EMA 4714 - Materials Selection and Failure Analysis Unit 2 Examination

Wednesday, April 9, 2008 Name _______________________

Open Book/Open Notes

1. [ 5 points] (a) In order to test the strength of a ceramic insulator, cylindrical specimens of length 25 mm and diameter 5 mm are placed in axial tension. The tensile stress s that causes 50% of the specimens to fracture is 120 MPa. Cylindrical ceramic components of length 50 mm and diameter 10 mm are required to withstand an axial stress s_c with a survival probability of 99%.

Given that m = 5, determine s_t.

Change Volume: s₂ = (V₁/V₂)^1/m s₁ = ( 1/ 8) ^0.2 (120 MPa) = 79 MPa

Change Probability: s₃ = [(ln P_f3) / (ln P_f2) ]^1/m s₂ = [ - 10^-6 / - 0.693] ^0.2 (79 MPa) = 5.4 MPa

[5 points] (b) What would be the expected mean strength if the larger cylinders were loaded in three-point flexure for the same P_f = 99%? In other words, how would this new request change the calculations above? [Show any work for partial credit.]

You should expect the mean strength value to increase by a factor of [2(m+1)²]^1/m.

s _{3 pt flex} = s _tension [2(m+1) ²] ^1/m = 5.4 MPa [ (72)^0.2] = 5.4 MPa * 2.4 = 12.7 MPa

2. [10 points] A company is faced with the decision of selecting injection molding machines manufactured by two different suppliers - both perform the same function equally well. Machine A costs $1400 with annual maintenance and operating costs of $200 for the first 10 years and $250 for the following 10 years. Machine B costs $2000 and requires annual operating and maintenance costs of $100 for the entire 20 year period. Assuming both machines to have zero salvage values at the end of their economic lives, by how much will the manufacturer of machine A have to adjust his purchase price in order for it to be competitive with machine B at a discount rate of 15%?

PV method

PV(A) = -$1400 - $200[P/A,15,10] -$250[F/A,15,10][P/F,15,10]

= -$1400 - $200*5.0188 - $250*5.0188*0.2472

= -$2713.92

PV(B) = -$2000 - $100[P/A,15,20] = -$2000 - $100*6.2593 = - $2625.93

3. [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 100 lb. If the cross section of the beam is rectangular with a 2” width, how deep should the beam be so that there is a 99.9% probability that its deflection at the free end is less than 2”?

Assume a normally distributed population of applied loads whose mean [m] is 500 lbs with

a standard deviation [s] of 100 lbs. Let x be a load which produces a beam deflection of 2

inches. The standard normal variable, z, for 99.9% [0.999] of applied loads producing

beam deflections of <2 inches is +3.090 and equals [x-m]/s; x = 809 lbs.

For d =FL³/3EI (cantilever beam), where I = bt³ / 12, t³ = [12 *809*(10*12)³/[3*2*30x10⁶ *2] = 46.6 in³, t = 3.6 inches

4. The last NASA Discovery mission discovered small pieces of metal on a stanchion for a solar panel on the space station. A bearing assembly moves “up and down” on this stanchion. The small particles were approximately 1 – 250 microns in size. It was discovered that the pieces were composed of a nitrided steel. The flakes were cracking within the nitrided layer in a shape reminiscent of cyclic fatigue.

a) [5 points] How would you go about determining the cause(s) of the damage? What characterization instruments would you suggest using? Why? What properties would you characterize?

Scanning Electron Microscopy can determine size. EDAX determines chemical content. Metallurgic analysis to examine microstructure. Microhardness indentations to find hardness. These methods would give a good idea of the composition of the debris. There are limited possibilities (there were four potential steel alloys present) and these can be distinguished by composition. Polishing striations could also be used by comparison with known material around stanchion.

b) [ 5 points] Who could be held liable for the damage? Why? Should there be a warranty for the product? If so, what should be covered? If not, why not?

The manufacturer of the part from which the particles are identified could be held liable if it is shown they did not meet specs and it was a composition difference or a heat treatment difference that caused unexpected behavior. Since it is a one-time product, there usually is not a warranty. The hold the government has over the manufacturer is that if they do not business with them for an extended period of time because the manufacturer does not repair the damage without additional cost, then there is a large loss of revenue for the manufacturer. That restriction is more important than a warranty.