Hello! Thank you very much for such a valuable course for static structural analysis with SW! BUT, pay attention that the lector used different materials while comparing the wrinkled beam's cases! In the first case aluminum alloy T6 and Titanium in the second! That's why we saw doubled difference in stress figures!
That's a good catch, Alexander, that the material changed on my Stepped Ends Simply Supported study. Must have been a mistake. BUT!...That has very little to do with the increased stress. This is a great example of the material properties having little effect on the STRESS result. I copied the study named "Stepped ends Simply Supported" and changed the material to the same Titanium (Ti-5AI-2.5Sn-) used with one end fixed study and the result was nearly identical to that for the 6061-T6 Aluminum: It increased from 786.5 MPa to 789.4 with the material change, less than 0.5%. So why is this? Well, what are the material properties used in a SW Sim linear static study? Young's Modulus and Poisson's Ratio. Modulus values in these two materials: 6061-T6 = 69,000MPa , Ti-5AI-2.5Sn- = 110,000MPa. Titanium is about 1.6 times stiffer than Aluminum. Poisson: 6061-T6 = 0.33, Ti-5AI-2.5Sn- = 0.31, similar range with Aluminum having slightly softer neck-down properties (larger value = softer). Compare to most steels at 0.27.
These two material differences mean that the Aluminum is more flexible than the Titanium. Under the same loading for identical geometry and boundary conditions, the Aluminum part will deform more than the Titanium and the local STRAIN will also be greater. BUT larger strain does not mean larger stress when comparing two models, in this case because the lower Modulus means a shallower slope of the stress-strain curve. So if one slides further to the right on the stress strain curve, with Aluminum you can get to a larger STRAIN than with Titanium before you reach the same STRESS value. Which would you expect to deform more? The Aluminum, of course. Max deformation for the Al study is 4.5mm while for the Ti it is 4.2mm, not as much difference as I expected, but still, the more flexible material does deform more. Now this all may be moot if the yield stress of the material is reached (which it certainly is for the 6061-T6 Aluminum). But remember that a linear/static FEA tool simply analyzes stress as a linear function of strain with no regard to the yield stress...in other words it will continue to calculate the stress way beyond material failure based on the linear stress-strain curve (Young's Modulus) and it is up to the user to observe that yield has been exceeded.
Thanks for the video. I was wondering that in case we pull the bar (the one in the video) from both sides, is there a good way to restrain it without causing errors in linear dynamic study?
Hi, Jamie, Sorry for the 1 year late reply but I must have missed the notification of your comment. If you're doing a linear dynamic study and you want to pull the right hole, say, to the right, you might get errors if there is nothing to prevent the model from drifting vertically (Y-direction) due to calculation round-off error. Now in this case I would not expect that, but it can happen. So let's say that the left hole is hinged, and we apply a force to the right hole pulling to the right. Or we might instead apply a prescribed displacement, say, 0.25mm to the face of the hole and move it to the right. A couple of things can insure convergence. You could restrain the right hole surface from vertical movement using a reference restraint in the Y direction and set the value to zero. That presupposes that the hole will stay exactly cylindrical which it might if it's glued to a super stiff cylindrical rod. But that's not usually the case and we might expect SOME deformation of that hole with a pull to the right. So we could instead go into our study options and turn on Soft Springs or Inertial relief. These prevent the model from drifting. Another option (a really GOOD option) might be to use a bearing force on the right hand hole. This requires a coordinate system to be created in the hole center (use a sketch). What a bearing load does is to apply the load in a realistic way to the hole, that is, with a sinusoidal or parabolic distribution around the 'pull' side of the hole (rather than to the entire hole). So you'll get uneven stresses around the hole (reality, right). And as a side effect, the bearing load should prevent vertical instability to the hole so your study converges. Hope this helps.
@@benrosow7997 thank you for your detailed explanation. I was working on this problem during my dissertation. It would have been great if UA-cam could notify you on time as your answer could save me lots of time :) As you suggested, I actually tried restricting the z axis movement, used bearing force with another coordinate system and a couple of other things (I don't remember). I got around 80% mass participation factor (or something like that), so that I think helped me solved the issue. If I have similar type of problem in the future, I will refer back to your answer and go through all methods you suggested. Thank you again for your answer!
@@jamiemitev When you say 80% Mass participation factor, that tells me you are running a dynamic analysis (frequency doain based) not a Non-linear dynamic study. Different animals. NL studies have the option of static or dynamic. A poor choice of words as static indicates use of pseudo time an dDynamic indicates boundary conditions with a time element and a true study time. Frequency domain studies (linear/dynamic) are essentially frequency studies with numerou bells and whistles. So I am not sure what your study intent was.
Very valuable... Thank you for sharing your knowledge.
Thank you, great lesson
very valuable, I wish the design model files in the videos were available to practice by our-self during videos...
Hello! Thank you very much for such a valuable course for static structural analysis with SW!
BUT, pay attention that the lector used different materials while comparing the wrinkled beam's cases! In the first case aluminum alloy T6 and Titanium in the second! That's why we saw doubled difference in stress figures!
That's a good catch, Alexander, that the material changed on my Stepped Ends Simply Supported study. Must have been a mistake. BUT!...That has very little to do with the increased stress. This is a great example of the material properties having little effect on the STRESS result. I copied the study named "Stepped ends Simply Supported" and changed the material to the same Titanium (Ti-5AI-2.5Sn-) used with one end fixed study and the result was nearly identical to that for the 6061-T6 Aluminum: It increased from 786.5 MPa to 789.4 with the material change, less than 0.5%. So why is this? Well, what are the material properties used in a SW Sim linear static study? Young's Modulus and Poisson's Ratio.
Modulus values in these two materials: 6061-T6 = 69,000MPa , Ti-5AI-2.5Sn- = 110,000MPa. Titanium is about 1.6 times stiffer than Aluminum.
Poisson: 6061-T6 = 0.33, Ti-5AI-2.5Sn- = 0.31, similar range with Aluminum having slightly softer neck-down properties (larger value = softer). Compare to most steels at 0.27.
These two material differences mean that the Aluminum is more flexible than the Titanium. Under the same loading for identical geometry and boundary conditions, the Aluminum part will deform more than the Titanium and the local STRAIN will also be greater. BUT larger strain does not mean larger stress when comparing two models, in this case because the lower Modulus means a shallower slope of the stress-strain curve. So if one slides further to the right on the stress strain curve, with Aluminum you can get to a larger STRAIN than with Titanium before you reach the same STRESS value.
Which would you expect to deform more? The Aluminum, of course. Max deformation for the Al study is 4.5mm while for the Ti it is 4.2mm, not as much difference as I expected, but still, the more flexible material does deform more.
Now this all may be moot if the yield stress of the material is reached (which it certainly is for the 6061-T6 Aluminum). But remember that a linear/static FEA tool simply analyzes stress as a linear function of strain with no regard to the yield stress...in other words it will continue to calculate the stress way beyond material failure based on the linear stress-strain curve (Young's Modulus) and it is up to the user to observe that yield has been exceeded.
29:38 >>> Use of Virtual Wall.
thanks
Thanks. Great Video.
For the press fit example, would the result be the same if we use foundation bolt instead of cylindrical constraint + virtual wall?
Thanks for the video. I was wondering that in case we pull the bar (the one in the video) from both sides, is there a good way to restrain it without causing errors in linear dynamic study?
Hi, Jamie, Sorry for the 1 year late reply but I must have missed the notification of your comment. If you're doing a linear dynamic study and you want to pull the right hole, say, to the right, you might get errors if there is nothing to prevent the model from drifting vertically (Y-direction) due to calculation round-off error. Now in this case I would not expect that, but it can happen. So let's say that the left hole is hinged, and we apply a force to the right hole pulling to the right. Or we might instead apply a prescribed displacement, say, 0.25mm to the face of the hole and move it to the right. A couple of things can insure convergence. You could restrain the right hole surface from vertical movement using a reference restraint in the Y direction and set the value to zero. That presupposes that the hole will stay exactly cylindrical which it might if it's glued to a super stiff cylindrical rod. But that's not usually the case and we might expect SOME deformation of that hole with a pull to the right. So we could instead go into our study options and turn on Soft Springs or Inertial relief. These prevent the model from drifting. Another option (a really GOOD option) might be to use a bearing force on the right hand hole. This requires a coordinate system to be created in the hole center (use a sketch). What a bearing load does is to apply the load in a realistic way to the hole, that is, with a sinusoidal or parabolic distribution around the 'pull' side of the hole (rather than to the entire hole). So you'll get uneven stresses around the hole (reality, right). And as a side effect, the bearing load should prevent vertical instability to the hole so your study converges. Hope this helps.
@@benrosow7997 thank you for your detailed explanation. I was working on this problem during my dissertation. It would have been great if UA-cam could notify you on time as your answer could save me lots of time :) As you suggested, I actually tried restricting the z axis movement, used bearing force with another coordinate system and a couple of other things (I don't remember). I got around 80% mass participation factor (or something like that), so that I think helped me solved the issue. If I have similar type of problem in the future, I will refer back to your answer and go through all methods you suggested. Thank you again for your answer!
@@jamiemitev When you say 80% Mass participation factor, that tells me you are running a dynamic analysis (frequency doain based) not a Non-linear dynamic study. Different animals. NL studies have the option of static or dynamic. A poor choice of words as static indicates use of pseudo time an dDynamic indicates boundary conditions with a time element and a true study time. Frequency domain studies (linear/dynamic) are essentially frequency studies with numerou bells and whistles. So I am not sure what your study intent was.
@@benrosow7997 My question above was on linear dynamic study (for frequency domain), and your initial answer is the correct way to solve it 😄
I would like more design, can send email ask you?