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### Transcript of Design of toggle jack

1. 1. L.D. college of engineering Subject: Machine Design and Industrial Drafting Topic: Design of toggle jack Year:2014-15 Sr.No. Name Enroll No. Roll No 1 Kuralkar Hemant Yogeshbhai 140283119009 419245 2 Shinde Kunal Bharatbhai 140283119024 419259 3 Sabalpara Nilesh 140283119023 419258 4 Rathod Jaydipsinh 140283119022 419257 5 Patel Sagar 140283119019 419254 6 Vaghela Kanu 140283119025 419260 7 Jadav Vipul 140283119007 419243
2. 2. Content: Mechanism of toggle jack Design of the screw Design of the Nut Design of the Pin Design of spanner Design of the Link
3. 3. Design of toggle jack
4. 4. Design of toggle jack- Assumed data Lifting load = 4KN Number of Link = 8 Length of the link = 110 mm Materials for the screw, Nut and pins = M.S. for M.S. = 100 MPa for M.S. = 50 MPa Limited bearing pressure = 20 MPa Pitch of the screw thread = 6 mm Co-efficient of the friction = 0.20
5. 5. Toggle Jack
6. 6. Design of screw
7. 7. Design of Square threaded screw Maximum load on screw occurs when the jack is in the bottom position. From figure cos = 10515 110 = 35.1
8. 8. Design of Square threaded screw Each nut carries half the total load on the jack Link CD is subjected to tension while the square threaded screw is under pull. F = 2 = 2 tan 35.1 = 2846N This similar pull acts on other nut,therefore total tensile load on square threaded rod 1= 2F =5692N
9. 9. Design of Square threaded screw Now considering tensile failure of the screw 1= 4 * 2 * 5692= 4 * 2 *100 = 8.5 mm say 10 mm Since the screw is also subjected to shear stress, therefore let = 14 mm
10. 10. Design of Square threaded screw Outer Diameter of screw = + P = 14+6 = 20 mm. Mean Diameter of screw d = - 2 = 20 3 d = 17 mm
11. 11. Checking of the screw for principal stress tan = = 6 17 = 0.1123 Effort required to rotate the screw P = 1 *tan( + ) P = 1( tan + tan 1tan tan ) = 5692( 0.1123+0.20 10.11230.20 ) P = 1822N
12. 12. Checking of the screw for stress Torque required to rotate the screw T = P* 2 = 1822*17/2 T = 15487 N.mm Shear stress due to torque = 16 T ( 3 ) = 1615487 143 = 28.7 N/2 Direct Tensile stresses in the screw = 1 4 2 = 5692 0.7855142 = 37 N/2
13. 13. Checking for maximum principal stress Maximum Principal stress () = 2 + 1 2 2 + 42 ()= 37 2 + 1 2 372 + 4 28.72 = 52.6 N/2 Maximum shear stress = 1 2 2 + 42 = 1 2 372 + 4 28.72 = 34.1 N/2 Since the maximum stresses are within limit, therefore design of square threaded screw is safe.
14. 14. Design of Nut
15. 15. Number of threads Let n is the number of the threads, which can be find by considering bearing failure of nut. =20= 1 4 2 2 = 5692 4 202142 n = 1.776 Inorder to have good stability and to prevent rocking of the screw let n = 4
16. 16. Dimensions of the Nut Thickness of nut t = n*p = 4*6 t = 24 mm. Width of the nut b = 1.5* =1.5*20 b = 30 mm
17. 17. Length of the screw To control the movements of the nuts beyond 210mm , rings of 8mm thickness are fitted on the screw with the help of set screw. length of screwed portion = 210 + 24 + 2*8 = 250 mm Since screw is operated by spanner, therefore extra 15 mm length both sides are provided. Total length = 250 + (2*15) = 280 mm
18. 18. Length of spanner Assuming that a force of 150 N is applied by each person at each end of the rod, T = 150*2*Length of spanner 15487 = 150*2*L Length of spanner = 51.62 mm We shall take length of spanner as 200 mm in order to facilitate the operation
19. 19. Design of the pins in the Nuts Let 1 is diameter of the pin. Considering double shear of the pin. F = 2* 4 1 2 * 2846 = 2* 4 1 2 *50 1 = 6.02 mm 8mm
20. 20. Design of the Links
21. 21. Load acting on Link Load on the link = F/2 = 2846/2 = 1423 N Assuming factor of safety = 5 = 5*1423 = 7115 N
22. 22. Dimensions of the link Let 1 = thickness of the link Let 1 = width of the link Assuming 1 = 31 . cross sectional area of the link ,A A = 31 *1 = 31 2 Moment of inertia of the cross section ,I I = 1 12 *1 *31 3 = 2.251 4 Radius of gyration k K = = 0.8661
23. 23. Buckling of the link in vertical plane In buckling in vertical plane link is considered as hinged. Therefore, L = l = 110 mm Rankines constant a = 1 7500 According to Rankines formula for column, = 1+( )2
24. 24. Buckling of the link in vertical plane 7115 = 100 31 2 1+ 1 7500 ( 110 0.8661 )2 1 2 = 23.7+ 23.72+451 2 = 25.7 1 = 5.07 mm 6mm and 1 = 3*6 = 18 mm
25. 25. Buckling of the link in plane perpendicular to vertical plane I = 1 12 *31 *1 3 = 0.251 4 A = 31 *1 = 31 2 K = = 0.29 1 Since the buckling of the link in plane perpendicular to the vertical plane, the ends are considered as the fixed. L = l/2 = 110/2 = 55mm
26. 26. Buckling of the link in plane perpendicular to vertical plane According to Rankine's formula = 1+( )2 = 100 31 2 1+ 1 7500 ( 55 0.291 )2 Taking 1 = 6 mm = 9532 N. Since buckling load is more then the calculated value, link is safe in design. So dimension of the link 1 = 6mm and 1 = 18 mm