The document summarizes the design of a toggle jack. It includes calculations to determine the appropriate dimensions for the square threaded screw, nuts, pins, and links. The screw diameter is determined to be 14 mm to withstand both tensile and shear stresses. Nut dimensions and a 4-thread design are selected for stability. Link thickness is calculated as 6 mm based on buckling load considerations. Overall, the design process involves analyzing each component to withstand appropriate stresses and loads from the 4 kN lifting force.

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.
• 𝞼 𝑡 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

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 Ѳ =
105−15
110
→ Ѳ = 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 =
𝑊
2 𝑡𝑎𝑛Ѳ
=
𝑊
2 tan 35.1
= 2846N
• This similar pull acts on other
nut,therefore total tensile load
on square threaded rod
→𝑊1= 2F =5692N

9. Design of Square threaded screw
• Now considering tensile failure
of the screw
→𝑊1=
𝞹
4
* 𝑑 𝑐
2
*𝞼 𝑡
→5692=
𝞹
4
* 𝑑 𝑐
2
*100
→𝑑 𝑐 = 8.5 mm say 10 mm
Since the screw is also subjected
to shear stress, therefore let
→𝑑 𝑐 = 14 mm

10. Design of Square threaded screw
• Outer Diameter of screw
→𝑑 𝑜 = 𝑑 𝑐 + P = 14+6
→𝑑 𝑜= 20 mm.
Mean Diameter of screw
→d = 𝑑 𝑜 - 𝑝
2 = 20 – 3
→d = 17 mm

11. Checking of the screw for principal stress
• tan α =
𝑝
𝞹 𝑑
=
6
𝞹∗17
= 0.1123
• Effort required to rotate the
screw
P = 𝑊1 *tan(α + Ø)
P = 𝑊1(
tan α+ tan Ø
1−tan α tan Ø
)
= 5692(
0.1123+0.20
1−0.1123∗0.20
)
P = 1822N

12. Checking of the screw for stress
• Torque required to rotate the
screw
→T = P*
𝑑
2
= 1822*17/2
→T = 15487 N.mm
Shear stress due to torque
→𝞽 = 16х
T
(𝞹∗𝑑 𝑐
3
)
=
16∗15487
𝞹∗143
→𝞽 = 28.7 N/𝑚𝑚2
• Direct Tensile stresses in the
screw
→𝞼 𝑡 =
𝑊1
𝞹
4
∗ 𝑑 𝑐
2 =
5692
0.7855∗142
→𝞼 𝑡 = 37 N/𝑚𝑚2

13. Checking for maximum principal stress
• Maximum Principal stress
→𝞼 𝑡(𝑚𝑎𝑥) =
𝞼 𝑡
2
+
1
2
𝞼 𝑡
2 + 4𝞽2
→𝞼 𝑡(𝑚𝑎𝑥)=
37
2
+
1
2
372 + 4 ∗ 28.72
→𝞼 𝑡 𝑚𝑎𝑥 = 52.6 N/𝑚𝑚2
• Maximum shear stress
→𝞽 𝑚𝑎𝑥 =
1
2
𝞼 𝑡
2 + 4𝞽2
→𝞽 𝑚𝑎𝑥 =
1
2
372 + 4 ∗ 28.72
→𝞽 𝑚𝑎𝑥 = 34.1 N/𝑚𝑚2
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.
𝑝 𝑏=20=
𝑊1
𝞹
4
𝑑 𝑜
2
−𝑑 𝑐
2
𝑛
=
5692
𝞹
4
202−142 𝑛
→ 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*𝑑 𝑜=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 𝑑1 is diameter of the pin.
• Considering double shear of the
pin.
→ F = 2*
𝞹
4
∗ 𝑑1
2
*𝞽
→2846 = 2*
𝞹
4
∗ 𝑑1
2
*50
→𝑑1 = 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
→𝑊𝑐𝑟 = 5*1423 = 7115 N

22. Dimensions of the link
• Let 𝑡1 = thickness of the link
• Let 𝑏1 = width of the link
• Assuming 𝑏1 = 3𝑡1 .
• cross sectional area of the link ,A
→A = 3𝑡1 *𝑡1 = 3𝑡1
2
• Moment of inertia of the cross
section ,I
→I =
1
12
*𝑡1 *3𝑡1
3
= 2.25𝑡1
4
• Radius of gyration k
→K =
𝐼
𝐴
= 0.866𝑡1

23. Buckling of the link in vertical plane
• In buckling in vertical plane link
is considered as hinged.
Therefore, L = l = 110 mm
Rankine’s constant a =
1
7500
According to Rankines formula for
column,
𝑊𝑐𝑟 =
𝞼 𝑐 ∗𝐴
1+𝑎(
𝐿
𝑘
)2

24. Buckling of the link in vertical plane
• 7115 =
100 ∗3𝑡1
2
1+
1
7500
(
110
0.866𝑡1
)2
• 𝑡1
2 =
23.7+ 23.72+4∗51
2
=
25.7
• 𝑡1 = 5.07 mm≈ 6mm and
• 𝑏1 = 3*6 = 18 mm

25. Buckling of the link in plane perpendicular to vertical plane
• I =
1
12
*3𝑡1 *𝑡1
3 = 0.25𝑡1
4
• A = 3𝑡1 *𝑡1 = 3𝑡1
2
• K =
𝐼
𝐴
= 0.29 𝑡1
• 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 𝑊𝑐𝑟 =
𝞼 𝑐 ∗𝐴
1+𝑎(
𝐿
𝑘
)2
→ 𝑊𝑐𝑟 =
100 ∗3𝑡1
2
1+
1
7500
(
55
0.29𝑡1
)2
Taking 𝑡1 = 6 mm
→ 𝑊𝑐𝑟 = 9532 N.
Since buckling load is more then the calculated value,
link is safe in design.
So dimension of the link
• 𝑡1 = 6mm and
• 𝑏1 = 18 mm