Module 1 Activity 1 – Failure Distribution Curve

1. Introduction
   • Purpose

This report proposes describes the testing and analysis of paper clips to determine the fatigue failure of different materials such as metals or plastics.

• Problem

The study of the strength of materials is extremely important in our day-to-day lives. This is because the knowledge of this can save us from very hazardous consequences in an event some materials should break due to fatigue. Hence, this experiment will serve as a good reference for both plastic and metal paper clips.

1.3 Scope

This experiment will not cover all the metal, ceramic or polymer materials. It is limited to the study of metal and plastic paper clips.

2. Background

2.1 Theory

Scientist analyzes materials such as metals, ceramics or polymers and engineers to reveal their mechanical properties to determine what use the materials may have. An example of a mechanical property that is tested for is the amount of stress a material can handle before it breaks. The stress a material can take before it breaks measures the strength of the material. Also, as a material gets older, it can handle less stress, which can cause it to fail at much lower stresses.

2.2 Research

Fatigue is a very common mode of failure for materials and has been studied for centuries. Fatigue occurs every day in objects that you are familiar with. An example, airplane wings fatigue thousands of cycles on every flight and bridges fatigue every time a car drives over them. However, just because a material is undergoing fatigue does not mean that it will always break. Engineers run careful experiments so that they can be sure that things will not break due to fatigue while you are using them.

3. Tests and Evaluation

• Apparatus

As provided in the assignment sheet, the apparatus that was are:

1. Small metal paper clips
2. Small plastic coated paper clips
3. Metric ruler
4. Protractor

Figure 1¹. Paper Clip

• Procedure

Also provided in the assignment sheet, the procedure is as follows:

1. Using your hands, open up the inner loop of the paperclip so that it makes 180-degree angle with the outer loop.
2. Bend the paperclip back to its original position. This counts as one loading cycle.
3. Repeat steps 1 and 2 until the paperclip breaks. Record the number of loading cycles that elapse. During the actual experiment, record all observations (i.e., changes in surface finish, color, etc.).
4. After completing the experiment for 180-degree angle, make a hypothesis below:
5. Repeat steps 1 through 3 for 45, 90, and 270-degree angles.
6. Record the following for each paperclip in Table A.
7. Look at the fracture surface and write down any observations in the Table B.
8. Plot your results on the graphs shown below.
9. Repeat entire experiment for a paperclip of a different size or material (such as plastic).

The tables below show the numerical results of the experiment.

Table A

Table B

4.2 Interpretation

From the data provided above, the plastic paper clip had more strength than the metal paper clip because it took more rotations before it broke. A possible explanation for this could be due to the fact that metals are not supposed to be bent; their main function will be to carry more weight. As opposed to the plastic paper clip that is more flexible.

5. Conclusion

5.1 Assessment

There was a linear relationship between the angle of rotation and the cycles to failure in both materials of paper clips. As the angle of rotation decreased, the number of cycles increased for both materials. It was also seen that metal paper clips could not take too many cycles before failure due to its rigidity.

5.2 Recommendation

More materials should also be studied. This will bring about more diversity in the experimentation. Also, it would be wise to have more angles of rotation to provide more accurate results.

Module 1 Activity 2 – Identifying Failure Sources

Failure Distribution Curve

The initial procedure

1. Use your thumb and pointer finger of each hand and pull the two end loops of the clip apart as far as possible so the center bend straightens up.
2. Bend the clip back and forth from fully open through to touching the fingers.
3. While counting the number of cycles repeat the open-close cycle until the wire breaks. Record the cycle count.

Break 20 paper clips and chart the results on a scatter plot.

Second procedure for improving the reliability of the paper clips.

1. To Open the Paper Clip, one hand will be holding the clip by the longer, outer loop.
2. With the thumb and forefinger of the other hand, grasp the smaller, inner loop.
3. Pull the smaller, inner loop out and down 90 degrees so that a right angle is formed
4. Push the smaller inner loop up and in 90 degrees so that the smaller loop is returned to the original upright position in line with the larger, outer loop this completes one cycle.
5. Repeat until the paper clip breaks and Count and record the cycles-to-failure for each clip.
6. Repeat the above procedure using the other 19 paper clips and record the cycles to failure for each clip and chart the results on a scatter plot.

(C ) Third  procedure for improving the reliability of the paper clips.

To Open the Paper Clip, one hand will be holding  the clip by the longer, outer loop.

2. With the thumb and forefinger of the other hand, grasp the smaller, inner loop.
3. Pull the smaller, inner loop out and down 45 degrees so that a right angle is formed
4. Push the smaller inner loop up and in 90 degrees so that the smaller loop is returned to the original upright position in line with the larger, outer loop this completes one cycle.
5. Repeat until the paper clip breaks and Count and record the cycles-to-failure for each clip.
6. Repeat the above procedure using the other 19 paper clips and record the cycles to failure for each clip and chart the results on a scatter plot.

Module 1 Activity 3 – Cumulative Failure and Reliability Plots

DATA OBTAINED.

Cycles-to-failure from the first procedure. 4,4, 5,5, 5, 5.5,5.5,5,5.5, 6, 6,6.5,6,6.5,6.5,6,4,4.5,6.5,5. cycles.

Cycles-to-failure when the paper clips are placed at 90°: 16,17 17,16, 18, 21, 22, 23, 16,23,22,21,18, 16,17,19,23,21,22,18. cycles.

Cycles-to-failure when the paper clips are placed at 45°: 58, 59,61,63, 65,67,69, 72,73, 78,81,83,85,85, 86 ,63,65,59,72,65. cycles.

Observations and analysis.

It is observed that the when the paper clips is at 180 degrees the number of cycles to failure are minimum. Reducing of angle degrees is an acceleration factor for the failure of the paper clips.

Module 1 Activity 5 – Separate Failure Mode Distribution Plot

Module 1 Activity 4 – Failure Distribution and Weibull Plot

Module 1 Activity 6 – Pump Reliability Improvement Business Case

Objective:

The objective of this project was to design a system to cool tomatoes to preserve freshness prior to being placed in refrigeration trucks for shipping. The design will consist of a hydro cooler that will effectively cool water that is being recirculated to cool tomatoes on a conveyer belt. The system described in the report will effectively cool the tomatoes to the correct temperature, conserve water, and maintain a clean cooling environment.

Theory:

Hydro cooling is a process of slowing the ripening of produces after harvesting by immersing in cold water. For cooling the water, a hydro cooler consists of a vapor-compression refrigeration system. The system consist of four components: a compressor, condenser, expansion valve, and evaporator. The compressor starts the refrigeration cycle; it takes in low-pressure, low-temperature refrigerant in gas form and compresses it into a high-pressure, high- temperature gas. The condenser takes in the gas from the compressor and water flows over the coils to remove heat from the refrigerant. As the refrigerant loses heat, it will begin to condense until all of the gas has condensed into a liquid. The expansion value takes in liquid from the condenser and restricts the flow of the refrigerant. When the high-pressure liquid goes through the expansion valve it enters the evaporator. The evaporator makes the refrigerant start to evaporate back into gas which causes the refrigerant to become very cold and absorb a lot of heat. This is where water interacts to essentially chill itself. The heat from the water is transferred to the refrigerant and the process starts over. The length of the process depends on the size of the produce, and the time can range from minutes to hours. With the use of a hydrocooler, it is essential to maintain clean water to prevent contamination so the produce doesn’t get introduced to bacteria.

For determining hydrocooling rates for many fruits and vegetables, many studies have been conducted in order to calculate how long it would take to cool certain produce to a desired temperature. For this to be determined, a Decimal Temperature Difference (DTD) and approximate size of the produce is needed. To determine DTD, the following equation is used:

Eq. 1

where T is the target temperature, W is temperature of the water, and P is the starting temperature of the produce. For determining the approximate size of the produce, the following table is used and is separated into seven categories depending on size:

Table 1: Size chart for cooling vegetables/fruits.

With the DTD and size category determined, the following graph is used to determine the active minutes needed to effectively cool the produce:

Figure 1: Graph for determining cooling time.

A horsepower (HP) is a unit of measurement of power, or the rate at which work is done. For many motors, the basis of output is horsepower. For this design, the power output is to be expressed in kilo-watts (kW) and the conversion from HP to kW is expressed as the following equation:

Eq. 2

Similarly, to calculate the amount of power in kW while given operating Voltage (V) and operating Amperage (A), Eq. 3 is used:

Eq. 3

With power in kW, it is possible to calculate cost based off the operation time. Assuming the cost of operation is $0.10 per kilo-watt-hour, the total operating cost can be expressed as follows:

Eq. 4

where T is the allotted time of operation in hours (hr).

Design:

Figure 2: The schematic of the tomato cooling system.

The system consists of a conveyer belt that will carry tomatoes driven by a motor. As the tomatoes are slowly transported, they are sprayed with cool water. This water collects in a catch pan located under the produce and conveyer belt. Since the water is now contaminated, it is sent through a pump which then goes through a sand filter to remove dirt and disinfector to assure drinking quality. The clean water is then collected in a clean water reservoir. From the clean water reservoir, the water is pumped to the chiller, which cools the water to the required temperature, and is then sent to the sprayers to cool off the tomatoes. The process repeats as necessary.

Pumps:

In an engineering based system occasionally a pump is seen in the schematic of it. Pumps are a key factor in systems where liquid or gas substances are being moved from one location to the next to enter the next phase in the system. A pump is essentially the motor that produces pressure in hoses or pipes to force water or another substance to the next location in a system. In addition to the prior statement, pumps are needed between each significant component in a system. For this specific tomato cooling system two pumps would be required.

One of the pumps would push water from the catch pan under the conveyor to the reservoir. This pump would be relatively small since it would not be moving a large amount of water. It would only make a maximum 55 pounds per square inch (psi) of pressure and move no more than 3 gallons of water per minute (GPM). The specific pump that was desired for use was found to draw 12 Volts (V) and 7.5 Amperes (A). Using these values for the pump and Equation 3 a value was calculated for the power required to run the pump to be 0.09kW.

A second pump would have to be implemented to move the water back to the conveyor once it made it through the reservoir. Though, this pump would need to be able to produce a pressure between 500 and 1000 psi to produce the mist at the nozzles that is sought after. It would also need to be capable of passing at least 5 GPM. This pump is registered at 6.0 HP, so converting to kW yields a value of 4.4742kW required to run this pump.

Conveyor Belt:

In industry conveyor belts are very useful and efficient. They are used to transport an object from point A to point B, or in this case used to slowly transport tomatoes from one side of the nozzle filled enclosed tunnel to the next. Higher quality conveyor systems will have adjustable belt speed settings.

A conveyor belt specific to a tomato cooling system would need to move slowly enough for the tomatoes to take the full effect of the cooling water. Employing a 15 foot long conveyor that is 4 foot wide and dividing the length by the required minutes of active cooling to drop the temperature of the tomato from 75° F to 40° F yields a required speed of 0.32 feet per minute (ft/min) for the belt. A specific conveyor motor was found to use 1HP or 0.7457kW. This motor also specified operating at 3,450 revolutions per minute (RPM). In addition, the use of a gearbox would assist in achieving the required angular velocity of 0.32ft/min.

Nozzles:

Cooling tomatoes using fresh cool water is ideal when one thinks of keeping tomatoes fresh. To accomplish this said freshness the tomato needs to endure spraying of cold water mist via hoses and nozzles. In fact, for this specific design 5 hoses for each foot of width that run the full 15 foot length of the belt and 8 nozzles each hose every 2 feet of length. A specific nozzle was found to output 0.028GMP. Since 5 hoses would be used with 8 nozzles on each one, multiplying the output value of 0.028 by a total 40 nozzles yields a total flow rate of 1.12GPM.

Chiller:

Another key factor to a tomato cooling system is the cooling system alone. For water to be cold enough to cool the freshly picked warm tomatoes a water-cooler chiller would be implemented between the filtration system and the reservoir to ensure all water travelling to the reservoir is at the required temperature.

This chiller is another basic appliance that is powered by electricity. Specifically, a chiller was found to be

Module 1 Activity 7 – Conveyor Failures Timeline

λ = n / T = 40/400 = 0.1/hr

P(0<t<2) = e^-λt₁ – e^-λt₂ = e^-λ*0 – e^-λ*2 = 1 – 0.08187 = 0.1813

Calculating Expected Failures freuquency :

F_exp(0<t<2) = N * P(0<t<2) = 40 * 0.1813 = 7.252

Similarly doing the calculation to all the values

After that χ² statistic = (Observed failure – Expected failure)²/ Expected failue

Total χ² Stat, W = 1.197

Rejection region, R = χ_(1-α) ² (k-1-m)

Where k = number of data intervals = 6

m = number of parameters of null hypothesis distribution estimated from the data

α = level of significance.

R = χ_(0.95)² (6 – 1 – 1) = 2.3719 * 4 = 9.4877

As W < R –

do not reject the null hypothesis at 95% confidence level implying the above problem follows exponential distribution.

The following time-to-failure data with the ranked value of ti. Test the hypothesis that the data fit a normal distribution. (Use the Kolmogorov test for this purpose.)

Mean of the sample data,

= / 10

= 140/10

= 14

Calculating Variance of the sample data,

S² =

Calculating Z =

S – sample standard deviation.

S_n(t_i) = i/n

Where n =10

Module 1 Activity 8 – Conveyor Reliability

From the above table we have observed the max K-S statistic from the columns

[|F(ti) -Sn(ti)|and |F(ti) -Sn(ti-1)|]

The max K-S statistic is 0.2294.

Now we must find the rejection region R at 5% significance level.

D_n(α) = 0.409 from tables of Critical value for the Kolmogonov-Smirnov statistic

Rejection region, R ≥ D_n(α) is defined.

From the values : K-S statistic value < D_n(α)

Therefore we accept the null hypothesis i.e, the above model follows normal distribution at 95% confidence level.

1. Mean of time to failure : = 235

Standard deviation of the time to failure =

Z = (X- )/ σ

P(T<200) = 0.05

Z = (200-235)/σ

But P(Z) = 0.05 => Z = -1.64

There σ = 35/1.64

= 21.28

Module 1 Activity 9 – Risk Ranking

1. P(L<t) = 0.01
   1. For 0.01 the value of Z = -2.326

Using Z = (X- )/ σ

-2.326 = (L_0.01 – 235) / 21.28

L_0.01 = 185.5hrs

1. P(T>250 | T>200) = =
   1. As this model follows normal distribution
   2. P(T<250) = P(Z<(250-235)/21.28) = P(Z<0.704) = 0.75804
   3. P(T<200) = P(Z<(200-235)/21.28) = P(Z<-1.644) = 0.505
   4. P(T>250 | T>200) =

µ_y = 12000, σ_y = 2000

µ_{t = ln}  = 9.379

                σ_t = _ln = 0.1655

P(B>12000)

First we have find Z values for which from the tables we get P(B<12000).

Z =  =    = 0.0827

P(Z=0.0827) = 0.532

P(B>12000) = 1 – P(B<12000) = 1 – 0.532 = 0.468

P(L=12000 | L>10000)

=

P(L > 10000)

Following above procedure

Z =  =    = -1.0187

P(Z) = 0.1538

P(L>10000) = 1 – P(L<10000) = 1 – 0.1538 = 0.846

P(L=12000 | L>10000)  = 0.468 * 1/ 0.846 = 0.55324