Figure Random vibration analysis of a simply supported beam

Transcription

1 Example [Spectrum Analysis] Determine the first two natural frequencies and the response of a damped two degree of freedom system subject to a random acceleration with a spectral density function as shown in the figure below. Apply the load at the base and assume 2 % constant modal damping. k 1 = lbf/in, k 2 = lbf/in, m 1 = 0.5 lbm, and m 2 = 1.0 lbm. Use the combination element COMBIN40. Solution: Figure Random vibration analysis of a simply supported beam This problem can be modeled with LINK1 and MASS21 elements, but here another element COMBIN40 is used. Note this element uses nodal coordinate system for results. Preferences 1. Structural & h-method Preprocessor 2. Element Type > Add/Edit/Delete > Add > select Combination 40 (COMBIN40) > OK > Element Types / Options > enter as in the dialog below. The following shows the COMBIN40 element. The degree of freedom is in y-direction and Mass location at J-th node (i.e., second node in the element).

2 3. [Watch the units here] Real Constants > Add > OK > select COMBIN40 > enter the spring constant (42832 lbf/in = 42832*32.2*12 lbm/s 2 ) and mass (0.5 lbm) for element 1 > OK > Add > Type 1 (same element type is used) > enter the spring constant and mass (32416 lbf/in = 32416*32.2*12 lbm/s 2 and 1 lbm) for element 2 > OK > Close 4. Modeling / Create > Nodes > On Working Plane > enter the node 1 at the bottom as (0,0) > CR > node 2 as (0,1) > CR > node 3 as (0,2) > CR >OK]. The actual lengths of elements are immaterial here. 5. Create / Elements > Elem Attributes > Element type number = COMBIN40 > OK > Auto Numbered / Thru Nodes > pick on node 1 and 2 > OK 6. Now, change the attributes for element 2 by Elem Attributes > change real constant number to 2 > OK > create second element Solution 7. Analysis Type / New Analysis > Modal > OK 8. Analysis Options > enter number of modes to expand, and check Elcalc option as shown below > OK 9. For modal analysis dialog, just accept the default > OK 10. Apply > Structural / Displacement > On Nodes > pick the bottom node > select UY > OK 11. Load Step Opts a. Output Ctrls / DB Results File > check Every substep for file write frequency > OK b. Time/Frequency > Damping > enter constant damping ratio 0.02 (i.e., 2%) > OK 12. Solve the problem [Current LS > OK] General Postprocessing 13. [List Results > Results Summary]. Two frequencies are listed as SET TIME/FREQ LOAD STEP SUBSTEP CUMULATIVE

3 Solution 14. Now, get both frequency data by [Parameters in utility menu > Get Scalar Data > Results data / Modal Results > OK >enter f1 for Name of parameter, Mode number 1, and select Frequency, FREQ > Apply > OK (for modal results data) > name = f2, mode number 2, and FREQ > OK] 15. Parameters > Scalar Parameters > see just defined scalars 16. Analysis Type / New Analysis > Spectrum >OK 17. Analysis Options > check PSD option, enter 2 for number of modes for solution, and check Elcalc option as shown below >OK 18. Define Loads / Apply > Structural / Spectrum > Base PSD Excitation / On Nodes > pick the bottom nodes > OK > Direction = Y > OK 19. Load Step Opts a. Output Ctrls > DB Results File > check Every substep >OK b. Time/Frequency> Damping > enter 0.02 for the constant damping ratio DMPRAT and STLOC Starting location N = 1 > OK

4 c. Spectrum > PSD i. Settings > set up the dialog as below >OK The PSDUNIT defines the unit for PSD input so that g2 / Hz can be used as unit and the significance threshold (SIGIF) combines only those modes whose significance level exceeds the SIGNIF threshold. Note that new Ansys versions may not have MCOMB any longer and the Significance threshold value has been moved to another place [Solution > Load Step Opts > Spectrum / PSD > Mode Combine]. ii. PSD vs. Frequency > Table no = 1 > OK > fill the frequency table as below >OK Note that the frequency must be positive. iii. Calculate PF (participation factor) > default dialog as below >OK iv. Calculate Controls > choose as below >OK

5 20. Solve the problem [Current LS >OK] General Postprocessing 21. See the results summary [List Results > Results Summary] The load step 2 has the PSD results data. 22. Read in data set for PSD [Read Results / By Load Step > Load step number = 2 and Substep number =1 > OK] 23. Load Case (LC) a. Create Load Case > check Results file > OK> enter as below > OK > repeat it for substeps nos. = 2 with LCDEF = 2

6 b. Calc Options / Scale Factor > enter as below > Apply > repeat this for LCNO = 2 c. Read Load Case (LCASE) > Ref no. of load case = 1 > OK 24. List Results > Nodal Solution > DOF Solution / Y-Component > OK. The 1σ acceleration responses of mass 1 and 2 are shown in the results data below. LOAD STEP= 1 SUBSTEP= 1 FREQ= LOAD CASE= 1 THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE NODAL COORDINATE SYSTEMS NODE UY E E Go back to Load Case above and use substep no = 2 to obtain the second result as LOAD STEP= 1 SUBSTEP= 2 FREQ= LOAD CASE= 1 THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE NODAL COORDINATE SYSTEMS NODE UY E E Try to store these data into scalar parameters and view them. a. Parameters in utility menu > Get Scalar Data > Results data / Nodal results > OK > enter parameter name like m1std, Node number = 2, DOF solution = UY > Apply > repeat this for node 3 b. See the values of 1 PSD acceleration responses by [Parameters / Scalar Parameters] as shown below.

7 27. For displacement, go back 19 iv and choose Displacement > 19 iii > Solve / Current LS > follow the rest of steps with scale factor, FACT = 1 to obtain the 1σ displacement responses of mass 1 and 2 LOAD STEP= 1 SUBSTEP= 1 FREQ= LOAD CASE= 1 THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE NODAL COORDINATE SYSTEMS NODE UY Go back to Load Case above and use substep no = 2 to obtain the second result as LOAD STEP= 1 SUBSTEP= 2 FREQ= LOAD CASE= 1 THE FOLLOWING DEGREE OF FREEDOM RESULTS ARE IN THE NODAL COORDINATE SYSTEMS NODE UY Try to store these data into scalar parameters and view them. TimeHistory Postprocessing 30. Store Data > accept the default resolution >OK 31. Define Variables > Add > Nodal DOF result > enter node number 2 and select UY > Apply > repeat this for node Calc Resp PSD > fill the dialog as below >OK

8 33. The PSD response graph is shown below [Graph Variables > enter Reference number = 4 > OK] Example [Ansys WB / Response Spectrum] A cantilever beam of length l has a square section 0.1 x 0.1 in 2 and is subjected to a vertical motion of support by displacement spectrum of 1 in. Two point weights of 10 lbf each are attached at mid-point and end point of the beam. Determine the displacement responses of two weights. Material of beam has density = 0.01 lbm/in 3 and E = 10 6 psi. Solution: Modal Analysis Geometry 1. Create > Point > Definition = Manual Input > Generate > repeat this at X=5 and 10 in 2. Concept > Lines From Points > drag select all three points > Generate 3. Right click the points > rename the second to m1 and third point to m2 4. Add the section to linebody Model 5. Right click Geometry > Insert > Point Mass > pick the second point in middle > Mass = 10, Behavior = Rigid > repeat this for other point at tip 6. Add a fixed support at the first point

9 7. Displacement > pick both masses > specify Z = 0 (for 2D analysis) Solution 8. Solve 9. Right click Solution / Tabular Data > Select All > right click > Create Mode Shape Results > right click the mode shapes in Outline > Evaluate All Results > see results of displacement response spectrum for two modes. Response Spectrum Analysis 10. Back in Project, right click Solution > Transfer Data to New > Response Spectrum 11. Double click Setup of Response Spectrum analysis 12. Click Response Spectrum in Outline > RS Base Excitation > RS Displacement > Boundart Condition = All BC Supports, Load Data = Tabular Data, enter 1 for tabular data, Direction = Y Axis 13. Add Directional Deformation > Orientation = Y Axis 14. Pick both masses > add Directional Deformation > Orientation = Y Axis 15. Solve and see deformation below Calculation 16. Let s use participation factors in vibration and calculate the displacements of both masses 17. First, obtain the mode shapes > pick both point masses > add Directional Deformation / Orientation = Y Axis > Duplicate it

10 18. In multi-window, pick the first window > first Directional Deformation > pick second window > second Directional Window > change Mode = 2 > right click Directional Deformation > Retrieve This Result > see the displacements of two masses 19. Thus, the mode vectors are obtained as V 1 = (0, , ) T and V 2 = (0, 5.917, ) T 20. Define mass vector w = (0, 10, 10) T > determine mass participation by M k = ( j w jv kj ) 2 w j V 2 as M 1 = j kj and M 2 = This results in M 1 + M 2 = indicating that two masses contribute almost 100% to total displacement spectrum. 21. Mode participation factor is defined by M k = j w jv kj w j V 2 resulting in MPF 1 = and MPF 2 j kj = Individual displacement spectrum is given by DS k = j MPF j V kj resulting in DS 1 = (0, , ) T and (0, , ) T 23. Adding individual displacement spectrum by SQSS (i.e., square root of sum of squares) for each mass yields and This results compare well with Ansys WB as shown above. Random Vibration 24. In Project, right click Solution in Modal Analysis > Transfer Data to New > Random Vibration 25. Double click Setup 26. Click Random Vibration > Environment / PSD Base Excitation > PSD G Accelertion > enter G Acceleration=1 at Frequency=10 and 100 in table > Orientation = Y Axis 27. Add Directional Deformation with Direction = Y Axis 28. Solve > see Directional Deformation as

11 29. User can add Directional Acceleration at two masses to see accelerations 30. In Details, Scale Factor = 1 Sigma and Probability = 68.3%. This means that probability of time in which the structure will be at or less than Y displacement = is 68.3%. Thus, only (3 Sigma probability) 0.3% of time, structure will exceed 3 x = σ results 1 σ results are generally used for: First passage limit calculations for example, the probability that the displacement at a DOF will exceed a displacement limit in a given time period limit calculations the statistics shows that the level of a variable like displacement or stress is at or below 1 σ 68.2% of the time, between 1 σ and 2 σ 27.2% of the time ( ), and between 2 σ and 3 σ 4.3% of the time ( ), and above 3 σ less than.3% of the time.