Hello, Can I ask that how can I add spring element at the middle of the beam? As When I add spring elements in the middle of the beam it doesn’t affect my simulation
thank you for your help I want to make simulation for high performance concrete but i dont know what is the parameters that i need it to make simulation to HPC , Can you explain it to me ?
Hi, I really love your work and how your present it. I was wondering how you would go about finding the force in the spring that you used? Im doing a project where I need to find that force in relation to a buckling load. Many thanks
@@TMsChannels Hello, thanks for the quick reply. If I were to replace a, let's say a roller support or a fixed support with a spring, how do I represent it in Abaqus?
It appears that you made a little mistake of writing triangle load equation. The equation of the triangle load is -120e3*X + 600e3 = 5y => y = -24e3*X + 120e3, so your input at 11:18 must've been -24e3*X + 120e3. Hope to hear from you soon. Thank you
Hi, I went and checked it and believe mine is correct. The 120 kN/m value given is the slope of the line. So you have y=-120e3X+c and you know that the point (5,0) lies on the line. By substituting that back you get c = 600e3, hence y = -120e3X + 600e3.
Hi, I asked my instructor about this problem, He said if we set the center of the global coordinate system at the left end of the beam and we use it to write the equation of the line load, then we have the line load passing through two point (0,120) and (5,0). Generally, we have the equation of a line passing through two points A(xA,yA) and B(xB,yB) as follows: (x - xA)/(xA - xB) = (y - yA)/(yA - yB) In this case, we have the equation passing through (0,120) and (5,0) as: (x - 0)/(0 - 5) = (y - 120)/(120 - 0) => y = -24x + 120 Hope you recheck again. Thank you.
Hi, you are indeed correct. My assumption that 120 was the slope is totally wrong. Its been a while since I looked at detailed analysis. I will add a note to the video and thanks for pointing it out :)
Hi, Thanks for the comment. I chose 2 as the second degree of freedom corresponds to displacement in the global Y-direction (which in this case is the direction our springs are orientated in). 1 would be for displacement in the global X direction and 6 corresponds to the rotation about the global Z-axis. Hope this helps :)
amazing video please slab reinforcee under static load please
Great work man!
These videos are awesome!
It is imposible.why did the end of beam go up?
Hello, Can I ask that how can I add spring element at the middle of the beam? As When I add spring elements in the middle of the beam it doesn’t affect my simulation
thank you for your help
I want to make simulation for high performance concrete but i dont know what is the parameters that i need it to make simulation to HPC , Can you explain it to me ?
Thank you soo much
Hi, I really love your work and how your present it. I was wondering how you would go about finding the force in the spring that you used? Im doing a project where I need to find that force in relation to a buckling load. Many thanks
thank you so much from india...
How do you calculate the spring stiffness?
Hi, its given in the problem statement. In real life typically a spring has a given stiffness which is available from the manufacturer.
@@TMsChannels Hello, thanks for the quick reply. If I were to replace a, let's say a roller support or a fixed support with a spring, how do I represent it in Abaqus?
It appears that you made a little mistake of writing triangle load equation.
The equation of the triangle load is -120e3*X + 600e3 = 5y => y = -24e3*X + 120e3, so your input at 11:18 must've been -24e3*X + 120e3. Hope to hear from you soon. Thank you
Hi, I went and checked it and believe mine is correct. The 120 kN/m value given is the slope of the line. So you have y=-120e3X+c and you know that the point (5,0) lies on the line. By substituting that back you get c = 600e3, hence y = -120e3X + 600e3.
Hi, I asked my instructor about this problem, He said if we set the center of the global coordinate system at the left end of the beam and we use it to write the equation of the line load, then we have the line load passing through two point (0,120) and (5,0). Generally, we have the equation of a line passing through two points A(xA,yA) and B(xB,yB) as follows:
(x - xA)/(xA - xB) = (y - yA)/(yA - yB)
In this case, we have the equation passing through (0,120) and (5,0) as:
(x - 0)/(0 - 5) = (y - 120)/(120 - 0) => y = -24x + 120
Hope you recheck again. Thank you.
Hi, you are indeed correct. My assumption that 120 was the slope is totally wrong. Its been a while since I looked at detailed analysis. I will add a note to the video and thanks for pointing it out :)
Hello !
Do we need to keep "e3" when you type the expression? I typed: 120e3-24*X
Thank you !
@@minhly650 is the line passing thru (0,600) and (5,0) ?
Thank you!
i have a simple question, when you chose degree of freedom for spring, why you chose 2 degree instead of 1? Thanks for your video.
Hi, Thanks for the comment. I chose 2 as the second degree of freedom corresponds to displacement in the global Y-direction (which in this case is the direction our springs are orientated in). 1 would be for displacement in the global X direction and 6 corresponds to the rotation about the global Z-axis. Hope this helps :)
Thank you for this tutorial
could you please simulate beam on elastic foundation in abaqus
Hi, It's my pleasure. I shall put in on my list of videos :)
Thank you very much in advance
Thank u very much
👍🏻
I think equation of line used is wrong.Equation of line must be y=-24x+120 where x varies from 0 to 5 mtrs
You're right :) I will add a note to the video
south african much?