Question

Baskı silindiri 1.2 KN 20° 60 mm 300 mm O 60 mm 100 mm Düz dişli In the figure, the movement to the printer spindle on the 12mm floor is from the vertex of the spur gear positioned C position. 1.2kN force applied by the gear will be applied. Between and between B bearings 4N/mm uniformly distributed load(up true). Press spindle servicing at 350 rpm for 6 hours a day, 300 days a year and 2 years It is desirable. i) Plot cut and moment diagrams for each force plane of the spindle. ii) Find the resultant forces to the A and B roller bearings. iii) Select suitable bearing for Spindle Bearings according to maximum resultant force. iv) By ignoring the torsional moment that will occur on the shaft, Check the safety of von Mises according to the static damage criteria. (Yield strength for shaft yield strength=820MPa) Note: Free body diagrams and equations of the shaft should be clearly drawn and written. No bearing calculations will be made statically for this question.

Answer 1

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solution: given F-2 plane, F2 = F cos 20° de A c RB2 2 1 RAZ : F2 = RAZT RB2 F cosao° 12 Cos 20 = 1.127 KN Taking Moment @ A, FZ (AC) = RBZ CAB) 1.127 x 0.52 : RB2 0.42 : RB2 = 1:39 KN given, Fe=RAzt & B2 1.127 - RAZ + 1.39 .. Raz = -0.263(1) = + 0.263 In Section 1-1, -0.263- : Mac - RAZ x = 26=0.42 = OC=0M= -0263 (0) = 0 M= -0.263 x 0-42 = -0.11 kN-m In Section 2-2 - Raz. x + RB2 (x- 0.42) Moc =

At, 26= 0.42 , M = -0.11 KN-m a=0.52 M= -0.263 x 0.52 + 1.396.52-0.42) = -0.1367 + 0.139 = 2.x 103 KN-m → Moment diagram will be : 2.3x103 K Nam (-0.11 KN-m → For y-xc plane : Fy - Fsin 20° 50.41 KN 1 A 1 4 KN/mm 2 4 RBY RAY Ray + "By (4x300) + Fy - 1200+ 0.41 = 1200.41 KN Taking moment @ A, -- 1200* (+0.06) - PBy CAB) + Fry (AC) = 0 : 1200 x 0.21 +0.41 X 0.52 = Rby (0.42) & By = 600.5 KN

RAY + Roy = 1200.41 .. RAY 1200.41- 600.5 = 599.91 KN In Section 1-1, -Ray 599.91 a . Ma = ut oc=0, Mx =0 =) ( = 0.06 Ma = - 599.9180.06 = -35.99 KN-m In section 2-2, + 4 (4 -0.05) x (0-0.06) MaL - Ray x 2 3-599.91 x + 2 (21-0.06)2 at x = 0.36 Mx = -599.91 x 0.36 + 2 (0.36= 0.06) 2 1 2-215.78 KN-m In Section 3-3, Moca 599.91 ou + 1200 (x-10:15+0.06)) est oc= 0.42, . MOC = -599.91 (0.42) + 1200 (0.42 - 0.21) 0.037 KN-m In Section 4-4, Moc = 1200 (OL-0.21) - Pay. IC – R BY (0-0.42) 9t 20= 0.52, 599.91(0.52) + 1200 (0.52-0.21) – 600.5 (0.52-0:42) . Moc -3 : Moc = -3.2 x 10 kN-m

→ The Moment Diagsam will be +0.037 KN-m B A -3 -3.2x10 KN-M -35.99 KN-m -215.78 KN-m (ii) Forces At n, fty = 599.91 KN, PA25 0.263 KN 600.5 KN IR BZ = 1.39 KN RBY RA - VR Ango+FAZZ Recytant ... At A, 1599.9,² +0.263² 599.91 KN - RA Resultent at B, RB = V Rey² +RB22 V 1600.5+ (1.39² 1 =600.5 KN (iii) Tortal Working hours, Lo = 6x 300X2 = 3600 hes 6 10 The Life, L10= (&) ²x 6ON Where, p=3 For bull bearings, : p= Fmax = Maximum load on beusings,

And C= dynamic capacity C : 3600 = 600.5 x 103 3 6 10 60x350 C= 2539. 14 KN → Under this condition SAE 52100 M12 deep grove ball bearing is suitable. (iv) If we plot the sheus force in the y-a plane, : RAY Е ву = 599.91 KN = 600.5 KN .: at 0-36 m from A, F- Ray - 4x300 2 599.91- 1200 -0.41 KN - 600.09 KN → Sheus Force Diggrum will be 599,91 KN 600.5 KN in А -0.41 KN - 600.09 KN → The Dommant plane is yox plane, so x-2 plane neglected. Mux. Force = 600.5 KN Shear stress Force Area

Is = 600.58103 Is = 600.5X103 Area : Area = TX (12 x 10-3)? 4 m2 = 1.13 x 10 1.13. 10-4 Is = 5.314 x 100 N/m2 Is = 531.4 MPa) MP4, GAK Factor of safety IS 820 531.4 FOS 1:543