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!
