Screw Jack Design 5mz60

Ministry of Higher Education Kabul University Faculty of Engineering Department of Mechanical Engineering

Screw Jack Design

Researcher/Designer Ahmad Murtaza Ershad (8173)

Course Instructor Professor Abdul Hamid Layan

Project Design, 10th Semester Kabul, 2010

i

Screw Jack Design

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Table of Contents Introduction………………………………………………….1 1. Power Screws………………………………………………..1 2. Screw Jack Design Procedure……………………………….3 3. Calculations………………………………………………….5 3.1. Design of the screw and torque calculation…………5 3.2. Design of the nut…………………………………....7 3.3. Design of the various diameters…………………….8 3.4. Design of the handle………………………………..9 3.5. Buckling of the screw………………………………10 3.6. Design of the body…………………………………11 3.7. System efficiency…………………………………..12 4. Conclusion………………………………………………….13 5. Recommendation…………………………………………...13 Reference…………………………………………………...14 Appendix…………………………………………….……..15

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Introduction The project design course is an essential course for the mechanical engineering undergraduate program. In this course, students are asked to design different projects which are either given by the instructor or students select themselves. The first project is to design a mechanical screw jack. The course requirement is only to determine the power screw dimensions, the required torque and the efficiency of the system. Students don’t have to worry about the manufacturing process of their design. The design process starts with specifying the amount of load which is going to be raised or lowered which in our case it is determined by the instructor. Next, some important data is gathered for the design process like the height of the lift, factor of safety, etc. After this, the actual calculation is started and different parts of the power screw are designed. It is a good experience for us to learn how to consider efficiency and cost effectiveness in our real projects once we graduate. 1. Power Screws Power screws are used to convert rotary motion in to translational motion. It is also called translational screw. They find use in machines such as universal tensile testing machines, machine tools, automotive jacks, vises; aircraft flap extenders, trench braces, linear actuators, adjustable floor posts, micrometers, and C-clamps. There are two kinds of power screws, hydraulic and mechanical power screws. A special case is screw jack which raises or lowers the load by applying a small force in the horizontal plane. A screw thread is formed by cutting a continuous helical groove around the cylinder. These grooves are cut either left hand or right hand. The majority of screws are tightened by clockwise rotation, which is termed a right-hand thread. Screws with left-hand threads are used in exceptional cases. For example, anticlockwise forces are applied to the screw (which would work to undo a right-hand thread), a left-hand-threaded screw would be an appropriate choice. Power screws are typically made from carbon steel, alloy steel, or stainless steel and they are usually used with bronze, plastic, or steel mating nuts. Bronze and plastic nuts are popular for higher duty applications and they provide low coefficients of friction for minimizing drive torques. There are important and figures that need to be understood before deg power screws: 1. Pitch: is the distance from a point on one thread to the corresponding thread on the next adjacent thread, measured parallel to the axial plane. 2. Lead: is the distance the screw would advance relative to the nut in one rotation. For single thread screw, lead is equal to pitch. 3. Helix Angle: is related to the lead and the mean radius by the equation below;

1

lead 2πrmean

tan α =

Figure 1: Common screw assembly ©1997-2005 Roton Products, Inc.

There are 3 types of screw threads used in power screws: 1. Square threads: a. Is used for power transmission in either direction b. Results in maximum efficiency and minimum c. It is employed in screw jacks and clamps 2. Acme threads: a. It is a modification of square thread b. Efficiency is lower than square threads c. The slope increases the area for shear d. It is easily manufactured 3. Buttress Thread: a. It is used when large forces act along the screw axis in one direction only. b. It has higher efficiency like square threads and ease of cutting like acme threads c. It is the most strong thread of all d. It has limited use of power transmission

Figure 2: Square thread

Figure 3: Acme threads

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Figure 4: Buttress thread

2. Screw Jack Design Procedure 2.1. Design Tools

1. 2. 3. 4. 5. 6.

Books Websites Stationary Calculator Laptop Printer

2.2. Design Objective • • •

• • •

State the problem and clarify what is expected from the design Specify design considerations such as factor of safety, material selection criteria, and etc. To study effects of stresses on the power screw parts o Direct tensile or compressive stress due to axial load o Torsional shear stress in the minimum cross section of the screw by the twisting moment o Shear stress at the threads of the screw at the room diameter and at the threads of the nut at the outside diameter due to axial loading o Bearing pressure at the thread surfaces of the screw and nut To determine the torque required to raise or lower the given load To determine the efficiency of the power screw To determine the dimensions of the different parts of the screw

2.3. Problem Statement

A mechanical power screw that can raise or lower 7 tons or 68.670 KN of load is intended to be designed. Different parts of the assembly such as the screw, the nut and the handle will be designed in an efficient and cost effective manner. 2.4. Design Considerations

1. Factor of safety for the assembly is taken 5 due to the nature of the design. Actually the factor of safety is taken 1.5 to 2 in static loading of ductile material. A higher factor of safety is considered due to the consequences of the failure. 2. Selection of Material for the screw and nut is of great importance. There are common materials used in the design of screw jacks like steel for the screw and cast iron, bronze or plastic for the nuts. Mild steel or hard steel is considered for different screw designs. In order to prevent friction cast iron or bronze is preferred for the design of the nut. Cup and frame are made of Grey cast iron which is cheap and has good mach inability. Material is selected as following:

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a. Screw:

Plane carbon steel (30C8 – IS: 1570-1978) is selected because screw is always under Torsional, bending and axial load. Carbon steel is chosen due to the strength issues. This steel is also used for the handle of the screw jack. (σ yield = 400 MPa, τ =240 MPa, E=207GPa) b. Nut:

In order to reduce the friction resistance between the screw and nut a softer material is selected for the nut. Phosphor Bronze (Grade 1-IS: 28-1975) is a proper material for nut construction because it acts very well against wear resistance and reduces torque to overcome friction. (σ ultimate = 190 MPa, σ MPa )

yield (tension)

=100 MPa, σ

yield (compression)

= 90 MPa, τ=80

c. Screw Jack Handle:

Plane carbon steel (30C8 – IS: 1570-1978) is selected for the handle of the jack because of the high strength it offers. (σ yield = 400 MPa, τ =240 MPa, E=207GPa) d. Frame:

Grey cast iron is used which is cheap and has good mach inability. 3. The effective lifting height is chosen to be 0.5m (500 mm). 4. Average coefficient of friction between the material soft steel and cast iron is taken 0.10 when it is lubricated. But for this specific design, it is taken 0.18 assuming it dry for safe operations.(1) 5. Limiting values for bearing pressure between steel and cast iron is taken 15.05 MPa.(2) 6. According to agronomists the force of the hand is about 150 to 200 N. In this design we assume that is the handle is rotated by two hands which give 400 N hand forces for the design of the handle.

(1) Table 17.6. Coefficient of friction, Chapter 17, Power screws, A Text of Machine Design, p. 642 (2) Table 17.7, Limiting values for bearing pressure, Chapter 17, Power Screws, A Text of Machine Design, p.646

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3. Calculations 3.1. Design of the screw

Procedure i. Core diameter of the screw is determined using allowable stress and the given load ii. Using the core diameter, the rest of the diameters and the pitch will be determined from the table iii. Torque will be determined using the mean diameter, coefficient of friction and the pitch iv. Principle stresses due to the shear and compression stresses will be studied v. The dimensions for the screw is safe if and only if the maximum stresses are less than the allowable stresses

σc =

P

π 4

dc =

dc

………………………………. (1) 2

P

π  σc  4  FS 

2

=

68670 N 6 π  400 *10 MPa   4 5  

2

= 0.03305m(33.05mm)

The next available diameter is 35 mm. For dc =35 mm, according to the Table 17.2 (Normal series) we have d o = 42mm(diameterouter ) Pitch = 7 mm d mean =

dc + do = 38.5mm 2

The torque required to rotate the screw:

T=

P * dmean  l + πµ dmean    2  πdmean − µl 

P * dmean T= * tan (α + φ ). 2

………….. (2, 3)

l= Lead=Pitch for single thread µ = Coefficient of friction α , φ = angles of helix and friction respectively

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68670 N * 0.0385 mm  0.007 mm + π 0.18 * 0.0385 mm    2  π 0.0385 mm − 0.18 * 0.007 mm  T = 320 N − m T=

Now, it is time to study principle stresses due to the combined stresses (compression and Torsional) and see if they are in limit for safe dimensions.

σc =

τ=

P

π

dc 4 T

π 16

σc =

...(1)

dc

2

68670 N

π 4

τ=

… (3) 2

(0.0385mm )

= 59.02MPa 2

320 N − m

π 16

(0.0385mm )

= 28.57 MPa 2

Principle stresses

σc

1 σc 2 + 4τ 2 2 2 ……………… (4, 5) 1 2 2 τ max = σc + 4τ 2

σ max =

±

σ max =

59.02MPa 1 + 59.02MPa 2 + 4(28.57 MPa) 2 = 70.5MPa( N / mm 2 ) Tension 2 2

σ max =

59.02MPa 1 − 59.02MPa 2 + 4(28.57 MPa) 2 = 11.56MPa( N / mm 2 ) Comp. 2 2

τ max =

1 59.02MPa 2 + 4(28.57 MPa) 2 = 41.06MPa( N / mm 2 ) 2

σyield

σallowable =

FS

σallowable =

τallowable =

……………………….. (6)

400 MPa = 80 MPa 5

τyield

… ……………………… (7) FS 240MPa τallowable = = 48MPa 5

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Criteria for safe design against principle stresses

τ allowable σ allowable

> τ

max

> σ

max

The design is therefore safe. 3.2. Design of the nut

Procedure i. ii. iii.

Number of threads in engagement is found Height of the nut is determined Shear stress produced at the threads of the screw at the core diameter and at threads of the nut at the major diameter is studied. For safe design, these shear stresses are compared with the allowable stresses

iv. Pb =

P

π

… ………………….. (8)

(do − dc )n 4 Pb= Bearing pressure n= Number of threads do= Outer diameter dc= Core diameter 2

15.05MPa =

2

68670 N

π 4

[(0.042m)

2

−

)]

(0.0385m) 2 n

n = 10.8

We take the number of threads n=12. The height of the nut is found from following equation: H = n* p H = 12 * 7 mm ………………………. (10) H = 84mm

The nut threads are subjected to crushing and shear. To check whether crushing is expected or not, P σcrushing = ………………… (11) π 2 2 (do − dc )n 4

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σcrushing =

68670 N

π

[(

)]

0.042m) 2 − (0.035m) 2 12 4 σcrushing = 13.53MPa σcrushing << σcy (bronze − 90MPa) From the above result, crushing is not expected because crushing stress is much smaller than the bronze yield stress at compression.

P , t = pitch / 2 π * do * n * t 68670 N τnut = …………… (12) π * 0.042m * 12 * 0.0035m τnut = 12.39MPa τnut << τyield (bronze − 80MPa)

τnut =

From the result above, the nut threads are safe against shear stress. P , t = pitch / 2 π * dc * n * t 68670 N τscrew = ……….. (13) π * 0.0385m *12 * 0.0035m τscrew = 13.52 MPa τscrew << τyield (bronze − 240MPa)

τscrew =

The screws are safe against the shear produced by the axial loading. 3.3. Design of the various diameters

Inner diameter of the nut collar (D1) P=

π 4

[(D )

2

1

π

2

]

− (d o ) σ t ................................(14)

[

2

(D1 ) − (0.042m ) 4 D1 = 0.078m(78mm)

68670 N =

2

100 * 10 6 Pa 5

]

Outer diameter of the nut collar (D2)

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P=

π

[(D ) 4

2

2

68670 N =

2

]

− (D1 ) σ c ............................(15)

π

[(D ) 4 2

6

2

− (0.078m )

2

]90 *105 Pa

D2 = 0.104m(104mm) Thickness of the nut collar (t1)

P = πD1t1τ ..............................................(16) t1 =

68670 N = 0.0175m(17.5mm) 80 * 10 6 Pa π 0.078m 5

Diameter of head at the top of screwed rod (D3) D3 = 1.7 * d o ............................................(17) D3 = 1.7 * 0.042m = 0.0714m(71.4mm) Diameter of pin which fits the cup (D4) D4 = D3 / 4...............................................(18) D4 = 0.0714 / 4 = 0.0178m(17.8mm) 3.4. Design of the handle

Ttotal = Force( hand ) * Lenght ( handle).............(19) Ttotal Lenght ( handle) = Force( hand ) Where; Ttotal = T 1 + T 2 ……………… (20) T1=torque required to rotate the screw; T1=320 N-m as calculated earlier T2=torque required to overcome friction at the top of the screw 2  ( R3 ) 3 − ( R4 ) 3  µP   ………………. (21) 3  ( R3 ) 2 − ( R4 ) 2  R3=radius of head R4=radius of pin T2 =

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T2 =

 (0.0714m / 2) 3 − (0.0178m / 2) 3  2 0.18 * 68670 N  2 2  3  (0.0714m / 2) − (0.0178m / 2) 

T 2 = 310.2 N − m Ttotal = 320 N − m + 310.2 N − m Ttotal = 630.2 N − m

Ttotal Force( hand ) 630.2 N − m Lenght ( handle) = = 1.57m 400 N Lenght ( handle) =

Note: If the length of handle is too large, an alternative is to place the handle centrally and apply the force.

Diameter of handle M (bending − moment ) =

π 32

σ b ( D) 3

M = Force( hand ) * Lenght ( handle) M = 400 N *1.57m = 628 N − m

σb =

σt FS

=

400MPa = 80MPa 5

M (bending − moment ) = D=3

……… (22)

π

σ b ( D) 3

32 M * 32 3 628 N − m * 32 = = 0.043m(43mm) πσ b 3.14 * 80 * 10 6 Pa

3.5. Buckling of the screw

Buckling is studied when the load is compressive and the uned length between the screw and the nut is long. When it is short, then it is assumed a column and buckling issue doesn’t rise. If the critical load is more than the load we have then our design is safe and there is no chance of buckling.

σY  (L / k )2  Pcritical = AC * σ Y 1 − 2  4Cπ E  ….... (23)

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Ac= Cross-sectional area of the screw core σ Y =Yield strength of the screw material L= effective length of the screw (L = screw lift height + ½(height of nut)) C= end fixity coefficient, in case of one end fix and one end free its value is 0.25 k=least radius of gyration, it is usually taken 0.25dc AC = π

(dc ) = π 4 (0.035m ) = 961.625 *10 −6 m 2 4 L = 0.5 + 1 / 2(0.084m) = 0.542m k = 0.25d C = 0.25 * 0.035m = 0.00875m 2

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σY  (L / k )2  Pcritical = AC * σ Y 1 − 2  4Cπ E  400 * 106 Pa   Pcritical = 961.625 * 10 − 6m 2 * 400 * 106 Pa 1 − (0.542m / 0.00875m) 2 9  4 * 0.25 * 3.14 * 207 * 10 Pa  Pcritical = 379.6kN

There is no chance of buckling because the critical load is much greater than the design load which is 68.670kN. 3.6. Design of the body

Height of the head

h = 2* D h = 2 * 0.043m = 0.086m(86mm)

.................... (24)

Diameter of the body at the top (D5) D5 = 1.5D2 = 1.5 * 0.104m = 0.156m(156mm) ..(25) Thickness of the body (t3) t 3 = 0.25d o = 0.25 * 0.042m = 0.0105m(10.5mm) … (26) Inside diameter at the bottom (D6) D6 = 2.25D2 = 2.25 * 0.104m = 0.234m(234mm) …(27) Inside diameter at the bottom (D7) D7 = 1.75D6 = 1.75 * 0.234m = 0.4095m(409.5mm) … (28) Thickness of the base (t2) 11

t 2 = 2t1 = 2 * 0.0175m = 0.035m(35mm) …. (29) Height if the body = Max. Lift Height + height of the nut+150mm = 500mm+84mm+150mm =734mm 3.7. System Efficiency TO ……………………… (30) Ttotal TO=Torque required to rotate the screw with no friction d TO = P tan(α ) mean 2 0.007m 0.0385m TO = 68670 N * * = 76.54 N − m π * 0.0385m 2 76.54 N − m Efficiency(η ) = = 0.121 630.2 N − m Efficiency(η ) = %12.1 Efficiency(η ) =

If the screw friction and collar is friction is neglected, the efficiency of the system is calculated as below:

η=

Effort Ideal ( P ' o ) P tan(α ) = Effort Actual ( P ' ) P tan(α + φ )

tan(3.3) * %100 tan(3.3 + 10.2) η = %24.02

η=

12

4. Conclusion

1. This project was the first machine element I designed. 2. Through this experience, I found the chance to apply my knowledge of previous courses like Mechanics of material, Kinematics, etc. 3. As a future design engineer, I learned to notice the importance of material selection and cost criteria in the design procedure. 4. This screw is a self locking jack which requires effort to lower the load. 5. The efficiency (%12) or (%24 in case of friction negligence) is very reasonable for a helix angle of 3.3 degrees. 5. Recommendation

1. When deg the system, the designer should make sure the material is cost effective, and durable. In the meantime, it should also be available in the market. 2. For the design of screw jacks, a higher factor of safety is recommended sue to the nature of the application. 3. Dimensions should be realistic. 4. The system should be tested fist for strength, and then sent for manufacturing.

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Reference

1- A Brief History of Screw Threads, Application Engineering, ©1997-2005 Roton Products, Inc. 2- Bhandari, Steinhoff, Dan (2007). Design of Machine Elements. 2nd Edition. McGraw Hill Publications. Pp. 224-232 3- Eggert, Rudolph (2004), Power Screws, McGraw Hill 4- Hall, S., Allen, Hollowenko, R., Alfred, Laughlin, G., Herman (1961). Power Screws and Threaded Fasteners, Scahuam’s Theory and Problems of Machine Design, pp. 145-153 5- Module 6, Lesson 2, Power Screws, Version 2 ME, IIT Kharagpur 6- Power Screws, Chapter 17, A Textbook of Machine Design 7- http://en.wikipedia.org/wiki/screws

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A Typical Screw Jack (Courtesy of Version 2 ME, IIT Kharagpur)

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