I study management, and I can outperform 98% of “mechanical engineers”!

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This post was last edited by 3983596_FPPZ on 2019-3-2 at 13:09. It was copied from the China Machinery Community: (To reiterate: this is not my opinion as the post owner or moderator; I just found this post to be quite controversial, so I shared it here for everyone to see: lol. The views expressed in it may not be correct, especially regarding how fitting tolerances should be calculated. In my opinion, there’s no need to learn or imitate such methods – we should focus on opinions and facts only. Please don’t target me: handshake. But as a mechanical engineer myself, it’s true that we should reflect on ourselves when being criticized by laypeople like this: L) My profile: . d, E8 K) h6 n’ S# x. X I graduated with a bachelor’s degree in management in 1990; I have almost no basic knowledge in science and engineering. From the perspective of the scientific and engineering knowledge I’ve learned, I’m nowhere near as knowledgeable as anyone else on this forum! ; O% _1 }, F6 o. u! g, L4 n 3 d. i% n1 R, A. @4 q8 G4 x I have been working for over 20 years; I have never dealt with machinery directly – I have always worked with CNC systems and in conjunction with mechanical components. I am responsible for the design, and every step in the process involves thorough reasoning and calculation. To be precise, each line of software code I write is the result of such reasoning and calculations. - o2 h7 @- l6 P# d9 a9 {( S" w! Therefore, in my imagination, machinery should also be like that; as a result, I have great trust in mechanical drawings. Moreover, due to the heavy workload associated with my own development tasks, I have never doubted the accuracy of those drawings. But: five years ago, by chance, I discovered that almost all the mechanical drawings I had seen lacked any analysis or calculations, and the data on those drawings were simply copied verbatim from manuals. 3 @* k/ {- ]+ R1 A* L Since then, I have started to pay attention to mechanical drawings; due to my position, I can access any such drawings. The following are the tasks I carry out regularly: rejecting mechanical design proposals, not based on authority but on data: * `1 ~8 N' A( W( B/ i 1. Obtain the drawings and gather the relevant designers together ； 2. Randomly select a marked value from the drawing ； 3. Question: Where does this data come from? How to calculate it? Of course, I don’t understand it at all and can’t do it ； / Q/ i! E5 P' q5 T/ {2 w0 E 4. So, I can only ask the designer: Where did this data come from? How should it be verified? 5. The designer will go check the manual temporarily and then tell me the calculation process ； " u7 H6 E& Y, j: ]# S% N3 t 6. Okay, I’ll go through the calculation using the method you provided ； 7. As a result, it’s obvious that the numbers on your drawing are incorrect – they’re off by a huge margin ； 8. The plan is rejected; redo it! / k5 P+ A m5 ^! c4 K p ....... # N2 k2 ^1 a' v8 N& ?/ V( ~9 \ This is the “Chinese phenomenon”: 98% of these robots are actually just draftsmen, and they aren’t even qualified to be draftsmen! Of course, I’m not qualified either, because I don’t even know CAD; how could I be capable of working as a draftsman? But please understand that I’m not a mechanical engineer, so I can’t draw mechanical drawings – that’s only natural! ' } k" \$ O0 o0 ` ! d j/ Y( e1 r# _1 G- I+ m Due to my limited knowledge level, it’s not possible for me to focus on too much mechanical expertise. Within my capabilities, I can only carry out simple calculations related to tolerance fitting in assembly, force analysis, torque, and power transmission. But with my poor skills – and I’m not bragging or being arrogant – I really feel like crying: ! {. y7 d1 u! x( }% R, h( f: u9 ^ I can actually outperform 98% of “mechanical engineers”! ; \1 }3 h3 Z) j# z4 I& ]) D; W; C, T % I8 w) ]% E( P: J3 a, K8 @- Naturally, many people in the forum are not convinced; if you’re not convinced, you need to use real evidence, not just rely on empty words! I’ll now reveal the basic tolerance matching problem I presented in the post “Why Do We Lack Argumentation Skills?” and describe the process for determining tolerances. You can use this to assess for yourself whether you truly “lack argumentation skills.” The original question is as follows (with the operational requirements mentioned in subsequent replies added): For a shaft-hole fit, the hole diameter is 50 mm; it is a through-hole with a depth of 10 mm (a hole drilled in 10 mm thick steel). Both the shaft and the hole are made of the most commonly used 45# steel. To simplify calculations, all other factors such as operational requirements are ignored – there is only one requirement: an interference fit. Please determine the specific dimensions of the shaft and the hole, that is, specify the tolerances for each! " ]9 X2 y0 l# K( n" S6 s! M' k, \ Isn’t this a too simple basic question? I have asked more than 100 people, and to date I have only encountered one person who was able to give a quick and accurate answer. Of course, his answer did not include specific numbers; it only described the process. I will repeat that process here: ) a- A8 S( g# k’ j! h7 _$ p: W0 t. There are many ways to achieve interference fitting (details omitted); I will use the pressing method as an example for calculation: + z y5 m* G” v5 V4 s) i) i. 1) Based on the strength data related to 45# steel, the maximum pressure that a 50mm shaft can withstand is calculated, denoted as P1. If the pressure exceeds P1, the shaft will be damaged ； 2) Based on the strength of the 10mm steel plate, the maximum pressure it can withstand is calculated: denoted as P2. If the pressure exceeds P2, the 50mm hole in the steel plate will be crushed ； 3) Take the smaller value of P1 and P2; this is the maximum allowable pressure value: P. 4) Based on this P value, and to achieve a penetration depth of 10 mm, it is possible to calculate the maximum allowable amount of interference, denoted as m1. 5) Using the 50 mm hole in a 10 mm thick steel plate, the maximum allowable amount of extrusion (which is related to the interference) can be determined, yielding the maximum allowable interference amount: m2. 6) Take the smaller value of m1 and m2 to obtain the ultimate allowable amount of interference: m. 7) Given the application requirements specified in the problem, such as no torque transmission, the minimum amount of interference does not need to be calculated; it can be directly set at 0%. 8) Thus, the range of allowable interference is from 0 to m. 9) Allocate the tolerances evenly between the shaft and the hole: for the hole, it is +0 to –m/2; for the shaft, it is +m/2 to +0. That’s all! 10) Meanwhile, based on the P-value, the parameters for the press can also be determined; taking 2 to 3 times the P-value gives the parameters for the press! % J- @9 i3 f- D( J" p7 H) R) H! E The answering is complete; although no data was provided, my evaluation can only be this: NB! ) R4 V’ ?$ U8 w - }. d! }3 ?; {& S, U6 Y. Of course, the values mentioned above are purely theoretical, or rather, the maximum values that can be used; actual applications require further optimization. I’m too lazy to go through each case and perform the calculations (those who want to calculate can find the methods in mechanical design manuals). Let’s assume a value and explain the optimization process: Assume m = 0.5; then as long as the interference amount remains between 0 and 0.5, any notation method is valid. For example: Shaft 50 +0.25 –0, Hole 50 +0 –0.25. It’s also possible to use: Shaft 50 +0.30 +0.05, Hole 50 +0.05 –0.20. 0.25 mm corresponds to 50 mm; according to mechanical design manuals, this represents the most generous allowable tolerance, meaning that it can be machined using any ordinary machine without any problems. z) x3 X0 d f$ o” R o% f; X, w& u. However, there is one small issue: 0.5 is based on the maximum allowable pressure. If workers in the workshop are not careful, they might compress the shaft or hole. To reduce the risk of damaging the parts during assembly, this value should be reduced slightly. ; J8 n3 At this point, the mechanical manual comes in handy again: consulting the tables shows that for IT9, the value is 0.074. We can use a slightly larger value, namely 0.075. In other words, we change 0.25 to 0.075, which reduces the pressing force significantly. There’s no longer any need to worry about the workpiece being damaged during assembly in the workshop. The precision requirement of IT9 results in little difference in the cost of part processing, while the cost of the press can be reduced considerably. Therefore, we choose IT9. Of course, you can also choose IT8; either way, as long as the interference fit falls within the range of 0–0.5, we simply consider both the costs of part processing and the press to select the better option :) ^ W; Z* A, L' s % g) @# H% ^5 ~1 x6 c8 j" T1 V Well, for the specific dimensions, you can just assign arbitrary values – any of the following combinations will work: Shaft: 50 +0.075/-0; Hole: 50 +0/-0.075. b" b9 Y8 x! O7 d+ |+ _, y6 n. Shaft: 50 +0.050/-0.025; Hole: 50 -0.025/-0.10. Shaft: 50 +0.075/+0.005; Hole: 50 -0.005/-0.075. (The minimum interference amount is 0.01.) 4 V- ?9 J; B# B' U3 h ......./ B' j' {# D- h # T6 ]1 a: s# O+ ` N/ G c Is H7/p6 necessary? Although this value is also within the 0–0.5 range, the processing costs for IT7 and IT6 are much higher than those for IT9. Is this necessary? * C% h3 \, D' i: Q4 W' f ' ]) U) \( G: m* F8 A: }! E 8 _$ x4 `6 S( }" N" U/ @+ {) |+ @ ( M9 q- q2 u/ R$ `; T/ _. S { : c, O d; f8 H5 q8 `' L' K Additional content (2013-7-8 18:00): There are quite a few “experts” in this forum! Try to knock me over? Write down any reasonable argumentative analysis process at random, and my ignorance will inevitably be exposed. 4 e# D5 _1 D! ?5 T- U. \ Here, the “expert” is very special: they use only knowledge of language and literature, and absolutely no mechanical expertise. & m& w7 d I/ V8 A( 98%, that’s 98%, and it will never change!