Torsion is the twisting of an object, caused

by a moment acting about the object’s longitudinal axis. It is a type of deformation. A moment which tends to cause twisting is

called torque. A common example of an object subjected to

torsion is the transmission shaft, which is used to transmit power by rotation. This could be the drive shaft and axles used

to transmit power from the engine of a car to the wheels, for example, or the shafts

used to transmit power from the blades of a wind turbine to its generator. Let’s explore what happens when we apply

torque to a circular bar. We can see that the applied torque causes

the bar to deform by twisting. An interesting thing we can observe is that

individual cross-sections of bar do not get distorted by the twisting. We can imagine that the bar is made up of

multiple individual disks, which rotate relative to each other when the torque is applied,

but do not deform. This is only true because the cross-section

of the bar is axisymmetric. A bar with a rectangular cross-section is

not axisymmetric, and so torsion results in warping of the bar cross-sections. This warping is complex, so in this video

we will keep things simple and only consider torsion as it relates to circular bars. Let’s fix our bar at one end, and track how

a line between point A and point B deforms as we apply a torque to the other end. The applied torque causes the free end of

the bar to rotate by an angle Phi. This is called the angle of twist. It varies linearly from zero at the fixed

end of the bar to Phi at the free end of the bar. We can calculate the angle of twist using

this equation. It is a function of four parameters – the

length of the bar L, the applied torque T, the shear modulus G, which is a material property,

and J, which is the polar moment of inertia. So, what is the polar moment of inertia? It defines the resistance of a cross-section

to torsional deformation, due only to the shape of the cross-section. The polar moment of inertia for a hollow bar

with an outer radius r.o and an inner radius r.i can be calculated using this equation. Setting the inner radius to zero gives us

the equation for a solid bar. One neat thing about the equation for the

angle of twist is that it gives us a way to determine a material’s shear modulus G experimentally. If we apply a known torque to a bar of known

length and cross-section, and measure the resulting angle of twist, we can use that

information to calculate the material’s shear modulus G. Torsion generates stresses and strains within

the bar, which we need to be able to calculate so that we can make sure our bar won’t fail. To figure out how to calculate these stresses

and strains, we can start by observing how a small rectangular element on the surface

of our bar deforms. The element is initially rectangular, but

when the torque is applied it gets distorted. Let’s take a closer look. Because the bar is axisymmetric, we know that

individual cross-sections will rotate but won’t get distorted. So the sides C-F and D-E of the element will

only move vertically along the lines shown here. After the torque is applied, the angles of

the element are no longer 90 degree angles. This gives rise to a shear strain, which corresponds

to the angle you can see here. We can calculate the shear strain by considering

only the geometry of the bar and the deformation. It corresponds to this angle between A-B and A-B’. We can use trigonometry to derive an equation

for shear strain. For small angles, Gamma will be approximately

equal to the tangent of Gamma, which makes it equal to the length B-B’ divided by

the length A-B. A-B is the length L of the bar. We can calculate the length B-B’ by realising

that it is the arc length of a circle with a radius R, covering an angle equal to the

angle of twist Theta. So the shear strain is equal to the radius

of the bar multiplied by the angle of twist, divided by the length of the bar. This is actually only an equation for the

shear strain on the surface of the bar. But what about inside it? It turns out that the shear strains increase

linearly with the distance from the centre of the cross-section. So if we define Rho as the radial distance

from the centre of the cross-section, we can replace r in this equation with Rho, to give

us an equation we can use to calculate shear strain due to torsion at any point within

the bar. That’s shear strains covered, but what about

shear stresses? Like the shear strains, shear stresses increase

linearly with the distance from the centre of the cross-section, with the maximum shear

stress occurring on the outer surface, as you can see here. This is true for a solid bar, but also for

a hollow bar. This is useful to know because it means that

hollow bars are way more efficient at carrying torsional loads, since the central part of

a solid bar is only resisting a small part of the total load. Let’s consider a small element within our

cross section that has an area equal to dA and is located at a distance Rho from the

centre of the cross-section. The internal force acting on this element

is equal to its area dA multiplied by the shear stress Tau. We can use this information to work out an

equation for calculating the shear stresses. The moments caused by the internal forces

acting on all of the elements within the cross-section must sum up to be equal to the torque T, otherwise

equilibrium is not maintained. We can represent that mathematically by this

integral. We know that the quantity Tau divided by Rho

is a constant, because the shear stress varies linearly with the distance from the centre

of the cross-section. So we can re-arrange the terms and move Tau

over Rho out of the integral. It turns out that the integral we now have

on the right is actually the definition of the polar moment of inertia, so we can replace

it with the letter J. And we can re-arrange this to get an equation

for shear stress. The shear stress is a function of the torque

T, the distance Rho from the centre of the cross-section, and the polar moment of inertia

J. It’s quite a simple equation! So we now have equations that allow us to

calculate the shear strains and shear stresses. We also have the equation for angle of twist

that we talked about earlier. These three equations tell us everything we

need to know about a circular bar which is under torsion. So far we have only talked about a uniform

bar fixed at one end with a single applied torque. But shafts are often loaded by multiple torques. This shaft for example, which is supported

by bearings at both ends, is driven by a gear at point B, and in turn drives two gears at

points A and C. It is loaded by three torques. Before we can use the equations for shear

stress, shear strain and angle of twist that we just developed, we need to figure out the

internal torque at each location along the shaft. The process for doing this is similar to calculating

the shear force along a beam, which I covered in a separate video. First we draw a free body diagram. Then we make imaginary cuts and use the concept

of equilibrium to determine the internal torque at different locations along the shaft. This will give us an internal torque diagram

that looks something like this. The maximum shear stress will occur in the

section of the shaft with the largest internal torque, and can easily be calculated using

the equation we derived earlier. I want to end the video by talking about failure

due to pure torsion. If we have two bars, one made of a ductile

material and one made of a brittle material, and we apply the same torque to both bars,

we will observe that they fail differently. The ductile bar fails at an angle perpendicular

to its axis, but the brittle bar fails at a 45 degree angle to its axis. We can explain this by remembering that ductile

materials tend to fail in shear, and so fracture along the plane of maximum shear stress. But brittle materials are weaker in tension

than in shear, and so tend to fracture along the plane of maximum tensile stress. Mohr’s circle for pure torsion looks like this. We can see that when our stress element is

oriented this way, the shear stresses are at their maximum values,

and we have no normal stresses. There is a 90 degree angle on Mohr’s circle

between maximum shear stress and maximum normal stress, which means that normal stresses are

at a maximum when our stress element is rotated by an angle of 45 degrees. This explains why brittle and ductile materials

fail in different ways due to pure torsion. That’s it for now. Thanks for watching! And don’t forget to subscribe if you haven’t

already!

1st comment 🤩

❤️❤️

Very Informative guys best ever explain Engineering funda

Dude, you're so amazing! I just found out about your channel and with my good graphic memory I now understand more about stress engineering, than I did before and I studied it in uni. I knew how to calculate it, but never fully understood it. That's no more the case, thanks to you!

If you'll do impacts and thermal stress, that would be awesome as well.

Explain Gear box – Sliding mess gear box and others pls make video on that

It's my personal request

Best ever explain- fun learn- Engineer rock

Thank you for this wonderful work ❤️✅😍

Thanks

Please made animation other topics regularly.

Thanks a lot.

It is very useful.

Could you please tell us what app do you use for this animation?

a picture worth a thousand words.

A video worth a thousand pictures

Great work, as always. Although, I've always wondered. Is bending moment the same as torque? What exactly is the difference between the two?

Thank you , Simple and elegant explanation

Great video. Just a dumb little nitpick because it's the Youtube comments section. The symbol used there is theta, not phi.

Normally i dont like studying this, but well, if i want my degree i have to and man, you really explain things in such a way that even i, who hate this, has come to kinda like it jajaj i have a exam this firday and you really saved me, really apreciate what you do, i hope more people find this channel so it can grow <3 much love

Your channel is fantastic, please keep posting more videos!

I got a torsion test tomorrow. Hope this helps

My man!

Need more friend

Excellent video.

Soo happy that I've found your videos!! Keep doing the amazing job!

Wow!! great video, this reaffirmed everything I just read in the book with visuals. I will be watching the other videos on your channel and subscribing.

Thank you so much! You’re really helping me with my course and these videos are a great visual representation!

Hope this channel gains traction, we engineers need it lol

The way you used to display the concept is really amazing and its very easy to understand the concept..keep doing this bro… I expect some more topics in structural analysis and soil mechanics😊