Hello Mr. Ameen. Thank you for the video. How can I duplicate and transfer a node and give constraint to one side of the spring&damper elements (to the duplicated node)?
I have explained this in the video already. Target mass of the block: 1.6kg Current mass: 8 kg. The target mass is smaller than the current mass. So, we need to multiply the mass by some factor (less than zero) to reduce it to 1.6 kg. So, 1.6/8 = 0.2 is the scaling factor. Since we cannot modify the mass directly, we change the density and this will change the mass.
@@AmeenTopa Hello, thankyou so much for sharing your knowledge on LS Dyna. Regarding the question, I like to get clarified that we might need to add more mesh elements to create two springs. since we used coarse mesh. I also like to add to know how to build a parallel spring system?
@@vspraneethdandibhotla3991 Spring elements are called Discrete elements, you need only two nodes. If say, you have three nodes which are colinear, then you made two springs. This is equivalent to a single spring system as long as the spring constant is same in all cases. For a parallel spring system, you need to create two spring elements which are parallel to each other, then connect their nodes at each side by using CONSTRAINED keyword.
hello. Can you please tell me, if a force of 10kN is directed to the cube (100 nods), how to adjust the DC constant damper so that a resistance of 5kN arises or moving velocity 10 m/s? Im have read vol2 manual and theory but can't understand DC constant unit measure and how this constant works P.s. many thanks for videos:)
Hello, Mr. Ameen, I employed a similar way to simulate spring by discrete elements, but the solution time became significantly long after the spring is introduced. I looked for some reference, they explained that timestep is sensitive to this kind of situation, so it is necessary to define mass for the spring part. But I noticed that you seem not consider the mass of spring while the solution time is acceptable. Could you please further introduce some knowledge about handle mass of spring? Thanks in advance.
Dear Evan, First of all, don't stress, stop screaming and enjoy life (referring to your profile picture) 😂😂 Regarding your questions, I am linking my spring to the two bodies. Thus, it will take the nodal mass data from the blocks. Each spring element consist of only two nodes. In my case, the nodes already have masses because they belong to the blocks. I hope this answers your question.
@@AmeenTopa Dear Ameen, That's a good advice, thank you! LOL It is enlightening for me, because, in my case, one node of the spring elements is directly restrained by Boundary_SPC, it means that node is massless. I will try to add some artificial mass to the node and see if it works! Thank you!😁
@@AmeenTopa That works, Mr. Ameen. It could be helpful for others, if you use discrete elements as spring and one of its nodes is not connected to any other element, please give it an artificial mass by element_mass otherwise the solution time will be ridiculous long.
The ACSII option is to generate extra data or more refined data. For example, you're d3plot intervals is 1 s. Then that will be the interval for all other output. So, if you use ASCII option and set DT for 0.1. Then you get the output at each 0.1 s. Some data are a not available in POST PROCESSING -> HISTORY output. Say, the forces between the parts, you cannot find it there. You need to use ASCII OPTION rcfoc for this output. I hope this answers your question.
Dear Ameen, in my modeling I use MAT_SPRING_ELASTOPLASTIC, I am getting confused what is the difference between Elastic stiffness and Tangent stiffness and how to get thess data? Thanks in advance 🙏
Video is amazing , thanks so much
You are welcome. Please share my videos if you find them useful.
وفقكم الله استاذ امين
آمين يا رب وإياك يا أخي العزيز
Hello, Thank you so much for the video. Just wanted to ask if it is possible to describe failure for the sprigs?
You are welcome. Springs are simplified elements. You can make it linear or non-linear but you cannot describe its failure (as far as I knw)
can you do incremental sheet metal forming using LS-DYNA ,please
th-cam.com/video/tg6LY7ab6_g/w-d-xo.html
Hi ,can you explain how to check FEA for vibrating screen in LS- DYNA
For this, it is recommended to use DEM approach.
Hello Mr. Ameen. Thank you for the video. How can I duplicate and transfer a node and give constraint to one side of the spring&damper elements (to the duplicated node)?
Transform - Translate, and tick on the copy node option.
*BOUNDARY SPC NODE to apply constraints.
شكرا على الفيديو استاذ امين, بس استفسار ليش قسمت الكتله على 8 لتغيرها ايش تعني
I have explained this in the video already.
Target mass of the block: 1.6kg
Current mass: 8 kg.
The target mass is smaller than the current mass. So, we need to multiply the mass by some factor (less than zero) to reduce it to 1.6 kg.
So, 1.6/8 = 0.2 is the scaling factor.
Since we cannot modify the mass directly, we change the density and this will change the mass.
@@AmeenTopa شكرا جزيلا لك
hallo dear how are you?
how to apply a parallel springs, equivalently?
The same way I made one spring, you can make two springs. Just make sure their length and stiffness are same.
@@AmeenTopa Hello, thankyou so much for sharing your knowledge on LS Dyna.
Regarding the question, I like to get clarified that we might need to add more mesh elements to create two springs. since we used coarse mesh.
I also like to add to know how to build a parallel spring system?
@@vspraneethdandibhotla3991 Spring elements are called Discrete elements, you need only two nodes.
If say, you have three nodes which are colinear, then you made two springs. This is equivalent to a single spring system as long as the spring constant is same in all cases.
For a parallel spring system, you need to create two spring elements which are parallel to each other, then connect their nodes at each side by using CONSTRAINED keyword.
thank you, do you say some thing about torsional spring?
Sorry I don't understand your question. Say what? The same thing?
ok, how we use it as a rotational spring or how we do rotational spring-damper element? thank you
could you answer please, as much as faster if you can
@@aklilualemneh9885 For rotational spring, you just change the value of DRO in SECTION DISCRETE. The rest of the steps should be the same.
@@AmeenTopa thanks so much, i got it
hello. Can you please tell me, if a force of 10kN is directed to the cube (100 nods), how to adjust the DC constant damper so that a resistance of 5kN arises or moving velocity 10 m/s?
Im have read vol2 manual and theory but can't understand DC constant unit measure and how this constant works
P.s. many thanks for videos:)
Hello, Mr. Ameen, I employed a similar way to simulate spring by discrete elements, but the solution time became significantly long after the spring is introduced. I looked for some reference, they explained that timestep is sensitive to this kind of situation, so it is necessary to define mass for the spring part. But I noticed that you seem not consider the mass of spring while the solution time is acceptable. Could you please further introduce some knowledge about handle mass of spring? Thanks in advance.
Dear Evan,
First of all, don't stress, stop screaming and enjoy life (referring to your profile picture) 😂😂
Regarding your questions, I am linking my spring to the two bodies. Thus, it will take the nodal mass data from the blocks.
Each spring element consist of only two nodes. In my case, the nodes already have masses because they belong to the blocks.
I hope this answers your question.
@@AmeenTopa Dear Ameen,
That's a good advice, thank you! LOL
It is enlightening for me, because, in my case, one node of the spring elements is directly restrained by Boundary_SPC, it means that node is massless. I will try to add some artificial mass to the node and see if it works!
Thank you!😁
@@AmeenTopa That works, Mr. Ameen.
It could be helpful for others, if you use discrete elements as spring and one of its nodes is not connected to any other element, please give it an artificial mass by element_mass otherwise the solution time will be ridiculous long.
@@Evanyali I don't know how to determine an artificaial mass? Can you tell me how much value we should assign that mass to one node?
Met mey,
Hint: you can measure the mass of the block. Use that value and you should get results similar to mine.
thanks for video. I've got a question. what does ASCII actually do? When do we need it?
The ACSII option is to generate extra data or more refined data. For example, you're d3plot intervals is 1 s. Then that will be the interval for all other output. So, if you use ASCII option and set DT for 0.1. Then you get the output at each 0.1 s.
Some data are a not available in POST PROCESSING -> HISTORY output. Say, the forces between the parts, you cannot find it there. You need to use ASCII OPTION rcfoc for this output.
I hope this answers your question.
Dear Ameen, in my modeling I use MAT_SPRING_ELASTOPLASTIC, I am getting confused what is the difference between Elastic stiffness and Tangent stiffness and how to get thess data?
Thanks in advance 🙏
Elastic stiffness is the initial stiffness whereas the tangent stiffness is the stiffness after yield point is reached.