Design of toggle jack

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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

Transcript of Design of toggle jack

Page 1: Design of toggle jack

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

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Content:

• Mechanism of toggle jack• Design of the screw• Design of the Nut• Design of the Pin• Design of spanner• Design of the Link

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Design of toggle jack

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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• 𝞽 for M.S. = 50 MPa• Limited bearing pressure = 20 MPa• Pitch of the screw thread = 6 mm• Co-efficient of the friction = 0.20

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Toggle Jack

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Design of screw

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Design of Square threaded screw

• Maximum load on screw occurs when the jack is in the bottom position.• From figure = → Ѳ = 35.1°

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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 = = = 2846N• This similar pull acts on other

nut,therefore total tensile load on square threaded rod

→= 2F =5692N

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Design of Square threaded screw

• Now considering tensile failure of the screw

→= * *→5692= * *100→ = 8.5 mm say 10 mmSince the screw is also subjected to shear stress, therefore let→ = 14 mm

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Design of Square threaded screw

• Outer Diameter of screw → = + P = 14+6→= 20 mm.Mean Diameter of screw→d = - = 20 – 3→d = 17 mm

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Checking of the screw for principal stress• = = = 0.1123• Effort required to rotate the

screwP = *) P = () = 5692()P = 1822N

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Checking of the screw for stress

• Torque required to rotate the screw

→T = P* = 1822*17/2 →T = 15487 N.mmShear stress due to torque→𝞽 = 16х = →𝞽 = 28.7 N/

• Direct Tensile stresses in the screw

→ = = → = 37 N/

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Checking for maximum principal stress• Maximum Principal stress→ = + + 4→=+→= 52.6 N/

• Maximum shear stress→ = + 4→ =→ = 34.1 N/

Since the maximum stresses are within limit, therefore design of square threaded screw is safe.

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Design of Nut

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Number of threads

• Let n is the number of the threads, which can be find by considering bearing failure of nut.

=20= =→ n = 1.776Inorder to have good stability and to prevent rocking of the screw let n = 4

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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

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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

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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 mmWe shall take length of spanner as 200 mm in order to facilitate the operation

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Design of the pins in the Nuts

• Let is diameter of the pin.• Considering double shear of the

pin.→ F = 2**𝞽→2846 = 2**50→ = 6.02 mm ≈ 8mm

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Design of the Links

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Load acting on Link

• Load on the link = F/2 = →2846/2 = 1423 NAssuming factor of safety = 5→ = 5*1423 = 7115 N

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Dimensions of the link• Let = thickness of the link• Let = width of the link• Assuming .• cross sectional area of the link ,A→A = * = • Moment of inertia of the cross

section ,I→I = ** = 2.25• Radius of gyration k→K = = 0.866

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Buckling of the link in vertical plane• In buckling in vertical plane link

is considered as hinged.Therefore, L = l = 110 mmRankine’s constant a = According to Rankines formula for column,

=

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Buckling of the link in vertical plane

• 7115 = • = = 25.7• = 5.07 mm≈ 6mm and• = 3*6 = 18 mm

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Buckling of the link in plane perpendicular to vertical plane

• I = *3* = 0.25 • A = * = • K = = 0.29 • 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

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Buckling of the link in plane perpendicular to vertical plane

• According to Rankine's formula = = Taking = 6 mm = 9532 N.Since buckling load is more then the calculated value,

link is safe in design.

So dimension of the link• = 6mm and• = 18 mm