Fatigue Problems Solution Problem 1. (a) G iven the values of
σ
(70 M Pa) and
m
σ
(210 MPa) we are asked asked t o compute σmax and σmin.
a
From Equation 1
σ
m
σ
=
max
+ σ
2
min = 70 M Pa Pa
Or, σ
max + σmin = 140 MPa
Furthermore, utilization of Equation 2 yields
σ
a
=
σ
max
− σ
2
min = 210 MPa MPa
Or, σ
max – σmin = 420 MPa
Simultaneously solving these two expressions leads to σ σ
(40,000 psi) psi) max = 280 MPa (40,000
20,000 ,000 psi) psi) min = − 140 MPa (−20
(b) Using Equation 3 the stress ratio ratio R is determined as follows:
R
=
min = −140 MPa = σ 280 MPa max σ
(c) The magnitude of the stress stress range
σ
r
= σ max
− σ
σ
−
0.50
is determined using Equation 4 as
r
2800 MPa MPa min = 28
−
(−14 1400 MPa MPa)) = 42 4200 MPa MPa (60,00 (60,0000 psi psi))
1
Problem 2. This problem asks that we det ermine t he minimum allowable bar diameter to ensure that fatigue
failure will not occur for a 1045 steel th at is subjected to cyclic loading for a load am plitude of 66,700 N (15,000 lbf ). From Fig . 1, the fatigue limit stress amplitude for this alloy is 310 MPa (45,000 psi). Stress is defined in as σ =
F A
. For a cylindrical bar
0
⎛ d ⎞2 0⎟ A = π ⎜ 0 2 ⎝ ⎠
Substitution for A0 into the Equation leads to σ =
F
=
A
0
4 F = πd 02 ⎛ d ⎞2 0 π⎜ ⎟ ⎝ 2 ⎠ F
We now solve for d 0, taking stress as the fatigue limit divided by the factor of safety. Thus 0 =
d
=
(4)(66,700 N) ⎛ 310 x 10 6 N /m2 ⎞ ⎟⎟ (π) ⎜⎜ 2 ⎝ ⎠
4 F ⎛ σ ⎞ π⎜ ⎟ ⎝ N ⎠
= 23.4 x 10−3 m = 23.4 mm (0.92 in.)
2
Problem 3.
We are asked to determine the fatigue life for a cy lindrical 2014-T6 aluminum rod given its diameter
(6.4 mm) and t he maximum tensile and compressive loads (+5340 N and –5340 N, respect ively). The fi r st thing that is necessary is to calculate values of σmax and σmin using stress equation. Thus F σ max = max = A 0
=
5340 N ⎛ 6.4 x 10−3 m ⎞2 ⎟⎟ (π) ⎜⎜ 2 ⎝ ⎠
−5340 N ⎛ 6.4 x 10−3 m ⎞2 ⎟⎟ (π) ⎜⎜ 2 ⎝ ⎠
max ⎛ d ⎞2 π⎜ 0 ⎟ ⎝ 2 ⎠
= 166 x 10 6 N/m2 = 166 MPa (24,400 psi)
σ min =
=
F
F
min ⎛ d 0 ⎞2 π⎜ ⎟ ⎝ 2 ⎠
= − 166 x 10 6 N/m2 = − 166 MPa (−24,400 psi)
Now it becomes necessary to compute the stress amplitude using Equation as
σa =
σ max − σ min 166 MPa − (−166 MPa ) = = 166 MPa (24,400 psi)
2
2
From Fig. 1, for the 2014-T6 alum inum, the number of cycles to failure at this stress am plitude is about 1 x 107 cycles.
3
Problem 4. This problem asks that we compute the maximum and minimum loads to which a 15.2
mm (0.60 in.)
diameter 2014-T6 al uminum alloy specimen may be subject ed in order to yield a fat igue life of 1.0 x 10 8 cycles; Fig. 1 is to be used assuming that data were taken for a mean stress of 35 MPa (5,000 psi). Upon consultation of Fig.1, a fat igue life of 1.0 x 10 8 cycles corresponds to a stress amplitude of 140 MPa (20,000 psi). Or, from Equation σ max − σ min = 2σa = (2)(140 MPa) = 280 MPa (40,000 psi)
Since σm = 35 MPa, then from Equation σ max + σ min = 2σ m = (2)(35 MPa) = 70 MPa (10,000 psi)
Simultaneous solution of these two expressions for σmax and σmin yields σmax = +175 MPa (+25,000 psi) σmin = –105 MPa (–15,000 psi)
Now, inasmuch as σ =
F A
(Equation ), and
0
max =
F
min =
F
σ max π d 20
4
σ min π d 20
4
=
(175 =
(−105
⎛ d ⎞2 0 ⎟ then A = π ⎜ 0 ⎝ 2 ⎠
x 10 6 N / m2 ) (π) (15.2 x 10−3 m) 4 x 10 6 N / m2 ) (π ) (15.2 x 10−3 m) 4
2
= 31,750 N (7070 lbf )
2
= − 19,000 N (−4240 lbf )
4
Creep Problems Solution Problem 1.
This problem asks that we determine the total elongation of a low carbon-nickel alloy that is exposed
to a tensile stress of 70 MPa (10,000 psi) at 427 °C for 10,000 h; the instantaneous and primary creep elongations are 1.3 mm (0.05 in.). From the 427 °C line in Fig .1, the steady state creep rate εÝ s is about 4.7 x 10 -7 h -1 at 70 MPa. The steady state creep strain, ε s, therefore, is just the product of εÝ s and time as ε
s
=
ε
s
x (time)
= (4.7 x 10-7 h-1) (10,000 h) = 4.7 x10-3 Strain and elongation are related as in Equation; solving for the steady state elongation,
∆l
s
∆l , s
leads to
= l 0 ε s = (1015 mm) (4.7 x 10-3) = 4.8 mm (0.19 in.)
Finally, t he total elongation is just t he sum of this
∆l and s
t he total of both instantaneous and primary creep
elongations [i.e., 1.3 mm (0.05 in.)]. Therefore, the total elongation is 4.8 mm + 1.3 mm = 6.1 mm (0.24 in.).
5
Problem 2. We are asked t o determine the tensile load necessary to
elongate a 635 m m long low carbon-nickel
alloy specimen 6.44 mm after 5,000 h at 538 °C. It is first necessary to calculate the steady state creep rate so that we may utilize Fig . 1, in order to determ ine the tensile stress. The steady state elongation, ∆l s, is just th e difference between the total elongation and the sum of the instantaneous and primary creep elongations; that is, ∆l = 6.44 mm − 1.8 mm = 4.64 mm (0.18 in.) s
Now the steady state creep rate, ε s is just ∆ l s
4.64 mm l ∆ε 0 = 635 mm = = εÝ s ∆ t ∆ t 5,000 h = 1.46 x 10 -6 h-1 Employing the 538 °C line in Fig .1, a steady state creep rate of 1.46 x 10 -6 h -1 corresponds t o a stress σ of about 40 M Pa (5,800 psi ) [since log (1.46 x 10 -6) = -5.836]. From this we m ay compute the tensile load using Equation as ⎛ d ⎞2 0⎟ F = σ A = σπ ⎜ 0 ⎝ 2 ⎠ ⎛ 19.0 x 10−3 m ⎞2 6 2 ⎟⎟ = 11,300 N (2560 lbf ) = (40 x 10 N/m ) (π) ⎜⎜ 2 ⎝ ⎠
6
Problem 3.
This problem asks us to calculate the rupture lifetime of a component fabricated from a low carbon-
nickel alloy exposed t o a tensile stress of 31 M Pa at 649 C. All that we need do is read from the 649 C line in °
°
Fig. 2 the rupture lifetime at 31 MPa; this value is about 10,000 h.
7
8-32
Problem. 4
We are asked in this problem to determine the maximum load that may be applied to a cylindrical low
carbon-nickel alloy component that must survive 10,000 h at 538 °C. From Fig. 2, the stress corresponding to 104 h is about 70 M Pa (10,000 psi ). Si nce stress is defined in Equation as σ = F/A0, and for a cylindrical ⎛ d ⎞2 specimen, A0 = π ⎜ 0 ⎟ , then ⎝ 2 ⎠ ⎛ d ⎞2 0⎟ F = σ A = σπ ⎜ 0 2 ⎝ ⎠ ⎛ 19.1 x 10−3 m ⎞2 6 2 ⎟⎟ = 20,000 N (4420 lb f ) = (70 x 10 N/m ) ( π) ⎜⎜ 2 ⎝ ⎠
8
Data Extapolation Problems Solution Problem. 2 We are asked in this problem to calculate the
temperature at which the rupture lifetime is 200 h when
an S-590 iron component is subjected to a stress of 55 M Pa (8000 psi). From the curve shown in Fig . 1, at 55 MPa, the value of the Larson-Miller parameter is 26.7 x 10 3 (K-h). Thus, 26.7 x 10 3 (K - h) =
Or, solving for T yields T = 1197 K (924 °C).
=
T [20
(20 +
T
log t r )
+ log(200 h)]
9
Problem. 3 This problem
asks that we determ ine, for an 18-8 Mo stainless steel, the tim e to rupture for a
component that is subjected to a stress of 100 M Pa (14,500 psi) at 600 C (873 K). From Fig. 2, the value of the Larson-Miller parameter at 100 MPa is about 22.4 x 10 3, for T in K and t r in h. Therefore, °
22.4 x 10 3 = T (20 + log t r ) = 873 (20 + log t r ) And, solving for t r 25.66 = 20 + log t r which leads to t r = 4.6 x 10 5 h = 52 yr.
10 .
Problem 4.
We are asked in this problem to calculate the stress levels at which the rupture lifetime will be 1 year
and 15 y ears when an 18-8 M o stainless steel co m ponent is subjected to a tem perature of 650 °C (923 K). It f irst becomes necessary to calculate the value of the Larson-Miller param eter for each tim e. The values of tr corresponding to 1 and 15 years are 8.76 x 10 3 h and 1.31 x 10 5 h, respectively. Hence, for a lifetime of 1 year
(20 +
log t r ) = 923 [20 + log (8.76 x 10 3)] = 22.10 x 10 3
(20 +
log t r ) = 923 [20 + log (1.31 x 105)] = 23.18 x 10 3
T
And for t r = 15 years T
Using the curve shown in Fig . 2, the stress values corresponding to the one- and fifteen-year lifetimes are approximately 110 MPa (16,000 psi) and 80 MPa (11,600 psi), respectively.
11 .
Problem 1. This problem asks that we com pute the maximum allowable stress level to give a rupture lifetim e of 20 days for an S-590 iron com ponent at 923 K. It is fi rst necessary to com pute the value of the Larson-Miller parameter as follows:
(20 +
T
log t r ) = (923 K){20 + log [(20 days)(24 h/day)]}
= 20.9 x 103
From the curve in Fig . 1, this value of the Larson-Miller parameter corresponds to a stress level of about 280 MPa (40,000 psi).
12
