Shear Stress Torsional Deflection Y RD2 X X

Example No. 2 A pulley of diameter 750 mm and a gear of 280

Reactions -Pulling forces of the belt should be resolved in the vertical and horizontal

Vertical Reactions 2120 N From Equilibrium: 900 N A B S M @ B

Horizontal Reactions From Equilibrium: S M @ B : 2120 N A B 375

Vertical Load Diagrams Horizontal 2120 N 900 N C A 375 D 375 1844

Critical Section at “C” is the critical one This section loaded by: 1. Shear

Shaft Design 1. Design Based on Bending Moment Mx =556. 5 N. m But,

Shaft Design 2. Design Based on twisting Moment Mt =127. 5 N. m And,

Shaft Design 3. Design Based on twisting & bending Moments Mt =127. 5 N.

Slides: 12

Download presentation

Shear Stress Torsional Deflection Y R=D/2 X X Y D L X Y Cross-section Stress Distribution Y X T

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/mm 2 and 71 N/mm 2 respectively. Determine the shaft diameter. 1670 N 450 A B 1330 N 375 900 N 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 2120 N From Equilibrium: 900 N A B S M @ B : 375 RA (375+500) = 375 500 (900 500)+2120 (375+500) 1250 RA = 450000+ 1855000 = 2305000 M @ A : RA RA = 1844 N RB (375+500) = 900 (375+375)+2120 375 1250 RB = 675000 + 795000 CKECK : RB = 1176 N FY = 0 : RA + RB = 1844+1176 = 2120 + 900 = 3020 Vertical reactions are right. RB

Horizontal Reactions From Equilibrium: S M @ B : 2120 N A B 375 HA (375+500) = 2120 (375+500) 375 500 1250 HA = 1855000 HA = 1484 N HA M @ A : HB (375+500) = 2120 375 1250 HB = 795000 HB = 636 N CKECK : FY = 0 : HA + HB = 1484 + 636 = 2120 Horizontal reactions are right. HB

Vertical Load Diagrams Horizontal 2120 N 900 N C A 375 D 375 1844 N + 276 2120 N B 500 S. F. D. 1176 + 1484 + 556. 5 588 N. m MT. D. 500 636 N 636 B. M. D. 127. 5 N. m 375 1484 N 1844 N _ B 375 1176 N + 619 A T =127. 5 N. m

Critical 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 But, My =619 N. m all = 71 N/mm 2 (given)

Shaft Design 2. Design Based on twisting Moment Mt =127. 5 N. m And, all = 57 N/mm 2 (given)

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 theories of elastic failure. (i. e. , Maximum shear or maximum principle stress theory).

EN D