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objective:To perform the explicit dynamics simulation for the 2 cases by changing the velocity of the cutting tool and then compare the results for both the cases. procedure: case1:cutting tool velcity of 20000mm/sec. 1.First edit the engineering data sources by adding the steel 1006 to the library and then import the…
Manohar SN
updated on 26 Feb 2021
objective:To perform the explicit dynamics simulation for the 2 cases by changing the velocity of the cutting tool and then compare the results for both the cases.
procedure:
case1:cutting tool velcity of 20000mm/sec.
1.First edit the engineering data sources by adding the steel 1006 to the library and then import the geometry and then assign the steel 1008 materail to the work piece and cutting tool to be structural steel and select the explicit dynamics.
2.Delete the contacts and then insert the mesh of automatic of tetrahedrons to the cutting tool and then sizing with the no of divisions as the 10 for the work piece.
Fig shows the mesh with the tetra hedrons and sizing of 10 divisions at the edges of workpiece.
3.In the analysis settings set the end time as the 7.5e-04sec and max no cycles is 1e+07 keep the remaing as it is and intial ,min time step, max time step to be program controlled.
4.Insert the velocity of 20000mm/sec in x direction with respect to global coordinates and remaining coordinates to be zero to the cutting tool and the work piece is to be fixed in all degrees.
Fig shows the velocity and the fixed support of the cutting tool and the work piece.
5.Add the solution of Equivalent Stress, Total Deformation insert a User defined result to calculate the temperature of the body and then click the solve after the solving of the problem the user defined result is defined for temperature of expression of temperatureall.
6. obtain the Equivalent Stress, Total Deformation and also insert a User defined result to calculate the temperature of the body.
Fig shows the total deformation of the min of omm and max of 34.311mm.
Fig shows the equivalent stress of min of 0mpa and max of 674.59mpa.
Fig shows the equivalent plastic strain of min of 0mm/mm and max of 1.3648mm/mm.
Fig shows the temperature of the cutting tool and work piece when it is conatct of min of 19.85 degree and max of 367.77 degree.
case2 :cutting tool velcity of 15000mm/sec.
Fig shows the total deformation of min of 0mm and max of 23.608mm.
Fig shows the equivalent stress of min of 0 mpa and max of 674.46mpa.
Fig shows the equivalent elastic strain of min of 0mm/mm and max of 1.4954mm/mm.
Fig shows the temperature as user defined result as the min of 19.85degree and max of 364.74degree.
RESULTS AND COMPARISION
CASES | Equivalent stress(mpa) | Total deformation(mm) | user defined result(temperature)(degrees) |
cutting tool velocity of 20000mm/sec |
Min :0 Max:674.59 |
Min:0 Max:34.311 |
Min:19.85 Max:367.77 |
cutting tool velocity of 15000mm/sec |
Min:0 Max:674.46 |
Min:0 Max:23.608 |
Min:19.85 Max:364.74 |
conclusion
In this challenge learnt about the explicit dynamics simualtion of the given geometry and then learnt about the cutting speed of the tool which can increases the temperature of cutting tool.
Note
In this challenge there is clear of generated data for the results and solution due to large file size.
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