Question

Question 2 (20 Marks) A 50 mm diameter shaft connected to 2 kW (kilowatt) electromotor is shown in Figure 2. The shaft is made from ductile AISI 1050 cold-drawn treated with ground surface finished. At service temperature 370°C, the shaft is subjected to load P-200N. The shaft rotates under sinusoidal time varying load from 2000 rpm to 3000 rpm. The shaft is designed for 10' cycles with manufacturing reliability of 99.9999% Assuming a miscellaneous factor of 1.0, determine factor of safety at the most critical point. State any assumptions made and you may also ignore the stress concerntration effect Connected to electromotor 200 mm 300 mm 320 mm Bearing are self-aligning so act as simple supports Figure 2

Answer 1

Dear student,

Here I solved the above problem.

Problem is combined load with fluctuating stress.

To get normal stress from combined loading(bending + torsion)

And Goodman ctmritera is used to find FOS

As load is very small and dimensions are large and required cycle is less . So FOS comes to be very huge

please share your feedback. Rate answer to work better. Thank you.

P=200N Zoomm 300mm 320mm. Dorom Efx=0; {fy= RA+ RB = 200 N. Emazo; (RB x 200) =(2000 x 300) RB=300N. RAZ loon. RA= (Toon downwards) BM @ D =0. Bm @ c to BM @ x, dist from A = FRAX X + (RB XC 200-x2)) BM @ 200 mm. fromA 200 x 200) + Rex (200-200). = - 20000 N.mm BM @ X, dise from A. = GRAX 2) + RB (22-200)

Tal Tmax - (Tmen! 2 d=somm J= T (44) alternating I torque. 19549 +6366 (904) Ta = 1957. Nomm J= 613592.31 mm I = I de Tm= - Tmih Tmax I 2 (584) = IF 306796.15mmy 9549 - 6366. 2 1591.5 N.m In= Ta = Ta XR J 7957.5k 61354 2.31 *) Ta= 0.32 42 N/m2 Tm= TM XR 15915 xlso 613592.31 Tm = 0.06484 N/mm2

B.M @ 200mm = 100*200 + 3008 2007 2007 4 I – 2000 o N mm e o ) = -100%Boo) + 300 ( 300-200) = 0 MB max = -20,000 N.mm * Even though Bending moment is considered to be Constant as the constant load P is acting, But due to shaft ration rotation, A Critical element experentes a compressine and Tensile stress alternatively * This can be considered to be completely reversed bending. To = Tomp = Gensile = M xy (As shage circular area) = x go 20000 306796.15 1.6297 N/mm2 -: max 6= Tmin= * - = 1.6297 N/mm² 1.6297 N/nom? sa= Gmax t amin 1 1.6297 + 1.6297) 2 1 = 1.6297 Nmme Jm= | Tmax & amin 116 297 -1.6297

combined ofera to loading, Usuary solved with von-missos get Normal stress Tam = V obmean + 3 Im? = Vot(0, 0848943) = Power Om= 0,123 N/mmy Ja = v vas + 372 = (1.62 97² + 3x 0.32423 Ja= 1.7237 N/mm2 (In general) AI SI loso Material Juts = 690 Mpa I Jy = 580 Mpa I propenites Te = 0.5 JUTS = 0.5x690 ae = 34s Мра Corrected endurance limit, For; 370°C.; K=1. d=50 K= leo97) 0.869 d . = 0.86995 (-0.097) 0.702 = 0. 8137. Preran 99.999%. cornered = ka. Kb. Kc. (Te) = 1x0-8687 X 0.702 (de) = 0.57) x oe 197 Mpa. se cometel

W.KiT) From Basanin equaton! St= A Nb. Sf = 0-9 out > 103 cycles Sf = decoreret lot cycles SF=? @ 104 Cycles. SF=0.9 out=621 Mpen. : 621= A (2000). log 621 = log A + 3b. log 62,-& b = logA. 2.793-36= cloga. - ③ Sf= Je conected 197 = A (06) log 197 = log A + 6b 2.294 – 6b= log A -@. so the eq 0 &0 -log A 736 = 2.793 log A +66= 2.294 36=-0.499 [b=-0. 1663. log A = 3.2910. TA= 1954.33. Sf @ 104 espur =) S6 = 1954.33 (104) TSE = 422.47 Mpa -0.1663

Goalmann Criteria Sm + Sa = 1 Junts Sf Fos حملهع ) 0.1123 + 1.7237 = 1 690. 422:47 - Fos I fos=2345