Design of knuckle joint (machine design & industrial drafting )
- 1. Sub :- MACHINE DESIGN AND INDUSTRIAL DRAFTING Shroff S.R. Rotary Institute of Chemical Technology Principle Supporter & Sponsor-United Phosphorous Ltd(UPL)/Shroff family Managed By Ankleshwar Rotary Education Society Approved by AICTE, New Delhi, Govt. of Gujarat & GTU Affiliated Prepared by :- 150990119013 :- MOIN MALEK 150990119014 :- TANMAY MANDALIYA 150990119015 :- PRATIJITSINH MANGROLA 150990119016 :- MILIND JAGTAP 150990119017 :- NISARG SANT Topic :- DESIGN OF KNUCKLE JOINT
- 2. KNUCKLE JOINT • A knuckle joint is a mechanical joint used to connect two rods which are under a tensile load, when there is a requirement of small amount of flexibility, or angular moment is necessary. There is always axial or linear line of action of load.
- 4. APPLICATIONS
- 5. Dimensions of knuckle joint are obtained as : 1. Diameter of rod (d) 2. Diameter of knuckle pin (dp) 3. Thickness of single eye (t) 4. Thickness of fork (t1) 5. Outside diameter of eye (D) 6. Failure of fork end in tension 7. Failure of fork end in shear 8. Failure of knuckle pin in bending 9. Standard propagation of knuckle joint
- 6. Diameter of rod (d): • The rod is subjected to a direct tensile stress and is given by, • From the equation, d can be obtained. Diameter of knuckle pin (dp) : • The knuckle pin is subjected to double shear. • The direct shear stress induced in a knuckle pin is, 4/2 d P t 4/)(2 2 p t d P
- 7. Thickness of single eye (t) : • The area of contact between the knuckle pin and the single eye is subjected to a crushing stress. • The crushing stress is induced by, • The minimum thickness of single eye is taken as, • “t” is taken as larger of two values. Thickness of fork (t1) : • the contact area between the knuckle pin and the fork is subjected to a crushing stress as, • Where the minimum thickness of the fork is taken as, • “t1” is taken as larger of two values. td P p t dt 25.1 12 td P p t dt 75.01
- 8. Outside diameter of eye (D) : • The single eye is subjected to direct tensile stress. • In addition, the single eye is subjected to double shear stress. • The outside diameter of the single eye “D” is calculated and the larger of the two values is adopted. Failure of fork end in tension and shear : Similar to the single eye, the fork end is subjected to direct tensile stress (σt) and double shear stress(τ) which is given as, tdD P p t )( tdD P p t )( 1)(2 tdD P p t 1)(2 tdD P p t
- 9. SUMMARY
- 10. Design a knuckle joint to connect two circular MS rods which are subjected to a tensile load of 63 kN. The allowable stresses are 80 MPa in tension, 56 MPa in shear and 80 MPa in crushing. Solution: Given: P = 63 kN = 63 x 10³ N σt = 80 N/mm² = 80 x 10³ τ = 56 N/mm² σc = 80 N/mm²
- 11. Diameter of rod (d) : The tensile stress induced in a rod is, d = 31.66 mm d = 32 mm Diameter of knuckle pin (dp) : The direct shear stress induced in the knuckle pin is, dp = 26.76 mm dp = 27 mm 4/2 d P t 4/ 10*63 80 2 3 d 4/)(2 2 p t d P 4/)(2 10*63 56 2 3 pd
- 12. Thickness of single eye (t) : The crushing stress is induced by, t = 29.16 mm t = 30 mm The minimum thickness of single eye is, = 40 mm so, t = 40 mm (taking larger value) Thickness of fork (t1) : t1 = 15 mm The minimum thickness of the fork is, t1 = 0.75d = 0.75*32 = 24 mm (taking larger value) so, t1 = 24 mm td P p t t*27 10*63 80 3 dlb 25.1 32*25.1t 12 td P p c 1 3 *27*2 10*63 80 t
- 13. Outside diameter of eye (D) : Tensile stress induced in a single eye, D = 46.88 mm or D = 47 mm Shear stress induced in a single eye, D = 55.32 mm or D = 56 mm tdD P p t )( 40*)27( 10*63 80 3 D tdD P p t )( 40*)27( 10*63 56 3 D
- 14. Failure of fork eye in tension : σt =45.26 N/mm² < 80 N/mm² Hence, fork end is safe against tensile failure. Failure of fork end in shear : τ = 45.26 N/mm² < 56 N/mm² Hence, fork end is safe against shear failure. 1)(2 tdD P p t 24*)2756(2 10*63 3 1)(2 tdD P p t 24*)2756(2 10*63 3