Question

Figure Q4 shows a complex multi-degree of freedom spring-mass system. a) Develop the equation of motion of the system. (6 Marks) b) If m - m - m - m and k, = kx - kyky+ ks = k, Determine the natural frequencies and mode shape of the system. (16 Marks) c) Estimate the largest strain that can occur to any of the spring in the system. State which spring in your answer. Marks) (8 ka ka ks ma m2 ܕܗ ܕܠ ܐ xa -x3 _Figure Q4

Answer 1

Answer: Ka A ki K₂ wwwm, mam wwwmg MW K5 HM Hida. (a) FBD of mass 'm; K K₂ (2,-X2) mi K K5 (21-13) mixi mix+ KiX, + K3 X– 32)+K5 (82- 13 ) = 0 mixi +(KitK, + K5) X, -RX, – K5 X3 = 0 - 0 FBD of mass m,: KAX2 K Kg (22-2₃) M₂ M₂ X₂ K2 (73-X m8a+ K, (X-X17 + K3 (%) – 73) + K4X2 : 0 m.), - Kz X, + (Ka+ K3 +Ka) Xy - k3 X3 = 0 -@

FBD Of mass mz : K3 ( 2₂ - H2) & m₃ X my K Kg [X₃ ~ 4 ) M₂ 23 + Kz (X3 - 2) + Ks (2₂ - 2) = 0 mziz - K5X, – K3 , + (kg+Ks ) X3 - 0 The equations of motron are mix+(K+ Kx + K5 ) X- Kz Y,- K5 13 = 0 m, 7, - K, X, + (ks+ką + Ka) X- Kz 12 = 0 ma üz- K52,- K₂X₂ + (Kzt Ks ) X₂ = 0 The equations of motion can be written in matrix form as mi ma X24 Kitkat kg -K₂ - kg Xto - ka Ky tkz tka - K₃ mai -kg -K} Kzt kB] 0 0 Given mie M, 2 m₂ = m Kiako = Kg = KA = ksek mo ♡ X TSK 4 +-k is -K 3R. -K 22 X3 ak m -k o

Finding natural frequencies K-W'MI = 0 3K-mu? -K -k 3K-mw? -K -K 2 komu? -k (3 K-mw? ) [C3K-mwo?)(K - mwo) – K2 ] +[-K12 K- mus?) KCK+K(3K-mw?)] = 0 - K? ] -mus+ 8 km² w 4 - 18m K²w² +9k3=0 let mw²=p p3 8 kp²+ 18 kipē ak3=0 P, = 4.3098K , P, : 06972 K, Pz : 3K 21.7321 m mu? = 4.3028k, mu? = 0.6972k, mw = 3k W, = 2.0743, K. Гk (К Wy 2 0.8350, 7 natural frequencies Mode shapes! W, : 20743 The w = 4.3028 K [rk-w'm77) 3K-4,3028 K -K X -K 3K-4.3028k -K -K 2K-4.3028 K 0 к X2

-KX, -1-3028KX, - KX3 = 0 - KX, - KX₂ - 9:3028 KXzco let X3 = 1 .. -X, - 1,3028 X, = +1 -X, X 2.3028 .: XX : +4.3025 X, :-6.6053 - 6.6053 Ø 4.3095 ☺ W, = 0.8350 ha W,-0.6972 K [K-WPM) x, وحدت X₂ х, X 2.3028K -K -K 2.3018 K -K -K X2 X₂ -k 6,3028K -KX, +2.3028 KX; K X3 20 - KX, - KX2 + 1.3098 KX3 = 0 let X3 = 1 -> -*, +2.3028X3! - X, -X, : - 1,3028 X-0.6056 X2 -0.6972

Ø : 0.6056 0.6972 * Wy: 1.7331 22. W: 3 ok ОК -K -k *** -k -K - KX₂ - KX₂ = 0 -KX, - KX₂ = 0 let X3 = 1 X₂ = -1 X, :- Mode shapes 4,3025 ist mode 2 -6.6053

and mode: 1 2 w 0.6056 0.6972 3rd mode: REA 3

- Flexibility matrix is [K] given by & the elements in Flexibility matter is called influence co-effercerent the + Influence co-effecvent (Sis) is deflection of ith node by giving unit force @ ;th nude, keeping all other of freedom) constrained dof (degrees
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C ЗК ( k ) : - -k -K 3k -K -k -k 2K [k]": Adj(K) IKI I KU : 3K (6K?K?) +* (2?-K?) - KCK'+3 **) 15 K 3 K ² AK² 863 AdJCKX + 8K

3K -K, 3K Adjka -K K 3K 2k-k -K -K 3K K " k ЗК 3K -K 5k? 3K² 4K² Adik : 3k? 5k? 4ke 8K2 AK 2 AK2 (kt] = Adik 4K² - 8K3 5k² .3 K2 3 K2 5K² AK? 4K? 4K2 8k? K ro.625 0.375 as 0.375 0.625 0.5 10.5 0.5 from the flexibility matrix chect the upper & lower triangle elements & Huke out the large influence concfficient 8,.- 832 be L let length of spring strain 632; (31 = 0.35 0.5 к Hos KL

The largest E311 0.5 KL strain is 22 with largest strain is [Kg & Kg The spring

Answer 2

Answer: Ka A ki K₂ wwwm, mam wwwmg MW K5 HM Hida. (a) FBD of mass 'm; K K₂ (2,-X2) mi K K5 (21-13) mixi mix+ KiX, + K3 X– 32)+K5 (82- 13 ) = 0 mixi +(KitK, + K5) X, -RX, – K5 X3 = 0 - 0 FBD of mass m,: KAX2 K Kg (22-2₃) M₂ M₂ X₂ K2 (73-X m8a+ K, (X-X17 + K3 (%) – 73) + K4X2 : 0 m.), - Kz X, + (Ka+ K3 +Ka) Xy - k3 X3 = 0 -@

FBD Of mass mz : K3 ( 2₂ - H2) & m₃ X my K Kg [X₃ ~ 4 ) M₂ 23 + Kz (X3 - 2) + Ks (2₂ - 2) = 0 mziz - K5X, – K3 , + (kg+Ks ) X3 - 0 The equations of motron are mix+(K+ Kx + K5 ) X- Kz Y,- K5 13 = 0 m, 7, - K, X, + (ks+ką + Ka) X- Kz 12 = 0 ma üz- K52,- K₂X₂ + (Kzt Ks ) X₂ = 0 The equations of motion can be written in matrix form as mi ma X24 Kitkat kg -K₂ - kg Xto - ka Ky tkz tka - K₃ mai -kg -K} Kzt kB] 0 0 Given mie M, 2 m₂ = m Kiako = Kg = KA = ksek mo ♡ X TSK 4 +-k is -K 3R. -K 22 X3 ak m -k o

Finding natural frequencies K-W'MI = 0 3K-mu? -K -k 3K-mw? -K -K 2 komu? -k (3 K-mw? ) [C3K-mwo?)(K - mwo) – K2 ] +[-K12 K- mus?) KCK+K(3K-mw?)] = 0 - K? ] -mus+ 8 km² w 4 - 18m K²w² +9k3=0 let mw²=p p3 8 kp²+ 18 kipē ak3=0 P, = 4.3098K , P, : 06972 K, Pz : 3K 21.7321 m mu? = 4.3028k, mu? = 0.6972k, mw = 3k W, = 2.0743, K. Гk (К Wy 2 0.8350, 7 natural frequencies Mode shapes! W, : 20743 The w = 4.3028 K [rk-w'm77) 3K-4,3028 K -K X -K 3K-4.3028k -K -K 2K-4.3028 K 0 к X2

-KX, -1-3028KX, - KX3 = 0 - KX, - KX₂ - 9:3028 KXzco let X3 = 1 .. -X, - 1,3028 X, = +1 -X, X 2.3028 .: XX : +4.3025 X, :-6.6053 - 6.6053 Ø 4.3095 ☺ W, = 0.8350 ha W,-0.6972 K [K-WPM) x, وحدت X₂ х, X 2.3028K -K -K 2.3018 K -K -K X2 X₂ -k 6,3028K -KX, +2.3028 KX; K X3 20 - KX, - KX2 + 1.3098 KX3 = 0 let X3 = 1 -> -*, +2.3028X3! - X, -X, : - 1,3028 X-0.6056 X2 -0.6972

Ø : 0.6056 0.6972 * Wy: 1.7331 22. W: 3 ok ОК -K -k *** -k -K - KX₂ - KX₂ = 0 -KX, - KX₂ = 0 let X3 = 1 X₂ = -1 X, :- Mode shapes 4,3025 ist mode 2 -6.6053

and mode: 1 2 w 0.6056 0.6972 3rd mode: REA 3

- Flexibility matrix is [K] given by & the elements in Flexibility matter is called influence co-effercerent the + Influence co-effecvent (Sis) is deflection of ith node by giving unit force @ ;th nude, keeping all other of freedom) constrained dof (degrees
​​​​​
C ЗК ( k ) : - -k -K 3k -K -k -k 2K [k]": Adj(K) IKI I KU : 3K (6K?K?) +* (2?-K?) - KCK'+3 **) 15 K 3 K ² AK² 863 AdJCKX + 8K

3K -K, 3K Adjka -K K 3K 2k-k -K -K 3K K " k ЗК 3K -K 5k? 3K² 4K² Adik : 3k? 5k? 4ke 8K2 AK 2 AK2 (kt] = Adik 4K² - 8K3 5k² .3 K2 3 K2 5K² AK? 4K? 4K2 8k? K ro.625 0.375 as 0.375 0.625 0.5 10.5 0.5 from the flexibility matrix chect the upper & lower triangle elements & Huke out the large influence concfficient 8,.- 832 be L let length of spring strain 632; (31 = 0.35 0.5 к Hos KL

The largest E311 0.5 KL strain is 22 with largest strain is [Kg & Kg The spring