What a great lacture on Som ...my all the doubt related topics have cleared.once again thanku sir 🙏🏻 .. keep it up Sir my request ... sir please please start fluid mechanics lacture also ..
What if in the first problem, Torque at junction is 5T clockwise instead of 2T. In that case as shown at 9:00 , shear stress for AB will exceed 80 MPa. Because in angle of twist formula resultant torque at AB is 4T (clockwise), so shear stress in AB will be 23.7 X 4 which is > 80 MPa. What to do in this case?
Professor please I don't get it. How it's possible two different rotations for the shear if you don't have other force to change that? I don't in last example how is it possible to add 400 rpm plus 1200 rpm it has given 800 rpm. In other wise results 1600 rpm. Sorry I don't understand just listen to. Maybe I get it if speak by words. Thank you very much.
At 18:52......just pause the video and take a look at the figure at bottom left. The torques acting at cross section B are 1200 clockwise and 400 counter clock wise, that means both of them are of an opposite nature. Hence we have subtracted them. The resultant torque i.e 1200-400 = 800 Nm shall have the same direction as that of 1200 Nm torque.....I would encourage you to revisit that small period of video once again and i am sure your doubts will vanish.....
Mr Manas, thank you so much for your videos, a month worth of lecture is summarized into 5 videos, It has helped me tremendously.
You are just a real world Thor for students 😌😌
What a great lacture on Som ...my all the doubt related topics have cleared.once again thanku sir 🙏🏻 .. keep it up
Sir my request ... sir please please start fluid mechanics lacture also ..
All the best
What if in the first problem, Torque at junction is 5T clockwise instead of 2T. In that case as shown at 9:00 , shear stress for AB will exceed 80 MPa. Because in angle of twist formula resultant torque at AB is 4T (clockwise), so shear stress in AB will be 23.7 X 4 which is > 80 MPa. What to do in this case?
Thank you so much sir, you provided so much clarity to me in this topic!!
Nicely explained sir thanks alot...
Sir please make video on machine drawing.It's a humble request plz sir... 🙏🙏
16:32 l1=1.1195l2
Sir, please solve the question
Draw the involute of a circular arc which subtends an angle @=90° at the center of the circle of dia=120mm
Thank you sir 😁
Thank you sir
Sir in the second problem if we find the length of shafts using the maximum stress given L1 is coming equal to L2
Can you please guide🙏🙏
Thank you sir ❤
Thankyou Sir
At 15.29 why angle of twist would be same
Good
Professor please I don't get it. How it's possible two different rotations for the shear if you don't have other force to change that? I don't in last example how is it possible to add 400 rpm plus 1200 rpm it has given 800 rpm. In other wise results 1600 rpm. Sorry I don't understand just listen to. Maybe I get it if speak by words. Thank you very much.
At 18:52......just pause the video and take a look at the figure at bottom left. The torques acting at cross section B are 1200 clockwise and 400 counter clock wise, that means both of them are of an opposite nature. Hence we have subtracted them. The resultant torque i.e 1200-400 = 800 Nm shall have the same direction as that of 1200 Nm torque.....I would encourage you to revisit that small period of video once again and i am sure your doubts will vanish.....
Hello Doctor, can I have the name of the source?
Problems are from the book Mechanics of Materials by RC Hibbler
@@ManasPatnaikofficial Thanks