Posts Tagged ‘vibration force’

Vibration Analysis in Mechanical Engineering, Part II

Sunday, March 14th, 2010

     Last week we began a new article on vibration analysis.  This week we’ll continue by looking at how to solve a vibration analysis problem.  

     Analysis of vibration in machine designs typically involves advanced knowledge of kinetics and higher level mathematics like calculus.  For this discussion, let’s consider a relatively simple balancing problem for a rotating system of masses.

     Suppose you have a ball with 10kg of mass on the end of a stick.  If you’ll recall from our discussion about kinetics, mass is the weight of an object divided by the acceleration of gravity.  For this problem, let’s say that only the ball has mass and not the stick.  Although this may seem nonsensical, it is necessary in order not to complicate our analysis well beyond the basic discussion we are having here.  Getting back to our ball and stick combo, the other end of the stick is attached to a rotating shaft, and the center of the ball is 0.5 meters (m) away from the center of the shaft.  This rotating system is shown in Figure 1.

Figure 1 – A Rotating System With One 10kg Ball


     When the shaft rotates at 60 revolutions per minute (RPM), the ball moves in a circle around the shaft via the stick as shown in Figure 2.  The centrifugal force created by the circular motion of the ball always points at a 90 degree angle to the ball’s path of movement.  The net result is an effect similar to that felt when you drive quickly around a sharp right-hand curve and you feel as though you are being pushed to one side.

Figure 2 – Side View of the Rotating System in Figure 1


     Getting back to Figure 2, since the centrifugal force is at a 90 degree angle to its movement, it keeps changing direction.  When the ball is at the top of its movement, the force points straight up as shown in (a).  When the ball is horizontal right, the force points right as shown in (b).  When the ball is at the bottom of its movement, the force points down as shown in (c). When the ball is horizontal left, the force points left as shown in (d).  This change in the direction of the centrifugal force pulls the rotating system up, to the right, down, and to the left as the shaft rotates, causing vibration in the whole system. The mechanics of this vibration are very similar to the unbalanced load in the washing machine during spin cycle that we talked about last week.

     To answer this question, let’s apply the formula for centrifugal force on a ball as it rotates in our system above:

F = (Mass) x ((RPM) x (1 min. / 60 sec.))2 x (4 π2) x

                                                        (Distance From Center of Rotation)

Plugging in values, the centrifugal force for the 10kg ball is calculated to be:

F10kg = (10 kg) x ((60 RPM) x (1 min. / 60 sec.))2 x (4 π2) x (0.5 m)

= 197.39 kg m/sec2 = 197.39 Newtons

In case you’re wondering, scientists got tired of talking about the metric units of force as “kg m/sec2,” so they decided to rename these units in honor of the great Sir Isaac Newton.

     In order for the vibration in our example to go away, the centrifugal force for the 10kg ball must equal the centrifugal force of the 6kg ball.  Knowing this, we can work backwards to calculate the distance from the center of rotation for the 6kg ball.  This distance will be key to balancing the system out because it acts to compensate for the unequal masses of the balls.

   F6kg = 197.39 kg m/sec2 = (6 kg) x ((60 RPM) x (1 min. / 60 sec.))2 x

                                             (4 π2) x (Distance From Center of Rotation6kg)

And the distance from the center of rotation becomes:

     Distance From Center of Rotation6kg =

                 197.39 kg m/sec2 /[(6 kg) x ((60 RPM) x (1 min. / 60 sec.))2 x (4 π2)]

                                                                        = 0.83 meters

     This tells us that we’d have to cut our stick so that the center of the 6kg ball is 0.83 meters from the center of the shaft in order to compensate for the vibration caused by the rotating 10kg ball.  See Figure 3.

Figure 3 – A Rotating System With One 10kg Ball and One 6kg Ball

     This concludes our basic look at vibration analysis.  Next week we’ll get into the last installment of our mechanical engineering series to see how everything we’ve learned throughout this series ties together in machine design.


Vibration Analysis in Mechanical Engineering, Part I

Sunday, March 7th, 2010

     Last week we wrapped up our discussion on heat transfer.  This week we’ll turn to a discussion on vibration analysis. 

     Vibration occurs when there is physical movement of a machine part when compared to a point of reference.  This physical movement can manifest in a straight line, a circle, or any way imaginable in three dimensions.  The point of reference can be a guideway for a sliding part or a shaft for a rotating part.

     As discussed in kinetics, machine parts have mass, and when mass moves it contains kinetic energy, so it makes sense that when a part moves within a machine, the energy created can result in forces that act upon the machine as a whole.  The net result is machine vibration which occurs in “sympathy” with the movement of the original mechanical part under discussion.

     Let’s talk about examples of vibration caused by straight line and circular motion.  And what figure conjures up a better image of up-and-down straight line vibration then a jackhammer, as shown in Figure 1.  This tool has a chisel which is attached to an air piston that moves up and down in a straight line, and the chisel and piston each have mass.  It is the rapid up and down movement of the total mass of the device that results in concrete-breaking force.  Unfortunately, those forces also vibrate back up the jackhammers shaft into the hands, arms and body of the construction worker operating it.

Figure 1 – A Jackhammer


     An example of vibration caused by circular movement can be found in your washing machine.  Ever notice what happens when you throw one heavy object, say a throw rug, into it, and it begins the spin cycle?  That “THUMP-THUMP-THUMP” sound that just won’t quit is due to the rug, now wet and congealed into a single heavy lump to one side of the agitator.  It continues to spin about the center of the agitator, creating an unbalanced outward centrifugal force that keeps changing direction around its central pivot point, the agitator.  This force makes the washing machine want to rock from side to side and front to back. Now imagine this situation taking place inside your washer day after day, hour after hour, every time you put a load to wash.  How long do you think your washer would last under this usage?  As this example illustrates, vibrations must be considered in machine design because if they are severe enough, they can cause machine parts to prematurely wear and fail.

     Speaking of wear and failure, you might have discovered how the tires on your car wore out prematurely and a mechanic said this happened because your shock absorbers or struts are bad.  Shock absorbers and struts help to safely control, or “dampen,” vibrations in the suspension system that result when your car’s wheels roll down the highway.  Without proper damping, the vibration forces can cause your tire to literally bounce down the road and grind on the pavement with each landing.  Oh, yes, if the vibration is strong enough, you can even lose control of your car and end up in a crash.

     Unfortunately, strong vibrations do not only affect machinery, but the people that come into contact with them, often causing physical discomfort and injury.  Ever heard of “white finger syndrome,” otherwise known as Raynaud’s syndrome?  This painful condition, which results when vibrations impair blood flow to the fingers, causes them to tingle, feel numb, and then turn white.  The longer a person uses a vibrating tool, and the faster the tool vibrates, the greater the risk.

     Well that’s our initial plunge into the world of vibrations.  Now that we know the basics of what vibrations are and what they can do, next time we can explore how to resolve a straightforward vibration problem which often presents itself in rotating mechanisms.