Solution Callister 8 40 8 50

Prévia do material em texto

1) Polish the surface to remove stress amplification sites.
2) Reduce the number of internal defects (pores, etc.) by means of altering processing and
fabrication techniques.
3) Modify the design to eliminate notches and sudden contour changes.
4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot
peening.
8.40 Creep becomes important at 0.4Tm, Tm being the absolute melting temperature of the metal.
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)
8.41 Schematic creep curves at both constant stress and constant load are shown below.
St
ra
in
Time
Constant Load
Constant Stress
With increasing time, the constant load curve becomes progressively higher than the constant
stress curve. Since these tests are tensile ones, the cross-sectional area diminishes as
deformation progresses. Thus, in order to maintain a constant stress, the applied load must
correspondingly be diminished since stress = load/area.
8.42 These creep data are plotted below
4 03 02 01 00
0.0
0.1
0.2
0.3
0.4
The steady-state creep rate ( / t) is the slope of the linear region as
t = 
0.230 - 0.09
30 min - 10 min = 7.0 x 10
-3 min-1
8.43 This problem asks that we determine the total elongation of a low carbon-nickel alloy that is
exposed to a tensile stress of 40 MPa (5800 psi) at 538 C for 5000 h; the instantaneous and
primary creep elongations are 1.5 mm (0.06 in.).
From the 538 C line in Figure 8.39, the steady state creep rate, 
.
s , is about 0.15
%/1000 h (or 1.5 x 10-4 %/h) at 40 MPa. The steady state creep strain, s, therefore, is just the
product of 
.
s and time as
s = 
.
s x (time)
= (1.5 x 10-4 %/h)(5000 h) = 0.75 % = 7.5 x 10-3
Strain and elongation are related as in Equation (6.2); solving for the steady state elongation,
ls, leads to
ls = lo s = (750 mm)(7.5 x 10
-3) = 5.6 mm (0.23 in.)
Finally, the total elongation is just the sum of this ls and the total of both instantaneous and
primary creep elongations [i.e., 1.5 mm (0.06 in.)]. Therefore, the total elongation is 7.1 mm
(0.29 in.).
8.44 We are asked to determine the tensile load necessary to elongate a 500 mm long low carbon-
nickel alloy specimen 3.2 mm after 10,000 h at 427 C. It is first necessary to calculate the
steady state creep rate so that we may utilize Figure 8.39 in order to determine the tensile
stress. The steady state elongation, ls, is just the difference between the total elongation and
the sum of the instantaneous and primary creep elongations; that is,
ls = 3.2 mm - 0.8 mm = 2.4 mm (0.10 in.)
Now the steady state creep rate, 
.
s is just
.
s = t = 
ls/lo
t = 
(2.4 mm)/(500 mm)
10000 h
= 4.8 x 10-7 (h)-1 = 0.048 %/1000 h
Employing the 427 C line in Figure 8.39, a steady state creep rate of 0.048 %/1000 h
corresponds to a stress of about 70 MPa (10,000 psi). From this we may compute the tensile
load using Equation (6.1) as
F = Ao = 
do
2
2
= (70 x 106 N/m2)( ) 10.0 x 10
-3 m
2
2
 = 5500 N (1250 lbf)
8.45 This problem asks us to calculate the rupture lifetime of a component fabricated from a low
carbon-nickel alloy exposed to a tensile stress of 60 MPa at 538 C. All that we need do is read
from the 538 C line in Figure 8.38 the rupture lifetime at 60 MPa; this value is about 36,000 h.
8.46 We are asked in this problem to determine the maximum load that may be applied to a
cylindrical low carbon-nickel alloy component which must survive 500 h at 649 C. From Figure
8.38, the stress corresponding to 500 h is 55 MPa (8000 psi). Since stress is defined in
Equation (6.1) as = F/Ao, and for a cylindrical specimen, Ao = ro
2, then
F = Ao = 
do
2
2
= (55 x 106 N/m2)( ) 12 x 10
-3 m
2
2
 = 6220 N (1570 lbf)
8.47 The slope of the line from a log 
.
s versus log plot yields the value of n in Equation (8.33);
that is
n = 
 log 
.
s
 log 
We are asked to determine the values of n for the creep data at the three temperatures in
Figure 8.39. This is accomplished by taking ratios of the differences between two log 
.
s and log
 values. Thus for 427 C
n = 
 log 
.
s
 log = 
log(10-1) - log(10-2)
log(85 MPa) - log(55 MPa) = 5.3
While for 538 C
n = 
 log 
.
s
 log = 
log(1.0) - log(10-2)
log(59 MPa) - log(23 MPa) = 4.9
And at 649 C
n = 
 log 
.
s
 log = 
log(1.0) - log(10-2)
log(15 MPa) - log(8.3 MPa) = 7.8
8.48 (a) We are asked to estimate the activation energy for creep for the low carbon-nickel alloy
having the steady-state creep behavior shown in Figure 8.39, using data taken at = 55 MPa
(8000 psi) and temperatures of 427 C and 538 C. Since is a constant, Equation (8.34) takes
the form
.
s = K2
nexp - 
Qc
RT = K2' exp - 
Qc
RT
where K2' is now a constant. (Note: the exponent n has about the same value at these two
temperatures per Problem 8.47.) Taking natural logarithms of the above expression
ln
.
s = ln K2' - 
Qc
RT
For the case in which we have creep data at two temperatures (denoted as T1 and T2) and their
corresponding steady-state creep rates (
.
s1
 and 
.
s2
), it is possible to set up two simultaneous
equations of the form as above, with two unknowns, namely K2' and Qc. Solving for Qc yields
Qc = - 
R ln 
.
s1
 - ln 
.
s2
1
T1
 - 1T2
Let us choose T1 as 427 C (700 K) and T2 as 538 C (811 K); then from Figure 8.39, at = 55
MPa,
.
s1
 = 0.01 %/1000 h = 1 x 10-7 (h)-1 and 
.
s2
 = 0.8 %/1000 h = 0.8 x 10-5 (h)-1.
Substitution of these values into the above equation leads to
Qc = - 
(8.31 J/mol-K)[ ]ln( )10-7 - ln( )0.8 x 10-5
[ ]1700 K - 1811 K
= 186,200 J/mol
(b) We are now asked to calculate 
.
s at 649 C (922 K). It is first necessary to determine the
value of K2' , which is accomplished using the first expression above, the value of Qc, and one
value each of 
.
s and T (say 
.
s1
 and T1). Thus,
K2' = 
.
s1
exp
Qc
RT1
= [ ]10-7 (h)-1 exp [ ]186200 J/mol(8.31 J/mol-K)(700 K) = 8.0 x 106 (h)-1
Now it is possible to calculate 
.
s at 922 K as follows:
.
s = K2' exp - 
Qc
RT
= [ ]8.0 x 106 (h)-1 exp [ ]- 186200 J/mol(8.31 J/mol-K)(922 K)
= 2.23 x 10-4 (h)-1 = 22.3 %/1000 h
8.49 This problem gives 
.
s values at two different stress levels and 1000 C, and the activation
energy for creep, and asks that we determine the steady-state creep rate at 850 C and 25 MPa
(3625 psi).
Taking the natural logarithm of Equation (8.34) yields
ln
.
s = ln K2 + n ln - 
Qc
RT
With the given data there are two unknowns in this equation--namely K2 and n. Using the data
provided in the problem we can set up two independent equations as follows:
ln[ ]1.0 x 10-4 (s)-1 = ln K2 + n ln(15 MPa) - 272000 J/mol(8.31 J/mol-K)(1273 K)
ln[ ]1.0 x 10-6 (s)-1 = ln K2 + n ln(4.5 MPa) - 272000 J/mol(8.31 J/mol-K)(1273 K)
Now, solving simultaneously for K2 and n leads to n = 3.825 and K2 = 466 (s)
-1. Thus it is now
possible to solve for 
.
s at 25 MPa and 1123 K using Equation (8.34) as
.
s = K2
nexp - 
Qc
RT
.
s = [ ]466 (s)-1 (25 MPa)3.825exp( )- 272000 J/mol(8.31 J/mol-K)(1123 K)
2.28 x 10-5 (s)-1
8.50 This problem gives 
.
s values at two different temperatures and 70 MPa (10,000 psi), and the
stress exponent n = 7.0, and asks that we determine the steady-state creep rate at a stress of
50 MPa (7250 psi)and 1250 K.
Taking the natural logarithm of Equation (8.34) yields
ln
.
s = ln K2 + n ln - 
Qc
RT
With the given data there are two unknowns in this equation--namely K2 and Qc. Using the
data provided in the problem we can set up two independent equations as follows:
ln[ ]1.0 x 10-5 (h)-1 = ln K2 + (7.0)ln(70 MPa) - 
Qc
(8.31 J/mol-K)(977 K)
ln[ ]2.5 x 10-3 (h)-1 = ln K2 + (7.0)ln(70 MPa) - 
Qc
(8.31 J/mol-K)(1089 K)
Now, solving simultaneously for K2 and Qc leads to K2 = 2.55 x 10
5 (h)-1 and Qc = 436,000
J/mol. Thus it is now possible to solve for 
.
s at 50 MPa and 1250 K using Equation (8.34) as
.
s = K2
nexp - 
Qc
RT
.
s = [ ]2.55 x 105 (h)-1 (50 MPa)7.0exp( )- 436000 J/mol(8.31 J/mol-K)(1250 K)
0.118 (h)-1
8.51 Three metallurgical/processing techniques that are employed to enhance the creep resistance
of metal alloys are 1) solid solution alloying, 2) dispersion strengthening by using an insoluble
second phase, and 3) increasing the grain size or producing a grain structure with a preferred
orientation.
Design Problems