Questions on Material Selection and Failure Analysis | EMA 4714, Exams of Materials science

Download Questions on Material Selection and Failure Analysis | EMA 4714 and more Exams Materials science in PDF only on Docsity! Unit II Examination EMA 4714 - Materials Selection and Failure Analysis KEY Monday, April 7, 2003 I. Statistics and Probability [20 points] 1. [10 points] A steel [E = 30x106 psi] cantilever beam of length 10 ft is required to support a concentrated load at its free end whose mean value is 500 lb and whose standard deviation is 50 lb. If the cross section of the beam is rectangular with a 2 inch width, how thick should the beam be so that there is a 99.9% probability that the deflection at the free end is less than 3 inches? ************************************************************************* Assume a normally distributed population of applied loads whose mean [µ] is 500 lbs with a standard deviation [σ] of 50 lbs. Let x be a load which produces a beam deflection of 3 inches. The standard normal variable, z, for 99.9% [0.999] of applied loads producing beam deflections of <3 inches is +3.090 and equals [x-µ]/σ; x = 654.5 lbs. For δ = FL3/3EI, where I = bt3, t3 = [4*654.5*(10*12)3/[3*2*30x106] = 25.13, t = 2.93 inches 2. [10 points] The mean and standard deviation for the lifetime [in hours] of computer hard drives A and B are A(bar) = 40,000 [σA = 6000] and B(bar) = 50,000 [σB = 8000]. Which drive is more likely to last at least 30,000 hours? Which one is more likely to last at least 60,000 hours? ************************************************************************* For PA(=30000), z = [30000-40000]/6000 = -1.6667 or 0.0475, 4.75% fail, 95.25% pass For PB(=30000), z = [30000-50000]/8000 = -2.5 or 0.0072, 0.72% fail, 99.38% pass, B better. For PA(=60000), z = [60000-40000]/6000 = 3.333 or 0.99957, 99.957% fail, 0.043% pass For PB(=60000), z = [60000-50000]/8000 = 1.25 or 0.8944, 89.44% fail, 10.56% pass, B better. II. Engineering Economics [10 points] Your client purchased a mainframe computer three years ago for $30,000. Now they are considering expanding their computing capability by either buying a new computer or adding components to the existing one. Additional components can be purchased for the present computer at a cost of $10,000. You estimate that the improved computer will have a life of 5 years with an annual operations, maintenance and repair [OMR] cost of $2500. It is not expected to have any salvage value at the end of the period. A second option is to buy a new ABC computer at a cost of $35,000. This unit will have an expected life of $15 years, an annual OMR cost of $1500, and a salvage value of $5000 [FW]. Another option is to purchase a new XYZ computer for $27,000. For this computer, the expected life is 10 years, with an annual OMR cost of $2000, and a salvage value of $3000. The existing computer has a trade-in value of $5000 if either of the new computers is purchased. All three computers have the same computational capacity. Assuming an interest rate of 10%, which alternative would you advise your client to choose? State all assumptions and conditions used in performing your analysis. *************************************************************************** Ignore $30,000 cost of mainframe computer - without upgrades, does it have value? Not for present owner, that is clear, but for someone else? Anyway, I would let this go..... PW(1) = -10.0-2.5(P/A,10,5) = -10-2.5(3.79079) = -$19.5k A(1) = -19.5(A/P,10,5) = -19.5/(P/A,10,5) = -19.5/3.79079 = -$5.14k PW(2) = -(35.0-5.00)-1.5(P/A,10,15)+5.0(P/F,10,15) = -30.0-1.5(7.60608)+5.0(0.23939) = -$40.2 A(2) = -40.2(A/P,10,15) = -40.2/(P/A,10,15) = -40.2/7.60608 = -$5.29k PW(3) = -(27.0-5.0)-2.0(P/A,10,10)+3.0(P/F,10,10) = -22.0-2.0(6.14457)+3.0(0.388554) = = -$33.1k A(3) = -33.1(A/P,10,10) = -33.1/(P/A,10,10) = -33.1/6.14457 = -$5.39 Option 1 best