Jordan University of Science and Technology
Engineering collage
Aeronautical Engineering Department
Solution Manual for: Material Science and and Engineering / 8th 8t h edition editio n
Ahmed Ahm ed Mustafa Mustafa El-Kha El-Khalili lili
CHAPTER 8
FAILURE
PROBLEM PROBLEM SOLUTIONS
Principl Principlee s of Fractur Fracturee Me chanics chanics
8.1 Wha t is th e magn magn itude of the maxim maximum um stress that exists exists at the tip of an intern al cra ck ha ving a r adiu s of curvature of 2.5
´
-4
-5
10 mm (10 in.) and a crack length of 2.5
´
-2
-3
10 mm (10 in.) when a tensile stress of 170
MPa (25,000 psi) is applied? Solution This problem asks t hat we compute the magnitude of th e max maxim imum um stres s th at exists exists at t he tip of an internal crack. Equation 8.1 is employed employed t o so lve this prob lem, lem, as
æ a ö1/ 2 sm = 2s0 ç ÷ è r t ø 1/2 é 2.5 2.5 ´ 10-2 mm ù ê ú 2 ú = 2404 = (2)( (2)(17 1700 MPa)ê 2404 MPa (354, (354, 000 psi) 2.5 ´ 10-4 mm ú ê 2.5 êë ûú
8.2 Estimate Estimate the theoretica l fra fra cture strength of a br ittle mater mater ial if it is known tha t fra fra cture occurs by the propagation p ropagation of an elliptically shaped surface crack of length 0.25 mm (0.01 in.) and having a tip radius of -3
-5
curvature of 1.2 1.2 ´ 10 mm (4.7 ´ 10 in.) when a stress of 1200 MPa (174,000 psi) is a pplied. Solution In order to estimate the theoretical fracture strength of this material it is necessary to calculate sm using Equation 8.1 8.1 given th at s0 = 1200 MPa, a = 0.25 mm, and r t = 1.2 ´ 10-3 mm. Thus Th us ,
æ a ö1/ 2 sm = 2s0 ç ÷ è r t ø 1/2 é 0.25 0.25 mm ù = (2)( (2)(12 1200 00 MPa)ê = 3.5 ´ 10 4 MP MPa = 35 GPa GPa ( 5.1 ´ 10 6 psi) -3 mm ú 1.2 1. 2 10 ´ ë û
2
8.3 If the specific surface energy for soda-lime soda-lime glass is 0.30 J/m , using data contained in Table 12.5, compute compute the critica l stress required for the pro pag atio n of a surface surface crack of length 0.05 mm. mm. Solution We may determine the critical stress required for the propagation of an surface crack in soda-lime glass XVLQJ ( TXD TXDW LRQ
W DNLQJ W KHYDO XHRI
* 3 D Ta ble 12.5 12.5)) as th e mod mod ulus lu s o f elas la s ticit ic ity y , we g et
é 2E g s ù1/ 2 sc = ê ú ë pa û
é ù1/ 2 9 N / m2 ) (0.30 ´ ( 2) 69 10 0. 30 N/m) ( ú = 16.2 ´ 10 6 N/m2 = 16.2 MPa =ê 3 ê ú (p) 0.05 0.05 ´ 10 m ë û
(
)
8.4 A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. 2
Determine the maximum allowable surface crack length if the the surface ener gy of polystyrene is 0.50 J /m (2.86 ´ 10 3
2
-
6
in.-lb f /in. ). Assume ssume a modulus of elasticit y of 3.0 GPa ( 0.435 ´ 10 psi). Solution 7 KH KHP P D[ LP XP DO O RZ DEO H VXUIDFH FUDFN O HQJ W K IRUS RUSRO\ VW \ UHQH QHP P D\ EH EHGHW HUP LQHG XVLQJ ( TXDW LRQ
3.0 3.0 GPa GPa as the modulus of elasticity, elasticity, and so lving lving for a , leads to
a
=
2 E g s (2) (3 ´ 109 N/m2 ) (0.50 0.50 N/m) N/m) = 2 2 p sc (p) (1.25 1.25 ´ 106 N/m2 ) = 6.1 ´ 10-4 m = 0.61 mm mm (0.024 in.) in. )
W DNLQJ
8.5 A specimen of a 434 0 steel a lloy having a plan e stra in fra cture tough ness of 45 MPa m ( 41 ksi in. ) is exposed to a stress of 1000 MPa ( 145,000 psi) . Will this specimen experie nce fra cture if it is known tha t the la rgest surface cra ck is 0.75 mm (0.03 in.) lon g? Why or why not? Assume that the pa ra meter Yha s a va lue of 1.0. Solution This problem asks u s to determine whether or no t the 4340 steel alloy specimen will fracture when expos ed to a stress of 1000 MPa, given the values of K Ic, Y, and the largest value of a in the material. This requires that we so lve for sc from Equation 8.6. Thus
sc =
K Ic Y
pa
=
45 MPa m (1.0) (p)(0.75 ´ 10-3 m)
= 927 MPa (133, 500 psi)
Therefore, fracture will most likely occur because this specimen will tolerate a stress of 927 MPa (133,500 psi) before fracture, which is less than the applied stress of 1000 MPa (145,000 psi).
8.6 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m (31.9 ksi in.). It has been determined that fracture results at a stress of 250 MPa (36,250 psi) when the maximum (or cr itical) interna l cra ck length is 2.0 mm (0.08 in.). For this same component and allo y, will fra cture occur at a stress level of 325 MPa ( 47,125 psi) when the maximum interna l cra ck length is 1.0 mm (0.04 i n.)? Why or why not? Solution We are asked to determine if an aircraft component will fracture for a given fracture toughness (35
MPa m ), stress level (325 MPa), and maximum internal crack length (1.0 mm), given that fracture occurs for the same component u sing the s ame alloy for another s tress level and internal crack length . It first beco mes nec ess ary to solve for the parameter Y, us ing Equation 8.5, for the cond itions under which fracture occurred (i.e., s = 250 MPa an d 2a = 2.0 mm). Therefore,
Y
=
K Ic
s pa
Now we will s olve for th e p roduct Y s
=
pa
35 MPa m = 2.50 æ 2 ´ 10-3 m ö (250 MPa) (p) ç ÷ 2 è ø for the other set of conditions , so as t o ascertain whether or not this
value is g reater than the K Ic for the alloy. Thus ,
Ys
æ 1 ´ 10-3 m ö p a = (2.50)(325 MPa) (p) ç ÷ 2 è ø = 32.2 MPa m
Therefore, fracture will not occur since this value
(29.5 ksi in.)
(32.3 MPa
m ) is less than the K Ic of the material, 35 MPa m .
8.7 Suppose that a wing component on an air craft is fab ricated from an a luminum alloy that ha s a plane strain fra cture toug hness of 40 MPa m (36.4 ksi in.). It ha s been determined that fra cture results at a stress of 365 MPa (53,000 psi) when the maximum interna l cra ck length is 2.5 mm (0.10 in.). For t his same component a nd alloy, compute the stress level a t which fra cture will occur for a critical interna l cra ck length of 4.0 mm (0.16 in.). Solution This problem asks u s to determine the st ress level at which an a wing componen t on an aircraft will fracture for a given fracture toughness
(40 MPa
m
)
and maximum internal crack length (4.0 mm), given that fracture
occurs for the s ame component us ing the same alloy at o ne s tress level (365 MPa) and anot her internal crack length (2.5 mm). It first becomes nece ss ary to solve for the parameter Y for the conditions under which fracture occurred us ing Equation 8.5. Th erefore,
Y
=
K Ic
s pa
=
40 MPa m = 1.75 æ 2.5 ´ 10-3 m ö (365 MPa) (p ) ç ÷ 2 è ø
Now we will so lve for sc us ing Equation 8.6 as
sc =
K Ic Y
pa
=
40 MPa m = 288 MPa (41, 500 psi) æ 4 ´ 10-3 m ö (1.75) (p ) ç ÷ 2 è ø
8.8 A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa m ( 50 ksi in. ) . If, during service use, the plate is exposed to a tensile stress of 200 MPa (29,000 psi), determine the minimum length of a surface crack that will lea d to fra cture. Assume a value o f 1.0 for Y. Solution For this p roblem, we are given values of K Ic
(55 MPa m ) , s (200 MPa), and Y(1.0) for a large plate and are
asked to determine the minimum length o f a su rface crack that will lead to fracture. All we need do is to so lve fora c XVLQJ ( TXDW LRQ
W KHUHIRUH 2 2 1 æ K Ic ö 1 é 55 MPa m ù ac = ç ú = 0.024 m = 24 mm (0.95 in.) ÷ = ê p è Y s ø p ë (1.0)(200 MPa) û
8.9 Calcula te the maximum interna l cra ck length allowa ble for a 7075-T651 aluminum alloy (Table 8.1) component that is loaded to a stress one half of its yield strength. Assume that the value of Yis 1.35. Solution This problem asks us to calculate the maximum internal crack length allowable for the 7075-T651 aluminum alloy in Table 8.1 given th at it is loaded to a st ress level equal to one-half of its y ield s trength. For this alloy,
K Ic
= 24 MPa m
(22 ksi
in. ) ; also, s = sy/2 = (495 MPa)/2 = 248 MPa (36,000 psi). Now so lving for 2a c using
Equation 8.7 yields 2 2 2 æ K Ic ö 2 é 24 MPa m ù 2ac = ç ú = 0.0033 m = 3.3 mm (0.13 in.) ÷ = ê p è Ys ø p ë (1.35)(248 MPa) û
8.10 A structura l component in the form of a wide plate is to be fabr icated from a steel alloy tha t ha s a pla ne stra in fra cture tou ghn ess of 77.0 MPa m (70.1 ksi in.) and a yield strength of 1400 MPa (205,000 psi). The flaw size resolu tion limit of the fla w detectio n a ppa ra tus is 4.0 mm (0.16 in .). If the design stress is one ha lf of the yield strength an d the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection. Solution This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (4.0 mm), the value of K Ic
(77 MPa
m ) , the design stress (sy/2 in
which s y = 1400 MPa), and Y= 1.0. We first n eed to co mpute th e value of a c XVLQJ ( TXDW LRQ
ac
W KXV
é ù2 2 1 æ K ö 1 ê 77 MPa m úú = ç Ic ÷ = ê = 0.0039 m = 3.9 mm (0.15 in.) æ 1400 MPa ö ú p è Ys ø pê ÷ êë (1.0) çè ø úû 2
Therefore, the critical flaw is not subject to detection since this value of a c (3.9 mm) is less than the 4.0 mm resolution limit.
8.11
After consultation of other references, write a brief report on one or two nondestructive test
techniques that a re used to detect an d measure intern al a nd/or surface flaws in metal a lloys.
The student should do this problem on his/her own.
Impact Fracture Testing 8.12 Following is tab ulated d ata tha t were gathered from a series of Char py impact tests on a ductile cast iron. Temperature (°C)
Impact Energy (J)
– 2 5
124
– 50
123
– 7 5
115
– 8 5
100
– 1 00
73
– 1 10
52
– 1 25
26
– 1 50
9
– 1 75
6
(a ) Plot the data as impact energy versus tempera ture. (b )
Determine a ductile-to-brittle transition temperature as that temperature corresponding to the
a verag e of the maximum a nd minimum impa ct ener gies. (c) Determine a ductile-to-brittle tra nsition tempera ture a s that temperature at which the impact energy is 80 J. Solution (a) The plot of impact energy v ersus temperature is sho wn below.
(b) The av erage o f the maximum and minimum impact energies from the data is
Average =
124 J + 6 J = 65 J 2
As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this criterion is about – 105°C. (c) Also, as n oted o n the p lot by the o ther set of dash ed lines , the ductile-to-brittle transition temperature for an impact en ergy of 80 J is a bout – 95°C.
8.13 Following is tabulated data that were gathered from a series of Charpy impact tests on a tempered 4140 steel alloy. Temperature (°C) 100 75 50 25 0 – 25 – 50 – 65 – 75 – 85 – 100 – 125 – 150 – 175
Impact Energy (J) 89.3 88.6 87.6 85.4 82.9 78.9 73.1 66.0 59.3 47.9 34.3 29.3 27.1 25.0
(a ) Plot the da ta a s impact energ y versus tempera ture. (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the a verag e of the maximum a nd minimum impa ct ener gies. (c) Determine a ductile-to-brittle tra nsition tempera ture a s that temperature a t which the impa ct energy is 70 J. Solution The plot of impact ene rgy versus temperature is sh own below.
(b) The av erage o f the maximum and minimum impact energies from the data is
Average =
89.3 J + 25 J = 57.2 J 2
As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this criterion is about – 75°C. (c) Also, as n oted o n the p lot by the o ther set of dash ed lines , the ductile-to-brittle transition temperature for an impact en ergy of 70 J is a bout – 55°C.
Cyclic Stresses (Fatigue) The S-N Curve
8.14 A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi). (a ) Compute the ma ximum and minimum stress levels. (b) Compute the stress ratio . (c) Compute the magnitude of the stress range. Solution (a) Given the values of
sm (50
MPa) and
sa
(225 MPa) we are asked to compute smax and smin . From
Equation 8.14
sm =
s max + s min = 50 MPa 2
Or,
smax + smin = 100 MPa Furthermore, utilization of Equation 8.16 yields
sa =
s max - s min = 225 MPa 2
Or,
smax – smin = 450 MPa Simultaneous ly solving these t wo express ions leads to
s max = 275 MPa (40, 000 psi) s min = - 175 MPa (-25, 500 psi) (b) Using Equation 8.17 the stress ratio R is determined as follows:
R =
s min -175 MPa = = - 0.64 s max 275 MPa
(c) The magnitude of the stress range sr is d etermined us ing Equation 8.15 as
s r = s max - s min = 275 MPa - (-175 MPa) = 450 MPa (65, 500 psi)
8.15 A cylindrical 1045 steel bar (Figure 8.34) is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 22,000 N (4950 lb f ), compute the minimum allowable bar diameter to ensure tha t fa tigu e fa ilur e will not occu r. Assume a factor of safety of 2.0. Solution From Figure 8.34, the fatigue limit s tres s amplitud e for this alloy is 310 MPa (45,000 psi). Stress is defined in F Equation 6.1 as s = . For a cylindrical bar A0
æ d0 ö2 A0 = p ç ÷ è 2 ø Subst itution for A0 into the Equation 6.1 leads to
s =
F A0
=
F
æ d0 ö2 p ç ÷ è 2 ø
=
4F pd02
We no w solve for d 0, taking stress as th e fatigue limit divided by th e factor of safety. Thus
d0
=
=
4F æ s ö pç ÷ è N ø
(4)(22,000 N) = 13.4 ´ 10-3 m = 13.4 mm (0.53 in.) æ 310 ´ 10 6 N / m2 ö (p) ç ÷ 2 è ø
8.16 An 8.0 mm (0.31 in.) diameter cylindrical rod fabricated from a red brass alloy (Figure 8.34) is subjected to reversed tensio n-compression l oa d cycling a long i ts axis. If the ma ximum tensile and compressive loads are +7500 N (1700 lb f ) and -7500 N (-1700 lb f ), r espectively, determine its fat igue life. Assume tha t the stress plotted in F igure 8.34 is stress amplitude. Solution We are asked to determine the fatigue life for a cylindrical red brass rod given its diameter (8.0 mm) and the maximum tensile and compressive loads (+7500 N and -7500 N, resp ectively). The first thing th at is neces sary is to calculate values of smax and smin us ing Equation 6.1. Thus
s max =
=
7500 N
æ 8.0 ´ 10-3 m ö2 (p) ç ÷ 2 è ø
Fmax A0
-7500 N æ 8.0 ´ 10-3 m ö2
(p) ç è
2
Fmax
æ d0 ö2 pç ÷ è 2 ø
= 150 ´ 10 6 N/m2 = 150 MPa (22, 500 psi)
s min =
=
=
Fmin
æ d ö2 pç 0 ÷ è 2 ø
= - 150 ´ 10 6 N/m2 = - 150 MPa (-22, 500 psi)
÷ ø
Now it becomes necessa ry to comput e the s tress amplitude us ing Equation 8.16 as
sa =
s max - s min 150 MPa - (-150 MPa) = = 150 MPa (22, 500 psi) 2 2
From Figure 8.34, f for the red brass , the nu mber of cycles t o failure at th is s tress amplitude is abou t 1 ´ 105 cycles.
8.17 A 12.5 mm (0.50 in.) diameter cylindrical rod fabricated from a 2014-T6 alloy (Figure 8.34) is subjected to a rep ea ted tension -compression lo a d cycling alo ng its axis. Compute the ma ximum a nd minimum load s that will be ap plied to yield a fatig ue life of 1.0
´
7
10 cycles. Assume tha t the stress plott ed on the vertica l
axis is stress amplitude, and da ta were taken for a mean stress of 50 MPa (7250 psi). Solution This problem asks that we compute the maximum and minimum loads to which a 12.5 mm (0.50 in.) diameter 2014-T6 aluminum alloy s pecimen may b e s ubjected in o rder to yield a fatigue life of 1.0 ´ 107 F\ FOHV ) LJ XUH
LVW R
be us ed as su ming that data were taken for a mean stres s of 50 MPa (7250 ps i). Upon con su ltation of Figure 8.34, a fatigue life of 1.0 ´ 107 cycles correspon ds to a s tress amplitude of 160 MPa (23,200 psi). Or, from Equation 8.16
s max - s min = 2s a = (2)(160 MPa) = 320 MPa (46, 400 psi) Since sm = 50 MPa, then from Equation 8.14
s max + s min = 2s m = (2)(50 MPa) = 100 MPa (14, 500 psi) Simultaneous so lution o f these t wo express ions for smax and smin yields
Now, inas much a s
Fmax
Fmin
s=
F A0
smax = +210 MPa
(+30,400 ps i)
smin = – 110 MPa
(– 16,000 ps i)
(Equation 6.1), and A0
æ d0 ö2 p = ç ÷ è 2 ø
then
s max p d 20 (210 ´ 10 6 N / m2 ) (p) (12.5 ´ 10-3 m) 2 = = = 25, 800 N (6000 lbf ) 4 4
s min p d 20 (-110 ´ 10 6 N / m2 ) (p) (12.5 ´ 10-3 m) 2 = = = - 13,500 N (-3140 lb f ) 4 4
8.18 The fatigu e da ta for a bra ss alloy ar e given as follows:
Stress Amplitude (MPa)
Cycles to Fa ilure
310
2 × 10
5
223
1 × 10
6
191
3 × 10
6
168
1 × 10
7
153
3 × 10
7
143
1 × 10
8
134
3 × 10
8
127
1 × 10
9
(a ) Make a n S–N plot (stress amplitude versus logarithm cycles to failur e) using these da ta. 5
(b) Determine the fatigu e strength a t 5 ´ 10 cycles. (c) Deter mine th e fa tigue l ife for 2 00 MPa . Solution (a) The fatigue data for this alloy are plotted b elow.
(b) As indicated by the “A” set o f dashed lines on the plot, the fatigue strength at 5 ´ 105 cycles [log (5 ´ 105) = 5.7] is ab out 250 MPa.
(c) As noted by th e “B” set of das hed lines, the fatigue life for 200 MPa is about 2 ´ 106 cycles (i.e., the log of th e lifetime is about 6.3).
8.19 Suppose that the fatigu e da ta for the b ra ss alloy in Pro blem 8.18 were taken from torsiona l tests, an d that a shaft of this alloy is to be used for a coupling that is attached to an electric motor operating at 1500 rpm. Give the maximum torsiona l stress a mplitu de po ssible for ea ch of the following lifetimes of the couplin g: (a ) 1 year , (b) 1 month, (c) 1 da y, an d (d) 2 hou rs. Solution For each lifetime, first compute t he nu mber of cycles, and th en read the co rresp onding fatigue s trength from the above plot. (a) Fatigue lifetime = (1 yr)(365 days /yr)(24 h/day )(60 min/h)(1500 cycles /min) = 7.9 ´ 108 cycles. The stress amplitude correspond ing to this lifetime is abou t 130 MPa. (b) Fatigue lifetime = (30 days)(24 h/day)(60 min/h)(1500 cycles/min) = 6.5
´
107 cycles.
The stress
amplitude corresponding to this lifetime is about 145 MPa. (c) Fatigue lifetime = (24 h)(60 min/h)(1500 cycles/min) = 2.2
´
106 cycles.
The stres s amplitude
105 cycles.
The stres s amplitude
correspond ing to this lifetime is about 195 MPa. (d) Fatigue lifetime = (2 h)(60 min/h)(1500 cycles/min) = 1.8 correspond ing to this lifetime is about 315 MPa.
´
8.20 The fatigu e da ta for a ductile ca st iron a re given as follows: Stress Amplitude [MPa (ksi)]
Cycles to Fa ilure
248 (36.0)
1 × 10
5
236 (34.2)
3 × 10
5
224 (32.5)
1 × 10
6
213 (30.9)
3 × 10
6
201 (29.1)
1 × 10
7
193 (28.0)
3 × 10
7
193 (28.0)
1 × 10
8
193 (28.0)
3 × 10
8
(a ) Make a n S–N plot (stress amplitude versus logarithm cycles to failu re) using th ese data . (b ) Wha t is the fat igue l imit for th is al loy? (c) Determine fatigu e lifetimes at stress amplitudes of 230 MPa ( 33,500 psi) a nd 17 5 MPa (25,000 psi). 5
(d ) Estimat e fa tigue strengths at 2 ´ 10 and 6
´
6
10 cycles.
Solution (a) The fatigue data for this alloy are plotted b elow.
(b) The fatigue limit is the s tress level at which th e curve becomes h orizont al, which is 193 MPa (28,000 psi).
(c) As noted by the “A” set of das hed lines, the fatigue lifetime at a stres s amplitude of 230 MPa is ab out 5
´
105 cycles (log N = 5.7). From the plot, th e fatigue lifetime at a st ress amplitude of 230 MPa (33,500 psi) is about
50,000 cycles (log N = 4.7). At 175 MPa (25,000 ps i) the fatigue lifetime is es sentially an infinite nu mber of cycles since th is s tress amplitude is below the fatigue limit. (d) As noted by the “B” set of dashed lines, the fatigue strength at 2 0 3D
´ 105 cycles (log N = 5.3) is about 240
SVL and according to the “C” set of das hed lines , the fatigue streng th at 6 ´ 106 cycles (log N = 6.78) is
about 205 MPa (30,000 psi).
8.21 Suppose that the fatig ue dat a for the cast iron in Problem 8.20 were taken for bending -rotating tests, and that a rod of this alloy is to be used for an automobile axle that rotates at an average rotational velocity of 750 revolutio ns per minut e. Give ma ximum lifetimes of contin uous driving tha t are a llowa ble for the following stress levels: (a) 250 MPa (36,250 psi), (b) 215 MPa (31,000 psi), (c) 200 MPa (29,000 psi), and (d) 150 MPa (21,750 psi). Solution For each stress level, first read the corresponding lifetime from the above plot, then convert it into the number of cycles. (a) For a stres s level of 250 MPa (36,250 ps i), the fatigu e lifetime is approximately 90,000 cycles. This translates into (9 ´ 104 cycles)(1 min/750 cycles) = 120 min. (b) For a s tres s level of 215 MPa (31,000 psi), the fatigue lifetime is approximately 2 ´ 106 cycles. This translates into (2 ´ 106 cycles)(1 min/750 cycles) = 2670 min = 44.4 h. (c) For a s tress level of 200 MPa (29,000 psi), the fatigue lifetime is ap proximately 1 ´ 107 cycles. This translates into (1 ´ 107 cycles)(1 min/750 cycles) = 1.33 ´ 104 min = 222 h. (d) For a stress level of 150 MPa (21,750 psi), the fatigue lifetime is essentially infinite since we are below the fatigu e limit [193 MPa (28,000 psi)].
8.22 Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one o f the maximum-minimum stress cycles listed b elow; the frequ ency is the same for al l thr ee tests.
Specimen
(MPa)
(MPa)
smax
smin
A
+ 450
– 3 50
B
+ 400
– 3 00
C
+340
– 3 40
(a ) Rank the fa tigu e lifetimes of these thr ee specimens from the longest t o the shor test. (b) Now justify this ranking using a schematic S–N plot. Solution In o rder to solve this problem, it is neces sary to co mpute b oth th e mean st ress a nd s tress amplitude for each sp ecimen. Since from Equation 8.14, mean s tres ses are the specimens are determined as follows:
sm =
s max + s min 2
s m (A) =
450 MPa + (-350 MPa) = 50 MPa 2
s m (B) =
400 MPa + (-300 MPa) = 50 MPa 2
s m (C ) =
340 MPa + (-340 MPa) = 0 MPa 2
Furthermore, us ing Equation 8.16, stress amplitudes are computed as
sa =
s max - s min 2
s a (A) =
450 MPa - (-350 MPa) = 400 MPa 2
s a (B) =
400 MPa - (-300 MPa) = 350 MPa 2
s a (C ) =
340 MPa - (-340 MPa) = 340 MPa 2
On the basis of these results, the fatigue lifetime for specimen C will be greater than specimen B, which in turn will be greater than sp ecimen A. This conclusion is bas ed upo n the following S-N plot on which curves are plotted for two s values. m
8.23 Cite five factor s tha t may lea d to scatter in fatig ue life da ta. Solution Five factors that lead to scat ter in fatigue life d ata are (1) specimen fabrication an d su rface preparation, (2) metallurgical variables, (3) specimen alignment in the test apparatus , (4) variation in mean st ress , and (5) variation in test cycle frequency.
Crack Initiation and Propagation Factors That Affect Fatigue Life
8.24 Briefly explain the difference between fatigue striations and beachmarks both in terms of (a) size and (b) origin. Solution (a) With regard to size, beach marks are normally of macroscop ic dimens ions and may be obs erved with the naked eye; fatigue s triations are of micros copic size and it is necess ary to obs erve them us ing electron micros copy. (b) With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation is correspond s to the adv ance of a fatigue crack during a s ingle load cycle.
8.25 List four measures tha t may be ta ken to increase th e resistance to fatigu e of a metal alloy. Solution Four measu res that may be taken to increase t he fatigue resistan ce of a metal alloy are: (1) Polish the su rface to remove s tress amplification sites . (2) Reduce the nu mber of internal defects (pores, etc.) by means of altering proces sing and fabrication techniques. (3) Modify the design to eliminate notches and s udden con tour changes . (4) Harden th e out er surface of the st ructure by cas e hardening (carburizing, nitriding) or shot peening.
Generalized Cre ep Be havior
8.26 Give the app roximate tempera ture at which creep deformatio n becomes an importa nt consideration for each of the following metals: nickel, copper, iron, tungsten, lead, and aluminum. Solution Creep becomes important at about 0.4Tm, Tm being th e abs olute melting t emperature of the metal. (The melting t emperatures in deg rees Celsius are found ins ide the front cov er of the bo ok.) For Ni, 0.4Tm = (0.4)(1455 + 273) = 691 K or 418°C (785°F) For Cu, 0.4Tm = (0.4)(1085 + 273) = 543 K or 270°C (518°F) For Fe, 0.4Tm = (0.4)(1538 + 273) = 725 K or 450°C (845°F) For W, 0.4Tm = (0.4)(3410 + 273) = 1473 K or 1200°C (2190°F) For Pb, 0.4Tm = (0.4)(327 + 273) = 240 K or -33°C (-27°F) For Al, 0.4Tm = (0.4)(660 + 273) = 373 K or 100°C (212°F)
8.27 The following creep data were taken on an aluminum alloy at 400°C (750°F) a nd a constant stress of 25 MPa (3660 psi). Plot the da ta as stra in versus time, then determine the steady-state or minimum creep rate. Note: The in itia l a nd insta nta neo us stra in is not included.
Time (min)
Strain
Time (min)
Strain
0
0.000
16
0.135
2
0.025
18
0.153
4
0.043
20
0.172
6
0.065
22
0.193
8
0.078
24
0.218
10
0.092
26
0.255
12
0.109
28
0.307
14
0.120
30
0.368
Solution These creep da ta are plotted below
The steady-state creep rate (De/Dt) is the slope of the linear region (i.e., the straight line that has been superimposed on the curve) as
0.230 - 0.09 De = = 7.0 ´ 10 -3 min -1 Dt 30 min - 10 min
Stress and Temperature Effects
8.28 A specimen 75 0 mm (30 in.) l ong of an S-590 a lloy (Figure 8 .31) is to be exposed to a tensile stress of 80 MPa (11,600 psi) at 815 °C (1500 °F). Determine its elonga tion after 5000 h. Assume that the total of both instanta neous a nd primary creep elonga tions is 1.5 mm (0.06 in.). Solution From the 815°C line in Figure 8.31, the steady state creep rate
es is about 5.5 ´
10-6 h -1 at 80 MPa. The
steady state creep s train, e , therefore, is just th e product of e s and time as s
es
= es x (time)
= (5.5 ´ 10-6 h -1) (5, 000 h) = 0.0275 Strain and elongatiRQDUHUHO DW HGDVLQ( TXDW LRQ
Dls
VROYLQJ IRUW KHVW HDG\ VW DW HHO RQJ DW LRQ Dls, leads to
= l0 es = (750 mm)
Finally, the total elongation is just the sum of this
(0.0275 ) = 20.6 Dls and
mm (0.81 in.)
the total of both instantaneous and primary creep
elongation s [i.e., 1.5 mm (0.06 in.)]. Therefore, the t otal elongation is 20.6 mm + 1.5 mm = 22.1 mm (0.87 in.).
8.29 For a cylindrical S-590 alloy specimen (Figure 8.31) originally 10 mm (0.40 in.) in diameter and 500 mm (20 in.) lo ng, what tensile loa d is necessary to produce a total elonga tion of 145 mm (5.7 in.) a fter 2,000 h at 730°C (1350°F)? Assume tha t the sum of instanta neous an d primary creep elonga tions is 8.6 mm (0.34 in.). Solution It is first necessary to calculate the steady state creep rate so that we may ut ilize Figure 8.31 in order to determine the tens ile stress . The stead y stat e elongation, Dls , is jus t the d ifference between the to tal elongat ion and the su m of the ins tantan eous an d primary creep elongations; that is,
Dls
= 145 mm - 8.6 mm = 136.4 mm (5.36 in.)
Now the stead y s tate creep rate, es is jus t
D ls es = .
De = Dt
l0 Dt
136.4 mm 500 mm = 2,000 h
= 1.36 ´ 10-4 h-1
Employing the 730°C line in Figure 8.31, a steady state creep rate of 1.36 ´ 10-4 h-1 corresponds to a stress s of about 200 MPa (or 29,000 ps i) [since log (1.36 ´ 10-4 ) = -3.866]. From this we may comput e the ten sile load using Equation 6.1 as
F =
= (200 ´
10 6 N/m2
sA0
=
æ d0 ö2 sp ç ÷ è 2 ø
æ 10.0 ´ 10-3 m ö2 ) (p ) ç ÷ = 15, 700 N (3645 lbf ) 2 è ø
8.30 If a component fabri ca ted from a n S-590 a lloy (Figure 8.30) is to be exposed to a tensile stress of 300 MPa ( 43,500 psi) at 650 °C (1200°F) , estimat e its rup ture l ifetime. Solution This problem asks us to calculate the rupt ure lifetime of a co mponen t fabricated from an S-590 alloy expos ed to a tensile stress of 300 MPa at 650°C. All that we need do is read from the 650°C line in Figure 8.30 the rupture O LIHW LP HDW
0 3D W KLVYDO XHLVDERXW
K
8.31 A cylindrical component constructed from an S-590 alloy (Figure 8.30) has a diameter of 12 mm (0.50 in.). Determine the maximum load tha t may be appl ied for it to survive 500 h a t 925°C ( 1700°F) . Solution We are asked in this problem to determine the maximum load that may be applied to a cylindrical S-590 alloy component that must survive 500 h at 925°C. From Figure 8.30, the stress corresponding to 500 h is abou t 50 MPa æ d ö2 (7,250 psi). Since stres s is defined in Equation 6.1 as s = F/ A0, and for a cylindrical specimen, A0 = p ç 0 ÷ , then
è 2 ø
F = sA0
= (50 ´
1 0 6 N/m2
æ d0 ö2 = sp ç ÷ è 2 ø
æ 12 ´ 10-3 m ö2 ) (p) ç ÷ = 5655 N (1424 lb f ) 2 è ø
8.32 From Equation 8.19, if the logarithm of es is plotted versus the loga rithm of σ, then a straight line should result, the slope of which is the stress exponent n. Using Figure 8.31, determine the value of n for the S-590 alloy at 925°C, and for the initial (i.e., lower-temperature) straight line segments at each of 650°C, 730°C, and 815°C. Solution The slope of the line from a log es versus log s plot yields the value of n LQ( TXDW LRQ
n=
W KDWLV
D log eÝs D log s
We are asked to determine the values of n for the creep data at the four temperatures in Figure 8.31 [i.e., at 925°C, and for the initial (i.e., lower-temperature) s traight line se gments at each of 650°C, 730°C, and 815°C].
This is
accomplished by taking ratios o f the differences b etween two log e s and log s values . (Note: Figure 8.31 plots log s versus log es ; therefore, values of n are equal to the reciprocals of the s lopes o f the st raight-line s egments.) Thus for 650°C
n=
D log eÝs log (10-1) - log (10-5 ) = 11.2 = D log s log (545 MPa) - log (240 MPa)
n=
D log eÝs log (1) - log (10-6 ) = 11.2 = D log s log (430 MPa) - log (125 MPa)
While for 730°C
And at 815°C
n=
D log eÝs log (1) - log (10-6 ) = 8.7 = D log s log (320 MPa) - log (65 MPa)
n=
log 10 2 - log (10-5 ) D log eÝs = 7.8 = D log s log (350 MPa) - log (44 MPa)
And, finally at 925°C
( )
8.33 (a ) Estimate the activation energy for creep (i.e., Qc in Equa tion 8.20) for the S-590 allo y having the steady-state creep behavior shown in Figure 8.31. Use data taken at a stress level of 300 MPa (43,500 psi) and temperatures of 650°C and 730°C . Assume that the stress exponent n is independent of temperature. (b) Estimate
es
at 600°C (873 K) and 300 MPa.
Solution (a) We are as ked to es timate the activation energy for creep for the S-590 alloy having the steady-state creep behavior shown in Figure 8.31, using data taken at s = 300 MPa and temperatures of 650°C and 730°C. Since s is a co ns tant, Equation 8.20 takes th e form
eÝs
= K 2s nexp
Q
ç- c ÷ = è RT ø
K 2' exp
Q
ç- c ÷ è RT ø
where K 2' is now a cons tant. (Note: the exponen t n has about the s ame value at the se t wo temperatures p er Problem 8.32.) Taking natural logarithms of the abov e expres sion
ln eÝs = ln K 2 '
Qc RT
For the case in which we have creep data at two temperatures (denoted as T and T ) and their corresponding steady-state creep rates ( es and 1
es
2
1
2
), it is pos sible to set up two simultaneous equations of the form as abo ve, with
two un knowns, namely K 2' and Qc. Solving for Qc yields
Qc =
-
R ç ln eÝs è 1
é1 ê êë T1
Let us choose T1 as 650°C (923 K) and T2 as 730°C (1
.
- ln eÝs ÷ 2 ø 1ù - ú T2 úû W KHQIURP ) LJ XUH
-5
DWs = 300 MPa, es = 8.9 ´ 10 1
-2
h -1 and es = 1.3 ´ 10 h-1. Subst itution of thes e values into the above equation leads to 2
[
]
(8.31 J / mol - K ) ln (8.9 ´ 10-5 ) - ln (1.3 ´ 10-2 ) Qc = é 1 1 ù êë 923 K - 1003 K úû
= 480,000 J/mol
(b) We are now asked to estimate es at 600°C (873 K) and 300 MPa. It is first n eces sary to determine the value of K 2' , which is acco mplished us ing the first express ion abov e, the value o f Qc, and one value each o f es and T (say es and T1). Thus, 1
K 2' = eÝs exp çç
Qc
÷
÷ è RT1 ø
1
é ù 480,000 J /mol = 8.9 ´ 10-5 h-1 expê = 1.34 ´ 10 23 h -1 ú ë (8.31 J /mol- K)(923 K) û
(
)
Now it is possible to calculate es at 600°C (873 K) and 300 MPa as follows:
eÝs
æ
Qc ö
è
RT ø
= K 2' exp ç -
÷
é ù 480,000 J/mol = 1.34 ´ 1023 h-1 expêë (8.31 J/mol- K)(873 K) úû
(
)
= 2.47 ´ 10-6 h -1
8.34 Steady-state creep rate da ta a re given below for nickel at 1000 °C (1273 K):
es (s – 1)
[MPa (psi)]
s
– 4
15 (2175)
– 6
4.5 (650)
10 10
If it is known that the activation energy for creep is 272,000 J/mol, compute the steady-state creep rate at a temperature of 850 °C (1123 K) a nd a stress level of 25 MPa ( 3625 psi). Solution Taking natural logarithms of b oth sides of Equation 8.20 yields
ln eÝs = ln K 2 + n ln s -
Qc RT
With the given data there are two unknowns in this equation--namely K 2 and n. Using the data provided in the problem st atement we can s et up two indep endent equations as follows:
(
)
272,000 J /mol (8.31 J/mol- K)(1273 K)
(
)
272,000 J /mol (8.31 J/mol - K)(1273 K)
ln 1 ´ 10-4 s-1 = ln K 2 + n ln (15 MPa) -
ln 1 ´ 10-6 s-1 = ln K 2 + n ln (4.5 MPa) -
Now, solving s imultaneous ly for n and K 2 leads to n = 3.825 and K 2 = 466 s -1 . Thus it is now pos sible to solve for es at 25 MPa and 1123 K using Equation 8.20 as
eÝs
= K 2s nexp
Q
ç- c ÷ è RT ø
é ù 272,000 J/mol = 466 s-1 (25 MPa)3.825exp êë (8.31 J/mol - K)(1123 K) úû
(
)
2.28 ´ 10-5 s -1
8.35 Steady-state creep data taken for a stainless steel at a stress level of 70 MPa (10,000 psi) are given a s follo ws:
es (s – 1)
T (K )
– 5
977
– 3
1089
1.0 × 10 2.5 × 10
If it is kno wn tha t the val ue of the stress exponent n for th is al loy is 7.0, compute t he stea dy-state creep rate a t 1250 K and a stress level of 50 MPa (7250 p si). Solution Taking natural logarithms of b oth sides of Equation 8.20 yields
ln eÝs = ln K 2 + n ln s -
Qc RT
With the given data there are two unknowns in this equation--namely K 2 and Qc. Using the data provided in the problem st atement we can s et up two indep endent equations as follows:
(
)
(
)
ln 1.0 ´ 10-5 s-1 = ln K 2 + (7.0) ln (70 MPa) -
ln 2.5 ´ 10-3 s-1 = ln K 2 + (7.0) ln (70 MPa) Now, so lving simultaneou sly for K 2 and Qc leads to K 2 = 2.55
´
Qc
(8.31 J/mol - K)(977 K) Qc
(8.31 J/mol - K)(1089 K)
105 s -1 and Q = 436,000 J/mol. Thu s, it is now c
pos sible to solve for es at 50 MPa and 1250 K using Equation 8.20 as
eÝs
= K 2s nexp
Q
ç- c ÷ è RT ø
436,000 J/mol = 2.55 ´ 10 5 s-1 (50 MPa) 7.0 expêë (8.31 J/mol - K)(1250 K ) úû
(
)
0.118 s -1
Alloys for High-Temperature Use
8.36 Cite thr ee metallu rgica l/processing techniques tha t ar e employed to enhance the creep resistance of metal allo ys. Solution Three metallurgical/proces sing tech niques that are employed to enh ance the creep resistance o f metal alloys are (1) solid solution alloying, (2) dispersion strengthening by using an insoluble second phase, and (3) increasing the g rain s ize or p roducing a grain s tructure with a preferred o rientation.
DESIGN PROBLEMS
8.D1 Each student (or group of students) is to obtain an object/structure/component that has failed. It may come from your home, an automobile repair shop, a machine shop, etc. Conduct an investigation to determine the cause an d type of failu re (i.e., simple fra cture, fatigu e, creep). In a ddition, propose measures tha t can be taken to pr event future incidents of this type of failu re. Fina lly, submit a report tha t a ddresses the a bove issues.
Each student or group of students is to submit their own report on a failure analysis investigation that was conducted.
Principles of Fracture Me chanics
8.D2 (a) For the thin-walled spherical tank discussed in Design Example 8.1, on the basis of critical crack size criterion [a s a ddressed in pa rt ( a) ] , ra nk the following polymers from longest to shortest critical crack length: nylon 6,6 (50% relative humidity), polycarbonate, poly(ethylene terephthalate), and poly(methyl methacrylate). Comment on the magnitude range of the computed values used in the ranking relative to those tabu lated for metal alloys a s provided in Table 8.3. For these computa tions, use data con tained in Tables B.4 and B.5 in Append ix B. (b) Now rank these same four po lymers relative to maximum allo wable pressure accor ding to the leak before- brea k cri terion , a s d escri bed in the (b) por tio n of Design Exa mple 8 .1. As ab ove, comment o n these val ues in rela tion to those for the metal a lloys tha t ar e tabulated in Table 8.4. Solution (a) This p ortion o f the problem calls for us to ran k four polymers relative to critical crack length in the wall of a sp herical pres su re vess el. In the dev elopment o f Design Example 8.1, it was n oted t hat critical crack length is proportional to the square of th e K Ic – sy ratio. Values of K Ic and sy as taken from Tables B.4 and B.5 are tabulated below. (Note: when a rang e of sy or K Ic values is given, the average value is us ed.)
Material
K Ic (MPa m )
sy (MPa)
Nylon 6,6
2.75
51.7
Polycarbonate
2.2
62.1
Poly(ethy lene tereph thlate)
5.0
59.3
Poly(methyl methacrylate)
1.2
63.5
On the bas is of these v alues, the four polymers are ranked per the s quares o f the K Ic – sy ratios as follows:
Material
æ K ö2 ç Ic ÷ çs ÷ è y ø
PET
7.11
Nylon 6,6
2.83
PC
1.26
PMMA
0.36
(mm)
Thes e values are smaller than th os e for the metal alloys given in Table 8.3, which range from 0.93 to 43.1 mm. 2 (b) Relative to the leak-before-break criterion, the K Ic
-
s y ratio
is us ed. The four polymers are ranked
according to values o f this ratio as follows:
Material
2 K Ic
sy
(MPa- m)
PET
0.422
Nylon 6,6
0.146
PC
0.078
PMMA
0.023
These values are all smaller than those for the metal alloys given in Table 8.4, which values range from 1.2 to 11.2 MPa-m.
Data Extrapolation Methods
8.D3 An S-590 a lloy component ( Figure 8.32) must ha ve a creep r upture lifetime of at least 100 days at 500 °C (773 K). Compute the maximum a llowa ble stress level. Solution This problem asks that we compute the maximum allowable stress level to give a rupture lifetime of 100 days for an S-590 iron co mponent at 773 K. It is first neces sa ry to compute th e value of the Larson-M iller parameter as follows:
T (20
+ log tr ) = (773 K){20 + log [(100 days)(24 h/day) ]} = 18.1 ´ 103
From the curve in Figure 8.32, this value of the Larson-Miller parameter corresponds to a stress level of about 530 MPa (77,000 psi).
8.D4 Consider an S-590 alloy component (Figur e 8.32) th at is subjected to a stress of 200 MPa (29,000 psi) . At wha t tempera tur e will th e ruptu re lifetime be 50 0 h? Solution We are asked in this problem to calculate the temperature at which the rupture lifetime is 500 h when an S590 iron component is subjected to a stress of 200 MPa (29,000 psi). From the curve s hown in Figure 8.32, at 200 MPa, the value of the Larso n-Miller parameter is 22.5 ´ 103 (K-h). Thus,
22.5 ´ 10 3 (K - h) = T (20 + log tr ) = T [20 + log (500 h) ] Or, solving for T yields T = 991 K (718°C).
8.D5 For an 18-8 Mo stainless steel (Figure 8.35), predict the time to rupture for a component that is subjected to a stress of 80 MPa (11,600 psi) a t 700°C (973 K). Solution This problem asks that we determine, for an 18-8 Mo st ainless steel, the time to rupt ure for a componen t that is subjected to a stress of 80 MPa (11,600 psi) at 700°C (973 K). From Figure 8.35, the value of the Larson-Miller parameter at 80 MPa is abo ut 23.5 ´ 103, for T in K and tr in h. Therefore,
23.5 ´ 10 3 = T (20 + log tr )
= 973(20 + log tr ) And, s olving for tr
24.15 = 20 + log t r which leads t o tr = 1.42 ´ 104 h = 1.6 yr.
