In a harmonic analysis, is the damping produced by MP,DMPR; TB,SDAMP; and DMPSTR constant vs frequen
October 4, 2017 at 12:20 pmVishal GanoreAnsys EmployeeNo. For a single mode, the ratio to critical is greater at frequencies below resonance and smaller at frequencies above resonance. This is documented in the Structural Analysis Guide Section 1.2.4. You can find the complete equation in the ANSYS Help documentation, under "Mechanical APDL -> Theory Reference -> 14.3.3 equation 14-22 parameter g, mj, gEj". The damping contribution g is related to the damping matrix by [C]=(2*g/Omega)*K. Omega is the excitation frequency. So (2*g/Omega) is a coefficient times the stiffness matrix like Beta damping. Beta = 2*eta/omega (omega is natural freq) while 2*g/Omega = 2*eta/omega: eta (ratio to critical) = g(omega/Omega) In a single degree of freedom test you will see the ratio to critical be higher at low excitation frequencies and lower at high excitation frequencies. A 1 dof test is provided below. Note that you can't measure the damping ratio by dividing the imaginary solution by the real solution. For example, at resonance, that is infinity, not any measure of damping. The only method I know is to compare the dynamic response to the static response. At resonance, the "dynamic load factor" will be 1/2*eta, where eta is the ratio to critical. The formula for the dynamic load factor at other frequencies is: 1/SQRT[(1-OMEGA^2/omega^2)^2+4(eta*OMEGA/omega)^2] Away from resonance, the harmonic response is not a strong function of damping, so the deviation from ratio to critical is small. /prep7 et,1,14 et,2,21,,,2 r,1,100*(2*3.1416)**2 r,2,1 n,1 n,2,1 e,1,2 type,2 real,2 e,2 d,1,all d,2,uy f,2,fx,1000 mp,dmpr,1,1.0 fini /solu antyp,static solve ux2=ux(2) ! static displacement = 0.253 fini /solu antype,harm harf,0,20 kbc,1 nsub,20 solve fini /post26 nsol,2,2,ux prcplx,0 prva,2
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