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Unformatted text preview: ?.9 Determine the critical crack length for a through crack contained within a thick plate of
?0?5T?51 aluminum alloy that is under uniaxial tension. For this alloych = 22.0 ksiqgin. and (If = 82.0 ksi. Assume Y= 1 [Ex 1 [22.0 ksiJin. Ir Yaf «5(321) ksi) 2
a = 1 = 0.0073 in. a"
For an internal through crack. the critical length is rte = 20 = 0.015 in. 110 What is the largest size (mm) internal through crack that a thick plate of ahuniniun allot,r TOTET651 can support at an applied stress of (a) threequarters of the yield strength and
{b} onehalf of the yield strength?} Assume I' = 1. From Table 5.1, for ?0?5T651,KK = 24.2 MPaJE and mm = 495 MPa. (a) Given J; = germ, we calculate, {If =i(495 MPa} = 37"1525 MPa. Assuming I": l, the crack length is, .121 ii =1 21.35xio3m=1.351mn
.rr ref 2? {1)(3TLZSMPa) Thus, for an internal through crack, ac = 2:: = 2.?0 mm {'5'} Given [If =%Uh =él£495 Puma): 24?.5 MPa, and assmning f: l__ . ' _ ,1 _
a:l zl m 23.04X10_"111=3.04nun
If I {If If (1)0415 MPa) Thus, for an internal through crack, ac = 2:: = 5.03 mm. Chapter 7, Problem 19 A fatigue test is made with a maximum stress of 25 ksi (1T2 MPa) and a minimum stress of 4.00 ksi (216 MPa).
Calculate (a) the stress range. (:5) the stress amplitude. (C) the mean stress. and (d) the stress ratio. Chapter 7, Solution 19
(a) The range of stress is 0",, = O'max —O'mm = 25 kSi — (—4 kSi) = 29 ksi (199.8 MP3). (b) The stress amplitude is O'a = =14.5 kSi [99.9 NIPEI).
0' +0'  25 k I—4k I .
(c) The mean stress is O'm — max f) mm — S: S] —'l 0.5 k8] (7’23 NIPEI).
 —4 k I
(d) The stress ratio is R = an“ = = —0.16.
0'mam 25 k3] Chapter 7, Problem 21 Describe the four basic structural changes that take place when a homogeneous ductile metal is caused to fail by
fatigue under cyclic stresses. Chapter 7, Solution 21 For a homogeneous ductile metal subjected to fatigue. four stages of structural changes are observed: 1.
2. {2.1 Crack initiation: Plastic deformation causes the onset and early development of fatigue damage. Siipbrind crack grout}! (Stage I): Slipband intrusions and extrusions are created on the surface of the metal
while damage along persistent slipbands occurs within the sample. As a result. cracks form at or near the
surface and propagate along planes subjected to high shear stresses. Crack growth on planes ofliig}? tensile stress iSmge H): The slow crack growth of Stage I is replaced by
rapid crack propagation as the crack direction shifts to a direction perpendicular to the direction of
maximum tensile stress. During this stage. striations are formed. Uiiiinaie dnciiiefailni'e: The crack achieves an area sufficient to cause the rupture of the sample by ductile fracture. Chapter 7, Problem 22 Describe the four major factors that affect the fatigue strength of a metal. Chapter 7, Solution 22 The four primary factors which affect the fatigue strength of a metal are: stress concentration. surface roughness.
surface condition and environment. Chapter 7, Problem 24 Same as Prob. T23 with tensile stress of T0 MPa. If the initial and critical crack lengths are 1.25 111.111 and 12 111.111 in
the plate and the fatigue life is 2.0 X 106 cycles. calculate the maximum tensile stress in MPa that will produce this
life. Assume m = 3.0 andA = 6.0 >< 10'13 in MP2: and 111ete1‘units. Assume Y= 1.20. Chapter ?, Solution 24 The maximum tensile stress is calculated as: a?” 2)+1 _ as (m 2)+1
0m — H.) mi? 7??? r
[—(HH H) +1]A:r } hf
3 3
0 012 111)—E+1— (0 0012< m)—3+1
03: 3( ' ' " =3.318x106
[—EH](6.0x10_13)(2rl'3)(1.2)3(2.0x'106) 0' = 149 NIPa Chapter 7, Problem 31 Equiaxed MARM 24? alloy is to support a stress of 2% MPa (Fig. 7'31). Determine the time to stress rupture at
850°C. Lalrsen—Mitier parameter. P = ITI°CJ + 273] I20 + IogirH >< I'D"1
22.2 23.3 24.4 25.5 26.7 27,8 28.9 30 3.l {190  l  l  IOU D5 MARM l4? hmgiludinul Ml‘li. ._ I2: Pm moroxmrcrs MAC. _ _ an
I 8? PUEU ha'AC. LR mm diameter production data;
4 e0 552 ' \ . 4 I4 : ‘ 
[)5 CM 24? LC. longitudinal.
I232°CI2 h r I260°CI2IJ MAC
932°C? th. 8? FUJI: th
3.11mi] 4. I rum LliiﬂllL‘lL‘l' specimens ___ 276 . i . _... .. .. _.. . . .. _ 40 E '3
2 hquiuxul MARM 24? NH); g _ 982’05 MAC + 311°0th MAC 5
2 lot and STITIEE} MAC nnly. _I 30 g
{7} LBInm diameter specimens "7 [)5 CM 347 LC longitudinal.
: mam: h + lasmm MACK H!’")°CH WAC. 8711320 MAC um] 123.7."0‘ 2 h I IZﬁU"C(2(I WAC.
HiﬁilDC'Il‘rh 5A1". RT'IT‘J'ZI] MAP.
3.3 and 4. mm diameter specimens 13% 4U 42 44 46 48 50 52 54 5E)
lumen—Miller purarncler. P = ITIEDF) + 46m [20 + logml X 10—" Figure 7.31 Chapter 7, Solution 31
From Fig. 7.31, for a stress of 276 MPa, the LM. parameter value for Equiaxed MARM 247 is 26 X 103 Kh. Thus, P=T(K)(20+log tr), T=850°C +273 21123 K
26,000 = (1123 K)(20+ log 1;.)
log fr 23.152
fr =1419h Chapter 7, Problem 35
Gamma titanium aluminjde is subjected to a stress of 50,000 psi. If the material is to be limited to 0.2% creep strain, how long can this material be used at 1100°F‘? Use Fig. 7.32. 1000 .I I I l_ — TitnAlJV
‘ — — 'I'IfaAlZSIIJZFOMH _   TI—GAl—ESI'I4ZI6Mo—Si
— [MI 829
  IMI 834    Alpha 2 [THAI] General
' m  r 4 Gzlmmul’TiAllccncrul
 INCONEL‘HX El IMI 829 ROC E I00 A TI—lSAlIllNhJIV—JMOTHAI ROC
E
[I] ll] 1 _ _ _ _ _ _ ._ _ . _ 28 30 3?. 34 36 33 4D
Larsen—Miller parameter. P = (460‘? + T°Fl[20 + log r [hrll X “3'3 Figure 7.32 Chapter 7, Solution 35
From Fig. 7.32, for a stress of 50 ksi, the L.M. parameter for 2% stress is P 2: 36,600. Thus, P = T(R)(20 Hog r02), 1" = 1 10001: +460 =1560°R 36,600 = (1560°R)(20 + log rm)
to 2 = 2894.0 11 :x 121 days ...
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This homework help was uploaded on 03/20/2008 for the course MSE 200 taught by Professor Wholedepartment during the Spring '08 term at N.C. State.
 Spring '08
 wholedepartment

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