Thank You. I would just like to query something which doesn't completely relate to your example: I have a hollow tube which is subjected to a 750N.m torque. Now I would like to determine the amount of torque resisted by a specific area on the tube. Here are the dimensions: Outer radius = 100mm, Internal radius = 25mm, and the area im interested in is 75mm from the centre to the Outside radius. After max shear stress was calculated,how do I then calculate the torque on that area?
very nice explanation, but it is way to basic i hope you make more advance tutorials since there is a wide range of problems in Mechanics of materials I hope you make tutorials about them. thanks
When it came to using the polar moment of inertia in the torsion equation did the power become 1.733 x 10^-6 when it was earlier worked out to be 1.733 x 10^6?
This is immensely helpful, thank you very much, I was wondering if you could help me with a problem I'm having though, I have a question which asks me to find the Maximum and Minimum shear stress on a shaft such as this one, and everywhere that I have looked only seems to refer to maximum stress, I'm finding virtually nothing on minimum stress. I can only assume that minimum stress would be taken on the radius of the internal diameter? is that correct?
Thank you very much for the very clear explanation,I loved it :-) I have one question, I have seen people using Torque =Force x Radius x Sin theta and sometimes just Force x Radius, can you explain when to consider angle theta and when not?
I think there is a mistake. =( at 5:40, you calculated torque (T) as being 29.670 Nm. Then at 7:35, you plugged torque in as 29670Nm (29,67 KN.m). That increased the shear force by 10^3. The answer would be 556.3 KPa, instead of MPa. The same goes for the angle, which happened to be 0.049 degrees. I made the calculations over and over. Perhaps, I've missed something. Can you clarify that, please?
best first 2 min illustration ever.hopefully this finds its way to schools
possibly the best strength of materials vid I have ever seen. Thanks heaps for such a real world example combining many different elements. Cheers :)
Wow, amazing, with an exam tomorrow it is so great to pass by a video like this.
Absolute life saver, just a note, we use the metric system in England too
@keropoklekor007 Since radius is half the diameter, J=pi/32* d^4 = pi/2 *r^4
I made some annotations on the video that I think clear things up. Towards the end, I mistakenly wrote J in m^4 when it should have been in mm^4
earned yourself a subscriber..this is good stuff
thanks for uploading this...i have xams tmr and i needed to study this topic :D
thanks for uploading such a best and helpfull video.
Thank You. I would just like to query something which doesn't completely relate to your example: I have a hollow tube which is subjected to a 750N.m torque. Now I would like to determine the amount of torque resisted by a specific area on the tube. Here are the dimensions: Outer radius = 100mm, Internal radius = 25mm, and the area im interested in is 75mm from the centre to the Outside radius. After max shear stress was calculated,how do I then calculate the torque on that area?
very nice explanation, but it is way to basic i hope you make more advance tutorials since there is a wide range of problems in Mechanics of materials I hope you make tutorials about them. thanks
thx! i looking forward to inderterminate prb about torsion in shaft.
Great visual! This really helps put things together. I can do the math fine for all this stuff, but I like to visualize what is really happening.
Awesome!! Thanks for sharing your knowledge!!
29670 N-m x 0.0325 m/1.733e-6 = 556,419,504 N/m^2
That is approximately 556.4 MPa
how about if the section is a cut hollow cylinder how does it affect the orientation of the shear center?
@shams2k No problem. I hope it helps. - RMF
Why use the radius r as 32.5 mm and not r = 32.5 mm - 10.5 mm = 22 mm? Because there will be no stress in the "hollow" part right?
Thank you. Can you please explain shear stress of a shaft with a transverse hole ( for a pin joint connection)?
oh and for the new J values, it should be mm^2 not m^2, I am pretty sure
When it came to using the polar moment of inertia in the torsion equation did the power become 1.733 x 10^-6 when it was earlier worked out to be 1.733 x 10^6?
Very helpful. Thanks!
we need more vedios in stress analysses please
dude I love you!
This is immensely helpful, thank you very much, I was wondering if you could help me with a problem I'm having though, I have a question which asks me to find the Maximum and Minimum shear stress on a shaft such as this one, and everywhere that I have looked only seems to refer to maximum stress, I'm finding virtually nothing on minimum stress. I can only assume that minimum stress would be taken on the radius of the internal diameter? is that correct?
use mohr's circle no?
great stuff
Thank you very much for the very clear explanation,I loved it :-) I have one question, I have seen people using Torque =Force x Radius x Sin theta and sometimes just Force x Radius, can you explain when to consider angle theta and when not?
When F is normal, then no need for sin theta. When it is inclined to an angle, the we require a perpendicular component, either sin or cos theta .
So at what T does the shaft buckle?
Hey changed units from mm^4 to m^4.
i learn to finding J is pie/2(c*4) not pie/32. whats the difference??
I think there is a mistake. =( at 5:40, you calculated torque (T) as being 29.670 Nm. Then at 7:35, you plugged torque in as 29670Nm (29,67 KN.m). That increased the shear force by 10^3. The answer would be 556.3 KPa, instead of MPa. The same goes for the angle, which happened to be 0.049 degrees. I made the calculations over and over. Perhaps, I've missed something. Can you clarify that, please?
How do u find a shaft diameter?
thank you man!!!
when you calculating J why dividing P by 32 and not 2 ?
He uses the diameter in stead of radius, which is why he divides by 2^4 = 16 times more:
1/32pie x d^4 = 1/2pie * (d/2)^4 = 1/2pie * r^4
Chrisjan Wust Thank you Chrisjan
and for the power i learnt that is equal to P=pie T n
Ive got 505.79 x 10^6 N.m^2 (505.79 MPa) :/
the method you are using to find the power and torsional constant is wrong. I pray for your teachings