Question

14. There is a two-degree-of-freedom system with no external force as shown in Figure 4. Here, kı=kz=k=10kN/m, ka=ks=2kN/m and m:=m2=2kg, answer the following. (25 points) 14-1. Find the equation of motion in matrix-vector form. 14-2. Find the natural frequencies W1, W2 (rad/sec) through the eigenvalue problem. 14-3. Find the eigenvectors corresponding to the eigenfrequencies through the eigenvalue problem, except that the first element is 1. X + ke ki 111; W ke Figure 4. Two degree of freedom model

Answer 1

Solution

FREE BODY DIAGRAM →X, >ta k5%, kg 64-22 kpp, M, -K-72) ma • Kgla Kuta 고 Using Newton's laws. -ka, -kgt, -k2(x,-*2) = myši - kuka tkq(x,-*2) - K342 = Mali Rearranging, mizi + (katky +ks)ą, - kale = 0

Maža + (ką tkz tkx) 22 - K24 = 0 Substituting known values, zäit 22x, 10 L2 = 0 aäg + 22x2 10 X, = 0 Expressing in matrix rector form, [M]{i} +[k]{x} = 0 22 - 10 3 + 10 Xg -10 ೩೩

Natural frequency 1 [k)- w [M] = 22x18 - Bus -10 -104 22x10² -20% (22x103 -Qwzje - 104 = 0 Take w²zd 4X² - 88x103 x + 386*100=0 Solving,

x=6000, 16000 .. Natural frequencies are 6000 = 77.4597 radls 126.4911 radys W, = √6000 Wa= 16000 Eigenvestors ([k} w\M]) {x} = 0 122x10 - Qw² Xi -10% 22x10²-26² 1) -104 X

Case 1, /S17 radما .11 = ا ا 04 0 - 10 X ا 16 Xa -to- 6 = يSolve : tox - tx ما ا (ox - lo X: X [!] eigen vector is Case 2, قهk۹ll r 126 دن

-10 4 -10 0 XI X2 -10 - 10% Solue: -15x, - 10x2=0 4 4 -10X1 = 10X2 X, - -Xa so eigencector is LIJ [:] and [-] . Eigenvectors are