FORCED VIBRATION UNDER HARMONIC LOADING 1 Response of Undamped SDOF System to Harmonic Motion In the forced vibration system, due to periodic force applied to the system, the vibrations continue as long as the periodic force is present. Periodic force which is harmonic in nature will be considered. π(π‘) = ππ π ππππ π‘ Where ππ is the peak amplitude and ππ is the frequency of the force in radians per second. Thus π π2 π₯ + ππ₯ = ππ π ππππ π‘ ππ‘ 2 So the differential equation of the periodic motions has a solution. π₯π = π₯(π‘) = π πΌπ‘ (π΄1 πππ π½π‘ + π΄2 π πππ½π‘) = π΄1 πππ ππ‘ + π΄2 π ππππ‘ Particular solution π₯π = ππ ππππ π‘ ππ₯π = πππ πππ ππ π‘ ππ‘ π2 π₯π = −πππ 2 π ππππ π‘ ππ‘ 2 Where π is the peak value of a particular solution Substituting −ππππ 2 π ππππ π‘ + πππ ππππ π‘ = ππ π ππππ π‘ −ππππ 2 + ππ = ππ π= ππ π − πππ 2 ππ ππ π π π= = πππ 2 ππ 2 1− 1− 2 π π Where π = ππ 2 π2 is the frequency ratio of the applied force frequency to natural frequency of vibration of the system. Therefore ππ π= π 2 1−π 1 Therefore the final solution will be ππ π₯(π‘) = π΄1 πππ ππ‘ + π΄2 π ππππ‘ + π 2 π ππππ π‘ 1−π Initial conditions are applied to obtain constantπ΄1 ππππ΄2 . (Recall: differential equations) In the case of initial condition Considering initial conditions π‘ = 0 π₯(0) = 0 π£=0 ππ π π΄2 = − 1 − π2 π π΄1 = 0, Thus ππ ππ π π ππππ‘ + π π πππ π‘ π₯(π‘) = − π 1 − π2 1 − π2 π ππ π₯(π‘) = − π 2 [π ππππ π‘ − ππ ππππ‘] 1−π As damping forces will always be present, the free frequency term will vanish eventually. Therefore the term (ππ ππππ‘) represents transient response of vibration. Transient response of vibration occurs to a system while it is approaching steady state. The forcing term is referred to as steady-state response is given by: ππ π₯(π‘) = π 2 π ππππ π‘, π ≠ 1 1−π When r =1, ππ = π t the amplitude of motions becomes infinitely large. This is resonance. The amplitude will increase gradually to infinite. The materials used in practice have strength limitation, thus structural failures occur long before extremely large amplitudes can be attained. Let ππ = π0 π π0 is the static deflection π= π0 1 − π2 π 1 = = πΉ(ππ ) π0 1 − π 2 π πΉ(ππ ) = π is called frequency response function. It gives the magnitude and he sign of steady state motion 0 as a function of frequency ratio “r” 2 The magnitudeπ·π , is called the steady-state or dynamic magnification factor, or gain as shown in the figure below π·π = |πΉ(ππ )| If r<1, the response is in phase with the excitation, since (1-r2) is positive. If r is more than 1.0, the response is 180o out of phase with the excitation, that is ππ π₯(π‘) = π 2 (−π ππππ π‘) 1−π π₯(π‘) = π0 (−π ππππ π‘) 1 − π2 When r is close to one but not equal to 1, the forced motion and natural motion alternatively reinforce each other and cancel each other giving the appearance of a beat phenomenon. 2 Response of Viscous-Damped SDOF System to Harmonic Motion Equation of motion π π2 π₯ ππ₯ +π + ππ₯ = ππ π ππππ π‘ 2 ππ‘ ππ‘ Solution π₯(π‘) = π₯π (π‘) + π₯π (π‘) 3 Where π₯π (π‘) is the complementary solution and π₯π (π‘) is the particular solution. For underdamped case where πΆ < πΆππ π₯π (π‘) = π −πππ‘ (π΄1 πππ ππ· π‘ + π΄2 π ππππ· π‘) Particular solution is given by π(π‘) = π ππππ π‘ ππ(π‘) ππ‘ π2 π(π‘) 2 ππ‘ = ππ πππ ππ π‘ = −ππ 2 π ππππ π‘ Gives two independent function π ππππ π‘ πππ πππ ππ π‘ π₯π = π1 π ππππ π‘ + π2 πππ ππ π‘ ππ₯π ππ‘ π2 π₯π 2 ππ‘ Substituting intoπ π2 π₯ ππ‘ 2 +π ππ₯ ππ‘ = π1 ππ πππ ππ π‘ − π2 ππ π ππππ π‘ = −π1 ππ 2 π ππππ π‘ − π2 ππ 2 πππ ππ π‘ + ππ₯ = ππ π ππππ π‘ π(−π1 ππ 2 π ππππ π‘ − π2 ππ 2 πππ ππ π‘) + π(π1 ππ πππ ππ π‘ − π2 ππ π ππππ π‘) + π(π1 π ππππ π‘ + π2 πππ ππ π‘) = ππ π ππππ π‘ −ππ1 ππ 2 π ππππ π‘ − ππ2 ππ 2 πππ ππ π‘ + ππ1 ππ πππ ππ π‘ − ππ2 ππ π ππππ π‘ + ππ1 π ππππ π‘ + kπ2 πππ ππ π‘ = ππ π ππππ π‘ (−ππ1 ππ 2 − ππ2 ππ + ππ1 )π ππππ π‘ + (−ππ2ππ 2 + ππ1 ππ + kπ2 )πππ ππ π‘ = ππ π ππππ π‘ The equations yields two algebraic equations −ππ1 ππ 2 − ππ2 ππ + ππ1 = ππ (π − π ππ 2 )π1 − πππ π2 = ππ −ππ2 ππ 2 + ππ1 ππ + kπ2 = 0 ππ1 ππ + (k − πππ 2 )π2 = 0 Divide through by k (1 − π 2 π ππ ππ )π1 − ππ π2 = π π π 4 π π π1 ππ + (1 − ππ 2 )π2 = 0 π π (1 − π 2 )π1 − 2πππ2 = ππ 2πππ1 + (1 − π 2 )π2 = 0 π π Where π = π = 2ππ π‘βπππππππ π = 2πππ ππ π1 = (1 − π 2 )ππ (1 − π 2 )2 + (2ππ)2 π2 = − (2ππ)ππ (1 − π 2 )2 + (2ππ)2 Therefore π₯π = (1 − π 2 )ππ (2ππ)ππ π ππππ π‘ − πππ ππ π‘ 2 2 2 (1 − π ) + (2ππ) (1 − π 2 )2 + (2ππ)2 π₯π = (1 − ππ [(1 − π 2 )π ππππ π‘ − (2ππ)πππ ππ π‘] + (2ππ)2 π 2 )2 Which can be written as π₯π = ππ √(1 − π 2 )2 + (2ππ)2 sin(ππ π‘−∝) Where ∝ is the phase angle defined by ∝= π‘ππ−1 ( 2ππ ) 1 − π2 Can be written in compact form as π₯π = ππ π½ sin(ππ π‘−∝) Where π½ is the magnification factor in the case of damped system. π½= π₯π 1 = ππ √(1 − π 2 )2 + (2ππ)2 Total response π₯(π‘) = π −πππ‘ (π΄1 πππ ππ· π‘ + π΄2 π ππππ· π‘) + ππ √(1 − π 2 )2 + (2ππ)2 sin(ππ π‘−∝) The constants π΄1 and π΄2 are determined from the initial conditions. The figure below gives the steady-state magnification factor ‘π½’ against frequency ratio ‘r’ and phase angle ∝ against frequency ratio “r” for various amounts of damping ratios π 5 FORCE 3 Transmission It is clear that by increasing the spring stiffness k and the damping coefficient c, the amplitude of vibration decreases. The increase in the stiffness and damping coefficients, however, may have an adverse effect on the force transmitted to the support. In order to reduce the force transmitted to the support, the stiffness and damping coefficients must be properly selected. Given the equation of motion π π2 π₯ ππ₯ +π + ππ₯ = ππ π ππππ π‘ 2 ππ‘ ππ‘ With particular solution π₯π = ππ π½ sin(ππ π‘−∝) (ππππππ? ) 6 ππ₯ The force transmitted to the support through the spring is kx and through the damping elementπ ππ‘ . Hence the total force transmitted to foundation ππ‘ = ππ₯ + π ππ₯ ππ‘ Therefore π₯π = ππ π½ sin(ππ π‘−∝) ππ₯π ππ‘ = ππ π½ππ cos(ππ π‘−∝) Therefore ππ‘ = π(ππ π½ sin(ππ π‘−∝)) + π(ππ π½ππ cos(ππ π‘−∝)) ππ‘ = ππ π½[π sin(ππ π‘−∝) + π ππ cos(ππ π‘−∝)] Given that πππ ) π π = π‘ππ−1 ( Then ππ‘ = ππ π½√π2 + (π ππ )2 sin(ππ π‘−∝ +π) ππ = ππ‘ = ππ π π½√π2 + (π ππ )2 sin(ππ π‘−∝ +π) ππ‘ = ππ π½√1 + ( π ππ π π ππ π 2 ) sin(ππ π‘−∝ +π) π Where π = π = 2ππ π‘βπππππππ π = 2πππ ππ 2πππππ ππ‘ = ππ π½√1 + ( ππ‘ = ππ π½√1 + ( π 2πππ π 2 ) sin(ππ π‘−∝ +π) 2 ) sin(ππ π‘−∝ +π) ππ‘ = ππ π½√1 + (2ππ)2 sin(ππ π‘−∝ +π) 7 Maximum force πΉπ‘ = ππ π½√1 + (2ππ)2 πΉπ‘ = ππ √1 + (2ππ)2 √(1 − π2 )2 + (2ππ)2 1 + (2ππ)2 √ πΉπ‘ = ππ (1 − π2 )2 + (2ππ)2 Transmissibility Tr is defined as the ratio between the amplitude of the force transmitted to the foundation and the amplitude of the force. Therefore ππ = πΉπ‘ ππ 1 + (2ππ)2 =√ (1 − π2 )2 + (2ππ)2 The transmissibility of motion from the foundation to the structure and the transmissibility of force from the structure to the foundation is given by the same expression. 4 Response to Support Motion There are cases where the foundation or support of a structure is subjected to time varying motion. Ground motion of earthquakes or other excitations such as explosion or dynamic action of machinery causes motion of structures. Harmonic motion π§(π‘) = π§π π ππππ π‘ Where π§π us the maximum amplitude and ππ is the frequency of the support motion. Given the equation of motion 8 π π2 π₯ ππ₯ ππ§ +π + ππ₯ = π + ππ§ 2 ππ‘ ππ‘ ππ‘ π§(π‘) = π§π π ππππ π‘ ππ§ = π§π ππ πππ ππ π‘ ππ‘ Substitution π π2 π₯ ππ₯ +π + ππ₯ = π§π (ππ ππππ π‘ + πππ πππ ππ π‘) 2 ππ‘ ππ‘ π π2 π₯ ππ₯ +π + ππ₯ = π§π √π 2 + (πππ )2 sin(ππ π‘ +∝π§ ) 2 ππ‘ ππ‘ π πππ 2 π2 π₯ ππ₯ +π + ππ₯ = π§π π√1 + ( ) sin(ππ π‘ +∝π§ ) 2 ππ‘ ππ‘ π π π2 π₯ ππ₯ +π + ππ₯ = π§π π√1 + (2ππ)2 sin(ππ π‘ +∝π§ ) 2 ππ‘ ππ‘ πππ ∝π§ = π‘ππ−1 ( ) = π‘ππ−1 (2ππ) π π2 π₯ ππ₯ This is same as π ππ‘ 2 + π ππ‘ + ππ₯ = ππ π ππππ π‘ except for the addition of angle ∝ in the sine function Therefore steady state response π₯π = π§π √1 + (2ππ)2 π½ sin(ππ π‘ +∝π§ −∝) Magnification factor/Transmissibility π₯π √1 + (2ππ)2 = π§π √(1 − π 2 )2 + (2ππ)2 Thus the expression for the relative transmission of the support motion to the oscillator is given by π₯π √1 + (2ππ)2 = × sin(ππ π‘ +∝π§ −∝) π§π √(1 − π 2 )2 + (2ππ)2 9 If r=0, Tr=1.0 If π = √2, Tr=1 Transmitted force In the case of the support motion, the force carried by the support can be obtained ππ₯ ππ§ πΉπ‘ = π ( − ) + π (π₯ − π§) ππ‘ ππ‘ πΉπ‘ = −π π2 π₯ ππ‘ 2 π₯π = π§π √1 + (2ππ)2 π½ sin(ππ π‘ +∝π§ −∝) ππ₯ = π§π ππ √1 + (2ππ)2 π½ cos(ππ π‘ +∝π§ −∝) ππ‘ π2 π₯ = −π§π ππ 2 √1 + (2ππ)2 π½ sin(ππ π‘ +∝π§ −∝) ππ‘ 2 Therefore πΉπ‘ = π. π§π ππ 2 √1 + (2ππ)2 π½ sin(ππ π‘ +∝π§ −∝) π πΉπ‘ = ( 2 ). π§π ππ 2 √1 + (2ππ)2 π½ sin(ππ π‘ +∝π§ −∝) π πΉπ‘ = π. π§π π 2 √1 + (2ππ)2 π½ sin(ππ π‘ +∝π§ −∝) 10 Mass motion relative to the support. The relative displacement denoted as z can be written as z=x-y π§=π₯−π¦ ππ§ ππ₯ ππ¦ = − ππ‘ ππ‘ ππ‘ π2 π§ π2 π₯ π2 π¦ = − ππ‘ 2 ππ‘ 2 ππ‘ 2 π2 π₯ ππ₯ Equation π ππ‘ 2 + π ππ‘ + ππ₯ = π§π (ππ ππππ π‘ + πππ πππ ππ π‘) can be written as π( π2 π§ π2 π¦ ππ§ + 2 ) = −ππ§ − π 2 ππ‘ ππ‘ ππ‘ Which can be written as π π2 π§ ππ§ π2 π¦ + π + ππ§ = −π = ππ§π ππ 2 π ππππ π‘ ππ‘ 2 ππ‘ ππ‘ 2 Therefore steady state response (particular solution) π₯π = π₯π = ππ§π ππ 2 /π √(1 − π 2 )2 + (2ππ)2 π§π π 2 √(1 − π 2 )2 + (2ππ)2 11 × sin(ππ π‘−∝) × sin(ππ π‘−∝)