Torsion in Shafts: Conceptual Doubt Explained

AI Thread Summary
The discussion centers on the conceptual understanding of torsion in shafts, particularly the treatment of one end as fixed while the other receives torque. It highlights that the fixed end is a simplification for analysis, as the actual rotation is relative between the two ends, not absolute. The torque in the shaft depends on the relative angular difference, not the fixed position of one end. When a motor drives the shaft, it induces torque necessary for power transmission, despite the shaft being free to rotate. Ultimately, the key takeaway is that treating one end as fixed is a convenient analytical approach rather than a strict physical requirement.
ramzerimar
Messages
178
Reaction score
23
I have some conceptual doubts about shafts subjected to torsion. When studying Strenght of Materials, to find stress and strain in power transmission shafts, we consider that one of the ends of the shaft is fixed, with all degrees of freedom restricted, and the other one is receiving torque. I'm having trouble to understand this fixed end. If it's a rotating shaft, how can it be fixed? But if it's not fixed, we can't apply equilibrium equations to solve the problem.
 
Engineering news on Phys.org
Suppose that the shaft is strain free (no twist) to begin. Make a mark on each end at the same angular position. Now, twist the shaft so that the angle between the marks is phi. There is now shear stress = T*c/J in the outer fibre, OK?

Now, suppose you roll both ends through an angle theta, maintaining the angular difference phi. The torque in the shaft is the same as it was before because it depends on the relative angle between the ends, not the absolute angle.

Where you choose the reference mark, at theta = 0, theta = pi/27, or something else, it is only the relative rotation, the twist, that matters as far as the induced torque. Treating one end as fixed may be convenient, but it is certainly not necessary.
 
Dr.D said:
Suppose that the shaft is strain free (no twist) to begin. Make a mark on each end at the same angular position. Now, twist the shaft so that the angle between the marks is phi. There is now shear stress = T*c/J in the outer fibre, OK?

Now, suppose you roll both ends through an angle theta, maintaining the angular difference phi. The torque in the shaft is the same as it was before because it depends on the relative angle between the ends, not the absolute angle.

Where you choose the reference mark, at theta = 0, theta = pi/27, or something else, it is only the relative rotation, the twist, that matters as far as the induced torque. Treating one end as fixed may be convenient, but it is certainly not necessary.

Suppose I have a shaft supported by a pair of bearings, one at each end, and there's a motor connected to the shaft. When we start the motor, the shaft will rotate with some angular velocity. Why would the shaft twist then, if it is free to rotate because of the bearings?
 
If it is truly free to rotate, there will be no torque in the shaft and no shear stress. Why would you do this?

The usual reason for turning a shaft with a motor is to enable the shaft to drive some machine, to do work, and that takes a torque in the shaft to transmit power to the driven machine.
 
Dr.D said:
Where you choose the reference mark, at theta = 0, theta = pi/27, or something else, it is only the relative rotation, the twist, that matters as far as the induced torque. Treating one end as fixed may be convenient, but it is certainly not necessary.

Okay. So when I treat one end as fixed is just a way of simplifying things. Actually, it's not really fixed, only in relation to the other end.
 
That's about the size of it.
 
How did you find PF?: Via Google search Hi, I have a vessel I 3D printed to investigate single bubble rise. The vessel has a 4 mm gap separated by acrylic panels. This is essentially my viewing chamber where I can record the bubble motion. The vessel is open to atmosphere. The bubble generation mechanism is composed of a syringe pump and glass capillary tube (Internal Diameter of 0.45 mm). I connect a 1/4” air line hose from the syringe to the capillary The bubble is formed at the tip...
Thread 'What type of toilet do I have?'
I was enrolled in an online plumbing course at Stratford University. My plumbing textbook lists four types of residential toilets: 1# upflush toilets 2# pressure assisted toilets 3# gravity-fed, rim jet toilets and 4# gravity-fed, siphon-jet toilets. I know my toilet is not an upflush toilet because my toilet is not below the sewage line, and my toilet does not have a grinder and a pump next to it to propel waste upwards. I am about 99% sure that my toilet is not a pressure assisted...
After over 25 years of engineering, designing and analyzing bolted joints, I just learned this little fact. According to ASME B1.2, Gages and Gaging for Unified Inch Screw Threads: "The no-go gage should not pass over more than three complete turns when inserted into the internal thread of the product. " 3 turns seems like way to much. I have some really critical nuts that are of standard geometry (5/8"-11 UNC 3B) and have about 4.5 threads when you account for the chamfers on either...
Back
Top