tractive force ( c) mr (cos θ si θ) This is maximum whe (cos θ si θ) is maximum, d or whe (cos si ) 0 d si θ cos θ 0 si θ cos θ ta θ θ 35 or 35 Whe θ 35 Maximum variatio i tractive force ( c) mr (cos 35 si 35 ) ( c) mr ( c) mr Whe θ 35 Maximum variatio i tractive force ( c) mr (cos 35 si 35 ) ( c) mr ( c) mr Thus, maximum variatio ( c) mr Darsha Istitute of Egieerig & Techology, Rajkot Page.3

38 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) III. Swayig Couple Ubalaced primary forces alog the lies of stroke are separated by a distace l apart ad Thus, costitute a couple. This teds to make the leadig wheels sway from side to side. Swayig couple momets of forces about the egie cetre lie l l ( c) mr cos ( c) mr cos(90 ) l ( c) mr (cos si ) This is maximum whe (cos θ + si θ) is maximum. d i.e., whe (cos si ) 0 dt si θ + cos θ 0 si θ cos θ ta θ θ 45 or 5 Whe θ 45, maximum swayig couple ( c ) mr l Whe θ 5, maximum swayig couple ( c ) mr l Thus, maximum swayig couple ( c ) mr l Example.3: A iside cylider locomotive has its cylider cetre lies 0.7 m apart ad has a stroke of 0.6 m. The rotatig masses per cylider are equivalet to 50 kg at the crak pi, ad the reciprocatig masses per cylider to 80 kg. The wheel cetre lies are.5 m apart. The craks are at right agles. The whole of the rotatig ad /3 of the reciprocatig masses are to be balaced by masses placed at a radius of 0.6 m. Fid the magitude ad directio of the balacig masses. Fid the fluctuatio i rail pressure uder oe wheel, variatio of tractive effort ad the magitude of swayig couple at a crak speed of 300 r.p.m. l B l C 0.6 m or r B r C 0.3 m; m 50 kg r A r D 0.6 m; m 80 kg; c /3 N 300 r.p.m. N rad / s Page.4 Darsha Istitute of Egieerig & Techology, Rajkot

39 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Equivalet mass of the rotatig parts to be balaced per cylider at the crak pi, m m B m C m + c.m kg 3 The magitude ad directio of balacig masses may be determied graphically as below: First of all, draw the space diagram to show the positios of the plaes of the wheels ad the cyliders, as show i Fig..7 (a). Sice the craks of the cyliders are at right agles, therefore assumig the positio of crak of the cylider B i the horizotal directio, draw OC ad OB at right agles to each other as show i Fig..7 (b). Fig..7 Tabulate the data as give i the followig table. Assume the plae of wheel A as the referece plae. Table. Cet.force Plae Agle Mass (m) kg Radius (r) m (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m A (R.P.) A m A m A 0 0 B C D D m D m D m D Now, draw the couple polygo from the data give i Table. (colum 7), to some suitable scale, as show i Fig.8 (a). The closig side c o represets the balacig couple ad it is proportioal to 0.9 m D. Therefore, by measuremet, 0.9 m D vector c o 94.5 kg-m m D 05 kg Darsha Istitute of Egieerig & Techology, Rajkot Page.5

40 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Fig..8 By measuremet, the agular positio of m D is θ D 50 i the aticlockwise directio from mass m B. I order to fid the balacig mass A, draw the force polygo from the data give i Table. (colum 5), to some suitable scale, as show i Fig..8 (b), The vector do represets the balacig force ad it is proportioal to 0.6 m A. Therefore by measuremet, 0.6 m A vector do 63 kg-m m A 05 kg By measuremet, the agular positio of m A is θ A 00 i the aticlockwise directio from mass m B. Each balacig mass 05 kg m 50 Balacig mass for rotatig masses, M m kg cm 80 Balacig mass for reciprocatig masses, M' m kg Balacig mass of 46.6 kg for reciprocatig masses gives rise to cetrifugal force. Fluctuatio i rail pressure or hammer blow M r Variatio of tractive effort Maximum variatio of tractive effort ( c) mr 46.6 x 0.6 x (3.4) 760 N ( ) (3.4) KN Page.6 Darsha Istitute of Egieerig & Techology, Rajkot

41 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Swayig couple Maximum swayig couple ( c) mr l ( ) (3.4) N.m Example.4: The followig data refers to two-cylider ucoupled locomotive: Rotatig mass per cylider 80 kg Reciprocatig mass per cylider 300 kg Distace betwee wheels 400 mm Distace betwee cylider ceters 600 mm Diameter of treads of drivig wheels 800 mm Crak radius 300 mm Radius of cetre of balace mass 60 mm Locomotive speed 50 km/hr Agle betwee cylider craks 90 Dead load o each wheel 3.5 toe Determie i. Balacig mass required i plaes of drivig wheels if whole of the revolvig ad /3 of reciprocatig mass are to be balaced ii. Swayig couple iii. Variatio i tractive force iv. Maximum ad miimum pressure o the rails v. Maximum speed of locomotive without liftig the wheels from rails. m 50 kg r B r C 0.3 m; m 80 kg r A r D 0.6 m; v 50 km/hr c /3 Dead load, W 3.5toe Fig..9 Total mass to be balaced per cylider, m B m C m P +cm kg Darsha Istitute of Egieerig & Techology, Rajkot Page.7

42 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Plae Agle Mass (m) kg Radius (r) m Table. Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m A (R.P.) A m A m A 0 0 B C D D m D m D m D Draw the couple polygo from the data give i Table. (colum 7), to some suitable scale, as show i Fig.0 (a). The closig side bo represets the balacig couple ad it is proportioal to 0.868m D. Therefore, by measuremet, 0.868m D vector bo 55. kg-m m D kg By measuremet, the agular positio of m D is θ D 48 i the aticlockwise directio from mass m B. Draw the force polygo from the data give i Table. (colum 5), to some suitable scale, as show i Fig.0 (b). The closig side co represets the balacig couple ad it is proportioal to 0.6m A. Therefore, by measuremet, 0.6m A vector co 0.78 kg-m m A kg By measuremet, the agular positio of m A is θ A 0 i the aticlockwise directio from mass m B. (a) Couple Polygo Fig..0 (b) Force Polygo Page.8 Darsha Istitute of Egieerig & Techology, Rajkot

43 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies v r 6 v 500 r / 5.43 rad/s Swayig couple ( c) mr l ( ) (5.43) N.m Variatio of tractive effort ( c) mr ( ) (5.43) N Balace mass for reciprocatig parts oly kg 480 Hammer blow mr x 0.6 x (5.43) 099 N Dead load 3.5 x 000 x N Maximum pressure o rails N Miimum pressure o rails N Maximum speed of the locomotive without liftig the wheels from the rails will be whe the dead load becomes equal to the hammer blow x 0.6 x rad/s Velocity of wheels r m / s km / hr 000 V km / hr Darsha Istitute of Egieerig & Techology, Rajkot Page.9

44 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Example.5: The three craks of a three cylider locomotive are all o the same axle ad are set at 0. The pitch of the cyliders is meter ad the stroke of each pisto is 0.6 m. The reciprocatig masses are 300 kg for iside cylider ad 60 kg for each outside cylider ad the plaes of rotatio of the balace masses are 0.8 m from the iside crak. If 40% of the reciprocatig parts are to be balaced, fid : (i) The magitude ad the positio of the balacig masses required at a radius of 0.6 m (ii) The hammer blow per wheel whe the axle makes 6 r.p.s. l A l B l C 0.6 m or r A r B r C 0.3 m ; m I 300 kg r r 0.6 m ; m O 60 kg c 40% 0.4 ; N 6 r.p.s. 6 π 37.7 rad/s Fig.. Sice 40% of the reciprocatig masses are to be balaced, therefore mass of the reciprocatig parts to be balaced for each outside cylider, m A m C c m O kg ad mass of the reciprocatig parts to be balaced for iside cylider, m B c m I kg Plae Agle Mass (m) Kg Radius (r) m Table.3 Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m A (R.P.) m m 0 0 B m m m C Draw the couple polygo with the data give i Table.3 (colum 7), to some suitable scale, as show i Fig.. (a). The closig side c o represets the balacig couple ad it is proportioal to 0.96 m. Therefore, by measuremet, Page.0 Darsha Istitute of Egieerig & Techology, Rajkot

45 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies 0.96 m vector c o 55. kg-m m 57.5 kg. By measuremet, the agular positio of m is θ 4 i the aticlockwise directio from mass m A. Draw the force polygo with the data give i Table.3 (colum 5), to some suitable scale, as show i Fig.. (b). The closig side co represets the balacig force ad it is proportioal to 0.6 m. Therefore, by measuremet, 0.6 m vector co 34.5 kg-m m 57.5 kg. By measuremet, the agular positio of m is θ 5 i the aticlockwise directio from mass m A. Hammer blow per wheel mr Fig x 0.6 x (37.7) N. Example.6: A two cylider locomotive has the followig specificatios; Reciprocatig mass per cylider 306 Kg Crak radius 300 mm Agle betwee craks 90 Drivig wheels diameter 800 mm Distace betwee cylider ceters 650 mm Distace betwee drivig wheel plaes 550 mm Determie (a) The fractio of reciprocatig masses to be balaced, if the hammer blow is ot to exceed 46 KN at 96.5 Km/hr. (b) The variatio i tractive force. (c) The maximum swayig couple. m 300 kg D.8 m orr 0.9 m r 0.3 m Hammer blow 46 kn v 96.5 km/h 6.8 m/s The mass of the reciprocatig parts to be balaced c.m 300c kg Darsha Istitute of Egieerig & Techology, Rajkot Page.

46 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Plae Agle Mass (m) kg Radius (r) m Fig..3 Table.4 Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m A (R.P.) A m A r A m A r A 0 0 B 0 300c c c C c c. 99c D D m D r D m D r D.55.55m D r D Now the couple polygo, to some suitable scale, may be draw the data give i Table.4 (colum 7), as show i Fig..4. The closig side of the polygo (vector c o ) represets the balacig couple ad is proportioal to.55 B.b. From the couple polygo,.55 D D (40.5 ) (99 ) 07 m r c c c m D r D 69c Agular speed, v/r Fig / rad/s Hammer blow m D r D c(9.8) c 0.75 Variatio of tractive effort ( c) mr ( 0.75) (9.8) 840 N Swayig couple ( c) mr l ( 0.75) (9.8) Nm. with Page. Darsha Istitute of Egieerig & Techology, Rajkot

47 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Example.7: The followig data apply to a outside cylider ucoupled locomotive: Mass of rotatig parts per cylider 360 kg Mass of reciprocatig parts per cylider 300 kg Agle betwee craks 90 Crak radius 0.3 m Cylider cetres.75 m Radius of balace masses 0.75 m Wheel cetres.45 m. If whole of the rotatig ad two-thirds of reciprocatig parts are to be balaced i plaes of the drivig wheels, fid: (a) Magitude ad agular positios of balace masses, (b) Speed i km/hr at which the wheel will lift off the rails whe the load o each drivig wheel is 30 kn ad the diameter of tread of drivig wheels is.8 m, ad (c) Swayig couple at speed arrived at i (b) above. m 360 kg r A r D 0.3 m m 300 kg r B r C 0.75 m c / 3. The equivalet mass of the rotatig parts to be balaced per cylider, m m A m D m + c.m kg Plae Agle (a) Positio of Plaes Mass (m) Kg Radius (r) m (b) Positio of Masses Fig..5 Table.5 Cet.force Distace from (mr) kg-m Ref. Plae (l) m Couple (mrl) kg-m A B (R.P.) B m B m B 0 0 C C m C m C m C D Darsha Istitute of Egieerig & Techology, Rajkot Page.3

48 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Draw the couple polygo with the data give i Table.5 colum (7), to some suitable scale as show i Fig..6(a). The closig side d o represets the balacig couple ad it is proportioal to.08 m C. Therefore, by measuremet,.08 m C 69.6 kg-m m C 49 kg By measuremet, the agular positio of m C is θ C 75 i the aticlockwise directio from mass m A. (a) Couple Polygo (b) Force Polygo Fig..6 Draw the force polygo with the data give i Table.5 colum (5), to some suitable scale as show i Fig..6(b). The closig side co represets the balacig force ad it is proportioal to 0.75 m B. Therefore, by measuremet, 0.75 m B kg-m m B 49 kg By measuremet, the agular positio of m B is θ B 74.5 i the aticlockwise directio from mass m A. Speed at which the wheel will lift off the rails Give : P 30 kn N D.8 m Agular speed at which the wheels will lift off the rails i rad/s, ad v Correspodig liear speed i km/h. Each balacig mass m B m C 49 kg cm 300 Balacig mass for reciprocatig parts, M kg. m P Mr rad/s (r r B r C ) v D/..8/ 9.08 m/s / km/h Page.4 Darsha Istitute of Egieerig & Techology, Rajkot

49 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Swayig couple at speed. rad/s Swayig couple ( c) mr l ( ) (.) N.m Example.8: The followig data refers to a four-coupled wheel locomotive with two iside cylider Pitch of cyliders 600 mm Reciprocatig mass/cylider 350 kg Revolvig mass/cylider 60 kg Distace betwee drivig wheels.6 m Distace betwee couplig rods m Diameter of drivig wheels.9 m Revolvig parts for each couplig rod crak 30 kg Egie crak radius 300 mm Couplig rod crak radius 40 mm Distace of cetre of balace mass i plaes of drivig wheels from axle cetre 750 mm Agle betwee egie craks 90 Agle betwee couplig rod crak with adjacet egie crak 80 The balaced mass required for the reciprocatig parts is equally divided betwee each pair of coupled wheels. Determie the (i) Magitude ad positio of the balace mass required to balace two-third of reciprocatig ad whole of the revolvig parts (ii) Hammer-blow ad the maximum variatio of tractive force whe the locomotive speed is 80 km/h m m 6 30 kg r r m 3 0 l 0. m r r m 4 90 l 3 0.5m r 3 r m 80 l 4.m 6 70 l 5.6m l 6.8m Leadig wheels: Balace mass o each leadig wheel mp cm m 3 m kg m r l 30 x 0.4 x (-0.) -6.4 m 3 r 3 l x 0.3 x m 4 r 4 l x 0.3 x Darsha Istitute of Egieerig & Techology, Rajkot Page.5

50 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) m 5 r 5 l 5 m 5 x 0.75 x.6. m 5 m 6 r 6 l 6 30 x 0.4 x Fig..7 mrl cos cos cos cos70 0 ad -6.4 si si si si70 0 Squarig, addig ad the solvig,.m ( 6.4cos cos cos cos70 ) ( 6.4si si si si70 ) 5 m5. (60.99) (64.9) 88.6 kg.mm m kg mrl si 64.9 ta mrl cos lies i the third quadrat (umerator is egative ad deomiator is egative) From symmetry of the system, m m kg ta lies i the third quadrat (umerator is egative ad deomiator is egative) Trailig wheels: The arragemet remais the same except that oly half of the required reciprocatig masses have to be balaced at the craks. m3 m kg 3 m 3 r 3 l 3 05 x 0.3 x Page.6 Darsha Istitute of Egieerig & Techology, Rajkot

51 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies m 4 r 4 l 4 05 x 0.3 x mrl cos cos cos cos70 0 ad -6.4 si si si si70 0 Squarig, addig ad the solvig,.m ( 6.4cos cos cos cos70 ) ( 6.4si si si si70 ) 5 m5. (.99) (.5) kg.mm m kg mrl si (.5) ta mrl cos lies i the secod quadrat (umerator is positive ad deomiator is egative) From symmetry of the system, m m kg.99 ta lies i the fourth quadrat (umerator is egative ad deomiator is positive) (iii) Hammer blow mr where m is the balace mass for reciprocatig parts oly ad eglectig m ad m 6 i the above calculatios. Thus, m r l m 6 r 6 l 6 0. m (5.75 cos cos 90 ) (5.75 si si 90 ) 5 m5. (5.75) (34.65) kg.mm m kg v r rad / s / Hammer blow 3.75 x 0.75 x (3.39) 305 N Variatio of tractive effort ( c) mr ( / 3) (3.39) 437 N Darsha Istitute of Egieerig & Techology, Rajkot Page.7

52 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690).9 Balacig of Multi Cylider Egie Balacig of Primary force ad couple The multi-cylider egies with the cylider cetre lies i the same plae ad o the same side of the cetre lie of the crakshaft are kow as I-lie egies. The followig two coditios must be satisfied i order to give the primary balace of the reciprocatig parts of a multi-cylider egie : (a) The algebraic sum of the primary forces must be equal to zero. I other words, the primary force polygo must close ad Primary force, F mr cos PF (b) The algebraic sum of the couples about ay poit i the plae of the primary forces must be equal to zero. I other words, primary couple polygo must close. Primary couple, F mrl cos The primary ubalaced force due to the reciprocatig masses is equal to the compoet, parallel to the lie of stroke, of the cetrifugal force produced by the equal mass placed at the crakpi ad revolvig with it. Therefore, i order to give the primary balace of the reciprocatig parts of a multi-cylider egie, it is coveiet to imagie the reciprocatig masses to be trasferred to their respective crakpis ad to treat the problem as oe of revolvig masses. Balacig of Secodary force ad couple Whe the coectig rod is ot too log (i.e. whe the obliquity of the coectig rod is cosidered), the the secodary disturbig force due to the reciprocatig mass arises. The followig two coditios must be satisfied i order to give a complete secodary balace of a egie : (a) The algebraic sum of the secodary forces must be equal to zero. I other words, the secodary force polygo must close, ad cos Secodary force, FSF mr (b) The algebraic sum of the couples about ay poit i the plae of the secodary forces must be equal to zero. I other words, the secodary couple polygo must close. cos Secodary couple, FSC mr l Example.9: A four crak egie has the two outer craks set at 0 to each other, ad their reciprocatig masses are each 400 kg. The distace betwee the plaes of rotatio of adjacet craks are 450 mm, 750 mm ad 600 mm. If the egie is to be i complete primary balace, fid the reciprocatig mass ad the relative agular positio for each of the ier craks. If the legth of each crak is 300 mm, the legth of each coectig rod is. m ad the speed of rotatio is 40 r.p.m., what is the maximum secodary ubalaced force? m m kg N 40 r.p.m N 40 r 300 mm 0.3 m 5.4 rad/s PC Page.8 Darsha Istitute of Egieerig & Techology, Rajkot

53 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies l. m (a) Positio of plaes Fig..8 (b) Primary crak positio Table.6 Agle Mass (m) Radius Cet.force Distace from Couple Plae Kg (r)m (mr) kg-m Ref. Plae(l) m (mrl) kg-m (R.P.) 336 m m m m m (a) Primary couple polygo Fig..9 (b) Primary force polygo Sice the egie is to be i complete primary balace, therefore the primary couple polygo ad the primary force polygo must close. First of all, the primary couple polygo, as show i Fig..9 (a), is draw to some suitable scale from the data give i Table.6 (colum 8), i order to fid the reciprocatig mass for crak 3. Now by measuremet, we fid that 0.5 m 3 96 kg-m m 3 87 kg. ad its agular positio with respect to crak i the aticlockwise directio, θ Now i order to fid the reciprocatig mass for crak, draw the primary force polygo, as show i Fig..9 (b), to some suitable scale from the data give i Table.6 (colum 6). Now by measuremet, we fid that Darsha Istitute of Egieerig & Techology, Rajkot Page.9

54 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) 0.3 m 84 kg-m m 947 kg. ad its agular positio with respect to crak i the aticlockwise directio, θ 68. Maximum secodary ubalaced force (a) Secodary crak positios Fig..0 (b) Secodary force polygo The secodary crak positios obtaied by rotatig the primary craks at twice the agle, is show i Fig..0 (a). Now draw the secodary force polygo, as show i Fig..0 (b), to some suitable scale, from the data give i Table.6 (colum 6). The closig side of the polygo show dotted i Fig..0 (b) represets the maximum secodary ubalaced force. By measuremet, we fid that the maximum secodary ubalaced force is proportioal to58 kg-m. Maximum Ubalaced Secodary Force, U.S.F. 58 (5.4) U.S.F. 58. / 0.3 U.S.F N Example.0: The itermediate craks of a four cylider symmetrical egie, which is i complete primary balace, are 90 to each other ad each has a reciprocatig mass of 300 kg. The cetre distace betwee itermediate craks is 600 mm ad betwee extreme craks it is 800 mm. Legths of the coectig rod ad craks are 900 mm ad 300 mm respectively. Calculate the masses fixed to the extreme craks with their relative agular positios. Also fid the magitudes of secodary forces ad couples about the cetre lie of the system if the egie speed is 500 rpm. m m kg r 300 mm 0.3 m l 0.9 m N 500 r.p.m N rad/s Page.30 Darsha Istitute of Egieerig & Techology, Rajkot

55 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Fig.. Positio of plaes Table.7 Cet.force Agle Mass (m) Radius Distace from Plae kg (r)m (mr) kg-m Ref. Plae(l) m (R.P.) 54 m m 0 0 Couple (mrl) kg-m m m m 4 (a) Primary couple polygo (b) Primary force polygo Fig.. Sice the egie is to be i complete primary balace, therefore the primary couple polygo ad the primary force polygo must close. First of all, the primary couple polygo, as show i Fig.. (a), is draw to some suitable scale from the data give i Table.7 (colum 8), i order to fid the reciprocatig mass for crak 4. Now by measuremet, we fid that 0.54 m kg-m m kg. ad its agular positio with respect to crak i the aticlockwise directio, θ Now i order to fid the reciprocatig mass for crak, draw the primary force polygo, as show i Fig.. (b), to some suitable scale from the data give i Table.7 (colum 6). Now by measuremet, we fid that Darsha Istitute of Egieerig & Techology, Rajkot Page.3

56 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) 0.3 m kg-m m 3.6 kg. ad its agular positio with respect to crak i the aticlockwise directio, θ (a) Secodary force polygo (b) Secodary couple polygo Fig..3 The secodary crak positios obtaied by rotatig the primary craks at twice the agle. Now draw the secodary force polygo, as show i Fig..3 (a), to some suitable scale, from the data give i Table.7 (colum 6). The closig side of the polygo show dotted i Fig..3 (a) represets the maximum secodary ubalaced force. By measuremet, we fid that the maximum secodary ubalaced force is proportioal to kg-m. Maximum Ubalaced Secodary Force, U.S.F (57.08) U.S.F / 0.3 U.S.F KN Now draw the secodary couple polygo, as show i Fig..3 (b), to some suitable scale, from the data give i Table.7 (colum 8). The closig side of the polygo show dotted i Fig..3 (b) represets the maximum secodary ubalaced couple. By measuremet, we fid that the maximum secodary ubalaced couple is proportioal to kg-m. Maximum Ubalaced Secodary Couple, U.S.C (57.08) U.S.C / 0.3 U.S.C KN.m Page.3 Darsha Istitute of Egieerig & Techology, Rajkot

57 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Example.: The craks ad coectig rods of a 4-cylider i-lie egie ruig at 800 r.p.m. are 60 mm ad 40 mm each respectively ad the cyliders are spaced 50 mm apart. If the cyliders are umbered to 4 i sequece from oe ed, the craks appear at itervals of 90 i a ed view i the order The reciprocatig mass correspodig to each cylider is.5 kg. Determie: (i) Ubalaced primary ad secodary forces, if ay, ad (ii) Ubalaced primary ad secodary couples with referece to cetral plae of the egie. r 60 mm l 40 mm m.5 kg N 800 r.p.m N rad/s Agle Agle Plae Mass (m) kg Radius (r) m Table.8 Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m Darsha Istitute of Egieerig & Techology, Rajkot Page.33

58 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Fig..4 Ubalaced primary forces ad couples The positio of the cylider plaes ad craks is show i Fig..4 (a) ad (b) respectively. With referece to cetral plae of the egie, the data may be tabulated as above: The primary force polygo from the data give i Table.8 (colum 6) is draw as show i Fig..4 (c). Sice the primary force polygo is a closed figure, therefore there are o ubalaced primary forces, Ubalaced Primary Force, U.P.F. 0. The primary couple polygo from the data give i Table.8 (colum 8) is draw as show i Fig..4 (d). The closig side of the polygo, show dotted i the figure, represets ubalaced primary couple. By measuremet, the ubalaced primary couple is proportioal to 0.09 kg-m. Ubalaced Primary Couple, U.P.C (88.5) Ubalaced secodary forces ad couples U.P.C N-m. The secodary crak positios, takig crak 3 as the referece crak, as show i Fig..4 (e). From the secodary force polygo as show i Fig..4 (f), it is a closed figure. Therefore there are o ubalaced secodary forces. Ubalaced Secodary Force, U.S.F. 0. The secodary couple polygo is show i Fig..4 (g). The ubalaced secodary couple is show by dotted lie. By measuremet, we fid that ubalaced secodary couple is proportioal to kg-m. Ubalaced Secodary Couple, (88.5) U S C 0.4 / 0.06 U.S.C N.m ( l / r) Page.34 Darsha Istitute of Egieerig & Techology, Rajkot

59 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Example.: The successive craks of a five cylider i-lie egie are at 44 apart. The spacig betwee cylider cetre lies is 400 mm. The legths of the crak ad the coectig rod are 00 mm ad 450 mm respectively ad the reciprocatig mass for each cylider is 0 kg. The egie speed is 630 r.p.m. Determie the maximum values of the primary ad secodary forces ad couples ad the positio of the cetral crak at which these occur. l 450 mm 0.45 m m 0 kg l / r 4.5 r 0. m N 630 r.p.m. N rad/s Fig..5 Cylider plae positio Table.9 Agle Agle Mass (m) Radius Cet.force Distace from Couple Plae Kg (r) m (mr) kg-m Ref. Plae (l) m (mrl) kg-m Ubalaced primary forces ad couples (a) Fig..6 Primary force polygo (b) Darsha Istitute of Egieerig & Techology, Rajkot Page.35

60 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) The positio of the cylider plaes ad craks is show i Fig..5 ad Fig..6 (b) respectively. With referece to cetral plae of the egie, the data may be tabulated as above: The primary force polygo from the data give i Table.9 (colum 6) is draw as show i Fig..6 (a). Sice the primary force polygo is a closed figure, therefore there are o ubalaced primary forces, Ubalaced Primary Force, U.P.F. 0. (a) (b) Fig..7 Primary couple polygo The primary couple polygo from the data give i Table.9 (colum 8) is draw as show i Fig..7 (a). The closig side of the polygo, show dotted i the figure, represets ubalaced primary couple. By measuremet, the ubalaced primary couple is proportioal to. kg-m. Ubalaced Primary Couple, U.P.C.. (65.97) U.P.C N-m. Ubalaced secodary forces ad couples (a) Fig..8 Secodary force polygo (b) Page.36 Darsha Istitute of Egieerig & Techology, Rajkot

61 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies The secodary crak positios are show i Fig..8 (b). From the secodary force polygo as show i Fig..8 (a), it is a closed figure. Therefore there are o ubalaced secodary forces. Ubalaced Secodary Force, U.S.F. 0. (a) (b) Fig..9 Secodary couple polygo The secodary couple polygo is show i Fig..9 (a). The ubalaced secodary couple is show by dotted lie. By measuremet, we fid that ubalaced secodary couple is proportioal to 3.4 kg-m. Ubalaced Secodary Couple, (65.97) U S C 0.45 / 0. U.S.C N.m ( l / r) Example.3: A four stroke five cylider i-lie egie has a firig order of The ceters lies of cyliders are spaced at equal itervals of 5 cm, the reciprocatig parts per cylider have a mass of 5 kg, the pisto stroke is 0 cm ad the coectig rods are 7.5 cm log. The egie rotates at 600 rpm. Determie the values of maximum primary ad secodary ubalaced forces ad couples about the cetral plae. l 0cm 0. m or r 0.05 m m 5 kg l / r 7.5/5 3.5 N 600 r.p.m. N rad/s Fig..30 Cylider plae positio Darsha Istitute of Egieerig & Techology, Rajkot Page.37

62 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Agle Agle Plae Mass (m) kg Radius (r) m Table.0 Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m Ubalaced primary forces ad couples (a) (b) Fig..3 Primary force polygo The positio of the cylider plaes ad craks is show i Fig..30 ad Fig..3(b) respectively. With referece to cetral plae of the egie, the data may be tabulated as above: The primary force polygo from the data give i Table.0 (colum 6) is draw as show i Fig..3 (a). Sice the primary force polygo is a closed figure, therefore there are o ubalaced primary forces, Ubalaced Primary Force, U.P.F. 0. (a) (b) Fig..3 Primary couple polygo Page.38 Darsha Istitute of Egieerig & Techology, Rajkot

63 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies The primary couple polygo from the data give i Table.0 (colum 8) is draw as show i Fig..3 (a). The closig side of the polygo, show dotted i the figure, represets ubalaced primary couple. By measuremet, the ubalaced primary couple is proportioal to 0.53 kg-m. Ubalaced Primary Couple, U.P.C (6.83) U.P.C N-m. Ubalaced secodary forces ad couples (a) Fig..33 Secodary force polygo (b) The secodary crak positios are show i Fig..33 (b). From the secodary force polygo as show i Fig..33 (a), it is a closed figure. Therefore there are o ubalaced secodary forces. Ubalaced Secodary Force, U.S.F. 0. (a) Fig..34 Secodary couple polygo (b) The secodary couple polygo is show i Fig..34 (a). The ubalaced secodary couple is show by dotted lie. By measuremet, we fid that ubalaced secodary couple is proportioal to 0.8 kg-m. Ubalaced Secodary Couple, U. S. C. 0.8 (6.83) U. S. C U.S.C. 03 N.m ( l / r) Darsha Istitute of Egieerig & Techology, Rajkot Page.39

64 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Example.4: I a i-lie six cylider egie workig o two stroke cycle, the cylider cetre lies are spaced at 600 mm. I the ed view, the craks are 60 apart ad i the order The stroke of each pisto is 400 mm ad the coectig rod legth is m. The mass of the reciprocatig parts is 00 kg per cylider ad that of rotatig parts 00 kg per crak. The egie rotates at 300 r.p.m. Examie the egie for the balace of primary ad secodary forces ad couples. Fid the maximum ubalaced forces ad couples. L 400 mm or r L / 00 mm 0. m l m N 300 r.p.m. m 00 kg N rad/s m 00 kg Fig..35 Positios of plaes of cyliders Table. Agle Agle Mass (m) Radius Cet. force Distace from Couple Plae kg (r) m (mr) kg-m Ref. Plae (l) m (mrl) kg-m Ubalaced primary forces ad couples (a) (b) Fig..36 Primary force polygo Page.40 Darsha Istitute of Egieerig & Techology, Rajkot

65 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies The positio of the cylider plaes ad craks is show i Fig..35 ad Fig..36(b) respectively. With referece to cetral plae of the egie, the data may be tabulated as above: The primary force polygo from the data give i Table. (colum 6) is draw as show i Fig..36 (a). Sice the primary force polygo is a closed figure, therefore there are o ubalaced primary forces, Ubalaced Primary Force, U.P.F. 0. (a) (b) Fig..37 Primary couple polygo The primary couple polygo from the data give i Table. (colum 8) is draw as show i Fig..37 (a). Sice the primary couple polygo is a closed figure, therefore there are o ubalaced primary couples, Ubalaced Primary Couple, U.P.C. 0 Ubalaced secodary forces ad couples (a) Fig..38 Secodary force polygo The secodary crak positios are show i Fig..38 (b). From the secodary force polygo as show i Fig..38 (a), it is a closed figure. Therefore there are o ubalaced secodary forces. Ubalaced Secodary Force, U.S.F. 0. (b) Darsha Istitute of Egieerig & Techology, Rajkot Page.4

66 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) (a) Fig..39 Secodary couple polygo (b) The secodary couple polygo is show i Fig..39 (a). The ubalaced secodary couple is show by dotted lie. By measuremet, we fid that ubalaced secodary couple is proportioal to 49 kg-m. Ubalaced Secodary Couple, U. S. C. 49 (3.4) U. S. C. 49 / 0. U.S.C N.m ( l / r) Example.5: The firig order i a 6 cylider vertical four stroke i-lie egie is The pisto stroke is 00 mm ad the legth of each coectig rod is 00 mm. The pitch distaces betwee the cylider cetre lies are 00 mm, 00 mm, 50 mm, 00 mm, ad 00 mm respectively. The reciprocatig mass per cylider is kg ad the egie rus at 3000 r.p.m. Determie the out-of-balace primary ad secodary forces ad couples o this egie, takig a plae midway betwee the cylider 3 ad 4 as the referece plae. L 00 mm or r L / 50 mm 0.05 m l 00 mm m kg N 3000 r.p.m. N rad/s Fig..40 Positios of plaes Page.4 Darsha Istitute of Egieerig & Techology, Rajkot

67 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Agle Agle Plae Mass (m) kg Radius (r) m Table. Cet.force (mr) kg-m Distace from Ref. Plae (l) m Couple (mrl) kg-m Ubalaced primary forces ad couples (a) (b) Fig..4 Primary force polygo The positio of the cylider plaes ad craks is show i Fig..40 ad Fig..4(b) respectively. With referece to cetral plae of the egie, the data may be tabulated as above: The primary force polygo from the data give i Table. (colum 6) is draw as show i Fig..4 (a). Sice the primary force polygo is a closed figure, therefore there are o ubalaced primary forces, Ubalaced Primary Force, U.P.F. 0. (a) (b) Fig..4 Primary couple polygo Darsha Istitute of Egieerig & Techology, Rajkot Page.43

68 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) The primary couple polygo from the data give i Table. (colum 8) is draw as show i Fig..4 (a). The closig side of the polygo, show dotted i the figure, represets ubalaced primary couple. By measuremet, the ubalaced primary couple is proportioal to kg-m. Ubalaced Primary Couple, U.P.C (34.6) U.P.C N-m. Ubalaced secodary forces ad couples (a) Fig..43 Secodary force polygo (b) The secodary crak positios are show i Fig..43 (b). From the secodary force polygo as show i Fig..43 (a), it is a closed figure. Therefore there are o ubalaced secodary forces. Ubalaced Secodary Force, U.S.F. 0. (a) (b) Fig..44 Secodary couple polygo The secodary couple polygo is draw as show i Fig..44 (a). Sice the secodary couple polygo is a closed figure, therefore there are o ubalaced secodary couples, Ubalaced Secodary Couple, U.S.C. 0 Page.44 Darsha Istitute of Egieerig & Techology, Rajkot

69 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies.0 Balacig of V Egie I V egies, a commo crak OA is operated by two coectig rods AB, ad AB. Fig..45 shows a symmetrical two cylider V cylider, the cetre lies of which are iclied at a agle α to the x axis. Let be the agle moved by the crak from the x axis. I. Primary force Primary force of alog lie of stroke Fig..45 OB mr cos( α) Primary force of alog x axis mr cos( α) cos α Primary force of alog lie of stroke OB mr cos( + α) Primary force of alog x axis mr cos( + α) cos α Total primary force alog x axis mr cosα *cos( α) + cos( + α)] mr cosα *(cos cosα + si siα) + (cos cosα si siα)] mr cosα cos cosα mr cos α cos Similarly, total primary force alog z axis mr siα *cos( α) cos( + α)] mr siα *(cos cosα + si siα) (cos cosα si siα)] mr siα si siα mr si α si Resultat primary force ( mr cos cos ) (mr si si ) mr ( cos cos ) ( si si ) It will be at a agle β with the x axis give by si si ta cos cos If α 90, resultat force mr ( cos 45 cos ) ( si 45 si ) mr si 45 si ta ta cos 45 cos Darsha Istitute of Egieerig & Techology, Rajkot Page.45

70 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) i.e., β or it acts alog the crak ad, therefore, ca be completely balaced by a mass at a suitable radius diametrically opposite to the crak such that m r r r mr. For a give value of α, the resultat primary force is maximum whe (cos cos ) ( si si ) is maximum (cos d d cos si si ) is maximum ( cos cos si si ) cos. cos si si. si cos 4 4 cos. si s. si 4 4 si ( si cos ) 0 0 i 0 As α is ot zero, therefore, for a give value of α, the resultat primary force is maximum whe is zero degree. II. Secodary force Secodary force of alog OB mr cos( ) Secodary force of alog x-axis mr cos( )cos Secodary force of alog OB mr cos( ) Secodary force of alog x-axis Total secodary force alog x-axis mr cos [cos( ) cos( )] mr cos( )cos mr cos [(cos cos si si ) (cos cos si si )] mr cos cos cos mr Similarly, secodary force alog z-axis si si si mr Resultat secodary force (cos cos cos ) (si si si ) si si si ta ' cos cos cos If α 90 or α 45 Page.46 Darsha Istitute of Egieerig & Techology, Rajkot

71 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies Secodary force mr si mr si ta ', ' 90 This meas that the force acts alog z-axis ad is a harmoic force ad special methods are eeded to balace it. Example.6: Reciprocatig mass per cylider i 60 V twi egie is.5 kg. The stroke ad coectig rod legth are 00 mm ad 50 mm respectively. If the egie rus at 500 rpm, determie the maximum ad miimum values of primary ad secodary forces. α 60 L 50 mm l 00 mm or r 50 mm m.5 kg N 500 rpm N rad / s Resultat primary force, F P mr ( cos cos ) ( si si ) mr ( cos 30 cos ) ( si 30 si ) 3 mr cos si 4 4 mr 9cos si.(i) The primary force is maximum, whe θ 0. Therefore substitutig θ 0 i equatio (i), maximum primary force, F P(max) mr (6.8) N The primary force is miimum, whe θ 90. Therefore substitutig θ 90 i equatio (i), miimum primary force, F P(mi) Resultat secodary force, F S mr (6.8) 570. N mr (cos cos cos ) (si si si ) mr (cos30 coscos60 ) (si30sisi60 ) Darsha Istitute of Egieerig & Techology, Rajkot Page.47

72 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) mr 3 3 cos si 3 mr Resultat secodary force, F S N (6.8) ( l / r) 0.5 / 0.05 Example.7: A V-twi egie has the cylider axes at right agles ad the coectig rods operate a commo crak. The reciprocatig mass per cylider is.5 kg ad the crak radius is 75 mm. The legth of the coectig rod is 0.3 m. Show that the egie may be balaced for primary forces by meas of a revolvig balace mass. If the egie speed is 500 rpm, what is the value of maximum resultat secodary force? α 90 r 75 mm m N 500 r.p.m. m.5 kg l 0.3 m N rad / s Primary force, F P mr ( cos cos ) ( si si ) mr ( cos 45 cos ) ( si 45 si ) cos si mr mr Sice the resultat primary force mr is the cetrifugal force of a mass m at the crak radius r whe rotatig at rad/s, therefore, the egie may be balaced by a rotatig balace mass. Secodary force, F S mr si This is maximum, whe si θ is maximum i.e. whe si θ ± or θ 45 or 35. Maximum resultat secodary force, F S(max) F S (max) mr (5.37) ( l / r) 0.3 / N (Substitutig θ 45) Page.48 Darsha Istitute of Egieerig & Techology, Rajkot

73 Dyamics of Machiery (690). Dyamics of Reciprocatig Egies. Balacig of Radial Egie A radial egie is a multi-cylider egie i which all the coectig rods are coected to a commo crak. The aalysis of forces i such type of egies is much simplified by usig the method of direct ad reverse craks. As all the forces are i the same plae, o ubalace couples exist. I a reciprocatig egie [Fig..46(a)], Fig..46 Primary force: mr cos θ (alog lie of stroke) I the method of direct ad reverse craks, a force idetical to this force is geerated by two masses i the followig way: A mass m/, placed at the crak pi A ad rotatig at a agular velocity i the give directio [Fig..46(b)]. A mass m/, placed at the crak pi of a imagiary crak OA' at the same agular positio as the real crak but i the opposite directio of the lie of stroke. This imagiary crak is assumed to rotate at the same agular velocity i the opposite directio to that of the real crak. Thus, while rotatig; the two masses coicide oly o the cylider cetre lie. Now, the compoets of cetrifugal force due to rotatig masses alog lie of stroke are Darsha Istitute of Egieerig & Techology, Rajkot Page.49

74 . Dyamics of Reciprocatig Egies Dyamics of Machiery (690) Due to mass at A m r cos m Due to mass at A r cos Thus, total force alog lie of stroke mr cos θ which is equal to the primary force. At ay istat, the compoets of the cetrifugal forces of these two masses ormal to the lie of stroke will be equal ad opposite. The crak rotatig i the directio of egie rotatio is kow as the direct crak ad the imagiary crak rotatig i the opposite directio is kow as the reverse crak. Secodary acceleratig force cos cos mr mr( ) 4 r m ( ) cos (alog lie of stroke) 4 This force ca also be geerated by two masses i a similar way as follows: A mass m/, placed at the ed of direct secodary crak of legth r/(4) at agle ad rotatig at a agular velocity i the give directio [Fig. 6.(c)]. A mass m/, placed at the ed of reverse secodary crak of legth r/(4) at agle - rotatig at a agular velocity i the opposite directio. Now, the compoets of cetrifugal force due to rotatig masses alog lie of stroke are m r mr Due to mass at C ( ) cos cos 4 m r mr Due to mass at C ' ( ) cos cos 4 m r mr Thus total force alog lie of stroke ( ) cos cos 4 Which is equal to the secodary force. Page.50 Darsha Istitute of Egieerig & Techology, Rajkot

75 3 Mechaical Vibratio Course Cotets 3. Itroductio to Mechaical Vibratios 3. Sigle Degrees of Freedom System (Liear ad Torsioal) 3.3 Two Degrees of Freedom System 3.4 Multi degree freedom systems ad aalysis (Free vibratios) 3.5 Vibratios of Cotiuous Systems (Free Vibratios) 3.7 Rotatig ubalace 3.8 Vibratio Measuremet Darsha Istitute of Egieerig & Techology, Rajkot Page 3.

76 3. Mechaical Vibratio Dyamics of Machiery (690) 3. Itroductio to Mechaical Vibratios Itroductio: Whe a elastic body such as, a sprig, a beam ad a shaft are displaced from the equilibrium positio by the applicatio of exteral forces, ad the released, they execute a vibratory motio, due to the elastic or strai eergy preset i the body. Whe the body reaches the equilibrium positio, the whole of the elastic or stai eergy is coverted ito kietic eergy due to which the body cotiues to move i the opposite directio. The etire KE is agai coverted ito strai eergy due to which the body agai returs to the equilibrium positio. Hece the vibratory motio is repeated idefiitely. Oscillatory motio is ay patter of motio where the system uder observatio moves back ad forth across some equilibrium positio, but dose ot ecessarily have ay particular repeatig patter. Periodic motio is a specific form of oscillatory motio where the motio patter repeats itself with a uiform time iterval. This uiform time iterval is referred to as the period ad has uits of secods per cycle. The reciprocal of the period is referred to as the frequecy ad has uits of cycles per secod. This uit of combiatio has bee give a special uit symbol ad is referred to as Hertz (Hz) Harmoic motio is a specific form of periodic motio where the motio patter ca be describe by either a sie or cosie. This motio is also sometimes referred to as simple harmoic motio. Because the sie or cosie techically used agles i radias, the frequecy term expressed i the uits radias per secods (rad/sec). This is sometimes referred to as the circular frequecy. The relatioship betwee the frequecy i Hz (cps) ad the frequecy i rad/sec is simply the relatioship. π rad/sec. Natural frequecy is the frequecy at which a udamped system will ted to oscillate due to iitial coditios i the absece of ay exteral excitatio. Because there is o dampig, the system will oscillate idefiitely. Damped atural frequecy is frequecy that a damped system will ted to oscillate due to iitial coditios i the absece of ay exteral excitatio. Because there is dampig i the system, the system respose will evetually decay to rest. Resoace is the coditio of havig a exteral excitatio at the atural frequecy of the system. I geeral, this is udesirable, potetially producig extremely large system respose. Degrees of freedom: The umbers of degrees of freedom that a body possesses are those ecessary to completely defie its positio ad orietatio i space. This is useful i several fields of study such as robotics ad vibratios. Cosider a spherical object that ca oly be positioed somewhere o the x axis. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.

77 3. Mechaical Vibratio Dyamics of Machiery (690) This eeds oly oe dimesio, x to defie the positio to the cetre of gravity so it has oe degree of freedom. If the object was a cylider, we also eed a agle θ to defie the orietatio so it has two degrees of freedom. Now cosider a sphere that ca be positioed i Cartesia coordiates aywhere o the z plae. This eeds two coordiates x ad y to defie the positio of the cetre of gravity so it has two degrees of freedom. A cylider, however, eeds the agle θ also to defie its orietatio i that plae so it has three degrees of freedom. I order to completely specify the positio ad orietatio of a cylider i Cartesia space, we would eed three coordiates x, y ad z ad three agles relative to each agle. This makes six degrees of freedom. A rigid body i space has (x,y,z,θ x θ y θ z ). I the study of free vibratios, we will be costraied to oe degree of freedom. Types of Vibratios: Free or atural vibratios: A free vibratio is oe that occurs aturally with o eergy beig added to the vibratig system. The vibratio is started by some iput of eergy but the vibratios die away with time as the eergy is dissipated. I each case, whe the body is moved away from the rest positio, there is a atural force that tries to retur it to its rest positio. Free or atural vibratios occur i a elastic system whe oly the iteral restorig forces of the system act upo a body. Sice these forces are proportioal to the displacemet of the body from the equilibrium positio, the acceleratio of the body is also proportioal to the displacemet ad is always directed towards the equilibrium positio, so that the body moves with SHM. Figure. Examples of vibratios with sigle degree of freedom. Note that the mass o the sprig could be made to swig like a pedulum as well as boucig up ad dow ad this would be a vibratio with two degrees of freedom. The umber of degrees of freedom of the system is the umber of differet modes of vibratio which the system may posses. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.3

78 3. Mechaical Vibratio Dyamics of Machiery (690) The motio that all these examples perform is called SIMPLE HARMONIC MOTION (S.H.M.). This motio is characterized by the fact that whe the displacemet is plotted agaist time, the resultig graph is basically siusoidal. Displacemet ca be liear (e.g. the distace moved by the mass o the sprig) or agular (e.g. the agle moved by the simple pedulum). Although we are studyig atural vibratios, it will help us uderstad S.H.M. if we study a forced vibratio produced by a mechaism such as the Scotch Yoke. Simple Harmoic Motio The wheel revolves at radias/sec ad the pi forces the yoke to move up ad dow. The pi slides i the slot ad Poit P o the yoke oscillates up ad dow as it is costraied to move oly i the vertical directio by the hole through which it slides. The motio of poit P is simple harmoic motio. Poit P moves up ad dow so at ay momet it has a displacemet x, velocity v ad a acceleratio a. Figure The pi is located at radius R from the cetre of the wheel. The vertical displacemet of the pi from the horizotal cetre lie at ay time is x. This is also the displacemet of poit P. The yoke reaches a maximum displacemet equal to R whe the pi is at the top ad R whe the pi is at the bottom. This is the amplitude of the oscillatio. If the wheel rotates at radia/sec the after time t secods the agle rotated is θ t radias. From the right agle triagle we fid x R Si(t) ad the graph of x - θ is show o figure 3a. Velocity is the rate of chage of distace with time. The plot is also show o figure 3a. v dx/dt R Cos(t). The maximum velocity or amplitude is R ad this occurs as the pi passes through the horizotal positio ad is plus o the way up ad mius o the way dow. This makes sese sice the tagetial velocity of a poit movig i a circle is v R ad at the horizotal poit they are the same thig. Acceleratio is the rate of chage of velocity with time. The plot is also show o figure 3a. a dv/dt - R Si(- R) The amplitude is R ad this is positive at the bottom ad mius at the top (whe the yoke is about to chage directio) Sice R Si(R) x; the substitutig x we fid a - x This is the usual defiitio of S.H.M. The equatio tells us that ay body that performs siusoidal motio must have a acceleratio that is directly proportioal to the displacemet ad is always directed to the poit of zero displacemet. The costat of proportioality is. Ay vibratig body that has a motio that ca be described i this way must vibrate with S.H.M. ad have the same equatios for displacemet, velocity ad acceleratio. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.4

79 3. Mechaical Vibratio Dyamics of Machiery (690) FIGURE 3a FIGURE 3b Agular Frequecy, Frequecy ad Periodic time is the agular velocity of the wheel but i ay vibratio such as the mass o the sprig, it is called the agular frequecy as o physical wheel exists. The frequecy of the wheel i revolutios/secod is equivalet to the frequecy of the vibratio. If the wheel rotates at rev/s the time of oe revolutio is ½ secods. If the wheel rotates at 5 rev/s the time of oe revolutio is / 5 secod. If it rotates at f rev/s the time of oe revolutio is / f. This formula is importat ad gives the periodic time. Periodic Time T time eeded to perform oe cycle. f is the frequecy or umber of cycles per secod. It follows that: T / f ad f / T Each cycle of a oscillatio is equivalet to oe rotatio of the wheel ad revolutio is a agle of π radias. Whe θ π ad t T. It follows that sice θ t; the π T Rearrage ad θ π / T. Substitutig T / f, the πf Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.5

80 3. Mechaical Vibratio Dyamics of Machiery (690) Equatios of S.H.M. Cosider the three equatios derived earlier. Displacemet x R Si(t). Velocity v dx/dt R Cos(t) ad Acceleratio a dv/dt - R Si(t) The plots of x, v ad a agaist agle θ are show o figure 3a. I the aalysis so far made, we measured agle θ from the horizotal positio ad arbitrarily decided that the time was zero at this poit. Suppose we start the timig after the agle has reached a value of φ from this poit. I these cases, φ is called the phase agle. The resultig equatios for displacemet, velocity ad acceleratio are the as follows. Displacemet x R Si(t + φ). Velocity v dx/dt R Cos(t + φ). Acceleratio a dv/dt - R Si((t + φ). The plots of x, v ad a are the same but the vertical axis is displaced by φ as show o figure 3b. A poit to ote o figure 3a ad 3b is that the velocity graph is shifted ¼ cycle (90 o ) to the left ad the acceleratio graph is shifted a further ¼ cycle makig it ½ cycle out of phase with x. Forced vibratios: Whe the body vibrates uder the ifluece of exteral force, the the body is said to be uder forced vibratios. The exteral force, applied to the body is a periodic disturbig force created by ubalace. The vibratios have the same frequecy as the applied force. (Note: Whe the frequecy of exteral force is same as that of the atural vibratios, resoace takes place) Damped vibratios: Whe there is a reductio i amplitude over every cycle of vibratio, the motio is said to be damped vibratio. This is due to the fact that a certai amout of eergy possessed by the vibratig system is always dissipated i overcomig frictioal resistace to the motio. Types of free vibratios: Liear / Logitudial vibratios: Whe the disc is displaced vertically dowwards by a exteral force ad released as show i the figure 4, all the particles of the rod ad disc move parallel to the axis of shaft. The rod is elogated ad shorteed alterately ad thus the tesile ad compressive stresses are iduced alterately i the rod. The vibratio occurs is kow as Liear/Logitudial vibratios. Trasverse vibratios: Whe the rod is displaced i the trasverse directio by a exteral force ad released as show i the figure 5, all the particles of rod ad disc move approximately perpedicular to the axis of the rod. The shaft is straight ad beds alterately iducig bedig stresses i the rod. The vibratio occurs is kow as trasverse vibratios. Torsioal vibratios: Whe the rod is twisted about its axis by a exteral force ad released as show i the figure 6, all the particles of the rod ad disc move i a circle about the axis of the rod. The rod is subjected to twist ad torsioal shear stress is iduced. The vibratio occurs is kow as torsioal vibratios. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.6

81 3. Mechaical Vibratio Dyamics of Machiery (690) ROD ROD ROD DISC A O A DISC O B DISC A B FIGURE 5 O FIGURE 4 Oscillatio of a floatig body: FIGURE 6 B You may have observed that some bodies floatig i water bob up ad dow. This is aother example of simple harmoic motio ad the restorig force i this case is buoyacy. Cosider a floatig body of mass M kg. Iitially it is at rest ad all the forces actig o it add up to zero. Suppose a force F is applied to the top to push it dow a distace x. The applied force F must overcome this buoyacy force ad also overcome the iertia of the body. Buoyacy force: The pressure o the bottom icreases by p ρ g x. The buoyacy force pushig it up o the bottom is Fb ad this icreases by p A. Substitute for p ad Fb ρ g x A Iertia force: The iertia force actig o the body is Fi M a Balace of forces: The applied force must be F Fi + Fb -this must be zero if the vibratio is free. 0 Ma + ρ g x A ρag a x M This shows that the acceleratio is directly proportioal to displacemet ad is always directed towards the rest positio so the motio must be simple harmoic ad the costat of proportioality must be the agular frequecy squared. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.7

82 3. Mechaical Vibratio Dyamics of Machiery (690) f ρag M ρag M π π ρag M Example: A cylidrical rod is 80 mm diameter ad has a mass of 5 kg. It floats vertically i water of desity 036 kg/m3. Calculate the frequecy at which it bobs up ad dow. (As Hz) Pricipal of super positio: The pricipal of super positio meas that, whe TWO or more waves meet, the wave ca be added or subtracted. Two waveforms combie i a maer, which simply adds their respective Amplitudes liearly at every poit i time. Thus, a complex SPECTRUM ca be built by mixig together differet Waves of various amplitudes. The priciple of superpositio may be applied to waves wheever two (or more) waves travelig through the same medium at the same time. The waves pass through each other without beig disturbed. The et displacemet of the medium at ay poit i space or time, is simply the sum of the idividual wave dispacemets. Geeral equatio of physical systems is: m & x&+ cx& + kx F(t) - This equatio is for a liear system, the iertia, dampig ad sprig force are liear fuctio & x, x& ad x respectively. This is ot true case of oliear systems. m & x&+ φ( x& ) + f ( x) F( t) - Dampig ad sprig force are ot liear fuctios of x& ad x Mathematically for liear systems, if xis a solutio of; m & x&+ cx& + kx F ( t) ad x is a solutio of; m & x&+ cx& + kx F ( t) the x + ) is a solutio of; ( x m & x&+ cx& + kx F ( t) + F ( t) Law of superpositio does ot hold good for o-liear systems. If more tha oe wave is travelig through the medium: The resultig et wave is give by the Superpositio Priciple give by the sum of the idividual waveforms Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.8

83 δ 3. Mechaical Vibratio Dyamics of Machiery (690) 3. (A) Udamped Free Vibratios Udamped Free Vibratiois: NATURAL FREQUENCY OF FREE LONGITUDINAL VIBRATION Equilibrium Method: Cosider a body of mass m suspeded from a sprig of egligible mass as show i the figure 4. Let m Mass of the body W Weight of the body mg K Stiffess of the sprig δ Static deflectio of the sprig due to W sprig k Stiffess By applyig a exteral force, assume the body is displaced vertically by a distace x, from the equilibrium positio. O the release of exteral force, the ubalaced forces ad acceleratio imparted to the body are related by Newto Secod Law of motio. The restorig force F - k x (-ve sig idicates, the restorig force k.x is opposite to the directio of the displacemet x ) By Newto s Law; F m a Wkδ m W x kx m m Ustraied postio x O x FIGURE 7 d x F k x m dt The differetial equatio of motio, if a body of mass m is acted upo by a restorig force k per uit displacemet from the equilibrium positio is; k(δ+x) mg m d x k + x 0 dt m This equatio represets SHM d x k + x 0 for SHM dt m π m The atural period of vibratio is T π Sec k The atural frequecy of vibratio is f π k m cycles / sec From the figure 7; whe the sprig is straied by a amout of δ due to the weight W mg δ k mg Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.9

84 3. Mechaical Vibratio Dyamics of Machiery (690) Hece k m f g δ π g δ Hz or cps Eergy method: The equatio of motio of a coservative system may be established from eergy cosideratios. If a coservative system set i motio, the mechaical eergy i the system is partially kietic ad partially potetial. The KE is due to the velocity of mass ad the PE is due to the stai eergy of the sprig by virtue of its deformatio. Sice the system is coservative; ad o eergy is trasmitted to the system ad from the system i the free vibratios, the sum of PE ad KE is costat. Both velocity of the mass ad deformatio of sprig are cyclic. Thus, therefore be costat iterchage of eergy betwee the mass ad the sprig. (KE is maximum, whe PE is miimum ad PE is maximum, whe KE is miimum - so system goes through cyclic motio) KE + PE Costat d dt [ KE +PE] 0 () dx KE m v m -() We have dt Potetial eergy due to the displacemet is equal to the strai eergy i the sprig, mius the PE chage i the elevatio of the mass. PE x 0 x 0 ( Total sprig forece) ( mg + kx mg) dx kx (3) dx mg dx sprig k Stiffess m FIGURE 8 Static Equilibrium postio x Equatio () becomes d dt dx dt + kx d x dx m + kx 0 dt dt kx PE x kx d x Either m + kx 0 OR dt dx But velocity ca be zero dt d x m + kx 0 dt d x k + x 0 dt m dx 0 dt for all values of time. [ m& x + kx 0] Equatio represets SHM Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.0

85 3. Mechaical Vibratio Dyamics of Machiery (690) Time period T π m k sec Natural frequecy of vibratio f ad π k m cycles/sec (The atural frequecy is iheret i the system. It is the fuctio of the system parameters 'k' ad 'm' ad it is idepedet of the amplitude of oscillatio or the maer i which the system is set ito motio.) Rayleigh s Method: The cocept is a extesio of eergy method. We kow, there is a costat iterchage of eergy betwee the PE of the sprig ad KE of the mass, whe the system executes cyclic motio. At the static equilibrium positio, the KE is maximum ad PE is zero; similarly whe the mass reached maximum displacemet (amplitude of oscillatio), the PE is maximum ad KE is zero (velocity is zero). But due to coservatio of eergy total eergy remais costat. Assumig the motio executed by the vibratio to be simple harmoic, the; x A Sit x displacemet of the body from the mea positio after time 't' sec ad A Maximum displacemet from the mea positio x& ASit At mea positio, t 0; Velocity is maximum v max Maximum Maximum We kow dx dt P.E P.E max Maximum K. E x max ( KE ) ( PE ) max m m k x ka A max k A max x max A f π k m / k m Qδ k m g k g m δ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.

86 3. Mechaical Vibratio Dyamics of Machiery (690). Determie the atural frequecy of the sprig-mass system, takig mass of the sprig ito accout. Let l Legth of the sprig uder equilibrium coditio ρ Mass/uit legth of the sprig Mass of the sprig ρ l m s Cosider a elemetal legth of 'dy' of the sprig at a distace 'y' from support. Mass of the elemet ρ dy At ay istat, the mass 'm' is displaced by a distace 'x' from equilibrium positio. P E k x K E of the system at this istat, is the sum of (KE) mass ad (KE) sprig K E G m x& l + 0 m x& + m x& + m x& + ( ρ dy ) y l x& l x& ρ y dy l 0 3 x& l ρ l 3 m m x& s m s x& l y dy k Sprig stiffess m x k x ms + m + x& 3 Differeti atig with respect to ' t ' ; ms k x x& + m + x& && x 0 3 k && x + x 0 ms m + 3 f OR f π k ms m k ms m + 3 cps radias We kow that PE + KE Costat Differeti / sec d dt ( PE + KE ) [ ρ l m ] 0 al equatio s Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.

87 3. Mechaical Vibratio Dyamics of Machiery (690). Determie the atural frequecy of the system show i figure by Eergy ad Newto's method. Whe mass 'm' moves dow a distace 'x' from its equilibrium positio, the ceter of the disc if mass m moves dow by x ad rotates though ad agle θ. k x/ x r θ x& θ& r KE ( KE ) Tr + ( KE ) rot θ x/ Io m Disc x/ PE d dt G ( KE + PE ) 3 m x& && x + m x& && x + 8 m + 3 m && x + 8 or m x& + m x& + m x& + k f f x π x& m + 4 m x& + m x& k 8 m + 3 m k 8 m + 3 m k x& x kx x& m r 4 r I & o θ 0 rad/ cps sec x m F Newto's Method: Use x, as co-ordiate or mass m; m && x F for disc m; && x kx m F + F I && o θ Fr F r Substituti g () i () ad (3) ad replace && x kx m m && x + F && x Io m && x r F r r && x Io m && x F r 3 && && x θ by r Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.3

88 3. Mechaical Vibratio Dyamics of Machiery (690) Addig equatios (4) ad (6) && x && x m + Io r kx m&& x { Io m r } 3. Determie the atural frequecy of the system show i figure by Eergy ad Newto's method. Assume the cylider rolls o the surface without slippig. a) Eergy method: Whe mass 'm' rotates through a agle θ, the ceter of the roller move distace 'x' m x m && + + 4m + kx 0 f f π k 3m + 8m rad / sec k cps or Hz 3m + 8m KE ( KE ) Tr + ( KE ) rot KE PE d dt ( KE + PE ) 3 Natural m x& + m x& k x m && x + m x& k x 0 0 frequecy I o θ& m r f x& r π k 3 m Roller/Disc Fr x Io m x r kx θ k Newto s method: ma F m&& x kx + Fr usig torque equatio mx && Fr mx && k x m&& x 3 mx && + k x 0 f π k cps 3m I && oθ Fr. r or f k 3m rad /sec Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.4

89 3. Mechaical Vibratio Dyamics of Machiery (690) 4. Determie the atural frequecy of the system show i figure by Eergy ad Newto's method. Eergy Method: Use θ or x as coordiate KE ( KE) + ( KE) KE m x& + m x& 4 PE k x d dt m x& + I & o θ x& m x& + m r r ( KE + PE) k x + Tr 0 k x&& x + m + 4 m + m Natural frequecy rot m x& && x 0 x& 0 f k m+ m rad / sec k m+ m G Newto s Method: Cosider motio of the disc with θ as coordiate. For the mass m : m & x& F. r π cps OR () π kr Io + m r For the disc : I & θ F. r kr θ () Substitute () i (): I && o θ m r && x k r θ I && o θ m r && θ k r θ && θ + k r θ 0 ( Io + m r ) f π k r Io + m r π k r Io + m r k m + m rad cps / sec cps or o & Hz r Io m r x θ m F Io m k Disc krθ cps orhz xrθ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.5

90 3. Mechaical Vibratio Dyamics of Machiery (690) 5. Determie the atural frequecy of the system show i figure 5 by Eergy ad Newto's method. Assume the cylider rolls o the surface without slippig. Eergy Method: KE ( KE ) + ( KE ) Tr rot k c θ k m x& + I & o θ m r & θ + m r & θ 3 KE m r & θ 4 PE k x k ( r + a) θ d dt ( KE + PE ) 3 m r & θ && θ + k 3 m r && θ + 4 k Natural 0 frequecy ( r + a ) ( r + a ) f θ & θ 0 θ 0 ( r + a ) 4 k ( r + a ) 4 k 3 m r rad / sec π 3 m r Newto s Method: Cosiderig combied traslatio ad rotatioal motio as show i Figure a. Hece it must satisfy: m&& x Force i x directio ad Natural m&& x F k ( x + aθ ) I && θ Torque about o F r k m r && θ F frequecy ( x + aθ ) k( r + a) ad m r && θ F r k Multiply equatio () by r 3m r && θ 4kr ( r + a) θ 3m r && θ + 4k[ r ( r + a) + 3m r && θ + 4k ( r + a) 0 f θ ' θ ' a ( r + a) aθ ad () by. The add 4k( r + a) ( r + a) a] θ ( r + a ) 4 k ( r + a ) 4 k 3m r aθ 0 () () rad / sec π a Io m FIGURE 5 o 3m r a (Appx) x r x o' c' Figure 'a' cps cps (x+aθ) +x or or F Hz k(x+aθ) Hz Refer PPT For more problems Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.6

91 3. Mechaical Vibratio Dyamics of Machiery (690) Itroductio: 3. (B) Damped Freee Vibratios Sigle DoF Dampig dissipatio of eergy. For a system to vibrate, it requires eergy. Durig vibratio of the system, there will be cotiuous trasformatio of eergy. Eergy will be trasformed from potetial/strai to kietic ad vice versa. I case of udamped vibratios, there will ot be ay dissipatio of eergy ad the system vibrates at costat amplitude. Ie, oce excited, the system vibrates at costat amplitude for ifiite period of time. But this is a purely hypothetical case. But i a actual vibratig system, eergy gets dissipated from the system i differet forms ad hece the amplitude of vibratio gradually dies dow. Fig. shows typical respose curves of udamped ad damped free vibratios. Types o dampig: (i) Viscous dampig I this type of dampig, the dampig resistace is proportioal to the relative velocity betwee the vibratig system ad the surroudigs. For this type of dampig, the differetial equatio of the system becomes liear ad hece the aalysis becomes easier. A schematic represetatio of viscous damper is show i Fig.. Here, F α x& or F cx&, where, F is dampig resistace, x& is relative velocity ad c is the dampig coefficiet. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.7

92 3. Mechaical Vibratio Dyamics of Machiery (690) (ii) Dry frictio or Coulomb dampig I this type of dampig, the dampig resistace is idepedet of rubbig velocity ad is practically costat. (iii) Structural dampig This type of dampig is due to the iteral frictio withi the structure of the material, whe it is deformed. Sprig-mass-damper system: Fig.3 shows the schematic of a simple sprig-mass-damper system, where, m is the mass of the system, k is the stiffess of the system ad c is the dampig coefficiet. If x is the displacemet of the system, from Newto s secod law of motio, it ca be writte m& x cx& kx Ie m & x + cx& + kx 0 () This is a liear differetial equatio of the secod order ad its solutio ca be writte as Differetiatig (), dx dt st x e () x& st se d x st & x s e dt st st st Substitutig i (), ms e + cse + ke 0 st ms + cs + k e ( ) 0 Or ms + cs + k 0 (3) Equatio (3) is called the characteristic equatio of the system, which is quadratic i s. The two values of s are give by s, c m ± c m k m (4) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.8

93 3. Mechaical Vibratio Dyamics of Machiery (690) The geeral solutio for () may be writte as st st x C e + Ce (5) Where, C ad C are arbitrary costats, which ca be determied from the iitial coditios. c I equatio (4), the values of s s, whe m c k Or, (6) m m Or c m, which is the property of the system ad is called critical dampig coefficiet ad is represeted by c c. Ie, critical dampig coefficiet cc m The ratio of actual dampig coefficiet c ad critical dampig coefficiet c c is called dampig factor or dampig ratio ad is represeted by ζ. c Ie, ζ (7) I equatio (4), s c m ca be writte as c c c m c c c k m cc ζ. m ζ. ± ζ. ζ ± ζ (8) Therefore,, ( ) [ ] The system ca be aalyzed for three coditios. (i) (ii) (iii) ζ >, ie, c > c c, which is called over damped system. ζ. ie, c c c, which is called critically damped system. ζ <, ie, c < c c, which is called uder damped system. Depedig upo the value of ζ, value of s i equatio (8), will be real ad uequal, real ad equal ad complex cojugate respectively. (i) Aalysis of over-damped system (ζ > ). I this case, values of s are real ad are give by s [ ζ + ζ ] ad s [ ζ ζ ] The, the solutio of the differetial equatio becomes x C e ζ + ζ t ζ ζ + C e t This is the fial solutio for a over damped system ad the costats C ad C are obtaied by applyig iitial coditios. Typical respose curve of a over damped system is show i fig.4. The amplitude decreases expoetially with time ad becomes zero at t. (9) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.9

94 3. Mechaical Vibratio Dyamics of Machiery (690) (ii) Aalysis of critically damped system (ζ ). I this case, based o equatio (8), s s - The solutio of the differetial equatio becomes st st x C e + Cte Ie, x t C e + C te t t Or, ( ) x C + C t e (0) This is the fial solutio for the critically damped system ad the costats C ad C are obtaied by applyig iitial coditios. Typical respose curve of the critically damped system is show i fig.5. I this case, the amplitude decreases at much faster rate compared to over damped system. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.0

95 3. Mechaical Vibratio Dyamics of Machiery (690) (iii) Aalysis of uder damped system (ζ < ). I this case, the roots are complex cojugates ad are give by [ ζ + j ζ ] [ ζ j ζ ] s s The solutio of the differetial equatio becomes x C e ζ + j ζ t + C e ζ j ζ t This equatio ca be rewritte as x e ζt Usig the relatioships j ζ t j ζ t C e + Ce () θ e i cosθ + isiθ θ e i Equatio () ca be writte as cosθ isiθ t x e [ C { ζ t j ζ t} C { ζ t j ζ t} ] ζ cos + si + cos si t Or x e [( C C ){ t} j( C C ){ t} ] ζ ζ ζ + cos + si () I equatio (), costats (C +C ) ad j(c -C ) are real quatities ad hece, the equatio ca also be writte as t x e [ A{ ζ t} B{ ζ t} ] ζ cos + si ζ t Or, x A e [{ si( ζ t + φ )} ] (3) The above equatios represet oscillatory motio ad the frequecy of this motio is represeted by ζ (4) d Where, d is the damped atural frequecy of the system. Costats A ad Φ are determied by applyig iitial coditios. The typical respose curve of a uder damped system is show i Fig.6. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.

96 3. Mechaical Vibratio Dyamics of Machiery (690) Applyig iitial coditios, equatio (3) becomes x x X o at t 0; ad x& 0 at t 0, ad fidig costats A ad Φ, X o ζ ζ t e si ζ + t ta (5) ζ ζ The term X o ζ e t ζ represets the amplitude of vibratio, which is observed to decay expoetially with time. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.

97 3. Mechaical Vibratio Dyamics of Machiery (690) LOGARITHMIC DECREMENT Referrig to Fig.7, poits A & B represet two successive peak poits o the respose curve of a uder damped system. X A ad X B represet the amplitude correspodig to poits A & B ad t A & t B represets the correspodig time. We kow that the atural frequecy of damped vibratio ζ rad/sec. Therefore, Hece, time period of oscillatio t From equatio (5), amplitude of vibratio X A d f d cycles/sec π B d π π t A sec (6) f d d ζ X o ζ e ζ t A X B X o ζ e ζ t B Or, X X A B e ζ ( t t ) ζ ( t t ) A B e B A Usig eq. (6), Or, X X A B log e X πζ ζ A e X B πζ ζ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.3

98 3. Mechaical Vibratio Dyamics of Machiery (690) This is called logarithmic decremet. It is defied as the logarithmic value of the ratio of two successive amplitudes of a uder damped oscillatio. It is ormally deoted by δ. Therefore, δ log X A e X B πζ ζ This idicates that the ratio of ay two successive amplitudes of a uder damped system is costat ad is a fuctio of dampig ratio of the system. For small values of ζ, δ πζ If X 0 represets the amplitude at a particular peak ad X represets the amplitude after (7) X cycles, the, logarithmic decremet δ log 0 X log X X e e X 0 X X e... X X X Addig all the terms, δ log X 0 Or, δ log (8) e X log e X X Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.4

99 3. Mechaical Vibratio Dyamics of Machiery (690) Solved problems ) The mass of a sprig-mass-dashpot system is give a iitial velocity 5, where is the udamped atural frequecy of the system. Fid the equatio of motio for the system, whe (i) ζ.0, (ii) ζ.0, (i) ζ 0.. Solutio: Case (i) For ζ.0 Over damped system For over damped system, the respose equatio is give by x C e ζ + ζ t ζ ζ t + Ce [ ] [ ] t 0.7 t Substitutig ζ.0, x C e + C e (a) Differetiatig, 0.7 t 3.73 t x& 0.7Ce Ce (b) Substitutig the iitial coditios x 0 at t 0; ad x& 5 at t 0 i (a) & (b), 0 C + C (c) C 3.73 C (d) Solvig (c) & (d), C.44 ad C Therefore, the respose equatio becomes [ 0.7] [ ] ( e ) t e t x.44 (e) Case (ii) For ζ.0 Critically damped system For critically damped system, the respose equatio is give by x t ( C + C t) e (f) & (g) t t Differetiatig, x ( C + C t) e + C e Substitutig the iitial coditios x 0 at t 0; ad x& 5 at t 0 i (f) & (g), C 0 ad C 5 Substitutig i (f), the respose equatio becomes t ( t) e x 5 (h) Case (iii) For ζ 0. uder damped system For uder damped system, the respose equatio is give by x A e [{ si ζ t + φ} ] ζ t 0. t Substitutig ζ 0., x A e [{ si( 0. t + φ )}] 98 (p) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.5

100 3. Mechaical Vibratio Dyamics of Machiery (690) Differetiatig, 0. t 0. t x& 0. A e [{ si( 0.98 t + φ )}] A e cos( t + φ ) (q) Substitutig the iitial coditios x 0 at t 0; ad A siφ 0 ad A cosφ 5. Solvig, A 5. ad Φ 0 Substitutig i (p), the respose equatio becomes x& 5 at t 0 i (p) & (q), [{ si( 0.98 t) }] 0. t x 5.e (r) ) A mass of 0kg is supported o two isolators as show i fig.q.. Determie the udamped ad damped atural frequecies of the system, eglectig the mass of the isolators. Solutio: Equivalet stiffess ad equivalet dampig coefficiet are calculated as k eq + k k C eq C + C Udamped atural frequecy k eq m f. 7cps π 0.74rad / sec Damped atural frequecy d ζ C ζ k eq eq m Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.6

101 3. Mechaical Vibratio Dyamics of Machiery (690) Or, d rad / sec 0.57 f d. 68cps π 3) A gu barrel of mass 500kg has a recoil sprig of stiffess 3,00,000 N/m. If the barrel recoils. meters o firig, determie, (a) iitial velocity of the barrel (b) critical dampig coefficiet of the dashpot which is egaged at the ed of the recoil stroke (c) time required for the barrel to retur to a positio 50mm from the iitial positio. Solutio: (a) Strai eergy stored i the sprig at the ed of recoil: P kx N m Kietic eergy lost by the gu barrel: T mv 500 v 50v, where v iitial velocity of the barrel Equatig kietic eergy lost to strai eergy gaied, ie T P, 50v 6000 v 9.39m/s (b) Critical dampig coefficiet C c km N sec/ m (c) Time for recoilig of the gu (udamped motio): k Udamped atural frequecy 4.49r / s m 500 π π Time period τ 0.59sec 4.9 τ 0.59 Time of recoil 0.065sec 4 4 Time take durig retur stroke: x C + C t e t Respose equatio for critically damped system ( ) & t t Differetiatig, x C e ( C + C t) e Applyig iitial coditios, x., at t 0 ad x& 0 at t 0, C., & C t Therefore, the respose equatio x ( t) e Whe x 0.05m, by trial ad error, t 0.0 sec Therefore, total time take time for recoil + time for retur sec The displacemet time plot is show i the followig figure. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.7

102 3. Mechaical Vibratio Dyamics of Machiery (690) 4) A 5 kg mass is restig o a sprig of 4900 N/m ad dashpot of 47 N-se/m i parallel. If a velocity of 0.0 m/sec is applied to the mass at the rest positio, what will be its displacemet from the equilibrium positio at the ed of first secod? Solutio: The above figure shows the arragemet of the system. Critical dampig coefficiet k 4900 Where 4r / s m 5 Therefore, c c c c m N sec/ m c 47 Sice C< C c, the system is uder damped ad ζ 0. c 700 c Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.8

103 3. Mechaical Vibratio Dyamics of Machiery (690) ζ t Hece, the respose equatio is x A e [{ si( ζ t + φ )} ] 0. 4t Substitutig ζ ad, x Ae [{ si( 0. 4t + φ )} ] x A e [{ si( 3. t + φ )}].94t 7.94t.94t Differetiatig, x&.94a e [{ si( 3.7t + φ )}] + 3.7A e cos( 3. t + φ ) Applyig the iitial coditios, x 0, at t 0 ad Φ 0 7 [{ si( φ )}] 3.7A ( ) A + φ cos Sice, Φ 0, A ; A x & 0.0m / s at t Displacemet at the ed of secod x e [{ si( 3.7) }] m 5) A rail road bumper is desiged as a sprig i parallel with a viscous damper. What is the bumper s dampig coefficiet such that the system has a dampig ratio of.5, whe the bumper is egaged by a rail car of 0000 kg mass. The stiffess of the sprig is E5 N/m. If the rail car egages the bumper, while travelig at a speed of 0m/s, what is the maximum deflectio of the bumper? m k c Solutio: Data m 0000 kg; k N/m; ζ. 5 Critical dampig coefficiet c c m k N sec/ m Dampig coefficiet 5 5 C ζ CC N sec/ m k Udamped atural frequecy 3.6r / s m 0000 Sice ζ. 5, the system is over damped. For over damped system, the respose equatio is give by x C e ζ + ζ t ζ ζ t + Ce Substitutig ζ.5, x [ 0.5] t [. 0] t C e + C e (a) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.9

104 3. Mechaical Vibratio Dyamics of Machiery (690) Differetiatig, Substitutig the iitial coditios 0.5t.0t x& 0.5 Ce. 0 Ce (b) x 0 at t 0; ad 0 C + C (c) C.0 C (d) Solvig (c) & (d), C 4.ad C -4. Therefore, the respose equatio becomes [.58 t ] [ 6.3 t ] x 4.( e e )m (e) x & 0m / s at t 0 i (a) & (b), The time at which, maximum deflectio occurs is obtaied by equatig velocity equatio to zero. 0.5t.0t Ie, x& 0.5 C e.0 C e 0.58t 6.3t Ie, 6.65e + 6.6e 0 Solvig the above equatio, t 0.9 secs. Therefore, maximum deflectio at t 0.9secs, [ ] [ ] Substitutig i (e), x 4.( e e )m,.99m. 6) A disc of a torsioal pedulum has a momet of iertia of 6E- kg-m ad is immersed i a viscous fluid. The shaft attached to it is 0.4m log ad 0.m i diameter. Whe the pedulum is oscillatig, the observed amplitudes o the same side of the mea positio for successive cycles are 9 0, 6 0 ad 4 0. Determie (i) logarithmic decremet (ii) dampig torque per uit velocity ad (iii) the periodic time of vibratio. Assume G 4.4E0 N/m, for the shaft material. Shaft dia. d 0.m Shaft legth l 0.4m Momet of iertia of disc J 0.06 kg-m. Modulus of rigidity G 4.4E0 N/m Solutio: The above figure shows the arragemet of the system. 9 6 (i) Logarithmic decremet δ log e log e (ii) The dampig torque per uit velocity dampig coefficiet of the system C. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.30

105 3. Mechaical Vibratio Dyamics of Machiery (690) We kow that logarithmic decremet δ πζ, rearragig which, we get ζ Dampig factor ζ δ π + δ 4π Also, ζ C, where, critical dampig coefficiet C k J C C Torsioal stiffess C GI p G πd π 0. 6 kt.08 0 N m / rad l l Critical dampig coefficiet CC kt J N m / rad Dampig coefficiet of the system C CC ζ N m / rad (iii) Periodic time of vibratio τ Where, udamped atural frequecy π Therefore, τ f d π d k t J π ζ sec t 44.6rad / sec 7) A mass of kg is to be supported o a sprig havig a stiffess of 9800 N/m. The dampig coefficiet is 5.9 N-sec/m. Determie the atural frequecy of the system. Fid also the logarithmic decremet ad the amplitude after three cycles if the iitial displacemet is 0.003m. Solutio: k 9800 Udamped atural frequecy 99r s m / Damped atural frequecy Critical dampig coefficiet c d c ζ c 5.9 Dampig factor ζ c c m 99 98N sec/ m Hece damped atural frequecy ζ d 98.99rad Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.3 / sec

106 3. Mechaical Vibratio Dyamics of Machiery (690) Logarithmic decremet δ πζ π ζ 0.03 X 0 Also, δ log ; if x e , X the, after 3 cycles, δ log ie, X m e X ie,0.88 log e X 3 X e 8) The damped vibratio record of a sprig-mass-dashpot system shows the followig data. Amplitude o secod cycle 0.0m; Amplitude o third cycle 0.005m; Sprig costat k 7840 N/m; Mass m kg. Determie the dampig costat, assumig it to be viscous. Solutio: X 0.0 Here, δ log loge e X πζ δ 0.33 Also, δ, rearragig, ζ 0. 0 ζ 4π + δ 4π Critical dampig coefficiet c c m k N sec/ m Dampig coefficiet C ζ CC N sec/ m 9) A mass of kg is supported o a isolator havig a sprig scale of 940 N/m ad viscous dampig. If the amplitude of free vibratio of the mass falls to oe half its origial value i.5 secods, determie the dampig coefficiet of the isolator. Solutio: k 940 Udamped atural frequecy 38.34r s m / Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.3

107 3. Mechaical Vibratio Dyamics of Machiery (690) Critical dampig coefficiet c c m N sec/ m ζ t Respose equatio of uder damped system x Ae [{ si( ζ t + φ )} ] Here, amplitude of vibratio A e ζ If amplitude X 0 at t 0, the, at t.5 sec, amplitude 0 Ie, Ae ζ X 0 Also,.5 A e ζ or A X 0 X 0 or t ζ X 0 e X 0 or e X 0 ζ ζ Ie, e, takig log, ζ ζ 0. 0 Dampig coefficiet C ζ CC N sec/ m Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.33

108 3. Mechaical Vibratio Dyamics of Machiery (690) Itroductio: 3. (C) Forced Vibratios I free u-damped vibratios a system oce disturbed from its iitial positio executes vibratios because of its elastic properties. There is o dampig i these systems ad hece o dissipatio of eergy ad hece it executes vibratios which do ot die dow. These systems give atural frequecy of the system. I free damped vibratios a system oce disturbed from its positio will execute vibratios which will ultimately die dow due to presece of dampig. That is there is dissipatio of eergy through dampig. Here oe ca fid the damped atural frequecy of the system. I forced vibratio there is a exteral force acts o the system. This exteral force which acts o the system executes the vibratio of the system. The exteral force may be harmoic ad periodic, o-harmoic ad periodic or o periodic. I this chapter oly exteral harmoic forces actig o the system are cosidered. Aalysis of o harmoic forcig fuctios is just a extesio of harmoic forcig fuctios. Examples of forced vibratios are air compressors, I.C. egies, turbies, machie tools etc,. Aalysis of forced vibratios ca be divided ito followig categories as per the syllabus.. Forced vibratio with costat harmoic excitatio. Forced vibratio with rotatig ad reciprocatig ubalace 3. Forced vibratio due to excitatio of the support A: Absolute amplitude Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.34

109 3. Mechaical Vibratio Dyamics of Machiery (690) B: Relative amplitude 4. Force ad motio trasmissibility For the above first a differetial equatio of motio is writte. Assume a suitable solutio to the differetial equatio. O obtaiig the suitable respose to the differetial equatio the ext step is to o-dimesioal the respose. The the frequecy respose ad phase agle plots are draw.. Forced vibratio with costat harmoic excitatio From the figure it is evidet that sprig force ad dampig force oppose the motio of the mass. A exteral excitatio force of costat magitude acts o the mass with a frequecy. Usig Newto s secod law of motio a equatio ca be writte i the followig maer. mx && + cx & + kx F o Si t Equatio is a liear o homogeeous secod order differetial equatio. The solutio to eq. cosists of complimetary fuctio part ad particular Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.35

110 3. Mechaical Vibratio Dyamics of Machiery (690) itegral. The complimetary fuctio part of eq, is obtaied by settig the equatio to zero. This derivatio for complemetary fuctio part was doe i damped free vibratio chapter. x x + x c p The complemetary fuctio solutio is give by the followig equatio. Equatio 3 has two costats which will have to be determied from the iitial coditios. But iitial coditios caot be applied to part of the solutio of eq. as give by eq. 3. The complete respose must be determied before applyig the iitial coditios. For complete respose the particular itegral of eq. must be determied. This particular solutio will be determied by vector method as this will give more isight ito the aalysis. Assume the particular solutio to be [ ] ξ t + φ 3 ζt xc Ae Si x p XSi ( t φ) 4 Differetiatig the above assumed solutio ad substitutig it i eq. m x& x&& p p XSi XSi t φ + XSi F Si t kxsi o π ( t φ + π ) ( t φ) c x t φ + ( t φ + π ) 0 5 π Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.36

111 3. Mechaical Vibratio Dyamics of Machiery (690) Fig. Vector represetatio of forces o system subjected to forced vibratio Followig poits are observed from the vector diagram. The displacemet lags behid the impressed force by a agle Φ.. Sprig force is always opposite i directio to displacemet. 3. The dampig force always lags the displacemet by 90. Dampig force is always opposite i directio to velocity. 4. Iertia force is i phase with the displacemet. The relative positios of vectors ad heir magitudes do ot chage with time. From the vector diagram oe ca obtai the steady state amplitude ad phase agle as follows Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.37

112 3. Mechaical Vibratio Dyamics of Machiery (690) X F [( ) ( ) ] k m + c 6 0 φ ta - [ ( )] c k m 7 The above equatios are made o-dimesioal by dividig the umerator ad deomiator by K. X φ [ ( ) ] + [ ξ ( )] ta - ξ X st ( ) ( ) 9 8 where, X st F o k is zero frequecy deflectio Therefore the complete solutio is give by x x c + x p x + A e ζ t Si [ ] ξ t + φ X stsit ( φ) [ ( ) ] + [ ξ ( )] 0 Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.38

113 3. Mechaical Vibratio Dyamics of Machiery (690) The two costats A ad φ have to be determied from the iitial coditios. The first part of the complete solutio that is the complemetary fuctio decays with time ad vaishes completely. This part is called trasiet vibratios. The secod part of the complete solutio that is the particular itegral is see to be siusoidal vibratio with costat amplitude ad is called as steady state vibratios. Trasiet vibratios take place at damped atural frequecy of the system, where as the steady state vibratios take place at frequecy of excitatio. After trasiets die out the complete solutio cosists of oly steady state vibratios. I case of forced vibratios without dampig equatio 0 chages to x A Si t [ + φ ] X Sit ( ) ( ) st + Φ is either 0 or 80 depedig o whether < or > Steady state Vibratios: The trasiets die out withi a short period of time leavig oly the steady state vibratios. Thus it is importat to kow the steady state behavior of the system, Thus Magificatio Factor (M.F.) is defied as the ratio of steady state amplitude to the zero frequecy deflectio. M.F. X X φ st ta - [ ( ) ] + [ ξ ( )] ξ ( ) ( ) 3 Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.39

114 3. Mechaical Vibratio Dyamics of Machiery (690) Equatios ad 3 give the magificatio factor ad phase agle. The steady state amplitude always lags behid the impressed force by a agle Φ. The above equatios are used to draw frequecy respose ad phase agle plots. Fig. Frequecy respose ad phase agle plots for system subjected forced vibratios. Frequecy respose plot: The curves start from uity at frequecy ratio of zero ad ted to zero as frequecy ratio teds to ifiity. The magificatio factor icreases with the icrease i frequecy ratio up to ad the decreases as frequecy ratio is further icreased. Near resoace the amplitudes are very high ad decrease with the icrease i the dampig ratio. The peak of magificatio factor lies slightly to the left of the resoace lie. This tilt to the left icreases with the icrease i the dampig ratio. Also the sharpess of the peak of the curve decreases with the icrease i the dampig. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.40

115 3. Mechaical Vibratio Dyamics of Machiery (690) Phase agle plot: At very low frequecy the phase agle is zero. At resoace the phase agle is 90. At very high frequecies the phase agle teds to 80. For low values of dampig there is a steep chage i the phase agle ear resoace. This decreases with the icrease i the dampig. The sharper the chage i the phase agle the sharper is the peak i the frequecy respose plot. The amplitude at resoace is give by equatio 4 c X X The frequecy at which maximum amplitude occurs is obtaied by differetiatig the magificatio factor equatio with respect to frequecy ratio ad equatig it to zero. Also o maxima will occur for r r X F st o p ξ ξ 4 5 ξ or ξ Rotatig ad Reciprocatig Ubalace Machies like electric motors, pumps, fas, clothes dryers, compressors have rotatig elemets with ubalaced mass. This geerates cetrifugal type harmoic excitatio o the machie. The fial ubalace is measured i terms of a equivalet mass m o rotatig with its c.g. at a distace e from the axis of rotatio. The cetrifugal force is proportioal to the square of frequecy of rotatio. It varies with the speed Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.4

116 3. Mechaical Vibratio Dyamics of Machiery (690) of rotatio ad is differet from the harmoic excitatio i which the maximum force is idepedet of the frequecy. Let m o Ubalaced mass e eccetricity of the ubalaced mass M Total mass of machie icludig ubalaced mo m o makes a agle t with ref. axis. M o e is the cetrifugal force that acts radially outwards. Equatio of motio is d x d o dt o dt Mx&& + cx& + kx m e ( M m ) + m ( x + esis) o si t kx c dx dt The solutio of followig equatio is give by mx&& + cx& + kx Fo si t x A e + ζ t Si [ ] ξ t + φ X stsi( t φ) [ ( ) ] + [ ξ( )] Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.4

117 Compare eq. with eq. the oly chage is Fo is replaced by m o e The trasiet part of the solutio remais the same. The oly chage is i the steady state part of the solutio. Therefore the steady state solutio of eq. ca be writte as The above equatio reduces to dimesioless form as The phase agle equatio ad its plot remais the same as show below ( ) o k c k M k m e where...x φ t Xsi x + o ξ M m e X + ξ ta φ Darsha Istitute of Egieerig & Techology, Rajkot Page No Mechaical Vibratio Dyamics of Machiery (690)

118 3. Mechaical Vibratio Dyamics of Machiery (690) Frequecy respose ad phase agle plots (Ubalace) At low speeds is small, hece all respose curves start from zero. At resoace /, therefore X m e o M Ad amplitude is limited due to dampig preset i the system. Uder these coditios the motio of mai mass (M-m o ) lags that of the mass m o by 90. Whe / is very large the ratio X/(moe/M) teds to uity ad the mai mass (M-m o ) has a amplitude of X moe/m. This motio is 80 out of phase with the excitig force. That is whe ubalaced mass moves up, the mai mass moves dow ad vice versa. Problem A couter rotatig eccetric weight exciter is used to produce the forced oscillatio of a sprig-supported mass as show i Fig. By varyig the speed ξ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.44

119 3. Mechaical Vibratio Dyamics of Machiery (690) of rotatio, a resoat amplitude of 0.60 cm was recorded. Whe the speed of rotatio was icrease cosiderably beyod the resoat frequecy, the amplitude appeared to approach a fixed value of 0.08 cm. Determie the dampig factor of the system. X m e o M ξ 0.6cm m e o X 0.08cm M Dividig oe by the other ξ (Aswer) Problem A system of beam supports a mass of 00 kg. The motor has a ubalaced mass of kg located at 6 cm radius. It is kow that the resoace occurs at 0 rpm. What amplitude of vibratio ca be expected at the motors operatig speed of 440 rpm if the dampig factor is assumed to be less tha 0. Solutio: Give: M 00 kg, m o kg, eccetricity e 0.06m, Resoace at 0 rpm, Operatig speed 440 rpm, ξ 0., X?. π N π N 3.43 rad / s op 50.79rad/ s r 0.65 X m e o M + ξ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.45

120 3. Mechaical Vibratio Dyamics of Machiery (690) Substitutig the appropriate values i the above eq. X mm (Aswer) However, if ξ is made zero, the amplitude X mm (Aswer) This meas if the dampig is less tha 0., the amplitude of vibratio will be betwee mm ad mm. (Aswer) Problem 3 A eccetric mass exciter is used to determie the vibratory characteristics of a structure of mass 00 kg. At a speed of 000 rpm a stroboscope showed the eccetric mass to be at the bottom positio at the istat the structure was movig dowward through its static equilibrium positio ad the correspodig amplitude was 0 mm. If the ubalace of the eccetric is 0.05 kg-m, determie, (a) u damped atural frequecy of the system (b) the dampig factor of the structure (c) the agular positio of the eccetric at 300 rpm at the istat whe the structure is movig dowward through its equilibrium positio. Solutio: Give: M 00 kg, Amplitude at 000 rpm 0 mm, moe 0.05 kg-m At 000 rpm the eccetric mass is at the bottom whe the structure was movig dowward This meas a there is phase lag of 90 (i.e., at resoace). At resoace. X moe M π N 60 ξ ξ rad/s (Aswer) φ ta φ ξ o (Aswer) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.46

121 3. Mechaical Vibratio Dyamics of Machiery (690) Problem 4 A 40 kg machie is supported by four sprigs each of stiffess 50 N/m. The rotor is ubalaced such that the ubalace effect is equivalet to a mass of 5 kg located at 50mm from the axis of rotatio. Fid the amplitude of vibratio whe the rotor rotates at 000 rpm ad 60 rpm. Assume dampig coefficiet to be 0.5 Solutio: Give: M 40 kg, m o 5 kg, e.05 m, ξ 0.5, N 000 rpm ad 60 rpm. Whe N 000 rpm π N 60 k M 04.67rad/s 5rad/s X m e o M X at 000 rpm 6.6 mm (Solutio) Whe N 60 rpm + ξ π N Usig the same eq. X at 60 rpm 4.9 mm (Solutio) 6.8rad/s Problem 5 A vertical sigle stage air compressor havig a mass of 500 kg is mouted o sprigs havig a stiffess of N/m ad a dampig coefficiet of 0.. The rotatig parts are completely balaced ad the equivalet reciprocatig parts have a mass of 0 kg. The stroke is 0. m. Determie the Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.47

122 3. Mechaical Vibratio Dyamics of Machiery (690) dyamic amplitude of vertical motio ad the phase differece betwee the motio ad excitatio force if the compressor is operated at 00 rpm. Solutio Give: M 500 kg, k N/m, ξ 0., m o 0 kg, stroke 0. m, N 00 rpm, X?. Stroke 0. m, i.e. eccetricity e stroke/ 0. m Usig the equatios X 0. mm ad φ 05.9 (Solutio) (a) c N s/m c c ξ c c c km N s/m 0.75 (Solutio) (b) X 0.5 cm ad φ 69 (Solutio) (c) (d) X m e o M N X r r + ξ φ ta k rad/s M rpm π moe 0.357cm ξ M π 400 c X kx N ξ 7.85N F ( c X ) + ( kx ).4N Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.48

123 3. Mechaical Vibratio Dyamics of Machiery (690) Coclusios o rotatig ad reciprocatig ubalace Ubalace i machies caot be made zero. Eve small ubalaced mass ca produce high cetrifugal force. This depeds o the speed of operatio. Steady state amplitude is determied for a machie subjected ubalaced force excitatio. For reciprocatig machies, the eccetricity ca be take as half the crak radius. Frequecy respose plot starts from zero at frequecy ratio zero ad teds to ed at uity at very high frequecy ratios.. Respose of a damped system uder the harmoic motio of the base I may cases the excitatio of the system is through the support or the base istead of beig applied through the mass. I such cases the support will be cosidered to be excited by a regular siusoidal motio. Example of such base excitatio is a automobile suspesio system excited by a road surface, the suspesio system ca be modeled by a liear sprig i parallel with a viscous damper. such model is depicted i Figure. There are two cases: (a) Absolute Amplitude of mass m (b) Relative amplitude of mass m (a) Absolute Amplitude of mass m m X K C YYsit Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.49

124 3. Mechaical Vibratio Dyamics of Machiery (690) It is assumed that the base moves harmoically, that is where Y deotes the amplitude of the base motio ad represets the frequecy of the base excitatio Substitutig Eq. i Eq. The above equatio ca be expressed as where Y[k + (c ) ] / is the amplitude of excitatio force. Examiatio of equatio 3 reveals that it is idetical to a Equatio developed durig derivatio for M.F. The solutio is: m&& x + cx& + kx Fo sit x Xsi ( t φ ) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.50

125 3. Mechaical Vibratio Dyamics of Machiery (690) X F o [ ( ) ] + [ ξ ( )] k x Xsi + ( t α - φ ) Therefore the steady state amplitude ad phase agle to eq. 3 is The above equatios ca be writte i dimesioless form as follows X Y + ξ + ξ φ ta ξ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.5

126 3. Mechaical Vibratio Dyamics of Machiery (690) φ α ta ξ ta ξ The motio of the mass m lags that of the support by a agle (φ α) as show by equatio 6. Equatio 5 which gives the ratio of (X/Y) is also kow as motio trasmissibility or displacemet trasmissibility. Fig. gives the frequecy respose curve for motio trasmissibility. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.5

127 3. Mechaical Vibratio Dyamics of Machiery (690). For low frequecy ratios the system moves as a rigid body ad X/Y.. At resoace the amplitudes are large 3. For very high frequecy ratios the body is almost statioary (X/Y 0) It will see later that the same respose curve is also used for Force Trasmissibility. (b) Relative Amplitude of mass m Here amplitude of mass m relative to the base motio is cosidered. The equatios are basically made use i the Seismic istrumets. If z represets the relative motio of the mass w.r.t. support, x y + z Substitutig the value of x i eq. z x y m z&& + cz& + kz m Y sit The above equatio is similar to the equatio developed for rotatig ad reciprocatig ubalaces. Thus the relative steady state amplitude ca writte as Z Y + ξ φ ta - ξ ( ) ( ) Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.53

128 3. Mechaical Vibratio Dyamics of Machiery (690) Thus eq. 7 ad eq.8 are similar to the oe developed durig the study of rotatig a reciprocatig ubalaces. Frequecy ad phase respose plots will also remai same. Problem The support of a sprig mass system is vibratig with a amplitude of 5 mm ad a frequecy of 50 cpm. If the mass is 0.9 kg ad the stiffess of sprigs is 960 N/m, Determie the amplitude of vibratio of mass. What amplitude will result if a dampig factor of 0. is icluded i the system. Solutio: Give: Y 5 mm, f 50cpm, m 0.9 kg, k 960 N/m, X? ξ 0., the X? k rad/s m 0.9 π f π 50/ rad/ s r.58 X Y + ξ Whe ξ 0, X mm (Solutio) Whe ξ 0., X.5 mm (Solutio) + ξ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.54

129 3. Mechaical Vibratio Dyamics of Machiery (690) Observe eve whe dampig has icreased the amplitude has ot decreased but it has icreased. Problem The sprigs of a automobile trailer are compressed 0. m uder its ow weight. Fid the critical speed whe the trailer is travellig over a road with a profile approximated by a sie wave of amplitude 0.08 m ad a wavelegth of 4 m. What will be the amplitude of vibratio at 60 km/hr. Solutio: Give: Static deflectio dst 0. m, Y 0.08 m, γ 4 m, Critical Speed?, X60?. Critical speed ca be foud by fidig atural frequecy. f k m g d st.576cps π rad/s V wavelegth f m/s Correspodig V.06 m/s 79.4 km/hr Amplitude X at 60 km/hr V m/s r.756 f velocity wavelegth π f 7.48rad/s Whe X dampig is zero X at 60km/hr 0.86 m (Solutio) Y + ξ + ξ.9cps Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.55

130 3. Mechaical Vibratio Dyamics of Machiery (690) Problem 3 A heavy machie of 3000 N, is supported o a resiliet foudatio. The static deflectio of the foudatio due to the weight of the machie is foud to be 7.5 cm. It is observed that the machie vibrates with a amplitude of cm whe the base of the machie is subjected to harmoic oscillatios at the udamped atural frequecy of the system with a amplitude of 0.5 cm. Fid (a) the dampig costat of the foudatio (b) the dyamic force amplitude o the base (c) the amplitude of the displacemet of the machie relative to the base. Solutio Give: mg 3000 N, Static deflectio dst 7.5 cm, X cm, Y 0.5 cm,, ξ?, Fbase?, Z? (a) Solvig for ξ 0.9 (c) X Y ( ξ ) ( ξ ) c ξcc ξ km N s/m Z Y Note: Z m, X 0.00m, Y m, Z is ot equal to X-Y due phase differece betwee x, y, z. Usig the above eq. whe, the + ξ Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.56

131 3. Mechaical Vibratio Dyamics of Machiery (690) relative amplitude is Z m (Solutio) (b) F d ( c Z ) + ( kz ) 3.65rad/s Fd 388.5N (Solutio) Problem 4 The time of free vibratio of a mass hug from the ed of a helical sprig is 0.8 s. Whe the mass is statioary, the upper ed is made to move upwards with displacemet y mm give by y 8 si πt, where t is time i secods measured from the begiig of the motio. Neglectig the mass of sprig ad dampig effect, determie the vertical distace through which the mass is moved i the first 0.3 secods. Solutio: Give: Time period of free vibratio 0.8 s., y 8 si πt, ξ 0, x at the ed of first 0.3 s.? mx&& + kx ky mx&& + kx kysi t where, Y 8 mm, ad π rad/s. The complete solutio cosists of Complemetary fuctio ad Particular itegral part. x x x x c p where, c + x Acos t + Bsi t Xsi ( φ α) p ( t + α φ) Y X 0, if o ( φ α) 80, if π π τ 0.8 π ad 0.8 φ α 0 Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.57

132 The complete solutio is give by Substitutig the iitial coditios i the above eq. costats A ad B ca be obtaied 0 α φ ad 0.8 π 0.8 π τ π si t Y x Hece, p si t Y Bsi t Acos t x + + Y B ad 0 A gives 0 t at 0, x 0 t at 0; x & Darsha Istitute of Egieerig & Techology, Rajkot Page No Mechaical Vibratio Dyamics of Machiery (690)

133 3. Mechaical Vibratio Dyamics of Machiery (690) Thus the complete solutio after substitutig the values of A ad B Y x si t si t whe t 0.3 s, the value of x from the above eq. is x 9. mm (Solutio) Coclusios o Respose of a damped system uder the harmoic motio of the base Review of forced vibratio ( costat excitatio force ad rotatig ad reciprocatig ubalace ). Steady state amplitude ad phase agle is determied whe the base is excited siusoidally. Derivatios were made for both absolute ad relative amplitudes of the mass. 4. Vibratio Isolatio ad Force Trasmissibility Vibratios developed i machies should be isolated from the foudatio so that adjoiig structure is ot set ito vibratios. (Force isolatio) Delicate istrumets must be isolated from their supports which may be subjected to vibratios. (Motio Isolatio) To achieve the above objectives it is ecessary to choose proper isolatio materials which may be cork, rubber, metallic sprigs or other suitable materials. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.59

134 3. Mechaical Vibratio Dyamics of Machiery (690) Thus i this study, derivatios are made for force isolatio ad motio isolatio which give isight ito respose of the system ad help i choosig proper isolatio materials. Trasmissibility is defied as the ratio of the force trasmitted to the foudatio to that impressed upo the system. m&& x + cx& + kx Fo sit x p XSi t ( φ) The force trasmitted to the base is the sum of the sprig force ad damper force. Hece, the amplitude of the trasmitted force is: Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.60

135 3. Mechaical Vibratio Dyamics of Machiery (690) Substitutig the value of X from Eq. i Eq. 3 yields Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.6

136 3. Mechaical Vibratio Dyamics of Machiery (690) Hece, the force trasmissio ratio or trasmissibility, TR is give by φ α ta ξ ta ξ Eq. 6 gives phase agle relatioship betwee Impressed force ad trasmitted force. Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.6

137 3. Mechaical Vibratio Dyamics of Machiery (690) Curves start from uity value of trasmissibility pass through uit value of trasmissibility at ( / ) ad after that they Ted to zero as ( / ). Darsha Istitute of Egieerig & Techology, Rajkot Page No.3.63