Since we are summing torques to zero (for equilibrium), then it actually doesn't really matter which way you assume as positive as long as you're consistent.
A steel shaft 50mm diameter and 0.5 m long is subjected to a twisting couple of 10³ Nm and the total angle of twist is 0.6°. Find the maximum shearing stress developed in the shaft and the modulus of rigidity. Plz solve this and give answer🙏
btw you are mistaken about the Torsion graph, right hand rule must be applied for internal torque not for external one so the graph should be upside down.
This is really great. My professor this semester is really disorganized and explains things rather poorly. This video really, really helped me. 😀
Still the best method I've found. Thanks!
So great! Thanks a ton
tnnx for giving the salutation of this numerical..i am so happy because tomorrow was my presentation in it ..
nice video very helpful, thanks
thanks a lot..i will be grateful to you forever
i was having trouble with my units thank you
Thank ya kindly for the vid!
Good job
Thank you
Wow nice
Great video and lovely accent!
can somebody tell me what to do if the shaft is fixed at both ends?, any relations with downwards and up wards force and all that?
Is this a torsional deformation?
thank you
What if we assume the shaft isnt connected to the wall, how will you know which is the positive direction
Since we are summing torques to zero (for equilibrium), then it actually doesn't really matter which way you assume as positive as long as you're consistent.
@@theryderproject5053 Thanks for your help!
A steel shaft 50mm diameter and 0.5 m long is subjected to a twisting couple of 10³ Nm and the total angle of twist is 0.6°. Find the maximum shearing stress developed in the shaft and the modulus of rigidity. Plz solve this and give answer🙏
So cute
btw you are mistaken about the Torsion graph, right hand rule must be applied for internal torque not for external one so the graph should be upside down.
Hi. It doesn't really matter which way you pick - just as long as you are consistent.