Thank you sir ..aapka lecture se kuch kuch n new sikhne ko milta hai..Numerical ka threat nhi hota ab ab numerical me mza aata hai .bcs of you and negi sir .
8:30pm ko clg ka online cls shuru hua toh beech me se chorr k jana para. Warna itna interesting class ko kon chorrna chahega. Because of you sir. I am seeing confidence in me towards concepts. Thankyou so much. ❤️ Special session jo 4pm ko hota h usme v jab jab me doubt puchta hu aap turant solve kr detey h. Agar aap solve naa krtey kuch conceptual doubt mere special class mey toh shayad aaj me doubt me he rehta. Thankyou so much sir. ❤️ Jitna v thanks bolu utna kam h. Oky see you in 4pm class tommorow ❤️ Aapkey jesa teacher se parhna student ka khuss naseebi h.
Sir, I have a doubt kindly reply to me Sir, you use scientific calculator. How do you find directly the final answer using the calculator without rearranging the terms.
In the 2nd question what is the logic behind assuming change in length as 0.8mm when the composite body has itself reached the bottom surface which then initiates the bottom normal force. After the lower reaction has been initiated, it becomes a compatibility equation where change in length will be zero. Also the top bar will expand by 0.8 mm and will only contribute to expansion resulting in the increase of its length from 1.2m to 1.2008m (the lower bar will just move due to the expansion because during reaction measurements, the lower bar didn't expand).... this seems more logical... please verify
Sir in last question agr ham log p2 ko opposite hi le to equation banega 2P1cos60°+P2=P..aur iss equation se to answer hi change ho jaa raha hai sir.. Isse P2 ka value 2P/3 aa raha hai.. And P1 ka P/3
Yes I think the same , sir's answer for P1 is -P , but by intuition only we can see that the particular bar with P1 will be in tension . So obviously his solution is wrong
Question no 2 me reaction ka jo unit hai usme mujhe confusion ho rhi hai Mujhe lag rha hai ke answer N hoga lekin sir aapne KN me liya hai mai confused hu please confusion dur kare meri
Thank you sir ..aapka lecture se kuch kuch n new sikhne ko milta hai..Numerical ka threat nhi hota ab ab numerical me mza aata hai .bcs of you and negi sir .
Last problem was lit❤
8:30pm ko clg ka online cls shuru hua toh beech me se chorr k jana para. Warna itna interesting class ko kon chorrna chahega. Because of you sir. I am seeing confidence in me towards concepts. Thankyou so much. ❤️ Special session jo 4pm ko hota h usme v jab jab me doubt puchta hu aap turant solve kr detey h. Agar aap solve naa krtey kuch conceptual doubt mere special class mey toh shayad aaj me doubt me he rehta. Thankyou so much sir. ❤️ Jitna v thanks bolu utna kam h. Oky see you in 4pm class tommorow ❤️ Aapkey jesa teacher se parhna student ka khuss naseebi h.
Thank you for the compliment 😊👍🏻
An amazing class with full content and concept..god bless you sir
Great content sir❤❤❤❤
due to your lectures i am improving in my conceptual clarity
Thank u so much sir for providing us this kind of amazing lectures👏🏻👏🏻🙏🙏
😊👍🏻
Sir if possible please try to solve ESE previous year conventional questions in class.
Sir your lectures are awesome. 🔥🔥
27:21 respected sir in this numerical why haven't we considered the load due to self-weight? Is it because they have not given specific wt?
amazing sir
Sir, I have a doubt kindly reply to me
Sir, you use scientific calculator. How do you find directly the final answer using the calculator without rearranging the terms.
In the 2nd question what is the logic behind assuming change in length as 0.8mm when the composite body has itself reached the bottom surface which then initiates the bottom normal force. After the lower reaction has been initiated, it becomes a compatibility equation where change in length will be zero. Also the top bar will expand by 0.8 mm and will only contribute to expansion resulting in the increase of its length from 1.2m to 1.2008m (the lower bar will just move due to the expansion because during reaction measurements, the lower bar didn't expand).... this seems more logical... please verify
Did you find the real answer of your question?
In last question, deflection of O, why we did not add force "P+P2".............. (P+P2)l/AE ??????
sir osm pdate ho aap
Strength of material ka bhagwan mil gaya mere ko❤️❤️❤️❤️❤️love you sir
Thanks sir
Sir ss rattan ka book kaisa hai for som. Because sir some of the example in this lecture are taken from that book .
Sir in last question agr ham log p2 ko opposite hi le to equation banega 2P1cos60°+P2=P..aur iss equation se to answer hi change ho jaa raha hai sir.. Isse P2 ka value 2P/3 aa raha hai.. And P1 ka P/3
Yes, exactly
Yes I think the same , sir's answer for P1 is -P , but by intuition only we can see that the particular bar with P1 will be in tension . So obviously his solution is wrong
Same doubt
Good morning sir 🌞🌞
Sir I'm unable to find special class on Unacademy can you put you tinyurl link here please.....
Best 👌👌👌
Is it help ful to ESE
Sir iske baad machine design start kariyega
Question no 2 me reaction ka jo unit hai usme mujhe confusion ho rhi hai
Mujhe lag rha hai ke answer N hoga lekin sir aapne KN me liya hai mai confused hu please confusion dur kare meri
❤️❤️❤️
Sir last question mein P2 ka value 2P/3 aa raha hai aur P1 ka value P/3.
9🥳🥳