Sir I got the same answer 8.65 but I rounded to 9 turns . But now I understood the concept.🤩🤩. Sir you are perfectly reading the mindset of aspirants 😍😍😍
Sir del to be same as 101.6 mm is permitted not more not less (as in question it is not mentioned that "not exceeding 101.6 mm" ), when N increases P decreases and P decreases then tao max is less than tao permissible then we should take N=9,( as tao can not greater than 124.105 MPa but can be less).
I think 9 is the right answer even as per your explanation.. because for a smaller spring to get deflected by same amount, more force would be needed.. which would result in stress.more than what is permitted
I am replying by watching concepts shown in video only! In first fomula, clearly, shear stress is proportional to load. And in second formula, to have same elongation with less number of turns, load needs to be increased.
Here we went for decreased value of deflection keeping max stress unchanged. If we go with decreasing value of stress keeping deflection unchanged, ans will be 9. Let we increase turns to 9. So to keep same deflection we have to reduce P. And we can reduce P by taking lower value of stress. In ques it is mentioned not to exceed the stress limit. We can reduce it. Why here reducing deflection is prioritized over reducing stress?
Good question. We are designing spring here and not designing the loading system. The approach you are proposing by changing the value of P is not correct. Loading is independent of number of coils and it can not be changed at our wish. If I give you a question to design a component and you will change the load applied and make the load 0, such designing will no work obviously. Apply the same concept here and you will get it.
@@aakashsingh7712 No bro, you were right.. Minimum 9 turns are required here.. Reason: Here an elongation of 101.6 mm is permitted which means that any time 101.6 mm of elongation may happen legally but for that amount of elongation if the total number of turns is 8, the generated shear stress will be more than 124.105 MPs, which does not meet the required condition and hence the correct answer of this question is 9 since for that amount of elongation the shearing stress will be less than 124.105 MPs which is permitted. So in integer, minimum 9 turns are required...
@@saikatkarmakar3040 Actually I know I m right. It was just that they cannot accept that they are wrong. So I wrote agreed. Ethically it is wrong. But those who understood, it doesn't matter whether I agree or not. Because these type of things needs to be understood as u did😊
@@aakashsingh7712 I paid ₹29k in installments for video course. But, because of the tremendous amount of wrong informations I did not continue the video course of exergic... and loosed my money🤣🤣... They are doing this kind of crap business for invaluable education... Never join exergic... they are crap... Great persons never show up themselves like Chandresh Mahajan does at the starting of his video...(underlining those words with a pen 🤣🤣)
Sir, in this same flt, there was a question on maximum stress on gear, Question no. 43 ( technical ) . There in velocity factor was given in the question and in solution velocity factor was multiplied to force induced to find effective load acting. But velocity factor is divided and service factor is multiplied. I am telling all this with reference to V. B. Bhandari book Either book or flt solution has to be wrong ?
sir i think this question has 2 sides , first , if we take the case in which we assume that we applied const load corresponding to max shear stress value than number of turns of 8 is ok as elongation is lesser than permitted value , as greatly explained by you, but if we consider that it is necessary to have a elongation of exactly 101.6 mm , then the load required to acheive it would be lesser and stress would be well below the limit of 124.105 MPa if number of turns are 9. it all comes down to its application i guess , wheather we are focussing on load bearing or want an exact amount of deflection . plz do reply sir if i am going in wrong direction
Sir, there is a mistake in solution of question no.51 thermo. It is solved assuming reversible refrigerator which is not given in question.Please correct.
Sir in shear stress we use Kw=( 4c-1)/4c-4) +0 .615/c c- dia ratio But since curvature effect is to be neglected so i put it to 1 ....but too Kw = diff. Formula not 1 ✍ i m confused.
to find out maximum shear stress induced in springs you have to use wahls factor...but we can use wahl stress factor only when there are curvature effects needed to be considered...since in question says curvature effects are neglected so we need to use the formula as mentioned in video...
Hello Argha, You can visit EXERGIC website for all course details. For any further queries, you can call on 9536651816 or mail at support@exergic.in Thank you.
GATE Course discounts, Ending SOON !!
For details, Visit 👉 www.exergic.in
no test series gave such a beautiful question this year you nailed it chandresh sir🙏🤟
Sir I got the same answer 8.65 but I rounded to 9 turns . But now I understood the concept.🤩🤩. Sir you are perfectly reading the mindset of aspirants 😍😍😍
Time to laugh at one of my friend who said that exergic mock test questions are formula based😂
🤣🤣🤣
@Navaneeth yupp
In answer 8.651 no. of full turns is 8 and 0.651 is a fraction of full turn.
Magnificently explained as always.
Proud to be a subscriber of exergic channel ...even I am from civil engineering background...I am always eager to watch ur lectures 🙏🙏
So nice of you!
Tq you sir.
Sir you are one, who gives more than expected...
Hope you can maintain at all...
Sir del to be same as 101.6 mm is permitted not more not less (as in question it is not mentioned that "not exceeding 101.6 mm" ), when N increases P decreases and P decreases then tao max is less than tao permissible then we should take N=9,( as tao can not greater than 124.105 MPa but can be less).
I've also similar opinion! 9 should be the right answer.
I think 9 is the right answer even as per your explanation.. because for a smaller spring to get deflected by same amount, more force would be needed.. which would result in stress.more than what is permitted
Please refer the basics of springs. There is a difference between how stress acts in a solid bar and a spring.
I am replying by watching concepts shown in video only! In first fomula, clearly, shear stress is proportional to load. And in second formula, to have same elongation with less number of turns, load needs to be increased.
Sir I missed the test please let us give the test on a later.... At least provide the pdf of questions of 2nd test
Here we went for decreased value of deflection keeping max stress unchanged. If we go with decreasing value of stress keeping deflection unchanged, ans will be 9.
Let we increase turns to 9. So to keep same deflection we have to reduce P. And we can reduce P by taking lower value of stress. In ques it is mentioned not to exceed the stress limit. We can reduce it.
Why here reducing deflection is prioritized over reducing stress?
Good question. We are designing spring here and not designing the loading system. The approach you are proposing by changing the value of P is not correct. Loading is independent of number of coils and it can not be changed at our wish.
If I give you a question to design a component and you will change the load applied and make the load 0, such designing will no work obviously.
Apply the same concept here and you will get it.
@@ExergicGATE
Agreed..
Thank u 😊
@@aakashsingh7712 No bro, you were right.. Minimum 9 turns are required here..
Reason:
Here an elongation of 101.6 mm is permitted which means that any time 101.6 mm of elongation may happen legally but for that amount of elongation if the total number of turns is 8, the generated shear stress will be more than 124.105 MPs, which does not meet the required condition and hence the correct answer of this question is 9 since for that amount of elongation the shearing stress will be less than 124.105 MPs which is permitted. So in integer, minimum 9 turns are required...
@@saikatkarmakar3040
Actually I know I m right.
It was just that they cannot accept that they are wrong. So I wrote agreed. Ethically it is wrong. But those who understood, it doesn't matter whether I agree or not. Because these type of things needs to be understood as u did😊
@@aakashsingh7712 I paid ₹29k in installments for video course. But, because of the tremendous amount of wrong informations I did not continue the video course of exergic... and loosed my money🤣🤣... They are doing this kind of crap business for invaluable education... Never join exergic... they are crap... Great persons never show up themselves like Chandresh Mahajan does at the starting of his video...(underlining those words with a pen 🤣🤣)
Sir, in this same flt, there was a question on maximum stress on gear, Question no. 43 ( technical ) .
There in velocity factor was given in the question and in solution velocity factor was multiplied to force induced to find effective load acting. But velocity factor is divided and service factor is multiplied.
I am telling all this with reference to V. B. Bhandari book
Either book or flt solution has to be wrong ?
Bro I didn't gave test but tell me if both values were more than 1 or not because if not it gets divided
Nicely explain ...😊🙏
Nice presentation skill of indirectly telling about video course
Good learning sir
But 1 doubt why we are limiting the elongation ? Why not shear stress?
If we limit shear stress then answer should be 9 right?
sir i think this question has 2 sides ,
first , if we take the case in which we assume that we applied const load corresponding to max shear stress value than number of turns of 8 is ok as elongation is lesser than permitted value , as greatly explained by you,
but
if we consider that it is necessary to have a elongation of exactly 101.6 mm , then the load required to acheive it would be lesser and stress would be well below the limit of 124.105 MPa if number of turns are 9.
it all comes down to its application i guess , wheather we are focussing on load bearing or want an exact amount of deflection . plz do reply sir if i am going in wrong direction
Beautiful explanation sir
Sir not any single flt test is provided with video solution?.all lessons are under construction?
It will be visible this week. In all.
I've missed the test, how can I at least watch the question paper ?plz reply
Sir please upload video of steam and gas turbine of applied thermo extra added course
Nhi hoga bhai ab 😂
Sir, there is a mistake in solution of question no.51 thermo. It is solved assuming reversible refrigerator which is not given in question.Please correct.
When nothing is given, it is taken as reversible. Just like if a gas is mentioned in a numerical, it is treated as ideal gas for calculation purposes.
What should one of cse student prefer to focus on gate or to prepare for placements ?
good analysis
In eoq model we get eoq for ex 232.55.. what we have to write in ans. In gate
Sir is there any course for gate chemistry students?
Sir in shear stress we use Kw=( 4c-1)/4c-4) +0 .615/c c- dia ratio
But since curvature effect is to be neglected so i put it to 1 ....but too Kw = diff. Formula not 1 ✍ i m confused.
to find out maximum shear stress induced in springs you have to use wahls factor...but we can use wahl stress factor only when there are curvature effects needed to be considered...since in question says curvature effects are neglected so we need to use the formula as mentioned in video...
@@v.venkatasainagakumar2644 ohh...i was not aware of this formula nd concept so far ...thanks 🤗🤗
Sir, can we get solutions of 2nd mock test?
Solution is available. Pls check here:
exergic.in/gate-gear-up/
Is there any course for ese mechanical?
Hello Argha,
You can visit EXERGIC website for all course details. For any further queries, you can call on 9536651816 or mail at support@exergic.in
Thank you.
Next mock is on 29th Jan sir ??
Yes Kanhu.
Is that free mock?
Yes bro.
Awesome 😍💋 💝💖♥️❤️
SIR PLZ UPLOAD LEAN MFG 456 VIDEOS
It will be uploaded today.
@@ExergicGATE THANQ SIR