I'm trying to do an incremental dynamic analysis using a harmonic function ( with 60cycles). the time duration is 23 sec, so 1sec cycle will be 2.6 So circular frequency will be = 2*pi*2.6=16.33 as I'm doing an IDA how can I increase the amplitude of harmonic load ?
Great video! Quick question, I'm trying to plot a maximum displacement per amplitude vs amplitude graph. Is there a way to plot the amplitude on the X axis and the output on the Y axis?
@@arsitech-ltd Sorry for the confusion. Please correct me if I am wrong, but I assume the amplitude is a load factor or multiplier to the input load. As a result, each amplitude produces a different load. For example, given a 100N load with amplitudes 1, 2, and 3, there would be loads 100N, 200N, and 300N as inputs. Ultimately, I am trying to plot the maximum displacements on the Y axis, and its respective applied amplitudes on the X axis.
Thank you for useful video! I have a question, when we use tabular for amplitude variations such as at t=0, amp=1 and t=1 and amp=0. this process is the same as deactivate load or not?
This is a great video. I have one question, though. I need to apply a sinusoidal load for fatigue simulation, and I know the fatigue life. It is around 10 million load cycles. How can I apply these many cycles of load? Is there a way to fix the time?
Dear, Decide time for 10 million cycle. Set circular frequency of load accordingly. For example 10 million (10,000,000) cycle you want to apply in 10,000 sec So in 1 sec cycle will be (10,000,000/10,000) = 1000 cycle So circular frequency will be = 2*pi*1000 I already explained this in video. However, for fatigue analysis you need to use Fesafe.
I wish you to guide me to achieve a better validation between experimental and Abaqus displacement graphs please, I got a good agreement regarding the peak value but I didn't get good agreement with respect to residual displacement (after peak value)
Thanks for sharing this valuable video. But I want to know how we can repeat a sine amplitude for more than 4 or 5 times? As you will have noticed, ABAQUS software consider only three cycles for each circular frequency.
Dear, In 3 sec you want w=8*pi It means in 3 sec you want to complete 4 cycles (or 8*pi). So how much cycle it will complete in 1 sec. That you have to feed into defination. In 1 sec it will complete (8*pi)/3
Respected sir, Can you please clarify....How to apply an amplitude of 20 microns with a frequency of 20000 hertz to a part and the amplitude is periodic. The part has to execute a periodic motion (vibration). Thank you.
Dear, You need to define amplitude as 20 micron (0.02 mm or 20e-6 m, depends on unit system u r considering) and instead of linear frequency you need to define circular frequency...that u can convert as per formula w= 2πf Here f is 20000 hz as you told. So define w = 125,663.706
@@arsitech-ltd Thank you sir.but I have done all of those mentioned by you.but when I apply the amplitude through velocity bc to the part,it is not vibrating or moving with the given velocity.can you suggest a solution. Thank you sir.
@@arsitech-ltd ok sir.i will try to solve it as per your suggestions.since it's a dynamic explicit problem,I need to check how I can change that increment size.. Thank you so much sir. And this video on periodic amplitude is really informative sir.
Dear, That is important when you have multiple steps in ur simulation. In case you have two situation 1. You want to carry out amplitude only for perticular step on which amplitude applied - step time. 2. You want to carry out amplitude for all steps - total time.
