Fatigue Problems Solution

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1 Fatigue Problems Solution Problem 1. (a) Given the values of σ m (7 MPa) and σ a (1 MPa) we are asked t o compute σ max and σ min. From Equation 1 Or, σ m σ max + σ min 7 MPa σ max + σ min 14 MPa Furthermore, utilization of Equation yields σ a σ max σ min 1 MPa Or, σ max σ min 4 MPa Simultaneously solving these two expressions leads to σ max 8 MPa (4, psi) σ min 14 MPa (, psi) (b) Using Equation 3 the stress ratio R is determined as follows: R σ min 14 MPa σ max 8 MPa.5 (c) The magnitude of the stress range σ r is determined using Equation 4 as σ r σ max σ min 8 MPa ( 14 MPa) 4 MPa (6, psi) 1

2 Problem. This problem asks that we determine the minimum allowable bar diameter to ensure that fatigue failure will not occur for a 145 steel that is subjected to cyclic loading for a load amplitude of 66,7 N (15, lb f ). From Fig. 1, the fatigue limit stress amplitude for this alloy is 31 MPa (45, psi). Stress is defined in as σ F A. For a cylindrical bar A π d Substitution for A into the Equation leads to σ F A F π d 4F πd We now solve for d, taking stress as the fatigue limit divided by the factor of safety. Thus d 4F π σ N (4)(66,7 N) (π) 31 x 16 N/m 3.4 x 1 3 m 3.4 mm (.9 in.)

3 Problem 3. We are asked to determine the fatigue life for a cylindrical 14-T6 aluminum rod given its diameter (6.4 mm) and the maximum tensile and compressive loads (+534 N and 534 N, respectively). The first thing that is necessary is to calculate values of σ max and σ min using stress equation. Thus σ max F max F max A π d 534 N (π) 6.4 x 1 3 m 166 x 16 N/m 166 MPa (4, 4 psi) σ min F min π d 534 N (π) 6.4 x 1 3 m 166 x 16 N/m 166 MPa ( 4, 4 psi) Now it becomes necessary to compute the stress amplitude using Equation as σ a σ max σ min 166 MPa ( 166 MPa ) 166 MPa (4, 4 psi) From Fig. 1, for the 14-T6 aluminum, the number of cycles to failure at this stress amplitude is about 1 x 1 7 cycles. 3

4 Problem 4. This problem asks that we compute the maximum and minimum loads to which a 15. mm (.6 in.) diameter 14-T6 aluminum alloy specimen may be subjected in order to yield a fatigue life of 1. x 1 8 cycles; Fig. 1 is to be used assuming that data were taken for a mean stress of 35 MPa (5, psi). Upon consultation of Fig.1, a fatigue life of 1. x 1 8 cycles corresponds to a stress amplitude of 14 MPa (, psi). Or, from Equation σ max σ min σ a ()(14 MPa) 8 MPa (4, psi) Since σ m 35 MPa, then from Equation σ max + σ min σ m ()(35 MPa) 7 MPa (1, psi) Simultaneous solution of these two expressions for σ max and σ min yields σ max +175 MPa (+5, psi) σ min 15 MPa ( 15, psi) Now, inasmuch as σ F (Equation ), and A A π d then F max σ max π d 4 (175 x 16 N/m ) (π)(15. x 1 3 m) 4 31,75 N (77 lb f ) F min σ min π d 4 ( 15 x 16 N/m ) (π)(15. x 1 3 m) 4 19, N ( 44 lb f ) 4

5 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 7 MPa (1, psi) at 47 C for 1, h; the instantaneous and primary creep elongations are 1.3 mm (.5 in.). From the 47 C line in Fig.1, the steady state creep rate ε Ý s is about 4.7 x 1-7 h -1 at 7 MPa. The steady state creep strain, ε s, therefore, is just the product of ε Ý s and time as ε s Ý ε s x (time) (4.7 x 1-7 h -1 )(1, h) 4.7 x1-3 Strain and elongation are related as in Equation; solving for the steady state elongation, l s, leads to l s l ε s (115 mm)(4.7 x 1-3 ) 4.8 mm (.19 in.) Finally, t he total elongation is just the sum of this l s and the total of both instantaneous and primary creep elongations [i.e., 1.3 mm (.5 in.)]. Therefore, the total elongation is 4.8 mm mm 6.1 mm (.4 in.). 5

6 Problem. We are asked to determine the tensile load necessary to elongate a 635 m m long low carbon-nickel alloy specimen 6.44 mm after 5, h at 538 C. It is first necessary to calculate the steady state creep rate so that we m ay utilize Fig. 1, in order to determine 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 s 6.44 mm 1.8 mm 4.64 mm (.18 in.) Now the steady state creep rate, ε Ý s is just ε Ý s ε t l s l t 4.64 mm 635 mm 5, h 1.46 x 1-6 h -1 Employing the 538 C line in Fig.1, a steady state creep rat e of 1.46 x 1-6 h -1 corresponds to a st ress σ of about 4 M Pa (5,8 psi ) [since log (1.46 x 1-6 ) ]. From this we m ay compute the tensile load using Equation as F σa σπ d (4 x 1 6 N/m )(π) 19. x 1 3 m 11,3 N (56 lb f ) 6

7 Problem 3. This problem asks us to calculate the rupture lifetime of a com ponent fabricated from a low carbonnickel alloy exposed to a tensile stress of 31 MPa at 649 C. All that we need do is read from the 649 C line in Fig. the rupture lifetime at 31 MPa; this value is about 1, h. 7

8 8-3 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 1, h at 538 C. From Fig., the stress corresponding to 1 4 h is about 7 M Pa (1, psi ). Since st ress i s defi ned i n Equat ion as σ F/A, and for a cylindrical specimen, A π d, then F σa σπ d (7 x 1 6 N/m )(π) 19.1 x 1 3 m, N (44 lb f ) 8

9 Data Extapolation Problems Solution Problem. We are asked in this problem to calculate the temperature at which the rupture lifetime is h when an S-59 iron component is subjected to a stress of 55 M Pa (8 psi). From the curve shown in Fig. 1, at 55 MPa, the value of the Larson-Miller parameter is 6.7 x 1 3 (K-h). Thus, 6.7 x 1 3 (K - h) T ( + log t r ) Or, solving for T yields T 1197 K (94 C). T [ + log( h) ] 9

10 Problem. 3 This problem asks that we determine, for an 18-8 Mo stainless steel, the time to rupture for a component that is subjected to a stress of 1 M Pa (14,5 psi) at 6 C (873 K). From Fig., the value of the Larson-Miller parameter at 1 MPa is about.4 x 1 3, for T in K and t r in h. Therefore,.4 x 1 3 T ( + log t r ) 873( + log t r ) And, solving for t r log t r which leads to t r 4.6 x 1 5 h 5 yr.. 1

11 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 component is subjected to a tem perature of 65 C (93 K). It first becomes necessary to calculate the value of the Larson-Miller parameter for each time. The values of t r corresponding to 1 and 15 years are 8.76 x 1 3 h and 1.31 x 1 5 h, respectively. Hence, for a lifetime of 1 year T ( + log t r ) 93[ + log (8.76 x 1 3 )].1 x 1 3 And for t r 15 years T ( + log t r ) 93[ + log (1.31 x 1 5 )] 3.18 x 1 3 Using the curve shown in Fig., the stress values corresponding to the one- and fifteen-year lifetimes are approximately 11 MPa (16, psi) and 8 MPa (11,6 psi), respectively.. 11

12 Problem 1. This problem asks that we com pute the maximum allowable stress level to give a rupture lifetim e of days for an S-59 iron com ponent at 93 K. It is fi rst necessary to com pute th e value of the Larson-Miller parameter as follows: T ( + log t r ) (93 K) { + log [( days)(4 h/day) ]}.9 x 1 3 From the curve in Fig. 1, this value of the Larson-Miller parameter corresponds to a stress level of about 8 MPa (4, psi). 1

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