Question

The shaft shown is made of AISI 1040 CD steel. It is machine finished and is subjected to a repeated bending stress of 15ksi. The diameter at the shoulder is 1.3in and will be used at a temperature of 400F. Estimate: 

(a) The endurance limit at 95% reliability 

(b) Endurance strength at 105 cycles and show on S-N plot 10 

(c) Plot the design region by Modified Goodman theory and determine if failure is by yield or fatigue  

(d) Find the factor of safety at the shoulder against fatigue at 10⁵ cycles using the Modified Goodman Criterion

Answer 1

solution: 0 0.098 rading 1.95 in lisin toe M for AISI 1040 CD Sut = 85 kpss Syt 71 kip] a) Specimen! andurance limit, 0.5885 0.5 sut = S Se こ 42.5 kpsi for machined Ginished d=1.3 b = -0.265 a= 2.70, ka factor Surface a sufb. 2.70 x (85) *0.265 ka 0855 T = 400°f R = 95% ht for bending momend, Load factor Kc =1

factor kd lo 018 Temprature factor Ke 0868 Reliability Endurance limit, se = kako ke ke ke se = 0.832x0.855 X 1 X. 1.01870.888 x 425 Se = 26.72 kpsi 95-7 Reliability is i. Endurance limit at 26.72 kpsi (0.9 Sut) b) => log, (0.9%85) log 10 10 1-88366 = = log, (se) log,086.42) 10 = 1: 4268 1 logiost A 1.88366 F E 1-4268 B C → 3 4- 5 6 Log Ny <10 from S-N diagram

DB Ynf (3) SE no (6-3) Y (188366->) (: -3) (1.88.266 1.4-2.68) 1'51908 1.57908 log Goe) 10 бе 37 94 kpsi las Endurance strength at Cycle is 31.94 kps ra og 2 om A se Goodlman line yield line ra B om m Syt Sut modified Goodman line diagram for Bending stress > Repeated stress.o=o. to isks; bending -Amplitude stress, 15-0 Ta 2

oa = 7.5 ksi 4 Mean stress 1.5 +0 = 2 m = 7.5ksi Tan o= 7.5 7.5 o = Tai (1) o 45 715 ksi within ABC region region, so ст = a Wi thin line it Then Goodman ict is will field by fatigue Goodman modified criteria. a) By using m a zh Sut Se 7.5 7.5 ) t 8.5 26.72 zt LL 2-71 N Na 2.71 The factor of satety is 2.71

Answer 2

solution: 0 0.098 rading 1.95 in lisin toe M for AISI 1040 CD Sut = 85 kpss Syt 71 kip] a) Specimen! andurance limit, 0.5885 0.5 sut = S Se こ 42.5 kpsi for machined Ginished d=1.3 b = -0.265 a= 2.70, ka factor Surface a sufb. 2.70 x (85) *0.265 ka 0855 T = 400°f R = 95% ht for bending momend, Load factor Kc =1

factor kd lo 018 Temprature factor Ke 0868 Reliability Endurance limit, se = kako ke ke ke se = 0.832x0.855 X 1 X. 1.01870.888 x 425 Se = 26.72 kpsi 95-7 Reliability is i. Endurance limit at 26.72 kpsi (0.9 Sut) b) => log, (0.9%85) log 10 10 1-88366 = = log, (se) log,086.42) 10 = 1: 4268 1 logiost A 1.88366 F E 1-4268 B C → 3 4- 5 6 Log Ny <10 from S-N diagram

DB Ynf (3) SE no (6-3) Y (188366->) (: -3) (1.88.266 1.4-2.68) 1'51908 1.57908 log Goe) 10 бе 37 94 kpsi las Endurance strength at Cycle is 31.94 kps ra og 2 om A se Goodlman line yield line ra B om m Syt Sut modified Goodman line diagram for Bending stress > Repeated stress.o=o. to isks; bending -Amplitude stress, 15-0 Ta 2

oa = 7.5 ksi 4 Mean stress 1.5 +0 = 2 m = 7.5ksi Tan o= 7.5 7.5 o = Tai (1) o 45 715 ksi within ABC region region, so ст = a Wi thin line it Then Goodman ict is will field by fatigue Goodman modified criteria. a) By using m a zh Sut Se 7.5 7.5 ) t 8.5 26.72 zt LL 2-71 N Na 2.71 The factor of satety is 2.71

Answer 3

For the given AISI 1040 Steel, the ultimate strength is obtained from manuals as 90ksi. The r/d and D/d is calculated for the given dimensions. The theoretical stress concentration factor is obtained for the values of r/d and D/d. Then Ka, Kb, Kc, and Kd values are obtained for the given data and material specification (Refer Shigley). Once Ka, Kb, Kc, Kd are known, the endurance strength can then be obtained. The calculated endurance strength will be for 10^6 cycles. The SN curve is prepared and the endurance strength corresponding to the 10^5 cycles is obtained. Using the modified Goodman theory, the variation of σa with σm is plotted. Using the modified Goodman Theory, the factor os safety is computed for the given mean stress and amplitude stress.

Answer 4

solution: 0 0.098 rading 1.95 in lisin toe M for AISI 1040 CD Sut = 85 kpss Syt 71 kip] a) Specimen! andurance limit, 0.5885 0.5 sut = S Se こ 42.5 kpsi for machined Ginished d=1.3 b = -0.265 a= 2.70, ka factor Surface a sufb. 2.70 x (85) *0.265 ka 0855 T = 400°f R = 95% ht for bending momend, Load factor Kc =1

factor kd lo 018 Temprature factor Ke 0868 Reliability Endurance limit, se = kako ke ke ke se = 0.832x0.855 X 1 X. 1.01870.888 x 425 Se = 26.72 kpsi 95-7 Reliability is i. Endurance limit at 26.72 kpsi (0.9 Sut) b) => log, (0.9%85) log 10 10 1-88366 = = log, (se) log,086.42) 10 = 1: 4268 1 logiost A 1.88366 F E 1-4268 B C → 3 4- 5 6 Log Ny <10 from S-N diagram

Answer 5

1. ·0.098racing Gasin 6 1.zin posin ban =) For AISI 1040 CD Sut = 85 kpsi Syt = 71 kips a) Speciment endurance limit, se = 0.5 Sut -0.5 x 85 Se = 42.5kppsi for machined finished, a = 2.70, b=-0.265 d = 1.3 = Surface factor, ka=asub 2.70 *[85-4-0.265 ka = 0.832, size factor, ko -0.107 - 0.879 d -- 0.879C1-33-0-107 kb.= 0.855 At, T = 400'F , R-95%. For bending moment, Load factor, ke=1 Temperatore factor, kd = 1.018 Reliability factor, ke = 0-868 - Endwance limit, se z Kakskekake se = 0.832 X0.855*1*1.01880.868XH25 Se = 26.72 kesi -

ei Endurance limitat 95%. Reliability is 26.72kpsi 2 b) log, 20.9 Sut J = log, (0.9%85) - 1.88366. => dog, (se) logio (26.72) Eloh268. log, ost 1.88366 A E 13 1.4268 D C 3 4 5 6 log, on From S.N diagram DB XAF FE AD (5-3) = 16 -3)*(1-88366->) (1.88366-), H268 X = 1,57908. log o cre) = 1.57908 bé = 3994 kpsi. Encionance strength at los cycles is 37.94 kpsi. C) sy • Goodman dine A se Ta 13 yield line e C um Tm syt sut Molifted Goodman linedogram for Bending stress.

3 =) Repeated bending stress, To=oto 15ksi Amplitude stress, Tá - 15-0 2 Ta = 7.5ksi Mean stress, The 15t0 2 m = 7.5kgi tand 7.5 7.5 z tani (1) فيا = م ܥ m = Ta - 7.5ksi within ABC region, so it is within Goodman line. Then it will kiled by fatigue. By using Modified Goodman criteria, um Sut + h Se 7.5 7.5 26.72 85 N x = 21 2.71 IN: 2.31 The factor of safety is 2.71.

Answer 6

Answer:

1. ·0.098racing Gasin 6 1.zin posin ban =) For AISI 1040 CD Sut = 85 kpsi Syt = 71 kips a) Speciment endurance limit, se = 0.5 Sut -0.5 x 85 Se = 42.5kppsi for machined finished, a = 2.70, b=-0.265 d = 1.3 = Surface factor, ka=asub 2.70 *[85-4-0.265 ka = 0.832, size factor, ko -0.107 - 0.879 d -- 0.879C1-33-0-107 kb.= 0.855 At, T = 400'F , R-95%. For bending moment, Load factor, ke=1 Temperatore factor, kd = 1.018 Reliability factor, ke = 0-868 - Endwance limit, se z Kakskekake se = 0.832 X0.855*1*1.01880.868XH25 Se = 26.72 kesi -

ei Endurance limitat 95%. Reliability is 26.72kpsi 2 b) log, 20.9 Sut J = log, (0.9%85) - 1.88366. => dog, (se) logio (26.72) Eloh268. log, ost 1.88366 A E 13 1.4268 D C 3 4 5 6 log, on From S.N diagram DB XAF FE AD (5-3) = 16 -3)*(1-88366->) (1.88366-), H268 X = 1,57908. log o cre) = 1.57908 bé = 3994 kpsi. Encionance strength at los cycles is 37.94 kpsi. C) sy • Goodman dine A se Ta 13 yield line e C um Tm syt sut Molifted Goodman linedogram for Bending stress.

3 =) Repeated bending stress, To=oto 15ksi Amplitude stress, Tá - 15-0 2 Ta = 7.5ksi Mean stress, The 15t0 2 m = 7.5kgi tand 7.5 7.5 z tani (1) فيا = م ܥ m = Ta - 7.5ksi within ABC region, so it is within Goodman line. Then it will kiled by fatigue. By using Modified Goodman criteria, um Sut + h Se 7.5 7.5 26.72 85 N x = 21 2.71 IN: 2.31 The factor of safety is 2.71.

Answer 7

solution: 0 0.098 rading 1.95 in lisin toe M for AISI 1040 CD Sut = 85 kpss Syt 71 kip] a) Specimen! andurance limit, 0.5885 0.5 sut = S Se こ 42.5 kpsi for machined Ginished d=1.3 b = -0.265 a= 2.70, ka factor Surface a sufb. 2.70 x (85) *0.265 ka 0855 T = 400°f R = 95% ht for bending momend, Load factor Kc =1

factor kd lo 018 Temprature factor Ke 0868 Reliability Endurance limit, se = kako ke ke ke se = 0.832x0.855 X 1 X. 1.01870.888 x 425 Se = 26.72 kpsi 95-7 Reliability is i. Endurance limit at 26.72 kpsi (0.9 Sut) b) => log, (0.9%85) log 10 10 1-88366 = = log, (se) log,086.42) 10 = 1: 4268 1 logiost A 1.88366 F E 1-4268 B C → 3 4- 5 6 Log Ny <10 from S-N diagram

DB Ynf (3) SE no (6-3) Y (188366->) (: -3) (1.88.266 1.4-2.68) 1'51908 1.57908 log Goe) 10 бе 37 94 kpsi las Endurance strength at Cycle is 31.94 kps ra og 2 om A se Goodlman line yield line ra B om m Syt Sut modified Goodman line diagram for Bending stress > Repeated stress.o=o. to isks; bending -Amplitude stress, 15-0 Ta 2

oa = 7.5 ksi 4 Mean stress 1.5 +0 = 2 m = 7.5ksi Tan o= 7.5 7.5 o = Tai (1) o 45 715 ksi within ABC region region, so ст = a Wi thin line it Then Goodman ict is will field by fatigue Goodman modified criteria. a) By using m a zh Sut Se 7.5 7.5 ) t 8.5 26.72 zt LL 2-71 N Na 2.71 The factor of satety is 2.71