<p> - For the case of linear damping where the energy loss is proportional to the square of the strain or amplitude (eg.4), the hysteresis curve is an ellipse. - When the damping loss is not a quadratic function of the strain or amplitude, the hysteresis in no longer an ellipse. </p><p>Equivalent Viscous Damping :</p><p>- The primary influence of damping on oscillatory systems is that of limiting the amplitude of response at resonance.</p><p>- The damping has little influence on the response in the frequency regions away from resonance.</p><p>- The equivalent damping Ceq is found by equating the energy dissipated by the viscous damping to that of the non-viscous damping force with assumed harmonic motion. 2 pCeq w X = w d (11)</p><p> where wd must be evaluated from the particular type of damping force.</p><p>- In the case of viscous damping, the amplitude at resonance, equation was found to be F X = 0 (10) Cwn</p><p>- For other types of damping, no such simple expression exists. It is possible, however, to approximate the resonant amplitude by substituting an equivalent</p><p> damping Ceq in the above equation. </p><p>Coulomb damping: </p><p>- It results from the sliding of two dry surfaces .</p><p>- The damping (friction) force is equal to the product of the normal force and the coefficient of friction m and is assumed to be independent of the velocity, once the motion is initiated. </p><p>- Hence, sometimes it is called a constant damping force. </p><p>- Since the sign of the damping force is always opposite to that of the velocity, the differential equation of motion for each sign is valid only for half-cycle intervals. </p><p>- Fig - To determine the decay of amplitude, we resort to the work-energy principle of equating the work done to the change in kinetic energy. - Choosing a half-cycle starting at the extreme position with velocity equal to zero</p><p> and the amplitude equal to X1 , the kinetic energy is zero and the work done on m is also zero:</p><p>1 2 2 k( X1- X- 1) - Fd ( X 1 + X - 1 ) = 0 2 1 or F= k( X - X ) d 2 1- 1 2k˙˙ or decay in amplitude X- X = d (12) 1- 1 k</p><p> where X -1 is the amplitude after the half-cycle as shown in the figure.</p><p>- Repeating this procedure for the next half-cycle a further decrease in amplitude of 2F d will be found, so that the decay in amplitude per cycle is a constant and equal k to </p><p>4F X- X = d 1 2 k</p><p>- The motion will cease, however, when the amplitude becomes less than D , at which position the spring face is insufficient to over-come the static friction, which is generally greater then the kinetic friction force.</p><p> k - The frequency of oscillation is w = , which is same as the undamped system m m</p><p>- It is assumed that under forced sinusoidal excitation the displacement of the system is with Coulomb damping is sinusoidal and equal to x= Xsinw t .</p><p>- The equivalent viscous damping can then be found , by noting that the work per</p><p> cycle by the Coulomb force Fd is equal to </p><p>Wd= F d (4 X ) 2 X average amplitude over a cycle pCeq w X= 4 F d X</p><p>4Fd or Ceq = pw X 2</p><p>The amplitude of forced vibration can be found by substituting Ceq (it should be noted that it contain X) in to the following eqn. F F X =0 = 0 22 2 2 2 4F (k- mw) + ( Ceq w) 2 骣 d (k- mw ) + 琪 2 桫p X</p><p>Solving for X, we obtain </p><p>2 2 2 骣4F 骣4Fd F - d 1- 琪 0 琪 p F 桫p F0 桫 0 X =2 = 2 ------(15) k- mw k 骣w 1- 琪 桫wn</p><p>X - It should be noted that unlike the system with viscous damping, , goes to d st</p><p> when w= wn .</p><p>4F - For numerator to remain real, the term d must be less than 1.0 p F0</p><p>Structural Damping :</p><p>- When the materials are cyclically stressed, energy is dissipated internally within the material itself. </p><p>- Experiments show that or most structural metals, such as steel or aluminum, the energy dissipated per cycle is independent of the frequency over a wide frequency range and proportional to the square of the amplitude of vibration.</p><p>- The internal damping fitting this classification is called solid damping or structural damping.</p><p>- With the energy dissipation per cycle proportion to the square of the vibration amplitude, the loss coefficient h is a constant and the shape of the hysteresis curve remains unchanged with amplitude and independent of the strain rate.</p><p>- Energy dissipated by structural damping may be written as </p><p>2 wd = a X (16)</p><p> where a is a constant with units of force/displacement .</p><p>- Using the concept of equivalent viscous damping, it gives 2 2 pCeq w X= a X a C = ------(17) eq pw</p><p>- Substitution of Ceq for C, the EOM can be written as </p><p>骣a mx˙˙+琪 x ˙ + kx = F( t) ------(18) 桫pw</p><p>Complex stiffness:</p><p>In the calculation of the flutter speeds of air plane wings and tail surfaces, the concept of complex stiffness is used. - It is arrived at by assuming the oscillation to be harmonic, which gives</p><p>骣 a jw t mx˙˙ +琪 k + j x = F0 e ------(19) 桫 p</p><p> jw t or mx˙˙ + k(1 + jg ) x = F0 e</p><p> a with g = ------(20) p k</p><p>(1+ jg ) Complex stiffness and g the structural damping factor.</p><p>- Using the concept of complex stiffness for problems in structural vibrations is advantageous in that one needs only to multiply the stiffness turns in the system by (1+ jg ) .</p><p>- The method is justified, however, only for harmonic oscillations.</p><p>- With the solution x= Xe jw t , the steady state amplitude from equation, becomes</p><p>F X = 0 2 ------(21) (k- mw) + i g k</p><p>- The amplitude at resonance is than </p><p>F X = 0 ------(22) w= wn g k</p><p>- Comparing this with the resonant response of a system with viscous damping </p><p>F X = 0 ------(23) 2Jk - We conclude that with equal magnitude at resonance, the structural damping factor is equal to twice the viscous damping factor.</p><p>Frequency Response with structural Damping :</p><p>- Starting with eqn. (21), the complex frequency response for structural damping can be shown to be a circle.</p><p> w - Letting = f , we get wn</p><p>2 Xk 1 (1- f ) -g H( r) = = = + j F 2 22 2 2 2 2 0 (1-f) + jg (1-f) +g( 1 - f ) + g</p><p>=x + iy ------(24)</p>