Design of the Toggle Jack

Scissor Jack Design Project GE 410 Fall 2005

Jim Ramirez David Hettinger

Instructor: Hall 12/05/2005

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ABSTRACT Scissor jacks are simple mechanisms used to drive large loads short distances. The power screw design of a common scissor jack reduces the amount of force required by the user to drive the mechanism. Most scissor jacks are similar in design, consisting of four main members driven by a power screw. In this report, a unique design of a scissor jack is proposed which is very easy to manufacture. Each member, including the power screw sleeves, is made of the common c-shape. This eliminates the need for machined power screw sleeves, which connect the four members and the power screw together. The manufacturability of the proposed scissor jack lowers the cost of production.

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TABLE OF CONTENTS Title Page…………………………………………………………………………………... i Abstract…………………………………………………………………………………….. ii Table of Contents…………………………………………………………………………... iii Introduction………………………………………………………………………………… Proposed Design…………………………………………………………………………… Figure 1: Labeled Scissor Jack Design…………………………………………………….. Table 1: Design Criteria……………………………………………………………………. Conclusions and Recommendations……………………………………………………….. Appendix A: Drawings…………………………………………………………………….. Appendix B: Calculations and Assumptions for Components 2, 4, 6, and 8……………… Appendix C: Calculations and Assumptions for Components 3 and 7……………………. Appendix D: Calculations and Assumptions for Components 1 and 5……………………. Appendix E: Calculations and Assumptions for Component 9……………………………. Appendix F: Calculations and Assumptions for All Pins………………………………….. Appendix G: Calculations and Assumptions for Crank Handle…………………………… Appendix H: ANSYS Force Analysis……………………………………………………...

4 5 5 9 10 11 12 13 14 15 16 17 18

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Introduction The most basic scissor jack design is truly engineering at its finest. With the power to magnify input forces, scissor jacks allow us to raise vast loads using only a fraction of the force ordinarily needed. Our goal in this project is to design an efficient scissor jack capable of raising a 2000lb load. As a screw-driven mechanical system, the jack will be manually operated and have at least 7 inches under load. The design will be transportable and storable, have a removable crank handle, and operate with a factor of safety of n = 2 using standard mechanical design methods for all components. The design itself has gone through multiple stages of development. We have taken several possible failure modes into account and are confident that our design is efficient and safe.

Proposed Design Summary The scissor jack design, shown in Figure 1, consists of four main lifting members, four connection members, a power screw and a crank. Members 1 through 8 are all primarily cshapes with ideal pin connections. Members 1 and 5 both have additional details to account for the contact surfaces. The power screw is single threaded with a collar at the member 3 connection. All members are 50 ksi strength steel with the exception of the rubber grip on the crank. The following is a summary of the design features for our proposed scissor jack. Details of the design specifications and failure criteria can be found in the attached appendices.

Figure 1: Proposed Scissor Jack Design with Labeled Members.

Main Lifting Members: These members are made from simple c-shapes. The web of the members is cut out near the pin connections to allow proper serviceability of the scissor jack at its maximum and minimum heights. Members 4 and 6 have ideal gear connections to balance the load between the left and right side. The flanges of the channels are to wrap around the flanges of the sleeve members. The lifting members are greater in length and are subjected to compression. Lifting member flanges on the outside of the sleeve flanges is to compensate for slenderness ratio by increasing the moment of inertia of the lifting members. Sleeve Members: The sleeve channels are to open inwards as shown in Figure 2. This is so the flanges are subjected to tension instead of compression. The bending moment from the power screw creates tension on the inner edge of the sleeve and compression on the outside edge. Tension along flanges on the inside prevents the possibility of localized bucking in the flanges from compressive forces.

Figure 2: Orientation of Sleeve Channels to Prevent Localized Buckling.

Additionally, the threaded sleeve section is to have additional thread surface area, shown in Figure 3. These additional threads safely transmit the stress from the power screw to the sleeve. Threading the thickness of the web of the channel would not be sufficient for reasonable power screw diameters. This addition is only made on the threaded sleeve section and not on the collared sleeve section. The collar transmits the stress safely to the c-shape.

Figure 3: Addition to C-Shape to Provide Adequate Threaded Area. Contact Members: The members that make contact with ground and the service load are members 1 and 5 respectively. Member 1 has additional flanges to provide a stable base for the mechanism while servicing the load. Member 5 has an attached plate atop to provide sufficient contact area. Most scissor jacks have ridges which lower the area of contact. This causes stress concentrations which can damage the underside of a car.

The Power Screw: The Power Screw is single threaded with a collar on the side in contact with Member 3. The collar is assumed to be frictionless and the power screw has been designed to be selflocking. The primary raising method is through the power screw’s hook coupling which is common to most scissor jacks. Incorporated into our proposed design is an option for a secondary raising method. The collar on the power screw doubles as a bolt with a hexagonal head. In a situation where the main hook coupling becomes inoperable, a standard socket wrench can be used on the hexagonal nut to raise the mechanism.

Design Criteria The design checks used in the design of the scissor jack are summarized in Table 1. The criteria are organized by failure mode with the applicable members identified. Table 1: Design Checks for Different Failure Modes of Members. Failure Member(s) Criteria Check Comments Mode Tr πd m4 τ max ≤ Torsion 9 J =  J 32 2 Pcr Cπ E Long Columns ≤ Buckling 2,3,4,6,7,8 2  (C=1) A (l / k ) 2 2 Intermediate Pcr S  l  1 = SY −  Y    Buckling 2,3,4,6,7,8 Length Columns  A  2π   k  CE (C=1) S Yielding Distortion Energy σ' ≤ Y 2,3,4,6,7,8  in C and T Theory n Selfπfd m > l 9  Locking AP = d B t plate is S σP ≤ Y n the projected area. Bearing All Pins  Check both S Y and FB Nominal σ P = SP

d B t plate

0.577 S Y n FB Nominal τ = At

τ≤

Shear

All Pins



Nominal includes threads in shear plane area At

Conclusion and Recommendations Our proposed design is similar to common scissor designs in some aspects, but also advantageous in others. Similar to others, our proposed design can safely raise a load of 2000 lbs to the required heights with relative ease on the user. Unique to our design, however, is the manufacturability of our design, which is much simpler. Since only c-shapes are utilized, bulk material can be more efficiently purchased and used. Also, less machining is required since there are no complex sleeves for the power screw. Only simple attachments which can be welded on are proposed. Therefore, when compared to similar scissor jack designs that perform equally as well, our proposed design is recommended for its manufacturability and lower cost.

Appendix A: Drawings Note: All Drawings are in Inches

Appendix B: Members 2, 4, 6, 8 Buckling Criteria (Long Columns):

Pcr Cπ 2 E ≤ A (n)( l / k ) 2

Pcr 3162 .28lbs = = 3162 .28 psi A 1in 2 Cπ 2 E (1)(π 2 )( 29 ,000 ,000 psi ) = = 508422 .58 psi ( n)( l / k ) 2 (2)( 6.32 in / 0.3767 in ) 2

3.1 6 2

k s i

≤ 5 0 8

.4 2 2

2

2

k s i

 P S  l  1 Buckling Criteria (Intermediate Length Columns): cr = S Y −  Y    A  2π   k  CE

Pcr 3162 .28lbs = = 3162 .28 psi A 1in 2 2

2

2

2

1 S  l  1  50000 psi   6.32in  SY −  Y    = 50000 psi −  = 49385 psi    2π    0.3767 in  (1)( 29000000 psi )  2π   k  CE

3.1 6 2

Yielding Using Distortion Energy Theory: σ ' ≤

k s i

≤ 4 9 .3 8 5



k s i

SY n

2 σ ' = σ X2 − σ X σ Y + σ Y2 + 3τ XY = (3162 .28 psi ) 2 − 0 + 0 + 0 = 3162 .28 psi

SY 50000 psi = = 25000 psi n 2

3.1 6 2

k s i

≤ 2 5 k s i

*See “Appendix B Support” for supporting calculations (not computer-generated)



**See ANSYS printouts in Appendix H: All force analysis support can be found there.

Appendix C: Members 3 and 7

Yielding Using Distortion Energy Theory: σ ' ≤

σY =

SY n

1000 lbs = 1641 psi  This calculation simplifies the cross-section of member 7. 0.609375 in 2 The addition of material inside the C-section (left) is added support for the power screw. It allows for increased shear strength. As a result, this calculation is conservative.

2 σ ' = σ X2 − σ X σ Y + σ Y2 + 3τ XY = 0 − 0 + (1641 psi ) 2 + 0 = 1641 psi

SY 50000 psi = = 25000 psi n 2

1.6 4 1

k s i

≤ 2 5 k s i



Combined Bending and Axial Compression: σ B =

Mc S Y ≤ I n

M = 3000 lbs ⋅ ( 2.38 in / 2) = 3570 lbs − in for Member 7 (see Drawings section)

σB =

Mc 3570 lbs − in ⋅ 0.35in = = 22718 psi = 22.7ksi I 0.055in 4

SY 50 ksi = = 25 ksi n 2

2 2 .7 k s i

≤ 2 5 k s i



For this particular section, we understand that combined bending and axial compression leads to eccentric loading that in turn magnifies the bending moment. However, given the small member size it is highly unlikely that the member would bend significantly enough to consider a change. Therefore, we just assume neglect the eccentric loading for our purposes.

*See “Appendix C Support” for supporting calculations (not computer-generated)

Appendix D: Members 1 and 5

Yielding Using Distortion Energy Theory: σ ' ≤

σX =

SY n

3000 lbs = 4000 psi 0.75in 2

2 σ ' = σ X2 − σ X σ Y + σ Y2 + 3τ XY = (4000 psi ) 2 − 0 + 0 + 0 = 4000 psi

SY 50000 psi = = 25000 psi n 2

 Note: The members are too short to consider buckling as a mode of failure.

4.0 k s i

≤ 2 5 k s i

Appendix E: Member 9

Yielding Using Distortion Energy Theory: σ ' ≤

SY n

Combined Tension and Torsion Assume a frictionless collar. τ=

Tr 16T 16 (50 lbs ⋅ 8in ) = = = 8547 .8 psi = 8.548 ksi J πd 3 π (0.62 in ) 3

σX =

FX 6000 lbs 6000 lbs = = = 19873 .67 psi = 19 .87 ksi 2 A (πd ) / 4 (π (0.62in ) 2 ) / 4

2 σ ' = σ X2 − σ X σ Y + σ Y2 + 3τ XY = (19874 psi ) 2 − 0 + 0 + 3(8548 psi ) 2 = 24782 .2 psi

SY 50000 psi = = 25000 psi n 2

2 4 .8k s i

≤ 2 5 k s i

 Self-Locking: πfd m > l

Assume coefficient of friction f =0.15, d m =0.62in and l =0.11in. (Referenced Table 8-3 Shigley) πfd m = π(0.15 )( 0.62 in ) = 0.292 in l = 0.11in

0 .2 9 2

 Bearing Stresses on the Collar: σ P ≤

σP =

FB A projected

=

FB SY where σ P = A projected n

6000 lbs 6000 lbs 6000 lbs = = = 20134 psi 2 2 Acollar − A powerscrew 0.6in − 0302 in 0.298 in 2

i n

> 0 .1 1 i n

SY 50 ksi = = 25 ksi n 2



2 0 .1 3 k s i

≤ 2 5 k s i

Appendix F: All Pins

Bearing Stresses: σ P ≤

σP =

FB SY where σ P = d B t plate n

FB 3.162 kips = = 16 .864 ksi 2d B t plate 2(0.375 in )( 0.25in )

SY 50 ksi = = 25 ksi n 2

1 6 .8 6 4

k s i



≤ 2 5 k s i

Note: We use two times the projected area because there are actually two identical projected areas through which the pin passes. Shear Stresses: τ ≤

τ=

FB 0.577 S Y where τ = At n

FB 3.162 kips 3.162 kips = = = 14 .315 ksi 2 2 At 2(πd / 4) 2(π (.375 in ) 2 / 4)

0.577 S Y (0.577 )50 ksi = = 14 .425 ksi n 2

1 4 .3 1 5

k s i

≤ 1 4 .4 2 5

k s i

 Note: We use two times the cross-sectional area because there are actually two shear planes through which the pin passes. We can assume that these hold true for all pins. The maximum axial force is 3.162 kips, so all of our designs assume a worst-case scenario.

Appendix G: Crank Handle Check of Raising Torque Requirement: Tapplied ≥TR Assume a human force of 50 lbs is applied to the crank handle. TR =

Fd m 2

 l + πfd m sec α  6000 lbs ⋅ 0.62in  0.11 + π (0.15)(0.62in ) sec 14.5    =   = 396 .7lb − in 2  π (0.62in ) − (0.15)(0.11) sec 14.5   πd m − fl sec α 

Tapplied = (50 lbs )( 8in ) = 400 lb −in

4 0 0

lb

− in ≥ 3 9 6

.7lb

− in



Appendix H: ANSYS Force Analysis Note: Power Screw is composed of members 9 and 10 in ANSYS report