Figure 3.4: Screw Nomenclature (Bhandari, 2010
The terminologies of the screw thread are defined as follows (Gupta, 2005):
(i) Pitch ()
The pitch is defined as the distance m easured parallel to the axis of the screw from a point on one thread to the corresponding point on the adjacent thread.
(ii) Lead ()
The lead is defined as the distance measured parallel to the axis of the screw that the nut will advance in one revolution of the screw.
For a single threaded screw =
For a double threaded screw =
It is the largest diameter of the screw. It is also called major diameter.
(iv) Core or Minor Diameter ()
It is the smallest diameter of the screw thread. =−
(v) Mean Diameter ()
=(+)/2 =−0.5
(vi) Helix Angle()
It is defined as the angle made by the helix of the thread with a plane perpendicular to the axis of the screw. The helix angle is related to the lead and the mean diameter of the screw.
Taking one thread of the screw and unwinding, one complete turn is developed. The thread will become the hypotenuse of a right-angled triangle with the base and height being equal to the lead .
Figure 3.5: Unwound thread
This right-angled triangle gives the relationship between the helix angle, mean diameter and lead, which can be expressed in the following form: tan= /
Where = ℎ ℎ ℎ ℎ.
The following conclusions can be drawn on the basis of the development of thread:
The screw can be considered as an inclined plane with as the angle of inclination.
The load always acts in the vertical downward direction. When the load is raised, it moves up the inclined plane. When the load is lowered, it moves down the inclined plane.
The load is raised or lowered by means of an imaginary force acting at the mean radius of the screw. The force multiplied by the mean radius (/2) gives the torque required to raise or lower the load. Force is perpendicular to load .
3.4 Torque Requirement - Lifting Load
The screw is considered as an inclined plane with inclination when the load is being raised. The following forces act at a point on this inclined plane:
Figure 3.6: Force diagram for lifting load Load : It always acts in the vertical downward direction.
Normal reaction : It acts perpendicular (normal) to the inclined plane.
Frictional force : Frictional force acts opposite to the motion. Since the load is moving up the inclined plane, frictional force acts along the inclined plane in downward direction.
Effort : The effort acts in a direction perpendicular to the load . It may act towards the right to overcome the friction and raise the load.
Resolving forces horizontally,
=cos+sin (3.0)
Resolving forces vertically,
=cos−sin (3.1)
Dividing equation (3.0) (3.1) we get:
=(cos+sin)(cos−sin) (3.2)
Dividing the numerator and denominator of the right hand side of e quation (3.2) by we get:
=(+tan)(1−tan) (3.3)
The coefficient of friction μ can be expressed as follows:
=tan (3.4)
Where
= the friction angle.
Substituting (3.4) into equation (3.3),
=(tan+tan)/(1−tantan) (3.5)
=tan(+) (3.6)
The torque required to raise the load is given by: =×/2 Whence
=[tan(+)]/2 (3.7)
3.5 Torque Requirement - Lowering Load
When the load is being lowered, the following forces act at a point on the inclined plane:
Load : It always acts in the vertical downward direction.
Normal reaction : It acts perpendicular (normal) to the inclined plane.
Frictional force : Frictional force acts opposite to the motion. Since the load is moving down the inclined plane, frictional force acts along the inclined plane in the upward direction
Figure 3.7: Force diagram for lowering load
Effort : The effort acts in a direction perpendicular to the load . It should act towards left to overcome the friction and lower the load.
Resolving horizontally,
= − (3.8)
Resolving vertically,
= + (3.9)
Dividing expression (3.8) by (3.9) we get as follows:
=(cos−sin)/(cos+sin) (3.10)
Dividing the numerator and denominator of the right hand side of e quation (3.10) by cos α:
=(−tan)/(1+tan) (3.11)
Substituting equation (3.4) into Equation (3.11),
=(tan−tan)/(1+tantan) (3.12)
Whence
= ( − ) (3.13)
The torque required to lower the load is given by, =×/2 Whence
=[tan(−)]/2 (3.14)
3.6 Over Hauling and Self-Locking Screws
From equation (3.14), we know torque re quired to lower load is given by: =[tan(−)]/2 18
Case 1: When <
The torque required to lower the load becomes negative. This indicates a condition that no force is required to lower the load and the load itself will begin to turn the screw and descend down, unless a restraining torque is applied. This condition is called overhauling of the screw.
Case 2: When >
The torque required to lower the load becomes positive. Under this condition, the load will not turn the screw and will not descend on its own unless effort is applied. This condition is called self- locking.
The rule for self-locking screw states that: A screw will be self-locking if the coefficient of friction is equal to or greater than the tangent of the helix angle.
For self-locking screw, tan≥tan
Or ≥/
Therefore, the following conclusion are made:
(i) Self-locking of the screw is not possible when the coefficient of friction (μ) is low. The coefficient of friction between the surfaces of the screw and the nut is reduced by lubrication. Excessive lubrication may cause the load to descend on its own.
(ii) The self-locking property of the screw is lost when the lead is large. The le ad increases with number of starts. For double-start thread, lead is twice of the pitch and for triple threaded screw, three times of pitch. Therefore, the single threaded screw is better than multiple threaded screws from self-locking considerations. Self-locking condition is essential in applications like screw jack (Naik, Apr 15, 2015).
3.7 Efficiency of the Square Threaded Screw
Referring to Figure 3.6: Force diagram for lifting the load ,the output consists of raising the load if the load moves from the lower end to the upper end of the inclined plane.
Therefore, = ℎ
=
The input consists of rotating the screw by means of an effortP. =
ℎ
= () The efficiency of the screw is given by,
= (3.15b)
This equation can also be expressed as:
=l/() (3.15c)
And
tan=/
Therefore
=tanα/ (3.15d)
Substituting for =tan (+) we get;
=tan/tan (+) (3.15e)
From the above equation, it is evident that the efficiency of the square threaded screw depends upon the helix angle and the friction angle. The following figure shows the variation of thefficiency of the square threaded screw against the helix angle for various values of the coefficient of friction. The graph is applicable when the load is being lifted
The efficiency of square threaded screw is given by (From equation 3.15e):
=tan/tan(+) For self-locking screw ≥
Substituting the limiting value ( = ) into the equation above
≤tan/tan(+)
Substituting for tan2 into the above expression,
≤tan/(1−2)tan(2) (3.16c)
Simplifying
≤1/2(1−2) (3.16d)
From the above expression we can deduce that the efficiency of self-locking square threaded power screw is less than 0.5 or 50%. If the efficiency is more than 50%, then the screw is said to be overhauling (Gupta, 2005).
3.9 Coefficient of Friction,
It has been found that the coefficient of friction () at the thread surface depends upon the
workmanship in cutting the threads and on the type of the lubricant used. It is practically independent of the load and dependent on rubbing velocity or materials. An average of 0.1 can be taken for the coefficient of friction when the screw is lubricated with mineral oil (Gupta, 2005
Chapter 4
Designing procedure for the screw jack
4.0 introduction
The generalized adopted design procedure for screw jack to rise a load of 3500kg, mini height=200mm, max height=400mm
4.1 Design for Screw Shaft
Material specification selected for the screw shaft is plain carbon steel to British Standard specification BS 970 080M30, Hardened and Tempered, whose properties are as shown in Appendix B and the
material yield strength is 700 MPa both in tension and pure compression and 450 MPa in shear. 4.1.1 Core Diameter
The core diameter is determined by considering the screw to be under pure compression. That is;
W=
c
AcWhere
c is pure compression stress=700mpaAc is cross sectional area of screw shaft=
/4(dc)
2Dc is core diameter
There for W=
c
/4(dc)
2
dc=
(4W/
c/)Taking factor of safety f.s=5
dc=
(4W/
c/5)
dc=
4
8583.75/
700/5)dc=0.008835m=8.835mm
For square threads of fine series, the following dimensions of screw are selected from Appendix D (Gupta, 2005) hence,,
The core diameter dc=10mm,do=12mm,pitch p=l=2mm
= tan
=0.1,
=5.71Section of screw spindle
4.1.2 Torque required to rotate the screw
When the torque required to rotate the screw is the same to torque required lift the load is given by
T1
P
dm
W
tan
dm
dm
do
dc
12
10
11mmtan
l
dm
2
11
0.05787Then T1=(8583.75tan(3.31227
5.71)0.011)/2=7.496Nm4.1.3 Screw Stresses
Compressive stresses duo to axial loads using the new c ore diameter is
c=W/Ac=W/(
dc
2/4)=4
8583.75/
0.01
2=109.29MpaThe shear stress due to this torque using the new core diameter is given
=T1dc/2J, where J is polar moments=
dc
4/32
=16T1/
dc
3=16
7.496/
0.01
3=38.177Mpa4.1.4 Principal Stresses
Maximum principal stress is as follows:
c max=
c
(
c)
(
)
c max=
.
.
4
3
.
7
=
c
max
=121.3068MpaAnd maximum shear stresses as follows:
max
=
c
max
=
/2
.
.
=66.66MpaDesign value of
is 450
=90MpaCheek; those maximum shear and compressive stresses ar e less than the permissible stresses which is safe design
.
4.2 Design for Nut 4.2.1 Height of the Nut
We find the height of the nut (h) by considering the bearing pb on the nut
Pb=W/
4
do
dc
n, where n isnumber of treads in contact with screwed spindle
Material specification for the nut is phosphor bronze which has tensile stress=150Mpa, compressive stress 125Mpa,shear stress=105Mpa specific bearing pressure not exceed 17Mpa and
=0.117
=
.
.
.
17
248.39
n
n
14.6Say n
15Then height of the nut is as follows;
h
n
p
15
2mm
30mmCheck:For a safe nut height ℎ ≤ 4dc
40mm4.2.2 Stresses in the Screw and Nut Shear stress in the screw is as follows
Screw
w/
n
dc
t where t is thickness of screw
p/2
2/2
1mm
Screw
8583.75/
15
0.01
0.001
18.215Mpa4.2.2 Stresses in the Screw and Nut Shear stress in the screw is as follow
Nut
W/
n
do
t ,8583.75/
15
0.012
0.001
15.179Mpa The given value of
is 105/5
21MpaCheck :These stresses are within permissible limit, hence, design for the nut is safe.
4.2.3 The outer diameter of Nut
Outer diameter of D1 is found by considering the tearing strength of the nut
t
W/
4
D1
2
do
2
t is tearing strength the nut
Tensile stress
t/f.s
150/5
30MpaThen we get D1 as follows
30
8583.75/
/4
D1
2
12
2D1
22.097mm,say D1
23mm4.2.4 The outside diameter of Collar
Outside diameter D2 is found by considering the cr ashing strength of the nut collar
c
W/
4
D2
2
D1
2
where
c is crushing strength of the nut
compressive strength
c
125/5
25MpaThen we get D2 as follow
25
8583.75/
4
D2
2
23
2
D2
31.083mm,sayD2
32mm4.2.5 Thickness of the Nut Collar
The thickness of nut collar t1 is found by considering shearing strength of the nut color
T1
W/
D1
is shearing strength of nut collar
105/5
21MpaT1
8583.75/
23
21
5.656mm,say t1
6mm4.3 Designs for Head and Cup
4.3.1 Dimensions of Diameter of Head on Top of Screw and for the Cup D3
ASSUMING
D3
1.75do
1.75
12mm
21mmThe seat for the cup is made equal to the diameter of the head and then chamfered at the top the cup prevents the load from rotating and is fitted with pine of diameter D4
D3/4D4
5.25mm,say D4
6mmSection of pin
Take length of pin to be 9mm.
Other dimensions for the cup are taken as:
Diameter at the top of the cup = Diameter of the head = 52mm Height of cup = 9mm
Thickness of cup = 3mm Fillet radii = 1mm
Figure 5.4: Section of Cup
4.3.2 Torque Required to Overcome Friction
We know that by assuming uniform pressure condition torque required to overcome friction is given as follows;
T2
W
D3
D4
D3
2
D4
2
Where D3
diameter of head
21mmD4
diameter of pin
6mmT2
0.1
8583.75
0.021
0.006
0.021
2
0.006
2
63.90Nm4.3.3 Total Torque Subjected to the Handle
Total torque to which the handle is subjected is given by
T
T1
T2
T
7.496Nm
63.90Nm
71.396Activity Professional use Domestic use
Pushing 200N (20.4kg) 119N (12.1kg)
Pulling 145N (14.8kg) 96N (9.8kg)
Table 4.2: Maximal Isometric Force by General European Working Population for Whole Body Work in a Standing Posture
Therefore taking the force of 96N in domestic use (J.J. Fereira, 2004) then the length of the handle required is
T
F
L
L
T
FL
T/F
71.396Nm/96N
0.7437m
743.7mm, say 744mmThe length of the handle may be fixed by giving some allowance for gripping 70mm
Therefore, the length of the handle/lever is 814mm
Section of Lever
4.3.4 Diameter of Handle/Lever
The diameter of the handle/lever, D may be obtained bending effects
M
32
c
D
While
b
t
c
700/5
140MpaM
Force applied
length of leverM
96N
0.7437m
71.395Nm71.395
32
140
10
6
D
D
0.0173m
17.3mm,say D
18mm4.3.5 Height of Head
The height of head is usually taken as t wice the diameter of handle.
H
2DH
2
18mm
36mm4.3.6 Design Check against Instability/Buckling
Effective length of screw, = + 1/2 ℎℎ
=1+ℎ/2 Leff
200
30/2
215mmWhen the screw reaches the maximum lift, it can be regarded as strut whose lower end is fixed and the load end is free. Therefore, buckling or critical load for this given condition is as follows (Gupta, 2005
Wcr
Ac.
y
y
4c
E
Leff
k
Wcr
13199.04N W
8583.75N4.4 Design of Body
4.4.1 Dimensions for the body of the screw
The dimension of the body may be fixed and given as in shown in the figure above (Gupta, 2005)
1. Diameter of the Body at the Top
D5
1.5D22. Thickness of the body
3. t2
0.25do, t2
0.25
12mm
3mm 3Inside Diameter at the BottomD6
2.25D2 D6
2.25
32
72mm4. Outer Diameter at the Bottom D7
1.75
D6D7
1.75
72mm
126 5. Thickness of Baset3
2t1, t3
2
6mm
12mm 6. Height of the BodyHeight of the body
max lift
height of nut
extra50mm
200mm
30mm
70mm
300mmFinally, the body is tapered in order to achieve stability of the jack.
4.5 Efficiency of the Screw Jack
Efficiency of screw jack is given as follows:
torque required to rotate screw with no friction
torque required out put
To
T
But To
W
tan
dm/2To
8583.75
0.05787
0.011
2 To
2.732NmAnd
T
71.396Nm
To/T
2.732/71.396
0.038
3.8
4.6 result and dissection
the results I find from my project are listed as follows to design individual parts of the screw jack
1 to design body(frame) 2 to design nut
3 to design handl (Tommy bar) 4 to designs the cup
5 to design set screw 6 to design washer 7 to design screw
Dissection
RESULTES OF NUMERICAL VALUE OF THE DESIGNE
Dc 10mm LPIN 9mm D5 =48mm
Do
12mm Dhead 52mm t2 =3mm
H
30mm H cup 9mm D6 =72mm
D1
23mm tcup 3mm D7 =126mm
D2
32mm Filit raduis 1mm t3 =12mm
t1
6mm Lhandl 814mm Hbody =300mm
D3
21mm Dhandl 18mm
D4
6mm H head 36mm
4.7 CONCULUTION
From my project I am concluded that from introduction part to design analysis we are seen clearly its working principle of the screw j ack and operation of the screw jack, efficiency of this designed screw jack, methods of increasing efficiency of the screw jack. A screw jack is an example of a power screw and referred to as a mechanical device that can increase the magnitude of an effort force. Screw jacks are used for raising and lowering platforms and they provide a high me chanical advantage in order to move moderately heavy and large weights with minimum effort. Based on my calculations and assumptions the designed values are safe.
4.8 Recommendation
From the case study, I concentrated on design of a simple mechanical screw jack where the nuts fixed in a cast iron frame and remains stationary while the spindle is being rotated by the lever. This design can only work for light loads hence when a screw jack is needed for heavy load application different designs required where the nut is rotated as the spindles moves. I therefore recommend design of a screw jack for the heavy loads. I recommended that the workshops and AutoCAD rooms open in order to practice more.
)
.1 Appendix A: Mechanical Properties of Cast Iron (Nyangasi, 18 December, 2006)
6.2 Appendix B: Mechanical Properties of Steels (Nyangasi, 18 December, 2006)
CD 63 510 650 10
Key:HR - Hot- Hot rolled and normalized CD - Cold drawn
H&T - Hardened and tempered
Appendix C: Safe Bearing Pressure for Power Screws (Gupta, 2005)
Type of power screw Material Safe bearing pressure,MPa
Rubbing speed m/s
screw Nut
Hand press Steel Bronze 17.0-24.1 Low speed ,well lubricated Screw jack Steel Cast Iron 12.0-17.0 Low speed <
2.5
Screw jack Steel Bronze 11.0-17.0 Low speed < 3 Hoisting screw Steel Cast Iron 4.0-7.0 Medium speed
(6-12)
Hoisting screw Steel Bronze 5.5-10.0 Medium speed (6-12)
References
1. Ashby, M. F., 2005. Material Selection in Mechanical Design. 3rd ed. New York: Pergamon Press .
2. Bhandari, V. B., 2010. Design of Machine Elements. Third Edition ed. New Delhi: Tata McGraw-Hill Education.
3. Collection, J., 2015. hubpages.com › Autos › Automobile History. [Online] Available at:
https://www.history of screw jacks.com [Accessed 11 November 2015].
4. Fasteners, C. o., 2005. Technical Reference Guide. Ninth Edition ed. Winona, Minnesota: Fastenal Industrial & Construction Supplies.
5. Gupta, R. K. &. J., 2005. Theory of Machines. Revised Edition ed. Punjab, India: S. Chand and Company