Question

A couple is applied to the ends of the rotating steel shaft as shown below with a force F= 350 lb, in the form of a repeated loading. The shaft is commercially grounded. The following properties are specified for the material Sut = 110ksi and Sy = 80ksi. The reliability specified is 95%. Find the factor of safety for the 8in shaft using the Modified Goodman Design Theory for infinite life. Ignore the stress concentration at the fillet. 

Answer 1

Answer 2

350 lb. # Given, Fmax ; min (: Repeated loading] Sut = lloksi sy = poksi ; Diameter of shaft (d) = 8 in. Realibility= 95% From the figure can see, Maximum Torque (Tmax) Fmax *45*2 (350 * 90) or max. = 31500 lb-in. 16 * 31500 T ²8 3 (Emin *90) -Also, Minimum torque min) From Torsion equation for circular shaft, 16* Tmax Imax 313.34 psi Td3 16* Tmin. Also, Tmin ad Mean stress (Tm) (Tmart Emin) (313.34+0) = 156.67 psi Alternating stress. (Ta) = (tmare - Tmin) 156.67 psi - 2 2 2 standard endurance limit ( sen) = 6.5* Sut). . (0.5*110) = 55ksi using Marin's equation, Sen x Ka*K* Ke *ų factor (het) b; where fut is in Mpa 0.085 for Grounded shaft, ka - 158*(1o86-894-0-085 sen where, ka = size a = 1.58 and ba 0.9

(6.859 -0.000873 xd Kb size factor where, dis diameter of shaft in mm So, Kb 0.859 0.000873 ( 8%25242 - 0.682 Also, Ke= Reliability factor = 0.868 for 95% reliability K = load factor 0.58 for Torsional load Sen Corrected endurance limit in ksi- 117 Kpi Sen So, = (55*0.9 x 0.682 x 0.868 X0.58) (Ans-) As per Maximum shear stress therry, shear endurance strength Ten = (0.54 Sen) = (0.5x17) 8.5 ksi shear ultimate strength (Tut) 0.5* Sut (o.sx 110) Sheer Yield Strength (ty) = (0.5 x Sy) - 40 ksi As per Goodman criteria, 55 ksi Im Ta + Eut W Ten 1 156.67 156.67 + 558 103 8.5X103 or, N = 47 Now using Langer criteria Im Ta + Ty Ty - 156.67 156.67 40 x 10² 40 x 103 So, Na < 127.66 11 2 You

To satisfy both the equation as per Modified Goodman, Factor of safety 47. (Ans--) So the shaft will never fail by fatigue. If static load acts then per Maximum shear Stress theory, Tmax Sy 2 N. 2) 313.34 $ 80 x 103 2 *N. or, N=127.65 So, the sh shaft will never fail by Yielding (Ans - -).

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Answer 3

Given repeated load, Fman = 35016 and Fminco (Understood) I Sut = 100ksi Tman Finane X 45X2 (For couple) 350 X 90 31500 lbstim d=8 in and Tmin=o.lb-im R = 95% torsional stress Eman = 16x Tmone 16X31500 Now, maximum Sy=80ksi Td3 TX 83 336 and Tmin=0. = 313 Shepsi Mean stress Tm mone + Emin = 156.67 Psi * stress Analysis Alternating stress Ta = (156.67 Psi standard For steel, endurance limit se = 0.5 Sut = 0.5X 110 = 55ksi Hence, Se= kako kc kase surface a (suto ka= factor since, the units are sI in VB bhandari, using the conversion For given grounded shaft, Iksi 6.894 N/mm² a=1.58, b = -0.085 & diameter & in 8 X 25.4 mm (Reference V.B Bhandari = 203- MD-Chapter 5) Hence, 51mm ccde 254 mm Ka= 1.58 (110X6.894) ka=0.9 kb=0.859-0. 000873d Ikb = 0.682 Size Factor mm -0.085 From, Se = 0.96 0.682 x 0.868 X 0.58 X 55 Now, ke= Reliability factor ke = 0.868 for 951 R & kueload Factor = 0.580 For torsional load Se = 17 FSI * Endurance Limit

To determine the type of failure cyield or Fatigue); For steel, from maximum shear stress theory, shear endurance strength Ten 0.5X Se 0.5X17 = -8.5ksi shear ultimate Strength Tut 0.5X Sut = 0.5 X 10 = 55 ksi Shear yield Zy 0.5% Sy = 40ksi From goodman Criteria, Ta s L 156.67 + 156.67 Ni 558 103 8.5X103 NI N = 47 From langer criteria, 156.67 + 156.67 SI N2 €127.667 40000 40000 Im Zut + N Ten en N2 es Hence from modified goodman theory, Fos (N = 4: and shaft cannot fail by fatigue To check against yielding, (For static load) Inan & Sy ZN 3362 3 < 80x100 2XN : 313.50000 NC 127.658 Hence shaft won't fail by yielding as well The design is safe ..

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Answer 4