Question

1 r s e e Thebe For a bolted by with both of each stitness of the members is = 25 MN perbeit. The of proof strength. Assume the allodialy M6 x 1 class 58 with rolled threads. Afici o with P = 60 N and P_ = 20 N (a) Determine the yielding factor of ty (b) Determine the overiod faca o sa (c) Determine the factor of safety based on joint sur (d) Determine the fatigue factor of styssing the Goodien ne heits ne = CP Pinux - Pmax Penin - Doin tp = SPA At = 20 lum? Sp= 380 MPa ; f = 0,75 At So M = Sp At-Fi plis i ša=c( Pine Preis mean = c(Peaux +P Emin & Fi 2 Az Sat = 520 MPa At Sé = 0,55 t ; ka = a Suts; a = 4 b = -0,265 ; Kf = 1 ; kc =0,85 ; Kf = 2,2 Se = kakekekake Kf. t.se ne SE ( Sut -si) Sat 5a + Se (sm-si) At

mechanical design

Answer 1

Solution Per bort . . 60, Pmax Pomax = - nooits Pomax = 60 Pomax = 7.5KN Pain Pomin - Mbolts Pbmiก - 20 Pbmin = 2.5kn KEI MN/mm 2.6 MN/mm SKb Kot km 1 +2.6 C =0,278 - AL -20.1 mm? Sp - 380 MPa F; -0.75 At SP Fi = 0.75 (20.1) 380610-3) Fi - 5.73 KN

The yielding factor of safety np = SPAL СР+Е; npe 380(20.11.1093 0.27867-5)+5.73 7.638 - 7.815 np np = 0.98 The yielding factor of safety is np = 0.98 b) The overload Factor of safety n = Sp AL-F; 1 CP ni z 380(20.1) 103-15.73) 0.278(7.5) nl = 1.908 2.085 me=0.915) The overload factor of safety is ne=0.9155 c) The factor of safety based on joint sepasation no = Fil pli-c) 5.73 no no = 5(1-0.278)

5.72 no 5.415 no = 1.06 The factor of safety based on joint separation is no=1.06 DjThe factor of Safety using Good man coiterion Ta = c(Pomax - Pb min) ZAL Ca 0.27217.5-2.5) 103 2x20.1 Ta - 1390 40.2 [5a = 34.6 MPa] on = c(Pbmax +Pomin) + Ei RAL 0.278(7.5+2.5) 100 5.73x113 2013 20.1 2120.1 om 2780 + 5730 m ū0.2 20.1 60 = 354.2 MPa) sut = 52ompa At G B48 5.73 X103 20.1

сі = 28S MPC se' = 0.5 Sut sé = 0.5(520) se = 260MPa a = 4.51 b = -0.265 Ka=asut Ra = 4.5 (s20 -0, 266 ) Ba = 0.86 0 Kb=1 Kc =0.85 kg = 2.2 ka K, Kc se /кс 0.86 x1 80.85260/2.2 Se=86.4 MPa ne : Se (sut -07) Sut at se com-oi) ne = 86.4 (520 - 285) 520034.6) +86.M(354-2-285) 20304 ng 23970.82 ne = 0.847] The factor of safety using Good man coiterion - ne =0.847