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!

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