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LS Dyna

LS Dyna

Topics related to LS-DYNA, Autodyn, Explicit STR and more.

to find max velocty of a bullet released from a pressure of 10psi from a cylindr

    • venkyvenkatesh374
      Subscriber

      Hello All., I really need help in solving this problem. My aim is to find the max velocity of a bullet. My initial conditions are pressure at 10PSI at back surface of bullet. Cylinder/barrel is fixed. contact is set to frictional. standard earth gravity is given.

      Initially i tried simulation at lower time frames (0.0001) i got a result with increasing velocity. as on when i crease the time frames, bullet velocity also increases.

      How do i find out max velocity in this case. will increasing more time frame gives me the desired result??. 

       

      have tested this practically and found the desired bulled reaches 8m( horizontal) @1.5 sec when fired at 90PSI.  i wanted to validate the same in ANSYS. I have set this problem in  transient structural. 

       

    • venkyvenkatesh374
      Subscriber

      Directional velocity is increasing

       

      In the above snip., I set pressure to 100Pa and simulated., the bullet is falling down because of earths gravity. but in graph the directional velocity in x axix is increasing constantly. this is confusing. 

    • peteroznewman
      Subscriber

      I don't know the size of your bullet or where you got 100 Pa for the pressure or the length of your barrel.

      Wikipedia has a page on the characteristics of a 9 mm cartridge, which includes a measurement of the exit velocity, which is 350 m/s.

      From that page, I got the mass and diameter of the bullet, and the length of the test barrel. Using those numbers, I created a set of equations to calculate the exit velocity assuming a constant average pressure on the back of the bullet for the length of the barrel. In reality, there is a pressure profile over the time the bullet is in the barrel. That Wiki page also lists the maximum pressure of 350 MPa.  I typed in numbers for the average pressure until I found that 67.5 MPa delivered the correct exit velocity.  At the exit time, there is no pressure to accelerate the bullet anymore. Instead there is a drag force on the bullet that gradually slows it down. You could model this using a 2 step solution where the pressure is 67.5 in step 1 till a time of 6.74e-4 then the pressure becomes some negative value that represents the drag force slowing the velocity down as it travels down range.

      Try using these equations with the mass of your bullet, or make a model using the values I used.

    • venkyvenkatesh374
      Subscriber

       

      Hello sir., Thanks for the detailed explanation. 

      I followed your suggestion by modelling in two step for the pressure values and could see the bullet speeding down. My mistake was giving constant pressure at the back surface of bullet. 

       

      I wanted to carry this analysis for 1.5sec ( this was time taken for the bullet to hit the ground fired horizontally from 1m height – Practicall tested). But if set time to 1.5sec the analysis is taking a very long time. Is there any chance i could carry this analysis in 2-3 steps. in first step i can set the time from 0-0.09secs, in 2nd step i’ll set the inirtial time 0.09 to 0.5secs and in 3rd i’ll set the time to 0.5 to 1.5secs. Is this possible sir?.

       

    • venkyvenkatesh374
      Subscriber

      any help on this please.??

    • peteroznewman
      Subscriber

      How did you get 0.09 s for the time in the barrel?  That seems too long.

      What was the exit velocity?

      What drag force are you using after the pressure on the back is gone?

      What are your analysis settings for Initial, Minimum and Maximum Time increment for each step?

      To save time, you could create a new Rigid Dynamics analysis with just one step. Apply an Initial Velocity to the bullet equal to the exit velocity and have the Standard Earth Gravity and the drag force. You can use a large Maximum Time increment to allow it to solve in a smaller number of time steps.

    • venkyvenkatesh374
      Subscriber

       

      Hello sir., thanks for the reply., 

      How did you get 0.09 s for the time in the barrel?  That seems too long.

      Assuming constant pressure throughout the barrel.,

      0.09 I just kept randomn value for timesteps. with a constant pressure of 90 PSI ( 620538 pa). The bullet is exiting the barrel at 0.008secs. refer snip 1 & 2

      What was the exit velocity? 

      The Bullet velocity is 37m/s at exit of barrel. refer snip 1 & 2

      What drag force are you using after the pressure on the back is gone?

      Air resistance of 0.189N (Drag force calculated) is given at front face of bullet

      What are your analysis settings for Initial, Minimum and Maximum Time increment for each step?

      Total step end time i set to 0.03secs and  I’ve turned off Auto time stepping and gave time step of 0.001secs. Refer snip 3

       

      I have practically done this and found bullet is hitting the ground at 8m away in 1.5secs when fired horizontally. I want to prove the same over analysis.

      So if i set the time to 1.5secs am getting solver error/Insufficient ram. how do i overcome this sir.

       

       

       

       

       

       

       

    • peteroznewman
      Subscriber

      In snip 1, I see that the pressure starts at 90 psi but ramps down linearly to 0 at 1e-3 seconds, so the average pressure over that 1 ms is 45 psi, not 90 psi.  You need to introduce a second step that ends when you want the pressure to end and keep the 90 psi at the start of that second step, then add a 3rd step that ends1e-5 seconds after the end of step 2 then in the Pressure Tabular data, you can type a 0 for the pressure in step 3.

      In the last snip, I see that you have turned Auto Time Stepping Off and the Time Step is 1e-3 s.  That means the solver is not updating the pressure between t=0 s and t=0.001 s so maybe you are getting the full 90 psi for the first 1 ms.  Perhaps this is why the results are changing when you use different Time Steps.  The proper way to do simulation is to reduce the time steps until the answers stop changing.

      Once the simulation reaches step 3, the velocity should change from increasing to nearly constant, except for the drag force, which I don't see in this simulation.

      In snip 2 I see that the velocity keeps increasing after the pressure has turned off. I don't see any load that is continuing to accelerate the bullet after the pressure turns off at 1e-3 seconds so I would say this simulation is producing wrong results.  It might start producing accurate results if you turn on Auto Time Stepping and use a much smaller Initial and Minimum Time step such as 1e-5 seconds.

    • venkyvenkatesh374
      Subscriber

       

      Got it sir., I will run this simulation as suggested.,  Also i wanted to know how would i measure how far this bullet has been travelled., As of now am using a geometry(pole) at 8m distance and seeing if bullet crosses at pole., Please let me know if there’s any probe or distance finder.

      Theorically am using time = 2h/g  to find time 

      and, Dist = velocity x time

       

    • peteroznewman
      Subscriber

      You don't need a pole at 8 m to measure distance. Simply add a Directional Displacement result in the X axis. That will tell you the position of the bullet from its starting point.

      To save computational time, you should Duplicate this analysis and convert it to a Rigid Dynamics analysis. That will change all the bodies to rigid bodies since you don't care about the stress and strain in the bullet or the barrel and you are incurring a large cost in computational time and space to store the results for every node when you don't care about anything other than the location of the center of mass.

      The equations you wrote are wrong. Look up the equations for Uniformly Accelerated Motion.  Gravity is an accelerationthat is pulling the bullet down.  You could add Standard Earth Gravity to your simulation to cause the bullet to drop down as it moves horizontally along the X axis.

    • venkyvenkatesh374
      Subscriber

      sure sir., I'll solve this rigid dynamics. also Directional displacement you mean as directional drformation??  I put a snip marking it can you please let me know if thats the one.

    • venkyvenkatesh374
      Subscriber
    • venkyvenkatesh374
      Subscriber

       

      Also In rigid dynamics., I’ve option to put my input conditions as acceleration, point load, std earth gravity, displacement. 

      I believe I’ve to put acceleration which i can obtain in transient analysis. also how do i put draf force acting on the surface??  Please correct me if am wrong sir.

       

    • peteroznewman
      Subscriber

      Yes, sorry, I meant Directional Deformation. Insert a second one for the Y (vertical) direction so you can see how far the bullet has fallen due to gravity.

      In Rigid Dynamics, under the Loads button is Remote Force which you can use with a + sign after you convert Pressure to Force and then with a - sign to use for the drag force. 

    • venkyvenkatesh374
      Subscriber

      Hello sir., Thanks and sorry for the late reply. 

      I've been working on it and for some reasons i couldn't match the experimental results with software results. Am just checking with boundary conditions and loads., I'll post the snips here once its complete. Thanks much again.

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