Supplementary Practice Problems for F23 Final Exam

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University of Michigan *

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250

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Mechanical Engineering

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Feb 20, 2024

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Transmissions Pulley A is driven counterclockwise with a speed of 1500 RPM. It is connected by a belt to Pulley B. The torque of pulley A is 28 Newton-meters. Pulley B and gear C are rigidly connected to the same shaft. Gear D and Sprocket E are also rigidly connected together on the same shaft. All gears have a diametral pitch of 20 teeth per inch. Assume 90% efficiency in the belt connecting pulleys A and B, 95% efficiency in each pair of gears that mesh, and 97% efficiency in the chain connecting sprocket E and F. A. (1 point) In what direction does Sprocket F rotate? B. (3 points) Find the Angular Velocity of Sprocket F ( RPM) C. (3 points) Find the torque output at Sprocket F.

SOLUTION: Transmissions A. (1 point) In what direction does Sprocket F rotate? Clockwise B. (3 points) Find the Angular Velocity of Sprocket F ( RPM) ω ? = ω ? ? ? ? ? ( ) ? ? ? ? ( ) ? ? ? ? ( ) ω ? = 600?𝑃? C. (3 points) Find the torque output at Sprocket F. ? ? = ? ? ? ? ? ? ( ) ? ? ? ? ( ) ? ? ? ? ( ) η ? η 𝑃 η ? ? ? = 58. 1 ??

Motors The specs of a metal gearmotor are provided below. Note that the provided torque and speed values are at the output shaft when the motor is being run at 6V. ● Stall Torque: 1.48 N-m ● No-Load Speed: 97 rpm ● Nominal Voltage: 6V For parts A – C, the motor is being powered by 3 AAA batteries, supplying 4.5V to the motor. A. The motor was selected for the application shown below, where an ME 250 student has proposed direct driving an arm with a holder at the end for a golf ball of mass 50 g. The arm also has a rope attached to it that is driving a pulley system somewhere else in their putt-putt course. They have modeled this as a 2 N force, labeled T. They added a 90 g counter-weight to help reduce torque on their metal gearmotor. The ball and counterweight can be approximated as point masses at the ends of the arm. The arm is 0.20 m long. Neglect the weight of the arm. Determine the load torque on the motor, in the position of the arm shown. B. Determine the operating speed of the motor at the maximum load torque. NOTE: If you were unable to determine a load torque from A, please use 0.25 Nm for the remainder of the problem. (0.25 Nm is not the correct answer from part A) C. The ME 250 student built a prototype to test the arm. They found that this operating speed was too fast. They would like to decrease it to 20 rpm. Please select an external gear ratio to help them achieve this new operating speed. Assume 85% efficiency for the gear mesh.

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SOLUTION - Motors The specs of a metal gearmotor are provided below. Note that the provided torque and speed values are at the output shaft when the motor is being run at 6V. ● Stall Torque: 1.48 N-m ● No-Load Speed: 97 rpm ● Nominal Voltage: 6V For parts A – C, the motor is being powered by 3 AAA batteries, supplying 4.5V to the motor. A. The motor was selected for the application shown below, where an ME 250 student has proposed direct driving an arm with a holder at the end for a golf ball of mass 50 g. The arm also has a rope attached to it that is driving a pulley system somewhere else in their putt-putt course. They have modeled this as a 2 N force, labeled T. They added a 90 g counter-weight to help reduce torque on their metal gearmotor. The ball and counterweight can be approximated as point masses at the ends of the arm. The arm is 0.20 m long. Neglect the weight of the arm. Determine the maximum load torque on the motor. ------------------------------------------------------------------------------------------------------------------------------- ? ? = ∑ ? ????? = ? 2 ∙? + 3? 4 • ? ???? − ? 4 • ? ?? ? ? = 0.2 ? 2 ∙2 ? + 3∙0.2 ? 4 • 50 𝑔 • 1 ?𝑔 1000 𝑔 • 9. 8 ? ? 2 − 0.2 ? 4 • 90 𝑔 • 1 ?𝑔 1000 𝑔 • 9. 8 ? ? 2 ? ? = 0. 2294 ?? +2 point for showing some form of moment analysis work +1 points for correct T L -------------------------------------------------------------------------------------------------------------------------------

B. Determine the operating speed of the motor at the maximum load torque. NOTE: If you were unable to determine a load torque from A, please use 0.25 Nm for the remainder of the problem. (0.25 Nm is not the correct answer from part A) ------------------------------------------------------------------------------------------------------------------------------- ? = ? ? − ?? => ? = ? ? ? ? = 1.48 ?? 97 ??? = 0. 01526 ?? ??? ? ?,4.5? = ? ??? ? ????? ? ?, 6? = 4.5 ? 6 ? 1. 48 ?? = 1. 11 ?? ? ?,4.5? = ? ??? ? ????? ? ?, 6? = 4.5 ? 6 ? 97 ??? = 72. 75 ??? ? ? = ? ?,4.5? − ?? => ? = ? ?,4.5? −? ? ? = 1.11 ??−0.2245 ?? 0.01526 ?? ??? = 57. 7 ??? OR: with the provided load torque ? = 1.11 ??−0.25 ?? 0.01526 ?? ??? = 56. 3 ??? ------------------------------------------------------------------------------------------------------------------------------- +1 point for correctly switching to 4.5 V values of stall torque and speed +1 point for correctly using the torque-speed equation. +1 point for correct answer. C. The ME 250 student built a prototype to test the arm. They found that this operating speed was too fast. They would like to decrease it to 20 rpm. Please select an external gear ratio to help them achieve this new operating speed. Assume 85% efficiency for the gear mesh. The key piece of information is to recognize that the load torque has not changed since we are still driving the same arm. Additionally, since we are at the same torque but want a lower speed, we can only achieve this with a speed reduction ratio, labeled M. ? ? = ? ?,??? − ? ??? ? ? ?,??? = η?? ?,4.5? & ? ?,??? = 1 ? ? ?,4.5? ? ??? = ? ?,??? ? ?,??? = η?? ?,4.5? 1 ? ? ?,4.5? = η? 2 ? ?,4.5? ? ?,4.5? = η? 2 ? ? ? = η?? ?,4.5? − η? 2 ?? ???

0. 2294 ?? (0. 25 ??) = 0. 85∙?∙1. 11 ?? − 0. 85∙? 2 ∙0. 01526 ?? ??? ∙20 ??? Simplified further: 0. 2594? 2 − 0. 944? + 0. 2294 = 0 is the desired speed of 20 rpm for the same load torque . Solving the quadratic equation for version ? ??? ? ? 1 yields M = 0.26 and 3.38. Identifying that 0.26 is not a speed reduction, 3.38 is the correct gear ratio. Note that 0.26 is also a valid solution, but it would run the motor inefficiently (close to its stall torque.) Note: this is NOT the same as 57.7 rpm / 20 rpm = 2.89. The gear ratio only tells you the ratio of the speeds. In part B, the motor shaft speed was 57.7 rpm. By adding the external transmission, we know the output shaft speed is 20 rpm. Writing 57.7 rpm / 20 rpm implies that the motor shaft is still moving at 57.7 rpm while the output shaft (driving the arm) is moving at 20 rpm. This is not correct. There is no constraint that implies the motor shaft would move at the same speed with / without an external gear ratio. We do know the constraint on the output shaft speed in B determined by the load torque, and in C given in the problem statement. We have to use those constraints to solve for the motor shaft speed. +1 point each for correct relationships of T s and n o with efficiency and gear ratio +1 point for correct relation for k new +2 points for correct T L equation +1 point for correct T L equation with correct numbers plugged in -1 point if they used the 6V torque/speed. +1 point for correct final answer

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Free-Body Diagrams / Materials Professor Austin-Breneman and Professor Umbriac have just opened a clown store in Pierpont Commons. They decide to place a sign outside of the store to help students find their new business venture. They attach a simply supported beam to the wall and secure it utilizing a massless wire from the end of the beam to the wall. They then hang the sign from the beam. Knowing they would like their students to do static analysis on their new setup, they measure the following: a. Draw the free body diagram of the beam. b. Write the 3 static equations of motion symbolically with the moment around point A for the beam. c. Calculate the internal tension in the cable. d. Calculate the reaction forces in the beam. e. The professors soon realize that the wire breaks above 430N of force. Identify a parameter the professors should change to reduce the internal force in the wire and in which direction (increase or decrease) should they change it?

SOLUTION: Free-Body Diagrams / Materials a. b. ∑ ? ? = 0 = ? ? + ? ? ?𝑖? ϑ − ? ? 𝑔 − ? ? 𝑔 ∑ ? ? = 0 = ? ? − ? ? ??? ϑ ∑ ? ? = 0 = ? ? ? ? ?𝑖? ϑ − ? ? 𝑔 ? ? 2 − ? ? 𝑔? ? c. ??? θ = 2? 4? →θ = ??? −1 ( 1 2 ) = 60° ? ? = ? ? 𝑔 ? ? 2 +? ? 𝑔? ? ? ? ?𝑖? ϑ = 25?𝑔*9.81 ? ? 2 * 2? 2 +40?𝑔*9.81 ? ? 2 *1.5? 2?*??? 60

= 245.25 ?𝑔? 2 ? 2 +588.6 ?𝑔? 2 ? 2 1.732? = 481? d. ? ? = ? ? ??? 60 = 240? ? ? = ? ? 𝑔 + ? ? 𝑔 − ? ? ?𝑖? 60 = 245. 25? + 392. 4? − 416. 925 = 221? e. Reduce the mass of the sign. Increase the wire length (To increase the angle at which the wire attaches to the beam). Reduce the mass of the beam. Reduce the length where the sign is attached. (Attach closer to the wall)

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Dimensioning & Tolerancing You are reviewing a design for a toggle clamp shown below in two positions. The partially-dimensioned drawings of the four parts are shown below, along with the General Tolerances table. Answer questions a through d from the following pages. Base Plate:

Handle: Connecting Link: Plunger:

General Tolerances for all four parts: a.) When the handle is in the position shown above (the components are in a straight horizontal line) what are the maximum and minimum values for dimension X , based on the drawings above? (When these parts are assembled to each other, assume that the centers of the holes are perfectly aligned between each part and the next one.) HINT: In order to calculate these values, you will need to reason through the additions and subtractions of the maximum and minimum values of the dimensions on these components. (6 points) b.) The amount of variation in dimension X, which you calculated in part a, is too large for the customer. Suggest a change to the dimensioning of the drawing of the base plate, that would reduce the variation of X. (2 points) c.) Suggest a change to the dimensioning of the drawing of the connecting link, that would reduce the variation of X. (2 points) d.) Given that the diameter of the plunger is 0.4988/.4981 inches (written as limit dimensions), what should be the maximum and minimum size of the hole in the base plate , to achieve an RC5 fit with the plunger? Refer to the table below. (2 points)

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SOLUTION: Dimensioning & Tolerancing Part a.) The values grouped in parentheses below represent the applicable dimensions of the plunger, connecting link, handle, and baseplate, respectively. X_max = (3.755 - .245) + (-.245 + 2.130) + (1.005) + (.205 - 4.48 + .130) = 2.255 X_min = (3.745 - .255) + (-.255 + 2.120) + (0.995) + (.195 - 4.52 + .120) = 2.145 RUBRIC: (6 pts total) ● +2 for applying BOTH of the General Tolerances correctly at least part of the time (for example, writing 4.48 or 4.52 as well as the dimensions which added .005) ● +2 for generally correct method of adding together the values. ● +2 for correct answers of 2.255 and 2.145 Part b.) Possible answers for the base plate include: ● Change the dimension of the center of the hole to be measured from the front of the boss on the LH side of the part. ● Dimension the hole from the datum on the LH side of the part, instead of the RH side. ● Make the overall length a 3-place dimension (not preferred, but acceptable as an answer to this question.) RUBRIC: (2 pts total) ● +2 for any one of the correct answers above Part c.) Possible answers for the connecting link include: ● Dimension from the center of one hole to the center of the other hole. Note: The distance between the holes is what contributes to X. Dimensioning the overall length of the connecting link to 3 decimal places would not improve the tolerance between the holes, because both of the holes are dimensioned from the same side of the part, not from opposite sides. RUBRIC: (2 pts total) ● +2 for the correct answer above

Part d.) The basic size (closest fraction) is .500, which is confirmed by the following reasoning: The .4988/.4981 size of the plunger (shaft) falls in the fourth row of the table, because it is in a size range of 0.40 - 0.71. For an RC5 fit, the shaft limits in the table are (-1.2, -1.9). The values in the table are written in thousandths of an inch. So the values in the table confirm that the basic size is .500, because .5000 - .0012 = .4988 , and .5000 - .0019 = .4981 . For this row and column in the table, the hole limits are (+1.0, 0). Again, these are written in thousandths of an inch. The answer for hole size is (.500, .501.) We will accept answers of those two values written in either order. RUBRIC: (2 pts total) ● +1 for using a hole tolerance of .001 based on the table. ● +1 for the correct answer of .501 / .500

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Manufacturing Processes Polymer Manufacturing (Sean) 4. Consider the Legos ® shown in the figure below. Mentalfloss.com a.) The Legos shown were made via injection molding. List three pieces of evidence that support this claim. b.) What is an alternative mass production method discussed in lecture to make a plastic part? Why do you think that method was not chosen for these Legos? c.) Why do you believe the company selected to use a polymer over a ceramic? Why do you believe the company selected to use a polymer over a metal? d.) The designed length L is 0.500”. What dimension should the mold cavity be? Assume the LEGO is made with ABS plastic. (Shrinkage of ABS= 0.006)

e.) What is the name of the feature circled in green below? What is one reason it was included in the design? This feature is also visible in the first figure. 27gen.com f.) A student entrepreneur interning at Lego has proposed the design feature shown below, with the logo molded through the side of the Lego for a cool new look. The intern claims this will cost the same to produce. Is this claim true? Why or why not? Do not consider the reduction in plastic material costs when answering this question. Note: the logo goes through the wall. g.) Consider the Lego Duplos shown below, which are much larger than regular Legos. Notice how the studs (the cylindrical features on top of the Duplos) are hollow instead of solid. What manufacturing

defect might you see if the studs were solid? Assume the thickness of the solid stud would be greater than the wall thickness of the part.

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SOLUTION: Manufacturing Processes 4. Consider the Legos ® shown in the figure below. (15 pts) Mentalfloss.com a.) The Legos shown were made via injection molding. List three pieces of evidence that support this claim. (3 pts) - +1 point for each acceptable response, up to 3 points ● Plastic work material ● Open shape with complex features ● Uniform wall thickness ● Thin wall thickness ● Tight dimensional tolerances ● Smooth surface finish ● No undercuts is an acceptable answer ● Ejector pin marks ● Nipple/witness marks are visible in the part e) image ● Ribs ● Fillets and drafts ● Inter-mold parting line (flash) ● High economic batch size b.) What is an alternative mass production method discussed in lecture to make a plastic part? Why do you think that method was not chosen ? (2 pts) - +1 point for method - +1 point for justification Blow molding ● Wrong shape - only for thin-walled, bottle-like shapes with a small opening Rotational molding ● Wrong shape - only for thin-walled, seamless, hollow shapes (no holes) Thermoforming

● Cannot simultaneously achieve the cylindrical bosses on the inside and outside ● Difficult to make the ribs, to the depth that is required for stiffness Extrusion ● Shape does not have a uniform cross-section Compression molding ● Injection molding is better to mass-produce small parts in batches ● Part thickness potentially too thin c.) Why do you believe the company selected to use a polymer over a ceramic? Why do you believe the company selected to use a polymer over a metal? (2 pts) - +1 point for acceptable answer for ceramic - +1 point for acceptable answer for metal Ceramics: ● Too brittle ● Heavier ● More expensive ● More difficult to manufacture with tight tolerance and surface finish Metals: ● Heavier ● More expensive ● More difficult to manufacture d.) The designed length L is 0.500”. What dimension should the mold cavity be? Assume the LEGO is made with ABS plastic. (3 pts) ? ? = ? ? 1−? ??? = 0.500" 1−0.006 = 0. 503" ?? ? ? (1 + ? ??? + ? ??? 2 ) = 0. 503" - +1.5 point for either equation - +1 point for correct answer - +0.5 points for units

e.) What is the name of the feature circled below? What is one reason it was included in the design? (2 pts) - +1 point for identifying correctly as a rib - +1 point for acceptable reason ● The labeled feature is a rib ○ Gusset is acceptable but not correct ● Increases stiffness of the cylindrical boss and of the wall ● Reduces warping in the wall. ● Allows plastic to more easily and/or quickly flow in the mold f.) A student entrepreneur interning at Lego has proposed the design feature shown below, with holes in the sides of the Legos for a cool new look. The intern claims this will cost the same to produce. Is this claim true? Why or why not? Do not consider the reduction in plastic material costs when answering this question. (2 pts)

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- +0.5 point for indicating it is not true - +1.5 point for acceptable justification ● The feature is an undercut, which would require side-action in the mold. This will drive up the cost and increase cycle time. ● 1/2: If they just said it would cost more because they’d need to buy a new mold. ● 1/2: If they indicated it would require post-processing like machining or laser-cutting g.) Consider the Legos shown below. Notice how the studs are hollow instead of solid. What defect might you see if the studs were solid? Assume the thickness of the stud is greater than the wall thickness of the part. (1 pt) - +1 point for acceptable answer ● Sink marks, voids, or shrinkage can appear for thicker cross-sections.

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Fasteners and Joining -- refer to the following charts for the next question

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Problem 1 Professor Austin-Breneman’s birthday is coming up, and Professor Umbriac wants to build him a custom-made machine for buffing (polishing) the hood of his 2014 Honda Odyssey. He has a preliminary design for the linkage shown in Figure 1. Figure 1 Unfortunately he finds that none of the links can make a complete rotation, so he cannot attach a motor to it. The locations of the baseplate’s joints cannot be adjusted. A. Professor Umbriac looks in his garage and finds 3 new links that he bought on his last trip to Links”R”Us. They have lengths (joint-to-joint) of 3.5 inches, 5 inches, and 8 inches. Which of these 3 new links can he use, and which old link will he replace with the new link, to allow a complete rotation? Show your calculations. (4 points) B. Professor Austin-Breneman loves the new machine. However, the metal tool (that attaches to the 4.5 inch link and holds the buffing pad) is creating dents in his car. Suggest a different family of material for this tool, and explain the reason for your choice. (2 points) C. Unfortunately the 17 horsepower motor, that Professor Umbriac has used to power the machine, is causing so much vibration that the 1/4-20 UNC bolts which mount the machine to Professor Austin-Breneman’s table are breaking. Professor Austin-Breneman has some 3/8-16 UNC bolts

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that he would like to use instead. What size should he drill the holes in his table, before using a 3/8-16 tap in those holes? (1 point) D. What size should he drill the new holes in the baseplate, for these bolts to go through? (He wants it to be easy to assemble, knowing that there are four holes that have to line up with the holes in the table.) (1 point) E. Professor Austin-Breneman’s aluminum table is 3/4 inch thick, and the holes will be tapped all the way through this thickness. The steel baseplate of the machine is 1/2 inch thick. The 3/8-16 UNC bolts are 1 inch long. Will the thread engagement be sufficient? Explain your answer. (3 points)

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Problem 1 Solution Solutions: 1. 4 points For complete rotation, the linkage must satisfy the Grashof Criterion s + l < p + q . ● The length between the baseplate joints = 9 - 3 = 6 inches and cannot be adjusted. ● In the preliminary design, 3 + 6 is not less than 4 + 4.5, so it can’t rotate. ● Directionally, we would want to make the 3” link shorter, but the available options are not shorter than 3”. ● Try replacing the coupler (4” link) with the 3.5” link: ○ 3 + 6 = 9 is not less than 3.5 + 4.5 = 8 can’t rotate ● Try replacing the coupler (4” link) with the 5” link: ○ 3 + 6 = 9 is less than 5 + 4.5 = 9.5 can rotate ● Try replacing the coupler (4” link) with the 8” link: ○ 3 + 8 = 11 is not less than 6 + 4.5 = 10.5 can’t rotate ● Other solutions might be possible, such as replacing the 4.5” link with the 8” link +1 point for using the Grashof criterion +1 point for showing the correct calculation +1 point for stating the correct new link +1 point for stating the correct old link to replace 2. 2 points The material families of Elastomers, or some Plastics would be a good choice. Their Young’s modulus (and also their hardness ) would be low enough that they could deflect before denting the car. (In the student’s answer, we would also accept the word “flexible” or “soft” or “lower stiffness,” even though stiffness is not an intrinsic property. Assuming the shape of the tool does not change, Young’s modulus correlates to stiffness). Ceramics and Glasses would dent the car, because of their high Young’s modulus and high hardness, so they would not be acceptable. They can also be brittle. We would also accept Foams as an answer -- although they might not be stiff enough if the tool is shaped like an arm, we did not specify the shape of the tool, so accept this answer. We would also accept Hybrids as an answer. Because hybrids are made from a combination of materials, the materials could be chosen to provide a low Young’s modulus and/or hardness. +1 point for choosing a correct family of material +1 point for a correct explanation 3. 1 point Use the tapdrill column in the table, to get an answer of 5/16” or .3125” +1 point for correct answer 4. 1 point Use the Free Fit Clearance hole column in the table, to get drill size 13/32” or .406” . Free fit holes provide fast and easy assembly.

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+1 point for correct answer 5. 3 points For the aluminum material of Professor Austin-Breneman’s table, we want the thread engagement to be 2 times the major diameter of the thread . 2 x 3/8 = 6/8 or 3/4 inch would be needed . The table is 3/4 inch thick, but the bolts are only 1 inch long, and 1/2 inch of that length is in the baseplate. (A similar case is clearly illustrated in the lecture slide about thread engagement.) Thus there is only 1/2 inch of thread engagement, which is not sufficient. To fix this problem, he should get longer bolts. +1 point for using 2x the major diameter of the thread +1 point for identifying that 3/4 inch of thread engagement is needed +1 point for stating that the thread engagement is not sufficient because of the bolt length

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Problem 2 1. You will be using a 6-32 screw to fasten two aluminum plates (Part A and Part B) together, as shown in the diagram below. (3 points) a. What size drill should you use for a free-fit hole in Part A? b. What size drill should you use for pre-drilling the hole in part B before tapping it? c. If the material for both plates was acrylic instead of aluminum, how would you design the joint differently, while still using a threaded fastener? Part A Part B 2. Under what circumstances would you want to use a needle roller bearing, compared to other types of bearings? (2 points) 3. In the following expression for thread, what do the values mean? (3 points) ¼”-20 UNC 1/4 __________________________ 20 __________________________ UNC ___________________________________ 4. When drilling a series of clearance holes, what is the primary reason for using free fit instead of close fit? (2 points)

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Problem 2 Solution 1. 3 points a. Use a #18 or 0.170” drill to make a free-fit hole in part A (+1 point) b. Use a #36 or 0.1065” drill to make the hole in part B before tapping (+1 point) c. If the material was acrylic instead of aluminum for both plates, bolts and nuts should be used instead of a screw since acrylic is too brittle to reliably tap threads. Washers should also be used under the head of the bolt and the nut to distribute the load better. (+1 point) 2. 2 points You would want to use a needle roller bearing when you have a high radial load and not enough space (radially) for a bearing. +1 point for mentioning high radial load +1 point for mentioning not enough space 3. 3 points ¼ - Major diameter (+1 point) 20 – threads per inch (+1 point) UNC – Unified National Coarse, or “thread series” or “type of thread” (shape) (+1 point) 4. 2 points The primary reason for using free fit instead of close fit is to make the joints easier to assemble.

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