Wednesday, 25 April 2018

solving standard problems in mechanical engineering(Task that I did recently at university)




The question:
You are supplied with output data from the CAD model of a prototype car brake disc. The data states the mass moment of inertia of the brake disc about the rotational axis. You are required to:

1. Determine the mass moment of inertia experimentally by oscillating the brake disc as a compound pendulum.
See the theory on the following page.
You should consider the design of your experiment including:
- How many swings should be counted (see Measurement and uncertainty Lab, simple pendulum part)
- Likely sources of error and their magnitude, think about the things you will need to measure and time. – See the reaction time exercise on Moodle.
2. Use theory from lectures and any other relevant sources to estimate the mass moment of inertia via hand calculation.
See question 2 from the Rotational Dynamics Tutorial (29th January- 4th February)
You will need to make assumptions, these should be clearly stated in your report.
3.    Develop/Design another experimental method that could be used to determine the mass moment of inertia for the object. This should include:
a.    The underpinning background theory for the method
b.    The method for gathering the required data
c.    The process for analysing the gathered data
Suggested methods you may wish to investigate are given below, this is not an exhaustive list:
i.      Falling mass (dynamics) (e.g. Question 5 of the Rotational Dynamics tutorial)
ii.     Torsional pendulum
ii.     Tri-filar suspension
When researching ideas we suggest you search for methods/labs for determining the mass moment of inertia of flywheels.
You do not need to carry out this experiment, it should be part of your report as an appendix.
4.    Present your findings, including a discussion of your results and the stated value from CAD.

 My answer to the question above...

❖ Abstract In this experiment we attempted to swing a brake disc twenty times and measure the time that it takes it to do that amount of swings using a timer.

The brake disc was set up on it inside ring to let the disc set freely then we moved it on it axes to either left/right side and let it swing, meanwhile one of the student in my group was holding a timer to get the most accurate results for the 20 swings.

We repeated the same process for four times to reduce the uncertainty for our own results, then by calculating the results and put it in the right equations we were able to calculate the Mass Moment of Inertia.

❖ Theory In order to find the mass moment of inertia I had to use many equations that been used for previous experiment such as the Pendulum experiment. By taking the average value from the provided data I was able to find the frequency f which I got it by dividing 20 by the average value of the time F = 𝑆𝑀𝑖𝑛𝑔𝑠 π‘‡π‘–π‘šπ‘’ π‘‘π‘Žπ‘˜π‘’π‘› , from that I got the time period which is equal to one over frequency. Then I calculated the angular velocity Ο‰, Ο‰ = 2 πœ‹F. Afterward I Had to find the value
of radius of gyration about the centre of gravity 𝐾𝐺, Ο‰ = √
𝑔 β„Ž 𝐾𝐺2+β„Ž2
 . I rearranged the
equation to calculate 𝐾𝐺, 𝐾𝐺2= (𝑔 β„Ž πœ”2
) - β„Ž2. In the end to get to calculate mass moment of inertia 𝐼𝐺 I used this equation 𝐼𝐺 = m 𝐾𝐺2. But I think all could be written in one equation which could make calculations easier by replacing 𝐾𝐺2 and Ο‰ to get: 𝐼𝐺 = π‘š 𝑔 β„Ž (2πœ‹π‘“)2 βˆ’ β„Ž2

❖ Experimental Methods The method was to put the brake disc on a small rod like shown in picture and move it to the side by small distance on it axes and let it go with initial speed of 0, and measure how long does it take for the disc to make 20 full oscillations, this experiment were repeated 4 times for more accurate measurement, using the same conditions and move it exact the same distance we were able to get four different measurement but close enough to be accurate of this experiment.


❖ Results Task 1: Ave value = 17.19+17.29+17.18+17.21 4 = 17.217β‰ˆ 17.22 s 𝑓 = 20 17.22 =  1.161 𝐻𝑍

𝑇𝑝 = 1 𝑓
 = 1/1.161 = 0.861 s πœ” = 2πœ‹π‘“ = 7.29 rad/s
Ο‰ = βˆšπ‘” β„Ž 𝐾𝐺2+β„Ž2

πœ”2 = 𝑔 β„Ž 𝐾𝐺2 + β„Ž2
 πΎπΊ2= (𝑔 β„Ž πœ”2 ) - β„Ž2 = 9.81Γ—68.5 Γ—10βˆ’3 7.292βˆ’ (68.5 Γ— 10βˆ’3)2 = 7.949Γ— 10βˆ’3
𝐾𝐺 = √ 7.949 Γ— 10βˆ’3

𝐾𝐺 = 89.161x10βˆ’3

𝐼𝐺 = m 𝐾𝐺2= 3.55 Γ— 7.949 Γ— 10βˆ’3 = 0.0282 Kg.π‘š2
 Uncertainty: Standard deviation = βˆšβˆ‘(π‘Ÿπ‘’π‘Žπ‘‘π‘–π‘›π‘”π‘–βˆ’π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’)2 𝑛 𝑖=1 π‘›βˆ’1 = √(17.19βˆ’17.22)2 4βˆ’1
 = 0.01732
Standard Uncertainty = π‘†π‘‘π‘Žπ‘›π‘‘π‘Žπ‘Ÿπ‘‘ π‘‘π‘’π‘£π‘–π‘Žπ‘‘π‘–π‘œπ‘› βˆšπ‘› = 0.01732 √4 = 8.66π‘₯10βˆ’3.

Task 2:  I broke the brake disc to three shapes the out-disc part(shape1), the ring which is the middle part(Shape2) and the small inside-disc(shape3). I didn’t include with my calculation the chamfers and the small holes, because their mass is too small to be considered in our experimental sample. The density 𝜌 = 7200 kg/ π‘š3 from the CAD drawing.

Shape1: π‘Ÿπ‘œ = 119.5π‘šπ‘š, π‘Ÿπ‘– = 68.5π‘šπ‘š and the thickness = 12mm 𝑉1 = πœ‹(π‘Ÿπ‘œ2 βˆ’ π‘Ÿπ‘–2)𝑑 = πœ‹(0.11952 βˆ’ 0.06852) Γ— 0.012 = 3.614Γ— 10βˆ’4 π‘š3 π‘š1 = 𝜌 Γ— 𝑣1 = 7200 Γ— 3.614 Γ— 10βˆ’4 = 2.6 Kg 𝐼1 = π‘š1 2 (π‘Ÿπ‘œ2 + π‘Ÿπ‘–2) = 0.0246 𝐾𝑔.π‘š2

Shape2: : π‘Ÿπ‘œ = 72.5π‘šπ‘š, π‘Ÿπ‘– = 68.5π‘šπ‘š and the thickness = 4mm 𝑉2 = πœ‹(π‘Ÿπ‘œ2 βˆ’ π‘Ÿπ‘–2)𝑑 = πœ‹(0.07252 βˆ’ 0.06852) Γ— 0.004 = 7.087Γ— 10βˆ’6 π‘š3 π‘š2 = 𝜌 Γ— 𝑣2 = 7200 Γ— 7.087 Γ— 10βˆ’6 = 51.02Γ— 10βˆ’3 Kg 𝐼2 = π‘š2 2 (π‘Ÿπ‘œ2 + π‘Ÿπ‘–2) = 0.2537 Γ— 10βˆ’3 𝐾𝑔.π‘š2

Shape3:  π‘Ÿπ‘œ = 68.5π‘šπ‘š, π‘Ÿπ‘– = 31.75π‘šπ‘š and the thickness = 8mm 𝑉3 = πœ‹(π‘Ÿπ‘œ2 βˆ’ π‘Ÿπ‘–2)𝑑 = πœ‹(0.06852 βˆ’ 0.031752) Γ— 0.008 = 9.259Γ— 10βˆ’5 π‘š3
π‘š3 = 𝜌 Γ— 𝑣3 = 7200 Γ— 9.259 Γ— 10βˆ’5=  0.6666 Kg  𝐼3 = π‘š3 2 (π‘Ÿπ‘œ2 + π‘Ÿπ‘–2) = 1.8999 Γ— 10βˆ’3 𝐾𝑔.π‘š2

Total: 𝐼 = 𝐼1 + 𝐼2 + 𝐼3 = 0.02675 Kg.π‘š2 β‰ˆ 0.027 Kg.π‘š2


❖ Discussion By looking at Task1 the answer was pretty accurate, but to get to the definitive answer I went through a lot of calculations and rearranged some equations too, I had to consider the difference in units and change some values from millimetres to metres.

In Task2 the answer wasn’t too accurate although calculation process was very slow and distractive so there was very high chance of making mistakes, some very small values like chamfer and the small holes was hard to calculate so I had to ignore it small value comparing to the disc value.

❖ Conclusion By comparing the results and the process I came to conclusion that CAD is faster and more accurate to provide with a good valuation of the mass properties, therefore its essential for an engineer to considerable CAD skills in order of saving much time and much effort.


❖ Appendix An alternative method I though of is by hanging the brake disc using a sting attached to it from the very top of the disc then by rotating the disc on it Y axes clockwise then release it when the string have enough tension to rotate the disc anticlockwise, at the release time a timer should start and then you count how many turns did the disc make until it start rotating clockwise again, then you count the total turns until the rest position of the disc.

By finding the string tension and the weight that affect the force diagram you will be able to find the force and from there you can find the acceleration and from the acceleration we can find the radius of gyration about the center of gravity and in the end we find the mass moment of inertia.

The End
My result came out to be 48% not quite sure where my mistakes was (since we don't get a feed back with where I made mistakes) but hopefully you find my assignment helpful and intersting enough for you.
Thank you

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