Thanks Dr. Sherif, you have a very well organized line of thought for the lectures made it very easy to understand for someone who's a mechanical engineer.
Very useful series of movies! Thanks! Here in part 3 at minute 9:46 the formula for "ye" is (A1*y1+A2*y2/sumA) maybe is better to move that bracket from the end before the division sign and write it (A1*y1+A2*y2)/sumA
Thank you Dr sherif for the great video. i have a question please. when we found that the collapse load is 0.7333 Mp. This collapse load corresponds to which plastic hinge ? they both occur at the same time or what ? 2nd question when i got the load factor to be 1.418 does that mean if. i multiplied the elastic load by 1.418 ill get the plastic load? 3rd question why did we take the 3 collapse mechanisms into consideration? the plastic hinge forms at maximum moment consecutively i think the 1st collapse is the only one that will happen in real life. fixed support will be hinge the under 2P will be another hinge so 3 hinges at st.line will cause failure
Than you for your interest and for your valuable questions. 1- The plastic hinges does not occur at the same time. If you have more than one plastic hinge, the first hinge will occure at the maximum moment then by increasing load another hing with form at the second maximum bending moment. The collabse load occure when you you hav number of plastic hinges that will result in unstability of the structure. 2- Yes. The load fator = Wp/We. So if you know two of them you get the third one.
3- The collapse mechanism depends on many things. A plastic hing may form under a smaller load if it is close to mid span than under a higher load close to support. It is also affected by the cross section of the beam in case you have beam with different cross sections. It is important to check all possible mechanisms to be sure. This will be clear in the coming video about Frame. It will come in a short time. Wait for it.
@@MrElgamal77 Professor, let me add this. For a statically underdetermined structure, you would need to perform the analysis of the elastic stage in order to determine in which sections are the maximum moments (matrix analysis, displacements or forces methods, Cross and so on...). After the first hinge is formed, again, you would need to see how the bending moments are redistributed in the elastic-plastic stage of the beam. And then find the maximum moment, where the next hinge is formed. Using the plastic hinges and the different possible collapse mechanisms, you do avoid the elastic analysis.
hi teacher / in ex:4 in 26:36 the n=r+1 r is equal to 3 my equation is this formula of n here is not right because n will equal 4 4 of plastic hing ? how did u equal n=3 ، the digree of indetemine is 3 ? can u explane
I have a question if you don't mind? In example 4: where did you get that there are 3 independent mechanisms? Nubmer of possibe hinges n = 3 Degree of redundancy r = 4-3 =1 Number of independent mechanisms = n-r = 3-1 = 2 Number of hinges for total collapse = r+1 = 1+1 = 2
Thank you so much for this video. However, I just have one question. What to do if we have uniform loading on beams? And a spring support is present? n=r+1 will still be applicable?
Thank you for your interest. The uniform load will not afect the number hinges. It will only affects the position of the hinge. It will be at the maximum positive moment. I do not understand what you mean. If you change the fixed support to a pin support, it will be a determinate beam and n= 0+1= 1
If you have a loading q(x) (could be uniform, but not necessarily), the virtual work associated to it, i.e, the external work, would be the integral, over the span where the load is acting, of q(x)*Area of the deformed beam after the plastic hinge was formed.
The maximum number is for but you may have a local failure by less number of plastic hinges. Foe ex. If you have an overhang, one plastic hinge will result in a collapse of the overhang part.
Thanks Dr. Sherif, you have a very well organized line of thought for the lectures made it very easy to understand for someone who's a mechanical engineer.
Very useful series of movies! Thanks!
Here in part 3 at minute 9:46 the formula for "ye" is (A1*y1+A2*y2/sumA) maybe is better to move that bracket from the end before the division sign and write it (A1*y1+A2*y2)/sumA
Well present of plastic and elastic structural behavior in details. Greatly appreciated Dr S El-Gamal.
Your explanation is a masterpiece Dr sherif ,keep it up 👏🏼
Thank you, I will. You may help me by sharing with your colleagues. Do not forget to "like, share, and subscribe".
This is so valuable and simply explained, great effort many thanks. million thumbs
Many thanks.
You deserve a lot by this episode mashaa Allah, I posted in my linked in www.linkedin.com/in/walidhassanmascpengpmp/detail/recent-activity/shares/
Great explanations throughout. Thank you!
The area moment of inertia for the elastic moment was supposed to be 1.5968 * 10^8 not 10^6
Well explained.
Thank you.
Thanks Dr....
in ex4, both ends with fixed supports, why there are no four plastic hinges please being (r+1)=4 ?
Thanks Dr. Sheriff, please is this the same as plastic method for steel and concrete structures...?
Thank you Dr sherif for the great video.
i have a question please. when we found that the collapse load is 0.7333 Mp. This collapse load corresponds to which plastic hinge ? they both occur at the same time or what ?
2nd question when i got the load factor to be 1.418 does that mean if. i multiplied the elastic load by 1.418 ill get the plastic load?
3rd question why did we take the 3 collapse mechanisms into consideration? the plastic hinge forms at maximum moment consecutively i think the 1st collapse is the only one that will happen in real life. fixed support will be hinge the under 2P will be another hinge so 3 hinges at st.line will cause failure
Than you for your interest and for your valuable questions.
1- The plastic hinges does not occur at the same time. If you have more than one plastic hinge, the first hinge will occure at the maximum moment then by increasing load another hing with form at the second maximum bending moment. The collabse load occure when you you hav number of plastic hinges that will result in unstability of the structure.
2- Yes. The load fator = Wp/We. So if you know two of them you get the third one.
3- The collapse mechanism depends on many things. A plastic hing may form under a smaller load if it is close to mid span than under a higher load close to support. It is also affected by the cross section of the beam in case you have beam with different cross sections. It is important to check all possible mechanisms to be sure. This will be clear in the coming video about Frame. It will come in a short time. Wait for it.
@@MrElgamal77 Professor, let me add this. For a statically underdetermined structure, you would need to perform the analysis of the elastic stage in order to determine in which sections are the maximum moments (matrix analysis, displacements or forces methods, Cross and so on...). After the first hinge is formed, again, you would need to see how the bending moments are redistributed in the elastic-plastic stage of the beam. And then find the maximum moment, where the next hinge is formed.
Using the plastic hinges and the different possible collapse mechanisms, you do avoid the elastic analysis.
hi teacher / in ex:4 in 26:36 the n=r+1
r is equal to 3
my equation is this formula of n here is not right because n will equal 4
4 of plastic hing ?
how did u equal n=3 ،
the digree of indetemine is 3 ?
can u explane
Thank you
Please do a video on inelastic zone length of fully plastic beam under a point load
Another fab video thanks
I have a question if you don't mind?
In example 4: where did you get that there are 3 independent mechanisms?
Nubmer of possibe hinges n = 3
Degree of redundancy r = 4-3 =1
Number of independent mechanisms = n-r = 3-1 = 2
Number of hinges for total collapse = r+1 = 1+1 = 2
Degree of redundancy in Eg 4 is not 1. It's an encastre beam, fixed on both sides. So degree of redundancy = 6-3 =3.
thank you very much
Thank you so much for this video. However, I just have one question. What to do if we have uniform loading on beams? And a spring support is present? n=r+1 will still be applicable?
Thank you for your interest. The uniform load will not afect the number hinges. It will only affects the position of the hinge. It will be at the maximum positive moment. I do not understand what you mean. If you change the fixed support to a pin support, it will be a determinate beam and n= 0+1= 1
If you have a loading q(x) (could be uniform, but not necessarily), the virtual work associated to it, i.e, the external work, would be the integral, over the span where the load is acting, of q(x)*Area of the deformed beam after the plastic hinge was formed.
Where can I found the table to determine the moment bending for elastic stage?
Supper! I really appreciate it. Bless you.
Thk doctor is there more videos in this field plastic analysis ?
Thanks you.
GREAT JOB
please can u make a video on torque of resistance at plastic hinges?
Thank you so much.
You're welcome.
Do you have any videos about the residual stress and load deflection diagram in plasticity ?
❤️
Thank doc
can you pls link the slides (Gdrive or slideshare ) in the description that would be of great help
ليه طريقة حساب ال N.A بتختلف فى ال elastic zone عنها فى ال plastic zone ؟
At 27:05 since the number oh hinges is n+1, why isn’t the hinges on the fixed to fixed beam 4? Instead it’s 3
The maximum number is for but you may have a local failure by less number of plastic hinges. Foe ex. If you have an overhang, one plastic hinge will result in a collapse of the overhang part.
Sir how do i get ymax ??
Check plastic analysis pPart 2.
21:18
Thank you Dr Sherif
After you have finsh Bs code go to ACI318-2019 if you can
Thank you. I will do my best.
Please do a video on inelastic zone length of fully plastic beam under a point load