Question

A couple is applied to the ends of the steel shaft as shown below with a force F= 150 lb, the shaft is commercially grounded. The following properties are specified for the material S_ut = 90ksi and S_y = 65ksi. The reliability specified is 95%. Find the factor of safety for the Sin shaft using the Modified Goodman Design Theory. 

Answer 1

Answer 2

sol: Given that force (F) = 150 lb Sut - 90 ksi Sy 6528i Relaibility = 95.7 95 =0.95 100 Now, For Modified Goodman Design Design fluctuating force is applied. So, (F) min=0, (max = 150 16 - Maximum Torque = Frase (90) = 150 X 90 (T) Mar = 13500 16f-in -> (min = Frin (90) =0 From Table As per b in fillet weld ( from figure) Theoretical stress concentration factor [ke) = 1.6 from figure chart, r= 1/2 in x but = 90 ksi notch Sensitivity (9) = 0.85 concentration factor (kts) Fatigue stress - 1+q (k-1)

Tuara min substituting : kps It Digs (1.6-1) = 1.51 16k4s Than - Maximum Shearstress (Tman ): Td 3 ll 16 x 1.51 X 13500 17 (8)3 Yuar - 202.8 psi 16 kq, Tuin I min -0 [ : Trino 113 Ymin -o Now, Traxt Ymin Mean Shear stress (Tu) = 2 20218+o n 2 = 101.4 psi am = 0.1014 ksi Amplitude stress (Ta) = 2 202.80 2 101.4 psi = 0.1014 ksi Ma

then, Endurance limit (se) = kako ke ka sé ka= a (sut) b Here, a=1.34, b = -0.085 (For ground) -0.085 Surface factor (ka) = 1.34 (90) 0.914 -0.157 size factor (ko) 0.9160) = Kb = 0.656 [ 22d210 In] koading factor Cke) = 0.59 for Torsion Reliability factor (14) = 0.868, CR=957] 0.5 ( Sut) Se = 0.5X 90 に 45 ksi Nold he get Endurance limit (se) Se= 10.914 ) x (0.656) x10.59)x145) Se s 15. 918 k5i By using Modified goodman theory : faitigue Streugth (same ) = 0.67 (Set) Torsional Ssu = 0.67X 90 に 60.3 KSI

Now, mp = Ya ) - TM Ssu Se m 01014 0.1014 :)+ 15.418 60.3 n 1 21.06 Thor fore, of safety (Fos) 8 In Facta Shabt is "121006 [ It's very large due to darge Shaft diameter

Answer 3