11hw2-21jan-soln - EML 4321 –HW#2 Solution Below I will...

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Unformatted text preview: EML 4321 –HW#2 Solution Below I will have a couple of solutions‐‐‐one from a TA and one from a student. I will comment in italics. JKS 1. Explain in a few sentences in your own words what the difference is between MRP and MRPII. From the TA: Early manufacturing software systems were designated MRP (Materials Resource Planning) programs. The basic functions of MRP systems included inventory control, bill of material processing and elementary scheduling. With the later addition of scheduling functionality (such as business planning, production planning and scheduling, capacity requirement planning, job costing, financial management and forecasting, order processing, shop floor control, time and attendance, performance measurement, and sales and operations planning), MRP became MRP II. This is a good listing of the characteristics of the two systems. But the “difference” is maybe expressed better by the student: MRP plans around the availability of the materials, mainly focusing on ordering as little and as late as possible into the production schedule. MRP2 additionally considers the availability and quality of resources such as people and machines. MRP2 is broader scope approach. 2. Textbook question Q1.1 From the TA: Examples of machinery products made a) in mass production: automobiles and their parts, appliances, lawn mowers, TV tubes and some TV sets, cellular telephone sets, small electric motors, screw bolts and nuts, spark plugs, ball bearings; b) in medium lot productions: road building machines, machine tools, airplanes, jet engines, medium electric motors, steam turbines, stationary gas turbines; small lot productions: large machine tools, large electric motors, fire trucks, boats, special machine tools, special medical equipment, etc. This is a very good list. However, there are always be arguments as to what constitutes “mass”, “medium”, and “small”. The question is biggest about what constitutes “medium”. As a consequence, I can take almost any reasonable answer here. The student’s list is also very good: a) Mass Production: Cars, ball bearings, cameras b) Medium‐lot Production: Bulldozers, ships, drill bits c) Small‐lot Production: $1M+ yachts, Fighter Jets, Custom motorcycles 3. A cylindrical steel specimen for a tensile test has an initial gage length of 1” and a diameter of 0.25”. Under a load of 2500 lbs, the gage length is increased to 1.1”. (Assuming no necking at that point.) First we will have the TA’s solution. He got it almost right, except for part g as I will explain below. The first parts are simple, as shown: (a) What is the engineering stress? π F S= ; A0 = (0.25 in) 2 , F = 2500lbs A0 4 4 ⋅ 2500 S= = 50.93 ksi ( ANS .) π (0.25) 2 (b) What is the engineering strain? ΔL (1.1 − 1.0) e= = = 0.1 ( ANS .) 1.0 L For part c, some of you asked about Poison’s ratio. Yes, you would have to consider all that during small, elastic deflections. But most of the deflection here is plastic. So I think the constant volume is a better solution. (c) What is the new diameter? V = const. V0 = A0 L0 = π 4 (0.25) 2 (1.0) = V1 = A1 L1 = π 4 (d ) 2 (1.1) 1.0 ⋅ (0.25) = 0.238 in ( ANS .) 1.1 2 d= (d) What is the true stress? F π ; A = (0.238 in) 2 , F = 2500lbs A 4 4 ⋅ 2500 σ= = 56.19 ksi ( ANS .) π (0.238) 2 σ= (e) What is the true strain? ⎛L⎞ ⎛ 1.1 ⎞ ⎟ = ln ⎜ ⎟ = 0.0953 ( ANS .) L0 ⎠ ⎝ 1.0 ⎠ ⎝ ε = ln ⎜ Part f requires you to think about what is happening. First the material will be elastically strained. Stress will increase linearly until it yields. We’re assuming a very ductile, non‐strain‐hardening material. So the engineering stress‐engineering strain curve makes a sharp corner and is very flat. That corner is where the engineering strain times Young’s modulus is equal to the yield strength: (f) Assuming the current engineering stress equals the yield stress (no strain hardening), what was the gage length when plastic strain (yielding) started? For steel, E=30×106 psi F 2500 → 30 ⋅106 ⋅ ey = π A0 (0.25) 2 4 ∴ ey = 0.001697 S = Ee = ey = Ly − L0 L0 = Ly − 1.0 1.0 = 0.001697 Ly = 1.001697 in ( ANS .) The TA got part g WRONG. He should be subtracting 0.001697 from L1 to get 1.0983: (g) If the load is removed, what will be the gage length? L1 − Ly = 1.1 − 1.001697 = 0.0983 in ( ANS .) Note that the springback is very little compared to the plastic strain. The student also has the correct answers for a through e: Aeng= πd2/4 π0.252/4= 0.049in2 a) 2500/.049= 50929.6 psi b) ΔL/L = (1.1‐1)/1= .1 c) (π.252/4 )(1)= πd(new)2/4(1.1) .252 = 1.1 d(new)2 d(new)2=.05682 d(new)=.238 in d) Anew= πd(new)2/4 π0.2382/4=.0445in2 2500/0.0445=56194.8 psi Too many significant figures here! Especially the “8” on the end must be removed! It is easy to overlook the decimal point think it is half a million. I even suggest “56,200 psi”, but “56194” would be OK. e) ln(1+eng strain)=ln(1+.1)= .095 However, I do not agree with his approach for part f. If you have no other data, you can use the “rule of thumb” of e=0.002 dividing elastic and plastic. But in reality, that point will vary with yield strength of the material. In this case we have more information, so I consider this answer WRONG: f) When engineering strain reaches e=0.002, plastic strain begins. 0.002= ΔL/1 New gage length is 1.002 when plastic strain starts However, he gets part g impressively right: g) E=σ/ε 50929.6/.002= 2.55x107 psi If we assume spring‐back occurs resulting in the strain reducing in a linear fashion, such that the stress‐strain curve is in parallel with the original elastic curve of slope E, we can find the resting length of the plastically deformed specimen. ie ΔL=.1‐.002=.098 (since the plastic portion of the curve is horizontal). Or we can use y=mx+b 50929.6=2.55x107(.1)+b b= ‐2.496x106 y=2.55x107x‐2.496x106 0=2.55x107x‐2.496x106 x=.098 ΔL/L=.098 ΔL=.098 New length is 1.098 in The TA gets problem 4 correct: 4. Using Table 2.1, plot the ultimate tensile strength of steel (MPa) as a function of Rockwell C from Rc=20 to Rc=50. (You may plot by spreadsheet, MATLAB, any plotting routine, or by hand.) However, I have problems with the solution of the student. He connects the data points. Even if engineering theory supports a relationship between the points (and how does he know which mathematical relationship to fit?), he should not plot values between the points because we have no such data. I DON’T WANT GRAPHS AND RESULTS WHICH ARE MISLEADING! Also, he does correctly have units, but they are wrong! It is “MPa”, not “Mpa”. Be careful, many software package “correct” spelling and make it wrong. For the abscissa I would prefer either “Rockwell C Hardness” or “Hardness (RC) and for the ordinate I would prefer “Ultimate Tensile Strength (MPa)”. 5. A steel specimen has a BHN of 200. The TA and student got this problem correct. I included this problem to remind you that almost everything we do in this course is only approximately correct. Different methods can give different answers. However, to verify your knowledge of the methods, if I tell you how I want something done with one method on tests and homework, I want it done only that way. I will only include the TA’s solution because the student had the same: (a) What is the UTS in psi according to Table 2.1? UTS of steel which has a BHN of 200 93 ksi (b) What is the UTS in psi according to Equation 2.11? Eqn 2.11→ UTS (lb / in 2 ) = 500 BHN (kgf / mm 2 ) ∴ 500 ⋅ 200 = 100000 psi =100 ksi For Problem 6, the TA got the correct answer. He has a good approach. Always make a drawing first so you understand the problem. Also, be sure what the question is asking. It is not asking “how much potential energy does the hammer have after impact”. It is asking what the difference is between before and after. 6. A Charpy impact test is conducted according to Figure 2.5. Assume that the radius of the arm is 750 mm and the hammer has a mass of 2 kg. If the hammer is initially held horizontal before dropping, plot the energy (J) absorbed by the specimen as a function of the maximum angle of the hammer from the specimen after impact. Hammer: 2 kg 750 mm = 0.75 m ● PE0=mgh0 θ ● Specimen PE1=mgh1 m kg ⋅ m2 PE0 = mgh = 2 kg ⋅ 9.8 2 ⋅ 0.75 m = 14.7 = 14.7 J s s2 And the absorbed energy = PE0 ‐ PE1, The angle can be calculated with figure below, 0.75 − h1 ⇒ cos θ = → h1 = 0.75(1 − cos θ ) 0.75 0.75 0.75 – h1 m PE1 = mgh1 = 2 kg ⋅ 9.8 2 ⋅ h1 = 19.6 ⋅ 0.75(1 − cos θ ) θ s ∴ PE0 − PE1 = 14.7 − 19.6 ⋅ 0.75(1 − cos θ ) = Absorbed Energy h1 * You may need to check between radian and degree when you plot it The student made a good attempt, but made an error at the start that threw everything off, so I’ll not include it here. ...
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This note was uploaded on 09/06/2011 for the course EML 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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