ohhhhh MO you got me! you are 100% correct - i meant to say the moment M is acting about the center of rotation Cr and the force V is applied at the center of mass Cm the moment arm is the eccentricity e between Cm and Cr and thus creates a clockwise moment about Cr. You just got pinned to the top homie! very good catch. I ended up doing the rest of the problem correctly (thankfully) but graphically showed that error on my part. BIG FLEX Mo
mfgman get after it! good luck and let me and the team know how it goes. I didnt realize ME's take the SE. Is that normal or are you just straight flexin?
@@Kestava_Engineering it's more of a decision for me to ensure I keep my business strong. I own Quanta Engineering in Olathe, KS. I spend a lot of my time designing small wind turbine and telecom towers and small steel structural shelters. A lot of states require an SE stamp. So, for me, getting an SE keeps the outside guy from taking my role.
I went ahead and calculated the Shear on SW4. The direct shear was 30 (both using the equation AND taking 40 -10 to get 30, the torsional shear force was the same 8.57. Will it always be the same once you calculate it the first time?
If there were two shear walls instead of one where shear wall 4 is, would Why-Bar be the same for both of them? The distance to the center of RIgidity?
hey. could you please answer this query. how do rcc slabs act in real life structures as rigid or semi rigid diaphragms when lateral forces act on them.are they designed for any lateral forces ?
reinforced concrete slabs (as I understand it) can be conservatively analyzed as rigid diaphragms if they have sufficient in-plane stiffness. However, if required, a RCC diaphragm can be analyzed as semi-rigid, and if done correctly, should give you a more efficient and refined diaphragm. Most of the time a semi-rigid analysis is required by code or when you believe you will have in-plane deformation issues. I believe RCC slabs ideally act as semi rigid systems in real life to some degree but it depends of slab properties and geometry, but please double check me!
i guess it should probably be better to consider X and Y signs when you were arranging your table, since these signs will later affect on the value of the torsional shear force (second portion of the formula). like for shear wall 4, we can not use (+20 ft) in the second term of the formula, since we need to use -20 ft. you agree with that?
you could have forces on another story above with a dissimilar footprint that requires a thicker shear wall on the ground floor. thats just one example tho.
Just curious, why isnt the sum of the 2 shear components a Vector like with Bolt groups. i'd like to avoid the damn vector calcs on the bolt groups lol
Rich, if we want to dive into rigid diaphragms some more, what code book or reference material were you getting your equations from? Or where would you recommend we go to learn more? Please don't say Wikipedia.
Hi Eric - I have a structural engineering reference manual that has a great example. I would also point you to the SEAOC design manuals. they have A LOT of great content for diaphragm design. no WIKI up in here!
the torsional moment is counter clockwise which means the shear resulting from it in wall 2 acts against the base shear. So shouldn't you subtract the torsional moment from wall 2. Its wall 4 you would add the torsional shear since on that side its acting in the same direction as the shear.
For the torsional part, the unit dimensions don’t match. The way you did you should had kip*ft^2 as a resultant force due the torsional moments. I believe that you should divide the torsional moments by the lever arm, not multiplying it. Thus, you can get the right units.
I was thinking Ip has units of ft^2 even tho not written out. So that way the units do work out and you don't need to divide by lever arm as suggested.
@@Kestava_Engineering Please do a quick 5 minute addendum to this video to explain that. I thought the magnitude AND sign were the same value for both 2 and 4
If I figured all the shear walls were the same thickness and height but varied only by length, could I accurately state that R is equal to 1, so my Resistance value is simply the lengths of each of my shear walls? Joes told me to ask that one
@@Kestava_Engineering Hi Kestava! A struggling/ bad engineer here. But, I will still make a comment. As a shear wall resists through in plane bending, the increase in length of a shearwall adds enormously to the rigidity of a shearwall, based on the formula bh^3/12. So, if the length of a shearwall is increased twice, the rigidity will increase by 7 times. Hence, the distribution of load will not be proportional to the length of the shearwall.
Matt don't let Jose convert you to the DARK SIDE.... like the side of the auditorium with the light out...next to the popcorn machine where he sits. DONT DO IT
Thanks for sharing great content. Just a quick question. Isn't Mt about CR and shear force (V) acts at CM hence creating clockwise moment (Mt) ?
ohhhhh MO you got me! you are 100% correct - i meant to say the moment M is acting about the center of rotation Cr and the force V is applied at the center of mass Cm the moment arm is the eccentricity e between Cm and Cr and thus creates a clockwise moment about Cr. You just got pinned to the top homie! very good catch. I ended up doing the rest of the problem correctly (thankfully) but graphically showed that error on my part. BIG FLEX Mo
I passed the PE 6 months ago thanks to your video and I am 62 years old. I am trying to pass the SE and took it last week.
I'm in debt to you man, thank you so much
I’m rally stoked about watching this. Been struggling the past few days with an open front structure with a job at work. Thanks
Thanks for all the work you put into these. I'm a Mechanical Eng PE and I'm taking the SE exam in a few weeks.
mfgman get after it! good luck and let me and the team know how it goes. I didnt realize ME's take the SE. Is that normal or are you just straight flexin?
@@Kestava_Engineering it's more of a decision for me to ensure I keep my business strong. I own Quanta Engineering in Olathe, KS. I spend a lot of my time designing small wind turbine and telecom towers and small steel structural shelters. A lot of states require an SE stamp. So, for me, getting an SE keeps the outside guy from taking my role.
Very valuable guide! Before i watched that i did not know how to account for the potion of V due to Mt. Now i know
Yes, please do reinforce the concrete shear wall too. Thanks. The more real-life problems the better.
You rock Rich!
Jeremy you're making e blush, I do it all for the team.
All hail the Pen of Learning
Thanks Man!!
Do you think you can do a full example for a reinforced masonry wall using the TMS 402/ 602-16 code?
oh baby baby that sounds fun. gunna take me a liitle bit to get one together but think thats a great idea!
@@Kestava_Engineering that will be awesome!!
Thank you so much for the great explanation! Will the external force acts to Center of Mass or Center of Rigidity?
I went ahead and calculated the Shear on SW4. The direct shear was 30 (both using the equation AND taking 40 -10 to get 30, the torsional shear force was the same 8.57. Will it always be the same once you calculate it the first time?
Are regidity and stiffness the same thing?
Thank very much ❤️
What if center of mass and center of rigidity was found in void area in slab
Is that problem??
very nice, thanks.
If there were two shear walls instead of one where shear wall 4 is, would Why-Bar be the same for both of them? The distance to the center of RIgidity?
Thanks buddy!!!
Any time Filipi
hey.
could you please answer this query.
how do rcc slabs act in real life structures as rigid or semi rigid diaphragms when lateral forces act on them.are they designed for any lateral forces ?
reinforced concrete slabs (as I understand it) can be conservatively analyzed as rigid diaphragms if they have sufficient in-plane stiffness. However, if required, a RCC diaphragm can be analyzed as semi-rigid, and if done correctly, should give you a more efficient and refined diaphragm. Most of the time a semi-rigid analysis is required by code or when you believe you will have in-plane deformation issues. I believe RCC slabs ideally act as semi rigid systems in real life to some degree but it depends of slab properties and geometry, but please double check me!
i guess it should probably be better to consider X and Y signs when you were arranging your table, since these signs will later affect on the value of the torsional shear force (second portion of the formula). like for shear wall 4, we can not use (+20 ft) in the second term of the formula, since we need to use -20 ft. you agree with that?
Hello why would we need to make one of the shear wall thicker than the others?
you could have forces on another story above with a dissimilar footprint that requires a thicker shear wall on the ground floor. thats just one example tho.
Just curious, why isnt the sum of the 2 shear components a Vector like with Bolt groups.
i'd like to avoid the damn vector calcs on the bolt groups lol
Rich, if we want to dive into rigid diaphragms some more, what code book or reference material were you getting your equations from? Or where would you recommend we go to learn more? Please don't say Wikipedia.
Hi Eric - I have a structural engineering reference manual that has a great example. I would also point you to the SEAOC design manuals. they have A LOT of great content for diaphragm design. no WIKI up in here!
the torsional moment is counter clockwise which means the shear resulting from it in wall 2 acts against the base shear. So shouldn't you subtract the torsional moment from wall 2. Its wall 4 you would add the torsional shear since on that side its acting in the same direction as the shear.
For the torsional part, the unit dimensions don’t match. The way you did you should had kip*ft^2 as a resultant force due the torsional moments.
I believe that you should divide the torsional moments by the lever arm, not multiplying it. Thus, you can get the right units.
Yes I agree - I was definitely sloppier back then and really need to redo this video more clearly!
I was thinking Ip has units of ft^2 even tho not written out. So that way the units do work out and you don't need to divide by lever arm as suggested.
Great ! Can you help us and post video about how to design diaphragm horizontally please 🙏
Hi zozo - consider it added to the list!
I just want to highlight something, in some cases the shear from torsion effect shall be in negative value (wall 4 for example)
spot on - I need to redo a rigid diaphragm analysis
@@Kestava_Engineering Please do a quick 5 minute addendum to this video to explain that. I thought the magnitude AND sign were the same value for both 2 and 4
thank you sir.
you're very welcome!
If I figured all the shear walls were the same thickness and height but varied only by length, could I accurately state that R is equal to 1, so my Resistance value is simply the lengths of each of my shear walls?
Joes told me to ask that one
I believe the answer to that one is yes Matt!
@@Kestava_Engineering Hi Kestava! A struggling/ bad engineer here. But, I will still make a comment. As a shear wall resists through in plane bending, the increase in length of a shearwall adds enormously to the rigidity of a shearwall, based on the formula bh^3/12. So, if the length of a shearwall is increased twice, the rigidity will increase by 7 times. Hence, the distribution of load will not be proportional to the length of the shearwall.
Jose keeps saying "eye why why?!"
Matt don't let Jose convert you to the DARK SIDE.... like the side of the auditorium with the light out...next to the popcorn machine where he sits. DONT DO IT
jesus there is so many adverts
crap really!!? Im just leaning about ads now. Ill do my best to tailor it better. Its 100% about the content and information ALWAYS Michael!