OK, time to look into some applications that make use of contact. We have three examples
lines up for this lesson. The first one is a bowling
ball ramming into the bowling pins to demonstrate an impact type contact.
In the second example we’ll revisit the discussion on belt drive
and see how friction is crucial for that design.
And in the third example we’ll see how contact pressure can be used for assessing
the design of an O-ring. In this example, we’ll also investigate how a change in the design
would change the way the contact formulation used for solving it would make a difference.
Let’s start with bowling. Bowling is a recreational activity where a heavy ball
is hurled towards a set of pins that are arranged
in a format with the objective of the game being to dislodge all the pins.
This game is very old, its origins dates all the way back to 5300 BC
which is the Egyptian times. The key for winning this game is
to transfer momentum from the ball to all the pins so that they’re dislodged.
Since the ball and the pins are all individual object, there must be a
contact formulation defined between them to enable this transfer of momentum.
Let’s simulate this game, we arrange the pins in the prescribed format and
then impart an initial velocity to the ball and let it on to the pins.
When we define a contact interaction between them, the momentum is properly transferred
to the pins and we see them getting dislodged, which is as expected.
Not let’s simulate the same scenario and ignore the contact formulation this time.
We see that ball simply drifts through them and does not transfer momentum to the pins.
Of course, this is not physical, or we’ll always end up with the ball in the gutter.
But this shows how contact is not automatically accounted for
in the governing equations which is why we need to define the interaction.
This is a simple example that demonstrates the importance of contact.
Now, in the next example which is the belt drive we’ll learn the importance of friction.
Belt drives are seen in various mechanical systems, they’re popularly used as timing
belts in automobiles. These are used for transmitting power from one shaft to another.
In general, there’s a flexible belt that runs over two or more shafts.
When one of the shafts rotates, due to the friction between the shaft and the belt, the
belt starts moving too and as a result it starts pulling the other shaft which sets it in motion.
A successful design would have enough friction between the shaft and the belt,
so the belt remains in a sticking state with the shaft while it is running over it.
But how crucial is friction in this application? Let’s simulate it and find out.
In this example we setup a simple belt drive system with two shafts and a belt
that’s running over them. We first create stress in the belt by stretching it
so there’s enough contact pressure developed between the shafts and the belt.
Next, we start rotating the smaller shaft which is called as the driver. As a result,
if there’s friction present between the belt and the shaft, the power is transmitted and the
bigger shaft, called as the driven also start rotating. This is how a belt drive works.
Now, suppose we lubricate the shaft surfaces more that we should and end up with nearly frictionless
interaction between the shaft and the belt. Now if we start rotating the driver,
the belt simply slips over the shaft and it never transmits the power to the driven.
We can also observe this by plotting the angle of rotation
of both the drive and the driven shaft in both the cases as a function of time.
In case of friction, we see that over time,
the driven shaft starts rotating but in case of frictionless case, the power is
never transmitted to the driven and therefore its angle of rotation does not change with time.
This example shows why frictional force is important
and how it can used in designing mechanical systems.
Now let’s look at our final example which is an O-ring assembly.
O-rings are commonly used to seal the gaps between engaging parts in several assemblies such as
water filters, espresso machines. The performance of an O-ring is assessed
based on the contact pressure that is created at the interface during operation.
In this example we will explore the design of an O-ring system
which is designed to seal a pressurized fluid.
Under the action of the fluid the O-ring is expected to deform but support the applied
pressure. After deforming and getting to the final state, it should fill up and seal the gaps.
We’ll explore two different designs of the mating assembly
and see how the O-ring performs in both the designs.
In both the designs, we have the O-ring installed between two steel components.
After installation, we apply a fluid pressure
on the O-ring to see if it can withstand the applied pressure.
In this case a frictional contact is used between the O-ring and the steel components.
The quantity of interest is the contact pressure between the O-ring and the steel components.
In Design 1, the lower component has a flatter base and a fluid pressure of
50 MPa is applied on the O-ring. In Design 2, the lower component has
an inclined base and a fluid pressure of 70 MPa is applied on the O-ring.
In both the designs, we will also see how the choice of contact formulation
may or may not affect the accuracy of results. In design 1, let’s compare the contact pressure
calculated using both the penalty and the Lagrange contact formulation.
Here are the contour plots for the contact pressure
distributed around the contact regions. From the legend we can see that there’s isn’t
any major difference in calculations between the two formulations.
In such a case we may go with a penalty formulation as it’s more robust due to
relatively inexpensive calculations. Now, let’s shift our focus on the second design
which has this small inclination cut into its design.
Once again, we’ll look at contact pressure as calculated by both the formulations.
We notice that there is a big difference in the values calculated by both the formulations.
This is because when the O-ring deforms it starts flowing about this point which is pretty sharp.
Since a penalty formulation is more forgiving in terms of penetration, it allows the O0ring
material to penetrate into this section which results in lower contact pressure calculation.
The same design when solved using Lagrange formulation
results in nearly zero penetration and therefore trustworthy contact pressure calculation.
In design 2, the change in design of the base
introduces a sharp corner over which the material flows.
As a result, the penalty formulation would require large contact stiffness to avoid penetration.
A Lagrange formulation on the other had imposes a zero-penetration constraint
so it captures the flow of material over sharper corner more accurately.
This simple example shows when one formulation may be beneficial over another
depending on the design or application. Of course, there are various other scenarios
and examples that demonstrate more uses of contact but the three examples we saw today
covers the concepts that we discussed as part of the section on contacts.