University of Portland School of Engineering ME Failure Analysis Assignment 7, Fall 20 I0

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1 University of Portland School of Engineering ME 421 Failure Analysis Assignment 7 Fall 20 0 Case Study Read the Mars Climate Orbiter article (link on course web page) and the Deepwater Horizon NY Time article (link below) to gain some background as to the root cause of these accidents. n about onehalf page explain similarities between the accidents of Space Shuttle Columbia Mars Climate Orbiter and Deepwater Horizon. Discuss general lessons learned (decision "root cause" not technical details). Link to the NY Times article on the Deepwater Horizon and be sure to watch the 2 minute video clip BEFORE reading the article (it will help you understand the article): http: // /us/2blowout.html?pagewanted= 1& r= 1 Fracture Mechanics and Fatigue. Assume fatigue loading Smax= 20ksi and Smin = O. How many cycles are required for the following cracks to grow J!4 inch longer than they start out at and how many cycles are required for them"both to grow to a final length of inch? Assume there is an edgecrack and the plate is very wide (small a/w). Use the Paris Law with n = 4 and C = 1X0 8. Note that this will be very low cycle fatigue and Paris Law is not really valid. Using a spreadsheet plot crack length verses number of cycles for an edge crack for both (a) and (b) (on the same graph). a) original crack length of 0.10 inch b) original crack length of 0.25 inch c) Repeat the above except use integration rather than numerical method. Compare the answers. 2. n order to solve future problems (such as problem 3 below) you need to establish an equation governing the steady crack growth. You receive a sample of the same steel that failed in Problem 3 and conduct a test. The test parameters are : a) constant sinusoidal stress amplitude b) Smin is 3ksi and Smax is S ksi c) center cracked panel; W= 6 inches; B = 0.1 inches; yield stress = 48 ksi d) K1c = 80 ksi root(inch) The following are the results showing crack length (a) after a given number of cycles (N): a = 0.05 N=O a = 0.2 N=24000

2 a = 0.4 N=54000 a = 0.7 N=68000 a = 1.0 N =74000 a = 2.0 N=77000 determine the Paris Law equation for this material. Hint create a graph (Excel) of logfda/dn) vs. log (L.K). Throwaway any data that does not fit on the straight line (i.e. those that don't obey Paris Law). Fit a straight line through the remaining points. "n" is the slope of the line 10g(C) is the yintercept. Hint you should get values of n=4 and C = 3.21XO'O 3. You have received a fractured item from a field failure. You see that the primary mode of failure was high cycle fatigue which originated from a sharp notch 0.5 inches deep ( 1/2 inch edge crack perhaps a quench crack). This can be considered to be an edgecrack in a large wide plate. Using an SEM (scanning electron microscope) you measure the fatigue striations near the starter crack to be about 8X 0 6 inches/cycle. You know that the stress ratio R was zero and the stress amplitude was nearly constant. Determine the peak cyclic stress. You must complete Problem 2 before solving this problem. 4. The following fatigue test data was acquired for nconel X750 at room temperature (25 C). L1K = 18 MPa rootmeter da/dn = 9XO 6 mm/cycle L1K = 50 MPa rootmeter da/dn = XO3 mrnlcycle An SEM image of the fracture surface of nconel X750 tested at 650 C is shown below. The loading conditions for this specimen was 35 MPa rootmeter. Does the fatigue rate for this material increase decrease or stay the same at 650 C? llof'jl.

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4 z../ ++'::...L..=_ s: 0.) "1C a.: 0 c.3 rj = 21' Z 19 U J.J=Z5/21 O=.3'('{ s: _= 0 _'Z.S T:=. 0 o V=. 2 'f'7 { z. ( 7 7 _z.es 7<S 1.00 AJ = L.53' 2 t l( = 7 +' C:O O.5 S.: O. ( 0... { :1b) 4 ::: O. 2S C\.:.C>. "A./= 8'0 6 =0. "Z...;S = ( b V.z: ZO

'P... c:)j. Given: a delta a delta K C(dK)/l n delta N Ntot Smax 20 ksi 0.01 0 Smin 0 0.02 0.01 6 1.00E05 9.96E+02 996 n 4 0.03 0.01 7 2.26E05 4.43E+02 1439 C 1.00E08 0.04 0.01 8 4.01E05 2.

5 'P... c:)j. Given: a delta a delta K C(dK)/l n delta N Ntot Smax 20 ksi Smin E E n E E C 1.00E E E edge crack small alw E E Find: number of cycles to grow 0.25" more and number of cycles to reach acrit E E E E E E E E E E part a) aortq E E part b) aoriq E E E E E E E E E E E E Ntot E E E E E E r E E+OO E E E E E E E E E E E E E E E E E E j i E E E E E E E E E E E E E E E E E E

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7 L Co 0 'Pa 111.1'&0 1../ '" \..L..'\ CL a; il 11 1 ' 1 ; ACL 5 Sol fs D lo 6.J \/..3 de (4') +: <".:5 sf " er.6s. ""d :..JoS kclil:u l:. «rla... P. ill \oj S e.1s r...j.s.f...nc..:_k:j (1... \ " e: L.f 6 /l. _ 0/9 \1\== Lf. O '«e):: {.'{9"i ) / 0 r= s. Zc./o J J :

8 Prob2 a N da dn da/dn ave a dk log(daldn) fg(dk) log(da/dn) o E E daldn vs delta K o : 3 "C j ' Doesn't "fit'" < l 6..1 _ log(delta K) "linear" data points only i 0.00 O.Do C "C OO 6.00 log (delta K) Y '" x /./! Page 1

9 3. 6' "'«0 ;; L( c.= 3. Z K/O..b?t..s L... = 0 S"......L J..J r.. 1 " " " / \... 1"'"1..:: 0" ru / c.;:r... "d1j ::: f2=ch r:..j. < (' S)"'"=O). S Ass «.»?C&f... L Ve4 6eP _. \ LtFfv\ C <6. ol...j. J] V"") 0 0 (.!J.. loxv LiJ 5 ) )(T t.91'. r41l L L o ( )(. 0 C7' u. Tv... CCA ) 104 ( )......" ) b). ' f) = 1 (3.z. 'G) + Ll lo! ( AlL)...J". 10.) (At ) (; ) \ 0"..1 l'.. Ll({ /7. b e:: J.cr'1[)... A z:... 0 s. 7 )c.s.l

$"J r (\... 1 sa oj RcslU..l) K: J...O V = 9oC. A 1<.: S6 lu:: )C/O. 1'.. 'S %yc.l "V".. ci :.J. /.J G5C)c:. +L SO c J 1 t.$

10 "J r (\... 1 sa oj RcslU..l) K: J...O V = 9oC. A 1<.: S6 lu:: )C/O. 1'.. 'S %yc.l "V".. ci :.J. /.J G5C)c:. +L SO c J 1 t. _ (tl +t>::67:; t==='o 1f1J EnccA./ pfcr1 fo(' AK.= 35 HPJ:l L 7""""%)?:> L; \u.. w / r2 T el..:: X/D "f c. 3 )<ofb 14% 4=.:..:::k;;)_' o

11 RT Delta K da/dn (mm/cyc) log(delta K) log(da/dn) E E E Prob _ O. JO // r 5.00 j..: 6.00 ' log (delta K)

A L A BA M A L A W R E V IE W

A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N