Homework 3 - F23 Solution

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Feb 20, 2024

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ME250 | F23 | University of Michigan HW3: Lectures 9, 10, 11, and 12 (67 pts) Engineering Drawings & Mfg Plans, Dimensions and Tolerances, Electric Motors, Machine Elements Due Friday, October 20th by 11:59 pm on Canvas This is an individual assignment, and your solution must be entirely prepared by you. Homework assignments must be completed on your own (unless they are team assignments), however you are encouraged to discuss the problems with your classmates. Upload a PDF of your solution to the Assignments tab on Canvas. Problem 1: Dynamic Analysis and Motors (20 points) Figure 1: FBD of the setup A. Answer: 1. Sum the moments: 𝐹 ? ? ? + 𝐹 ??? ? ??? = τ ?????? (1.28 oz)(3”) + (0.488 oz)(1.5”) = 4.572 oz-in +1 point for sum of moments equation +1 point for correct answer (within ±0.05) 2. Apply Safety Factor: τ ?????? ? ?????? = τ ??????, ???? 4. 15 ??𝑖? ( ) 2 ( ) = 9. 144 ??𝑖? +1 point for applying safety factor +1 point for applying safety factor of 2 3. Motor Specs from here : Stall torque ( τ ? ) ?? 3𝑉 = 1. 2 ??𝑖? Stall torque ( τ ? ) ?? 6𝑉 = 2. 4 ??𝑖? +1 point for scaling the voltage

4. Apply loss of efficiency: τ ? = 0. 2γτ ? = 0. 2 ( ) 0. 3 ( ) 2. 4 ( ) = 0. 144 ??𝑖? +1 point for applying loss of efficiency NOTE: some students may use 0.4 efficiency, this will affect the answers following by a factor of 2, but should be treated as correct due to also applying the factor of safety in pt 2. 5. Calculate required gear ratio: 𝑀 = τ ? τ ? = 9.144 0.144 = 63. 5 +1 point for equation +1 point for calculating correct gear ratio required 6. The smallest gear ratio that satisfies the torque requirement is 80:1 and verify: τ ? ≤ 0. 2γ𝑀τ ? = 9. 144??𝑖? ≤ 0. 2 ( ) 0. 3 ( ) 80 ( ) 2. 4??𝑖? ( ) = 11. 52??𝑖? +1 point for picking correct gear ratio +1 point for correct units throughout problem Verifying if the torque requirement is met is good practice, but not required B. With an 80:1 gear ratio, is scaled by 80 and the no-load speed is divided by 80. The torque τ ? ? ? speed curves change as shown below (they do not need to show a plot): τ ? ≤ γ𝑀τ ? = 9. 144??𝑖? ≤ 0. 3 ( ) 80 ( ) 2. 4??𝑖? ( ) = 57. 6??𝑖? z To calculate the speed of the motor, use the torque speed curve where the torque is 57.6 oz-in. The equation to calculate the speed is where y is the torque and x is the angular velocity: ? ? τ ? τ ? − ? ( ) = ? For the 80:1 gear ratio: 170 57.6 57. 6 − 9. 144 ( ) = ? = 143 ???

NOTE: may vary based on the results of part a. A consistent answer should not be penalized τ ? twice. +1 point for finding T_s +2.5 points for showing any work for calculating the speed (a plot, equation, etc.) +1.5 points for getting the correct speed (within ±3 rpm) C. The highest gear ratio is 400:1, the angular velocity is: 34 288 288 − 9. 144 ( ) = ? = 32. 9 ??? +2 points for picking 400:1 gear ratio +3 points for calculating the correct speed correctly based on the gear ratio they put

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Problem 2: Power Screws (10 pts) A standard 3/8-24 UNF threaded rod is used as a power screw to raise and lower a load. It has a lead of 0.0417 inches, a pitch diameter of 0.3479 inches, and a coefficient of friction of 0.3. A. When the power screw is used to raise the load, the input torque is 2.4 inch-pounds. What is the weight of the load that is being raised? (4 pts) ? ? = 𝐹 𝐷 ? 2 ?+ µπ𝐷 ? π𝐷 ? − µ? ( ) 𝐹 = 2? ? 𝐷 ? ?+ µπ𝐷 ? π𝐷 ? − µ? ( ) F = 40.33 lbf +2 points for equation +2 points for correct force B. How much torque does it take to lower this same load using the power screw? (4 pts) ? ? = 𝐹 𝐷 ? 2 µπ𝐷 ? −? π𝐷 ? + µ? ( ) ? ? = 1. 82 ?? − 𝑖?. +2 points for equation +2 points for correct torque C. A lubricant is applied to the threads, how would this affect your answer to part B? (2 pts) NEW This would lower the coefficient of friction, hence decreasing the lowering torque. +2 pts for correct answer

Problem 3: Stackup Analysis and GD&T (13 pts) A. What is the nominal thread engagement? Does this meet the minimum thread engagement requirement? ( pts) L e, nominal = L b - a 1 - a 2 - a 3 = 1.25 - 0.2 - 0.3 - 0.1 = 0.65in Minimum engagement for aluminum: 2 * .25 = 0.5in Since 0.65 > 0.5, this does meet the minimum thread engagement requirement. Also, since a 4 is 0.5in, then 0.5in is the maximum thread engagement possible so it still does meet minimum thread engagement requirement. +1pts for correctly calculating L e, nominal (0.65in or 0.5in is acceptable) +1pts for correctly stating L e, nominal does meet the minimum requirement for thread engagement ( -0.25pts if minimum thread engagement calculation is not shown anywhere in parts A-B for comparison) B. Calculate the WCS stackup for thread engagement. Does this meet the minimum thread engagement requirement? (2 pts) WCS = L b tol + a 1 tol + a 2 tol + a 3 tol = 0.1 + 0.05 + 0.05 + 0.010 = 0.19in L e, WCS = 0.65 - 0.21 = 0.44in Since 0.44 < 0.5, this does not meet the minimum thread engagement requirement. +1pts for correctly calculating L e, WCS +1pts for correctly stating L e,WCS does not meet the minimum thread engagement requirement. C. Respond to and explain your answer to the following: a. Do you think the worst-case scenario (WCS) method used above is a good strategy for calculating tolerance stack ups? (1 pt) (+1 pt) Any answer with justification is acceptable here (ie, yes because it’s conservative, or no because it’s probably never going to actually happen unless manufacturing volume is very high) b. Is the WCS likely to ever occur? (1 pt) +1 pt for correctly answering NO. Intuitively, it’s more likely that an actual measurement would fall somewhere in between the high and low of the tolerance zone than the high or low themselves.

D. A different part in the same assembly has these two feature control frames. Translate each tolerance description into a feature control frame by filling in the provided blank feature control frames. (7 pts) a. Position of 0.005 at MMC to datum A, datum B at MMC, and datum C at LMC. +0.5 pts for each correct symbol above ( 4.5 pts total ) -0.5 points for each extraneous symbol (ie, adding MMC anywhere but on datum B) b. Profile of a surface of 0.01 to datum E at MMC and datum C. + 0.5 points .for each correct symbol above ( 2.5 pts total ) -0.5 points for each extraneous symbol (ie, putting diameter symbol, adding MMC anywhere but on datum E)

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Problem 4: Engineering Drawings (14 pts) A. Using the provided top and right views of the drawing below, draw the missing front view to scale on the provided sheet. (6 pts) +1 point if problem was attempted +0.5 points per centreline (1.5 points total) +0.25 points per hidden line (1.5 points total) +0.5 points for each proper height circled below (1.5 points total) + 0.5 points for cut-out shown below

B. With the three views you now have, create an isometric sketch of the part on the provided isometric grid paper. (8 pts) Note: Example solution is shown from right side angle, but left side angle is acceptable and should be graded the same. +2 points for correct scale (allow .25” of variation in lengths) +1 point for each hole (3 points total, be lenient on quality of drawings of holes) +1 point if front cut-out is drawn properly (see above) +2 points if general shape is correct

Problem 5: Fasteners and Joining Techniques (10 pts) Diagram for reference. A. To solve the fasteners questions, we need to reference the Drill and Tap Chart provided: (5 pts) Based on the Drill and Tap Chart, for a #18 free fit, we identify that this is for a #6 screw. Next, we are provided that the tapped hole uses a #33 drill. Looking at the tap size column, we see this corresponds to a 6-40 screw. The major diameter of a #6 screw is listed as 0.138”. The threads per inch is 40. The minimum required thread engagement is twice the major diameter when tapping aluminum, which is . 2×0. 138" = 0. 276" +1.5 points for correct major diameter +1.5 points for correctly threads per inch +2 points for correct minimum thread engagement.

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B. To solve the fasteners questions, we need to reference the Drill and Tap Chart provided: (3 pts) Based on the charts, for a #29 free fit, we are using a #4 screw. The #4 screw has two thread count options: 40, 48. Thus, 40 is the coarse thread count. The major diameter is listed as 0.112”, and the threads/inch is 40 threads/inch. +1.5 points for correct major diameter +1.5 points for correct threads/inch

C. To solve the fasteners questions, we need to reference the Drill and Tap Chart provided: (2 pts) Based on the chart, for a 9/32” size drill free fit, we identify this is a ¼” screw. Looking at the other chart, since it is a coarse thread fastener, we identify this is a ¼”-20. The major diameter for a ¼” fastener is 0.250”, and the coarse threads/inch is 20. +1 point for correct major diameter +1 point for correct threads/inch