knuckle joint.pptx

Information about knuckle joints

1. Design of Knuckle Joint Sunilkumar N Chaudhari Mechanical engineering Department BBIT V.V.Nagar Date :30/7/2020

2. Introduction Knuckle joint is used to connect two rods subjected to axial tensile loads. The joint permit slight angular movement between the rods and little offset between the axis of the rods. It is not suitable to connect rotating shafts which transmit torque.

3. Main parts of knuckle joint Single eye end(rod end) Double eye end (fork end) Knuckle pin Collar Taper pin(split pin)

4. Application of knuckle joint To connect valve rod and eccentric rod of steam engine . To connect structural members under tension. To connect links in some material handling equipment. In locomotive engine for connecting links for The air brake assembly.

5. Advantages of the knuckle joint Permits slight angular movement between the rods. Easy to assemble and disassemble. Permits little offset between the axes of the rods

6. Design procedure of knuckle joint

7. 1.Tensile failure of rod

8. Refer fig. P= A×σ t P=Л/4×d 2 × σ t

9. 2. Shear Failure of knuckle pin

10. As shown in fig. The pin may fail in double shear due to tensile load P=2× Л/4×d p 2 ×[τ] Where d p =pin diameter

11. 3.Tensile failure of the single eye-having hole for pin

12. Eye dimension are obtained by taking proportions. D 1 =2d p ; t=1.2 d p Checking the eye for tensile failure P=( D 1 - d p )×t× σ t If σ t ˂[ σ t ] the eye is safe for tensile failure

13. 4. Crushing failure of single eye/knuckle pin

14. P= d p ×t × σ c If σ c ˂[ σ c ] the eye / pin is safe for crushing failure

15. 5. Tensile failure of double eye end /forked end

16. P=2×C× t 1 × σ t Here, t 1 =t/2 C=2 t 1

17. 6. Tensile failure of double eye through the hole for the pin

18. P=2(D 1 - d p )t 1 × σ t If σ t ˂[ σ t ] the double eye is safe for tensile failure

19. 7. Double shear failure of the single eye/ double eye

20. P=2×[(D 1 - d p )/2] × t×τ If τ˂[ τ] the double eye end / single eye end is safe against shear failure.

21. 8. Bending failure of knuckle pin

22. In the knuckle joint, the clearance is provided between pin and the hole of single and double eye and hence the possibility of bending of the pin will be there. Pin is considered to be a beam subjected to UDL on its length in the single eye and varying load on the length in the double eye with maximum load at the inner most end and zero at outer end . Hear , as shown in fig. a = distance of line of action of load on one side of single eye end from the axis of the joint. b = distance of line of action of load on one side of double eye end from the axis of the joint. With this loading , maximum bending stress in the pin is determined and compared with allowable to check for the failure.

23. Bending moment , M b =P/2×b - P/2×a =P/2 (b-a) b = t/2 +t 1 /3 ; a=t/4 M b =P/2[t/2 +t 1 /3 - t/4] =P/2 [t/4 +t 1 /3] Taking , t 1 = t/2   M b = P/2 [t/4 +t/6] M b =5/24 P×t σ b = M b /Z Z= Л/32 d p 3 Here , If σ b ˂[ σ b ] the pin is safe for bending  

24. Problem 1 Design of knuckle joint to connect two m .s rods of equal diameter subjected to axial tensile load of 12kN [ σ t ]=75N/mm 2 ; [ σ c ] =140 N/mm 2 ; [τ]=55 N/mm 2

25. Solution 1. Tensile failure of rod Refer fig. P= A×σ t P=Л/4×d 2 × σ t 12000= Л/4× d 2 ×75 d 2 =203.71 d =14.27mm= 16mm

26. 2.Shear Failure of knuckle pin As shown in fig. The pin may fail in double shear due to tensile load P=2× Л/4×d p 2 ×[τ] Where d p =pin diameter 12000=2× Л/4×d p 2 ×55 d p =11.78 note : here ,pin is also subjected to bending and hence possibility of bending is more compared to shear. Hence the diameter of pin is increased by 25% to 30% to that obtained for shear failure. Generally for safety and simplicity in design , the pin diameter is taken same as rod diameter. d p =d=16mm

27. 3.Tensile failure of the single eye-having hole for pin Eye dimension are obtained by taking proportions. D 1 =2d p ; t=1.2 d p D 1 =2×16=32mm t =1.2 ×16 =19.2=20mm Checking the eye for tensile failure P=( D 1 - d p )×t× σ t 12000=( 32 – 16)×20× σ t σ t =37.5 N/mm 2 allowable stress [ σ t ]=75N/mm 2 as σ t ˂[ σ t ] The eye is safe for tensile failure

28. 4.Crushing failure of single eye/knuckle pin P= d p × t×σ c 12000= 16 ×20×σ c σ c =37.5 N/mm 2 here, σ c ˂[ σ c ] , hence the eye / pin is safe for crushing failure  

29. 5.Tensile failure of double eye end /forked end P=2×C× t 1 × σ t Here, t 1 =t/2 =20/2=10mm ; C=2t 1 =2×10=20mm 12000=2×20× 10× σ t σ t =30 N/mm 2 here, σ t ˂[ σ t ] the double eye is safe for tensile failure

30. 6.Tensile failure of double eye through the hole for the pin P=2(D 1 - d p )t 1 × σ t 12000=2(32- 16 )10× σ t σ t =37.5 N/mm 2 here,σ t ˂[ σ t ], hence double eye is safe for tensile failure

31. 7.Double shear failure of the single eye/ double eye P=2×[(D 1 - d p )/2] × t×τ 12000=2×[(32- 16 )/2] ×20×τ τ =37.5 N/mm 2 here , τ˂[ τ] , hence the double eye end / single eye end is safe against shear failure.

32. 8. Bending failure of knuckle pin In the knuckle joint, the clearance is provided between pin and the hole of single and double eye and hence the possibilityof bending of the pin will be there. Pin is considered to be a beam subjected to UDL on its length in the single eye and varying load on the length in the double eye with maximum load at the inner most end and zero at outer end . Hear , as shown in fig. a = distance of line of action of load on one side of single eye end from the axis of the joint. b = distance of line of action of load on one side of double eye end from the axis of the joint. With this loading , maximum bending stress in the pin is determined and compared with allowable to check for the failure.

33. Bending moment,M b =P/2×b - P/2×a =P/2 (b-a) b = t/2 +t 1 /3 ; a=t/4 M b =P/2[t/2 +t 1 /3 - t/4] =P/2 [t/4 +t 1 /3] Taking , t 1 = t/2   M b = P/2 [t/4 +t/6] M b =5/24 P×t σ b = M b /Z taking [ σ b ] =1.4 [ σ t ] [ σ b ]= 1.4×75 [ σ b ] =105N/mm 2   Z= Л/32 d p 3 = Л/32 ×(16) 3 Z =402.12 mm 3

34. M b =5/24×12000×20 M b =50000N.mm σ b = M b /Z σ b =50000/401.92 σ b =124.40 N/ mm 2 Here , σ b ˃[ σ b ] the pin is not safe in bending Let pin diameter, d p =20mm Z= Л/32 ×(20) 3 Z=785.39 mm 3 σ b = M b /Z σ b = 50000/785.39 σ b =63.66 N/ mm 2 Here , σ b ˂[ σ b ] the pin is safe in bending