This document describes the design and modeling of a screw jack. It includes analyzing the stresses on each component of the screw jack, which consists of a screw, nut, body, handle, cup and other parts. Materials were selected for each component based on their strength and stress requirements. Dimensions for each part were calculated using analytical methods to ensure the screw jack could safely lift a load of 250kN over a height of 270mm. 3D models were created for each component in Solid Edge software and then assembled to visually demonstrate the complete screw jack design.

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CONTENT
SI.NO TITLE PAGE.NO
1. INTRODUCTION 3-6
2. OBJECTIVES 7
3. METHODOLOGY (Problem) 8-13
4. FIGURE OF SCREW JACK (software) 14-16
5. CONCLUSION 17
6. REFERENCE 18

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ABSTRACT
The screw jack are used to covert rotary motion into translator motion. A screw jack is
an example of a power screw in which a small force is applied in horizontal plane is used
to raise a large load.
In this project we are calculating or finding the strength of screw jack. In other words
how much weight it can carry.
In screw jack there are 7 components. All the 7 components are made in the solid edge
software.

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Chapter:-01
1.1 INTRODUCTION
FIG 1.1
It is commonly used to lift moderately and heavy weights, such as vehicles to raise and
lower the horizontal stabilizers of aircraft and as adjustable supports for heavy loads,
such as the foundations of houses.
It is very common throughout the world, screw jacks are found wherever there is a need
to lift, position, align and hold load. Their high reliability and synchronization makes
screw jacks suitable for a wide variety of uses that alternative methods of handling
cannot achieve.

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This applications reaches across market sectors and includes Steel Works Equipment,
Water Processing, Medical and Laboratory Equipment, Packaging Equipment, Nuclear
and Aerospace and General Mechanical Handling. Additionally, screw jacks are
increasingly finding uses as alternatives to conventionally pneumatic and hydraulic
systems.
Today, screw jacks can be linked mechanically or electronically and with the advances in
motion-control, loads can be positioned to within microns. Car jacks usually use
mechanical advantage to allow a human to lift a vehicle by manual force alone.
1.2 ADVANTAGE, DISADVANTAGE & APPLICATION
1.2(a) ADVANTAGE :-
Screw jacks lift or raise the moderate heavy weights against gravity and uses very small
handle force that can be applied manually. Generally screw jacks have the following
advantages:
 They are self-locking.
 They are simple to design.
 They are large load carrying capacity.
 They can lift moderately loads like cars with very less force.

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1.2(b) DISADVANTAGE:-
The major disadvantage of the screw jack is that chances of dropping, tipping or
slipping of the load are high, they should always be lubricated, and they cannot be used
to lift or support very heavy loads and can cause serious accidents hence the device is
termed as not safe fail.
1.2(c) APPLICATION:-
 To raise the load, eg, screw-jack.
 To obtain accurate motion in machining operations, e.g. lead-screw of lathe.
 To clamp a work piece, e.g. vice.
 To load a specimen, e.g. universal testing machine.
1.2(d) COMPONENTS OF SCREW JACK
There are 7 components of screw jack they are :-
 Body (Frame)
 Screw rod (spindle)
 Nut and collar for nut
 Handle (Tommy bar)
 Cup
 Set screw
 Washer

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FIG1.2(a) FIG1.2(B)

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Chapter:-02
2.1 OBJECTIVE
The general objective of this project is to design screw jack with the details
drawing
 Parts to be designed
 To be designed Screw spindle.
 To be designed Nut and washer.
 To be designed Cup at the top of head for the load,
 To be designed Body of the screw jack.
 Forces to be considered
 To be considered Load applied
 To be considered Shear stress
 To be considered Bending stress

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Chapter:-03
3.1 METHODOLOGY
Problem statement (Determine the screw jack by
analytical method)
1. Design a screw jack to lift load of 250 KN and having a maximum lift
of 270 mm select proper material and draw a proportional sketch

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Selection of material for screw
The screw is under compression due to steady load but undergoes tension when
rotated. It is subjected to buckling also when the load is raised through the max lift.
Therefore from strength consideration, let us select steel as the material of the screw.
Let us take C40
Selection of FOS
As screw is subjected to both compressive as well as torsional shear stress and to
account for high-stress concentration in threads.
Hence consider higher FOS
Take n=4based on σy value
Permissible stresses for screw
Take σy=330 N/mm2
σt=σc=σyFOS=330/4=82.5 N/mm2
Take σc=82 N/mm2
By maximum shear stress theory
e=0.5×σyFOSe=0.5×σyFOS
e=41.25 N/mm2
Take e=41 N/mm2
Selection of material for nut
Considering nut material like cast iron which is easy to manufacture and cheap C.I also
wear rest.
Taking Grey C.I. i.e. GCI 20
σ4=200 N/mm2
Selection of FOS
As the material is brittle and to account for high-stress concentration threads, consider
high FOS=6 based on σu
Permissible stress for nut
σt=σuFOS=2006=33.33 N/mm2
Considering e=σt=33 N/mm2
As compressive strength of C.I. is more than tension strength
consider σc=2σt=66 N/mm2

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Bearing stress selection
From screw and C.I. nut combination
σbr=80 Kgf/cm2
=80×10 N/cm2=80×10 N/cm2
=800×10−2 N/mm2=800×10−2 N/mm2
=8 N/mm2=8 N/mm2
Design of screw
Consider the compressive failure of the screw
σc=WAc
Ac=2.50×10382Ac=2.50×10382
Ac=3.0487×103 mm2
Ac=π4d2cAc=π4dc2 where dc=62.3 mm
Above diameter is based on only direct compressive stress but the actual screw is also
subjected to torsional shear stress. Hence to take into account the effect of torsional
shear stress take a diameter 30% greater than the above value
∴dc=1.3×62.3=80.9 mm
∴Ac=π4d2c=5151.72 mm2
from selecting normal series we can provide a standard screw of the cover area Ac
Take Ac=5411 mm2
nominal diameter
do=95 mm
coverdia dc=23 mm
pitchdia p=12 mm
Mean dia dmdm = d0+dc2=39 mm
helix angle is selected by a single shot
tan(α)tan⁡(α) = pitchπdmpitchπdm
∴α=tan−1(12π×89) ∴α=tan−1⁡(12π×89)
α=2.45degree
Selecting the angle of friction
ϕϕ =6 to 8degree from ϕϕ >α, self-locking will be provided
∴∴ frictional torque acting on the
screw, T1=P×dm2=wtan(ϕ+α)dm2T1=P×dm2=wtan(ϕ+α)dm2
=250×103tan(6+2.45)892=250×103tan⁡ (6+2.45)892
T1=1.65×106 N−mm
∴ shear stress developed in screw '
e=16T1πdc3

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as TIp=er
Ip=π64d4+π64d4=π32d4
e=16×1.65×106π×832
e=14.89 N/mm2
∴∴the maximum principal stress in screw
σcmax = 12[σc+(σ2c+4e2)−−−−−−−−√]12[σc+(σc2+4e2)]
where σc is the direct compressive stress developed in the screw
∴σc=wAc = 250×1035411250×1035411
σc=46.20 N/mm2
∴σcmax=12[46.20+(46.20)2+4(14.69)2−−−−−−−−−−−−−√]∴σcmax=12[46.20+(46.20)2
+4(14.69)2]
σcmax=50.47 N/mm2
σcmax=50.47<σc=82 N/mm2
∴screw is safe in compressive stress Maximum shear stress developed in the screw
τmax = 12[(σ2c+4τ2)−−−−−−−−−√]12[(σc2+4τ2)]
τmax = 12[((46.2)2+4(14.69)2)−−−−−−−−−−−−−−−−√]12[((46.2)2+4(14.69)2)]
τmax=27.37 N/mm2<τ=41 N/mm2
∴ screw is safe in shear stress
Let us check screw for buckling
Assume the nut is a solid
Ratio Hdp= 2 where H = 2×dm=2×89=178 mm
Length of the screw for buckling purpose i.e. unsupport length of screw
∴L=∴L= lifting height +H/2
L=270+178/2=359 mm
Considering screw as a short column whose one end is fixed to nut and other end is free
The end fixing coefficient is n=0.25 from using Johnson's parabolic formula,
Pc=acσy[1−σy4nπ2E(tk)2]
where K=K= radius of gyration
K=0.25×dc=0.25×83=20.75 mm
E=2.1×105N/mm2
∴Pc=5411×330[1−3304(0.25)π2×2.1×105(35920.75)2]∴Pc=5411×330[1−3304(0.25)π2×
2.1×105(35920.75)2]
Pc=1.7×106 N>250×103 N
As value of buckling load is greater than the load at which screw is designed so there is
no chance of screw to buckle hence it is safe.

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Design of nut
We know nominal dia of screw is d0=95 mmd0=95 mm Taking d=d0+10 mm=105 mm
Considering crushing failure of nut column W=π4(D2−d2)×σEW=π4(D2−d2)×σE
D=125.88 mm
Let n be the no of threads in engagement
∴height of the nut ,
H=pitch×n
178=p×n=12×n
n=14.83take n=15
Let us check bearing stress developed in nut threads
W = π4(d02−dc2)×n×σbr
250×103=π4(952−832)×15×σbr
∴ Nut threads fails in bearing stress.
∴ Increase no of threads in engagement
Taking n=20
W = π4(d02−dc2)×20×σbr
∴σbr=7.45 N/mm2<σbr=8 N/mm2
∴ nut threads are safe
Let us check shear stress developed in nut threads
∴ W = π×d0×t1×n×Tπ×d0×t1×n×T induced
t1=pitch2t1=pitch2
∴ 250×103103 = π×95×6×20×TinducedTinduced
Tinduced=6.98 < Tnut =33 ∴ nuts and threads are safe
Let us check shear stress developed in the nut threads
∴W=π×d0×t1×n×Tinduced∴W=π×d0×t1×n×Tinduced
t1=pitch2=12/2=6mm
∴250×103=π×6×20×Tinduced
Tinduced=6.98 N/mm2ltTnut=33
∴ nuts and threads are safe in shear.
Now calculate 't’ by considering the shear failure of nut column
W=π×d×t×T
250×103=π×105× t× 33
t=22.93 mm
Take t=23 mm
Height of nut H=240 mm

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Design of handle
The diameter of the head (D1)on top of the screw is taken 1.5 to 1.7 of d0
∴D1 =1.6× d0 =1.6×95 =152 mm D2=0.5D1 =76mm
Then find the torque required to overcome friction at the top of the screw assuming
uniform pressing
T2=23μNR31−R33R21−R22T2=23μNR13−R33R12−R22
R1=D1/2=152/2=76
R2=D2/2=76/2=38
Assuming coefficient of friction μ=tanϕμ
μ=0.1
T2=23μNR31−R33R31−R32
T2=23×0.1×250×103[763−383762−382]
=1.47×106 N−mm
∴∴ Total torque to lift load
T=T1+T2=1.65×106+1.47×106 = 3.12×106 N−mm3.
Let Ln be the length of the handle T=F×Ln=T/F=7.8m
Generally length of handle is 1m
Maximum bending moment will be M=force×Length=400×1000
Consider material of handle C40 and FOS
σc=σbending=σyFOS=330/2=165 N/mm2
165=400×103π32×dh3
dh=29.12 mm

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3.2 3D MODELS OF SCREW JACK
SCREWM12 CAP
WASHER CUP
SCREW TOMMY BAR

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BODY
SCREW JACK ASSEMBLE

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CONCLUSION
The screw car jack is successfully designed so that it withstands all the mechanical
stresses acting on it. The stresses in above-mentioned conditions are found out and
thickness of various parts is selected such that the stresses produced in each member
are within the maximum allowable range. All the selected have been successfully verified
and hence the design of screw jack is safe.

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REFERENCES
1. Machine component design by “Robert Juvinall & kurt marshek.
2. Design of machine element by “ V.B. Bhandari”
3. Problem is solved by with the help of DDHB “ K. Madhavan”
4. 3D Design is done by “Solid edge software” Version “v-19”