Calculation of Shaft Diameter for Pulley and Gear Assembly
A detailed analysis is conducted to determine the shaft diameter required for a system consisting of a pulley and a gear on a steel shaft. Reactions, load diagrams, critical section identification, and stress distribution are assessed to calculate the dimensions effectively based on the applied forces and material properties.
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Presentation Transcript
Shear Stress 4 4 D 50 4 = = = = = = I 613281 mm p 32 32 M r 219000 25 = = t 2 = = = = 9 N / mm I 613281 p Torsional Deflection M G = = t L I r p Y R=D/2 Y X X D L T Y Stress Distribution Y Cross-section
M L = = t GI p 219000 450 = = = = 0 0016 . rad 4 9 95 . 10 613281 I p 360 = = = = = = 0 0061 . rad 0 0061 . 0 0925 . deg ree 2
Example No.2 A pulley of diameter 750 mm and a gear of 280 mm diameter are mounted on a steel shaft. The belt pulls on the pulley with 1670 N and 1330 N at 450. The gear tooth force is 900 N . The allowable shear stress and tensile stress for the shaft material are 57 N/mm2and 71 N/mm2respectively. Determine the shaft diameter. 1670 N 450 A B 1330 N 900 N 375 375 500
Reactions -Pulling forces of the belt should be resolved in the vertical and horizontal directions. i.e., Fv: Vertical tension component. Fh: Horizontal tension component. Fv = Fh = (1670+1330)sin 450 = 3000 sin 45 =2120 N
Vertical Reactions From Equilibrium: 2120 N 900 N A B M @ B : 375 375 500 RA (375+375+500) = (900 500)+2120 (375+500) 1250 RA= 450000+ 1855000 = 2305000 M @ A : RB RA RA = 1844 N RB (375+375+500) = 900 (375+375)+2120 375 1250 RB= 675000 + 795000 RB = 1176 N CKECK : FY= 0 : RA+ RB= 1844+1176 = 2120 + 900 = 3020 Vertical reactions are right.
Horizontal Reactions From Equilibrium: 2120 N A B M @ B : 375 375 500 HA (375+375+500) = 2120 (375+500) 1250 HA= 1855000 HA = 1484 N HB HA M @ A : HB (375+375+500) = 2120 375 1250 HB= 795000 HB = 636 N CKECK : FY= 0 Horizontal reactions are right. :HA+ HB= 1484 + 636 = 2120
Load Diagrams Vertical Horizontal 2120 N C 900 N 2120 N D B A B A 375 375 500 375 375 500 1844 N 1176 N 1484 N 636 N 1844 N 1484 + + S.F.D. _ 636 276 1176 B.M.D. + + 556.5 588 N.m 619 127.5 N.m MT.D. T =127.5 N.m
Critical Section Section at C is the critical one This section loaded by: 1.Shear force Qx=1484 N & Qy=1844 N (Neglected values) 2. Bending Moment Mx=556.5 N.m & My=619 N.m 3. Torque Mt= 127.5 N.m
Shaft Design 1. Design Based on Bending Moment Mx=556.5 N.m My=619 N.m 2 y x b equiv 2 = = + + M M ( ) M ( ) Mb= = 832 N m . 2 2 = = + + M 556 ( 5 . ) 619 ( ) b equiv But, all = 71 N/mm2 (given) 3 8 32 10 D 2 832000 32 = = = = b 4 3 D D M Y bx 64 = = b I 832000 32 x = = 71 3 D 832000 32 = = = = D 49 22 . mm 3 71 = = D 50 mm
Shaft Design 2. Design Based on twisting Moment Mt=127.5 N.m And, all = 57 N/mm2 (given) 3 127 5 . 10 D 2 M r = = 4 D = = t 32 I 127500 16 p = = 57 3 D 127500 16 = = = = D 22 49 . mm 3 57 = = D 23 mm
Shaft Design 3. Design Based on twisting & bending Moments Mt=127.5 N.m Mx=556.5 N.m My=619 N.m Design will be carried out according to one of the theories of elastic failure. (i.e., Maximum shear or maximum principle stress theory).
