I have a doubt. R2 will be generated only after touching the lower support. Untill then 0.8mm deformation is going to happen by both R1 and 40 kN only. Then how does it possible to frame an equation that's give a combined effect all the three forces. Pls clarify
@@althaf4432 in the question, situation should have been explained clearly. Until that this is a clumsy to understand. Based on imagination and an assumption only this is valid
@@ManasPatnaikofficial sir your teaching skills are excellent,a student dumb like me ,wants to complete som,thank you sir, It's a great privilege to have your videos here on UA-cam. Love and respect.
It is still unclear to me as to how are we applying the reaction force from the ground in the entire scenario in the second sum while it will come into effect only when the bar touches the lower support when the bars have already elongated by 0.8mm
These are conventional problems and these are going to help in both Stage 1 and Stage 2 of ESE. Anyway, the video on objective problems for both ESE and GATE will be last video in every topic. And Himanshu, if you have recently gone through the ESE 2019 paper, some questions were asked in the objective type which earlier had appeared in Conventional Paper. So stay connected, video making is a time taking process and i am sure you will be benefitted.
@@ManasPatnaikofficial I am following your channel since last 3 months and your efforts can be seen in presentation sir..keep uploading and I am preparing for ESE and gate. After selection I will also help you in every possible way. But for that first help us.
Thanks for pointing it out NAVEEN. It should have been 14.5 KN and 25.4 kN. And congrats you watched the video right till the end. There are similar queries from other students as well in the comments section.
No, sir did the sum wrong At first u will find a load till the bar touches the end that will be expressed as [p'×L2/A2×E] = 0.8 You will get the value of p' then You will get a net load of 40-p'
You're a great gem when it comes to engineering. What inspires me most is your hand writing. So clean. Thanks and much adoration
You deeply understand for what you teach and digest the nature of mechanic. Big Respect to you.
Sir R1= 25.46 KN not a 254.6 KN
R2= 14.53 KN not a 145.3 KN
Yes.....
yes
Yes
yes
Ha right hai
SUPER EXPLANATION SIR IN MECHANICS OF SOLIDS SUPER WHAT A GOOD EFFORT SIR
Sir your way of teaching is very good.please keep it up.And provides more lectures
What if we also have temperature changes in problem first i.e horizontal bar suspended from two brass and one steel bar.
I have a doubt. R2 will be generated only after touching the lower support. Untill then 0.8mm deformation is going to happen by both R1 and 40 kN only. Then how does it possible to frame an equation that's give a combined effect all the three forces. Pls clarify
untill then body 2 will be freely expanding once it touches rigid plate then stresses devolped in body 2
Same doubt for me too... I hope this shall be addressed.
@@althaf4432 in the question, situation should have been explained clearly. Until that this is a clumsy to understand. Based on imagination and an assumption only this is valid
Sir, can you make playlists of other mechanical engineering subjects ?
If it's possible it would be very helpful.🙂
That will gradually happen Pawan. Video making is a long process. Once MOS is complete we will start Fluid Mechanics.
@@ManasPatnaikofficial sir your teaching skills are excellent,a student dumb like me ,wants to complete som,thank you sir,
It's a great privilege to have your videos here on UA-cam.
Love and respect.
Nobody's dumb Pavan......
It's just that you are at the right place in the right time.
All the best.....
In the second problem the E wouldn't have to be 205.10^9Pa? Or it was meant to be KPa but it was written wrong?
My 1st like to your video
Sir I got Pb=25/36 Ps and as well all the value changes afterwards
Extra marks for good handwriting ☺️☺️
It is still unclear to me as to how are we applying the reaction force from the ground in the entire scenario in the second sum while it will come into effect only when the bar touches the lower support when the bars have already elongated by 0.8mm
Sir the content quality is good ...but we want Gate and ese oriented concepts and questions...thanks
These are conventional problems and these are going to help in both Stage 1 and Stage 2 of ESE. Anyway, the video on objective problems for both ESE and GATE will be last video in every topic. And Himanshu, if you have recently gone through the ESE 2019 paper, some questions were asked in the objective type which earlier had appeared in Conventional Paper. So stay connected, video making is a time taking process and i am sure you will be benefitted.
@@ManasPatnaikofficial I am following your channel since last 3 months and your efforts can be seen in presentation sir..keep uploading and I am preparing for ESE and gate. After selection I will also help you in every possible way.
But for that first help us.
Super sir 👌
21:05 .., you wrote wrong sir while converting r1 and r2 values to Kilo Newton's 😊
Thanks for pointing it out NAVEEN. It should have been 14.5 KN and 25.4 kN. And congrats you watched the video right till the end. There are similar queries from other students as well in the comments section.
@@ManasPatnaikofficial thanks for replying sir ♥️
Sir in problem 2 value of R2 is negative .so value of R1 add hoga na ....plz rply sir....
No, sir did the sum wrong
At first u will find a load till the bar touches the end that will be expressed as [p'×L2/A2×E] = 0.8
You will get the value of p' then
You will get a net load of 40-p'
I don't understand why r2 is pointing upwards. Could anyone pls clarify?
Bro it will be compressive that's why
sir R1 value will come in negative...ie...is -14493.33
bro please take the value of 40 in kn i.e. 40*1000
R1=40*1000-14.53*1000=25.47*1000
Sir,why aren't we considering elongation due to its own weight?
To make question easy. but in practical we consider everything it's may be self weight, thermal stress etc.
Sir please help me with this equation I can't solve I don't know why but I can't solve
2(2.85sigma al +80)(1200)+(2400sigma al) =400×10^3
Keep it up sir!👌👌👌
7th lecture completed
sir bohut jada ad ata hai jitna video nhi chalta hai hai usse jyada ad ata hai
Nice sir
Pb=25/54ps ?some budy explain calacution part