# Design of toggle jack

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14-Aug-2015Category

## Engineering

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### Transcript of Design of toggle jack

- 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. 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. Design of toggle jack
- 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. Toggle Jack
- 6. Design of screw
- 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. 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. 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. 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. 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. 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. 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. Design of Nut
- 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. 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. 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. 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. 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. Design of the Links
- 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. 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. 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. 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. 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. 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