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No matter how many times i read the subject on my own, your videos never fails to surprise me and makes me feel that i have to do it once again. Awesome😊😊😊

Man this dude has superb foundation ....make us see hidden concepts generally we don't even think of

Exceptional effort from your side sir.U have got air 37 being in college for a reason.. please keep on adding such more videos .. please provide us correct solutions of such previous years questions more.thank you.

This is called The Quality 💯✨ I always feel grateful after completing every video lecture because I always find something new even as a repeater.. 😊 Thank you so much sir.. Such a quality is damn needed... 😊 I have been watching you giving your best, now it's my turn to give my best..!

I didn't know those big formulas. I have solved it in simple geometrical way. My answer has been 5.01400718 unit. #My approach: 1st: I assumed that at first two normal uniform loads are applied on the element so that only the given two normal strains will be generated . 2nd: Now I calculated the lengths of the elongated sides of the rectangle ( e.g; elongated horizontal side's length= A=4+4×0.001 unit.) 3rd: Now after the normal strains took place (i.e; after equilibrium is reached), a uniform shear load is applied on the element so that it generates the given shear strain. 4th: now during shear strain, the previously elongated lengths of the two consecutive sides won't change . But the angle between them will be changed . After equilibrium is reached, the angle will be (π/2+r) radian, where, r=shear strain. 5th: Now the elongated side's length (which is asked as the answer of the question) is the length of a side of a triangle of which another two sides' lengths and angle between them are known. It can be found easily from geometry. Elongated side's length= √(A²+B²-2ABcos(π/2+r))= 5.01400718 unit. where, A=elongated length of PQ=4×(1+0.001) unit ; B= elongated length of QR= 3×(1+0.002) unit ; r= shear strain=0.003; #NB: try to solve it with opposite of the above approach (i.e; assume that first shear load is applied and then the normal loads will be applied so that the given strains would happen). This approach may clear some of your doubts providing some special knowledge. # Is my approach correct sir ?

Every time I watch your video, I learn something new. Your Calmness while explaining questions really pass on to us . Thank you for this video. I have already purchased test series and a question bank of exergic and they are really helpful and to the point and more importantly, it focuses on Gate level dificulty.

Now that's what we call "Strong Foundation" on the Concepts.

You are doing a great job, keep going and clarifying the doubts.

Amazing! 👍 My doubt got cleared 😊 Thank you Sir.

True sir. Many of us have doubt in the 'theta' value. You just hit the bullseye for clearing & strengthening the concept. Thank you. :) :) :) :)

Thank you so much sir Absolutely right said its a must needed video .. Thank you for explaining all these in a clear way

This is the concept I was searching for in this particular Question... finally found it in this video...Thank you so much...I was confused about this question for a long time... finally cleared 🙏

What I did was calculated normal strain at 36.87 and shear strain at a plane (90 + 36.87),as diagonal length changes because of both , changes in linear dimensions as well as angular dimensions , added both to get final diagonal as 5.0146 mm......

Just amazing to see the explanation of question which I had stucked in past and was confused a lot thank you for clearing the concept it gives motivation to us that what our mind ,basic concepts telling just proceed with that.

Even after calculating with your method we are getting 5.015 !!!

Thank you so much sir. My doubt got cleared. Sir kindly explain GATE 2019 Set 1 Question no: 41 ( UTM Experiment Question )

each and every time I watch your video one of my misconceptions get cleared thank you sir

Learned a New concept that no one discussed before Thank U Sir

I need more like buttons 😍

Bhai plz upload soon engineering mechanics video course

In transformation of strain it is common that elongation and the plane on which stress acting there is difference of 90 degree so we can take theta from horizontal counterclockwise in case of strain formulas...

Thank you sir🙌🏻

Sir it was always my confusion, You have cleared it nicely, Thank you so much.

I was doing it the same way, just took the inclination as -theta and calculated strain(y')(i.e; at (90+theta)) yielding the same result. Had a feeling that other sources were mistaken in some way. Thanks for the clarification :)

Thank you sir... very good explanation of concept

Excellent Sir...I have a doubt..if they asked for the other diagonal what value of theta should we take

Very good bring some more quality lectures

When i tried to calculate elongation on opposite diagonal by reversing shear strain direction ,i got different angle ,so different answer. But both cases looks very similar.

Thank you for the video sir...

Thanks sir for cleaning the misconception.

Are Exergic workbook questions of higher level than Exergic test series? or same?

Bohot bohot badhayi sir ji!! 😍😍 💍

great new concept... thank you for the free knowledge

Concepts got dam cleared

great explanation amazed

The concept is good. But we can also find the answer considering PR plane, just take Epsilon y'.

Thank u so much sir, I got the concept.

sir, please post lecture videos of SOM

Yes sir I observed this mistake

sir could u plz make some videos on RAC subject

Sir...will you explain what us the difference between torsional shear stress and normal shear stress

Thanks sir

I think answer is same but the concept was wrong from other sources as theta value taken same value

Sir why we take normal strain formula why we don't take shear strain formula shear strain is responsible for stretching the diagonal

Sir what about aptitude session

sir, in the very beginning for the inclined plane to have a stress or strain we have to consider any plane normal to that plane and taking an angle theta from the vertical but rather we have taken just the angle between the inclined plane and the normal .Can u please explain it once more.

Just a comment, while explaining the last concept if you have exposed the internal element at the plane, that will be mindblowing :). By the way awesome video #cheers

We already paying lakhs to universities but we want to pay even this exergic also but we cant do that we r unable to do that so the one thing is just loosing hopes so rather than that I just following your youtube videos even I can cover few models right so that's why so please upload all important models sir pls

Superb!

Thank you for the concept tho !

thnk u sir i was having always a dout in this concept.

VERY NICE

u use the same angle as the other source used then how the answer can vary and by which concept the answer is vary????

great work....how you clear your concept..??form where you get all this concept.???

Amazing sir.....

Sir do we have to remember the formulas of moment of inertia of all shapes?

Thanq soo much sir. Do more vedios sir

Sir please make vedio on deflection of beam

thanks a lot sir

Sir sab samaz aaya bt ye btaye plz answer different kaise hua sab source ne angle to same liya na aur formula bhi

Sir ,I have a trivial confusion left to be cleared. How the answer varies when every other source and you have taken the value of angle same in the formula?

Next to wowsome...

Thank you

sir if I change the direction of shear strain then elongated diagonal become QS

Sir, epsilon PR is being calculated by the epsilon thita formula, mentioned in the video, but my answer is coming negative nd not even matching. Plz help.

Hello sir please tell me from which subject I should start ur vedio lacture.

❤

Me to aise hi karta hu

Hello sir.. I've a question.. when should I start to learn C language?? I've just passed out 12th science with PCM

Sir the question bank consists mathematics ??

Sir how u decided the direction of shear stress as it is not told in question that they r acting towards P and R

Test series is good for civil ??

Sir can you please check the answer of this I have a similar doubt in the GATE civil 2018 morning paper in the 8th question about torsion in circular cross section ,cantilever beam with torque applied at end..we have to calculate tau(xy) at the midpoint of the beam ...the answer is given zero.. But if we visualize tau(xy) we have to take plane perpendicular to x direction and shear stress acting in y direction..right? If we look at the topmost point of the circular cross section there is such shear stress ...right? I asked another professor and he agreed with me.Am I wrong? Sir Please look into this.

I need more like buttons

I am getting 5.015

Sir, what about stresses in plane PR. For calculating them do we consider plane as PR or the plane perpendicular to PR?

I gave this gate. Got 77.63 Mark's but couldn't get this question😅

Sir i have purchased the question bank last year in October. Its validity is till October 2.Is their any provision to extend till Gate 2020 or else i have to purchase it again. I am even planning to get the test series of Exergic in this month end. Is their any discount or else provision to extend validity at discounted rate?

Sir thoda hindi me explain karo maine apka video bahot bar dekha par samaj me nahi ara hai

👏👏👏👏👏

Question bank having previous year qs ?

SK Mondal taught it wrong