Integrity, Professionalism, Entrepreneurship Study Materials

Integrity,Professionalism,& Entrepreneurship Mata Kuliah Kode SKS : Dinamika Struktur & Pengantar Rekayasa Kegempaan : CIV - 308 : 3 SKS Single Degree of Freedom System Forced Vibration Pertemuan - 3

Integrity, Professionalism, & Entrepreneurship TIU : Mahasiswa dapat menjelaskan tentang teori dinamika struktur. Mahasiswa dapat membuat model matematik dari masalah teknis yang ada serta mencari solusinya. TIK : Mahasiswa mampu menghitung respon struktur dengan eksitasi harmonik dan eksitasi periodik

Integrity, Professionalism, & Entrepreneurship Sub Pokok Bahasan : Eksitasi Harmonik Eksitasi Periodik

Integrity, Professionalism, & Entrepreneurship Undamped SDoF Harmonic Loading The impressed force p(t) acting on the simple oscillator in the figure is assumed to be harmonic and equal to Fo sin t , where Fo is the amplitude or maximum value of the force and its frequency is called the exciting frequency or forcing frequency. u(t) u(t) fS = ku k m Fosin t fI(t) = m p(t) = Fosin t

Integrity, Professionalism, & Entrepreneurship The differential equation obtained by summing all the forces in the Free Body Diagram, is : F ku u m o = + sin t (1) The solution can be expresses as : ( ) t ( ) t ( ) t = + u u u c p (2) Complementary Solution uc(t)= A cos nt + B sin nt Particular Solution up(t) = U sin t (3.b) (3.a)

Integrity, Professionalism, & Entrepreneurship Substituting Eq. (3.b) into Eq. (1) gives : + = 2 m U kU o F Or : F F k = = o o U (4) 1 2 2 k m Which represents the ratio of the applied forced frequency to the natural frequency of vibration of the system : = (5) n

Integrity, Professionalism, & Entrepreneurship Combining Eq. (3.a & b) and (4) with Eq. (2) yields : F k ( ) t = + + o u A cos t B sin t sin t (6) n n 1 ( ) 0 2 ( ) 0 = u = With initial conditions : u u u ( ) ( ) 2 0 F 1 k F k u (7) ( ) t ( ) 0 = + + o o u u cos t sin t sin t n n 2 1 n Steady State Response Transient Response

Integrity, Professionalism, & Entrepreneurship 3 2 1 ( )( u t ) k F / 0 o 0 0.5 1 1.5 2 2.5 3 3.5 4 -1 -2 -3 Total Response Steady State Response Transient Response ( ) 0 ( ) 0 = = and 0 u u F / k n o = 0 20 , n

Integrity, Professionalism, & Entrepreneurship Steady state response present because of the applied force, no matter what the initial conditions. Transient response depends on the initial displacement and velocity. Transient response exists even if In which Eq. (7) specializes to ( ) 0 ( ) 0 = u = 0 u F 1 k ( ) t ( ) t n (8) = o u sin t sin 2 It can be seen that when the forcing frequency is equal to natural frequency ( = 1), the amplitude of the motion becomes infinitely large. A system acted upon by an external excitation of frequency coinciding with the natural frequency is said to be at resonance.

Integrity, Professionalism, & Entrepreneurship If = n( = 1), the solution of Eq. (1) becomes : 1 F ( ) t ( ) t n = o u t cos t sin (9) n n 2 k 40 30 20 ( )( 10 u t ) k 0 F / 0 0.5 1 1.5 2 2.5 3 3.5 o -10 -20 -30 -40

Integrity, Professionalism, & Entrepreneurship Assignment 3 If system in the figure have initial condition And subjected to harmonic loading p(t) = 5000 sin ( t) Plot a time history of displacement response of the system, for t = 0 s until t = 0,25 s, if = 0%, and : = 0,1 ; 0,25 ; 0,60 ; 0,90; 1,00; 1,25 & 1,75 W = 2,5 tons EI ( ) 0 ( ) 0 = cm/s 20 = 40x40 cm2 3 cm u & u 3 m T2 (b) E= 23.500 MPa

Integrity, Professionalism, & Entrepreneurship Damped SDoF Harmonic Loading Including viscous damping the differential equation governing the response of SDoF systems to harmonic loading is : + + = m u c u ku F sin t (10) o u(t) u(t) c fD(t) = c m p(t) fI(t) = m Fosin t fS(t) = ku k

Integrity, Professionalism, & Entrepreneurship The complementary solution of Eq. (9) is : ( ) t ( ) t D (11) = + t u e A cos t B sin n c D The particular solution of Eq. (9) is : ( ) C t up = + sin t D cos t (12) + 2 1 F (13.a) Where : = o C 1 2 k 2 2 2 + 2 F (13.b) = o D 1 2 2 k 2 2

Integrity, Professionalism, & Entrepreneurship The complete solution of Eq. (9) is : ( ) ( cos A e t u c = ) + + + t t B sin t C sin t D cos t n D D (14) Transient Response Steady State Response F 2 o = A ( 1 ) ( ) 2 k 2 2 + 2 ( ) ( 1 ) ) 2 2 F 2 o = B ( 1 ) ( ) ( k 2 0 5 , 2 2 2 + 2 1

Integrity, Professionalism, & Entrepreneurship Response of damped system to harmonic force with = 0,2, = 0,05, u(0) = 0, (0) = nFo/k

Integrity, Professionalism, & Entrepreneurship The total response is shown by the solid line and the steady state response by the dashed line. The difference between the two is the transient response, which decays exponentially with time at a rate depending on and . After awhile, essentially the forced response remains, and called steady state response The largest deformation peak may occur before the system has reached steady state.

Integrity, Professionalism, & Entrepreneurship If = n( = 1), the solution of Eq. (10) becomes : 1 F (15) ( ) t = + t o u e cos t sin t cos t n D D n 2 k 2 1

Integrity, Professionalism, & Entrepreneurship Considering only the steady state response, Eq. (12) & Eq. (13.a, b), can be rewritten as : ( ) ( ) = t sin U t u (16) 2 F k Where : = = o U tan (17) ( 1 ) 2 ( )2 1 2 + 2 2 Ratio of the steady state amplitude, U to the static deflection ust(=Fo/k) is known as the dynamic magnification factor, D : 1 U = = D (18) ( 1 ) ( )2 u 2 + 2 2 st

Integrity, Professionalism, & Entrepreneurship

Integrity, Professionalism, & Entrepreneurship Exercise The steel frame in the figure supports a rotating machine that exerts a horizontal force at the girder level p(t) = 100 sin 4 t kg. Assuming 5% of critical damping, determine : (a) the steady-state amplitude of vibration and (b) the maximum dynamic stress in the columns. Assume the girder is rigid W = 6,8 tons EI 4,5 m WF 250.125 I = 4,050 cm4 (b) E = 205.000 MPa

Integrity, Professionalism, & Entrepreneurship Assignment 4 If system in the figure have initial condition And subjected to harmonic loading p(t) = 5000 sin ( t) Plot a time history of displacement response of the system, for t = 0 s until t = 5 s, if : = 5% ( ) 0 ( ) 0 = cm/s 20 = 3 cm u & u W = 2,5 tons EI = 0,1 ; 0,25 ; 0,60 ; 0,90; 1,00; 1,25 & 1,75 40x40 cm2 3 m T3 (b) E= 23.500 MPa