mechanical design project

1. By: 1- Omar Mari
2- Nathmi Shurafa
3- Omyr Wahbi
4- Ruqaya Turabi
5- Mariam Shehada

2. A jack is mechanical device use to lift heavy loads or apply great forces.
There are types of car Jack:
1-pneumatic or hydraulic.
2-mechanical hand.
The most common form is a car jack ,floor jack or garage jack which lifts
vehicles so that maintenance can be performed
Conventionally a screw jack is used for :
1- lifting a vehicle to change a tire.
2-maintenance and many material handling operations.

3.  Paper, press, printing industry
 Gypsum factories
 Sheet metal forming machinery
 Mechanical lifting applications
 Platform lifting applications
 Food processing machinery
 Construction sector
 Bridge jacks for road and bridge lifting
 Shipyards
 Opening and closing of penstocks
 Industrial process
 Roll form machinery
 Mining industry
 Defence industry
 Lift tables
 Stage setup applications

4. Hazards:
In screw jack applications, the hazards are dropping, tipping or slipping of
machines or their parts during the operation.
The main reasons
1. Jack is overloaded
2. Load is improperly secured on the jack
3. Jack not placed on hard and level surface
4. Jack is used for a purpose for which it is not designed

5.  1-Platform and base:
These members that make contact with ground and the service
load.
Base platform

6. 2-power screw:
The power screw is a single Acme threaded screw using to
convert turning power to transmission power.

7. 3- Joints(Trunion):
Joints are used to link two beams with power screw.

8. 4- Beams:

9. Rivet is used as fasteners and to link beam with base and platform
5-Rivets:

10. 6-Handle:
A cranck is used to turn the screw in the center of the jack

11.  The force analysis consideration is based on the assumption that the screw jack is
loaded vertically symmetrical and the weight on the screw jack equal 4000 N :
1-Member analysis:
 𝐹𝑋 = 0
 F1 sin(𝜃) – F2 sin(𝜃)=0 F1 = F2 =F

12.  𝐹𝑌 = 0
F1 cos(𝜃) + F2 cos(𝜃)=w F=
𝑊
2𝐶𝑂𝑆(𝜃)
 𝐹𝑌 = 0
 F1 cos(𝜃) – F3 cos(𝜃)=0 F1 = F3 =F
 𝐹𝑋 = 0
 F1 sin (𝜃) + F3sin(𝜃)= Fs FS = 2F*sin(𝜃)

13.  At maximum raising height of the jack when h= 26.58 and 𝜃 = 20°
 F=
𝑊
2𝐶𝑂𝑆(𝜃)
=
400
2𝐶𝑂𝑆(20)
=2128.4 N
 FS = 2F*sin (𝜃) = 2*212.84*sin (20)=1455.9 N
Figure(a):shows position of the jack at maximum height.

14. At minimum raising height of the jack when h= 9.58 and 𝜃 = 70°
F=
𝑊
2𝐶𝑂𝑆(𝜃)
=
400
2𝐶𝑂𝑆(70)
=5847.6 N
FS = 2F*sin (𝜃) = 2*5847.6*sin (70)= 10990 N
Since the maximum loading force will act at the minimum raising height of the jack, the
design stresses will be analyzed at that point (h=9.58 m,𝜃 = 70° ).
Figure(b):shows position of the jack at minimum height

15.  Shear in rivet:
 Arivet = π/4 * d^2 = π/4 * (7.2)^2 = 40.72 mm2
 𝜏=
𝑅
𝐴
=
2923.8
40.72
= 71.8 𝑀𝑝𝑎
 N=
𝑆𝑆𝑦
𝜏
=
0.5∗180
71.8
= 1.254
 Bearing in rivet:
 Ab = d*thickness = 7.2*(4) =28.8 mm2
 𝜎 =
𝑅
𝐴𝑏
=
2923.8
28.8
= 101.5𝑀𝑝𝑎
 N =
𝑆𝑦
𝜎
=
180
101.5
= 1.77
2- Rivet analysis:
Figure(c):show forces in load platform and its distribution in
member and Revit.

16.  Bearing in member :
 Ab = d*thickness = 7.2*(4) =28.8 mm2
 𝜎 =
𝑅
𝐴𝑏
=
2923.8
28.8
= 101.5𝑀𝑝𝑎
 N =
𝑆𝑦
𝜎
=
180
101.5
= 1.77
 Rupture in member : member in compression
 A = (10+10+35)*4= 220 mm2
 𝜎 =
𝐹1
𝐴
=
5847.6
220
= 26.58 𝑀𝑝𝑎
 N =
𝑆𝑦
𝜎
=
180
26.58
= 6.7

17. 3- ACME power screw analysis :
 D = 13.5 mm
 P = 3 mm
 Dm = D -
𝑃
2
= 13.5 -
3
2
= 12 𝑚𝑚
 Nt = 107 thread
 Dr = Dm -
𝑃
2
=12-1.5 =10.5 mm
 𝛼 = 29°/2 = 14.5°
 l=p=3mm
 f = 0.2
 TR =
𝐹𝑠∗𝐷𝑚
2
*(
𝑙+𝜋∗𝑓∗𝐷𝑚∗sec(𝛼)
𝜋∗𝐷𝑚−𝑓∗𝑙∗sec(𝛼)
) =
10990∗12
2
∗ (
3+3.14∗0.2∗12∗sec(14.5)
3.14∗12−0.2∗3∗sec(14.5)
)=19195.86 N.mm

18.  To find the force acting by person in the end of the handle:
 TR = Fh * 200 Fh = 19195.86/200 = 95.97 N
 Stresses on the body of the power screw:
 • Shear stress:
 From force :
 𝑐 = 𝐷𝑟/2=5.25 mm
 𝐽 =
𝜋
32
*𝐷𝑟
4
=
𝜋
32
*10.54
= 1192.71 mm4

 𝜏𝑇 =
𝑇𝑅∗𝑐
𝐽
=
19195.86∗5.25
1192.71
= 84.5 Mpa
 • Normal (axial compressive)stresses in ACME screw by Fs :
 A = π/4* 𝐷𝑟
2
=
3.14∗(10.5)^2
4
= 86.54 mm2
 𝜎 =
𝐹𝑠
𝐴
=
10990
86.54
= 127 𝑀𝑝𝑎

19.  Thread stresses in power screw :
 • Nominal thread stress(bearing stress) in power screw
 AB = 𝜋 ∗ 𝑑𝑚 ∗ 𝑛𝑡 ∗ 𝑝/2
 𝜎B =
𝐹𝑠
𝐴𝐵
=
2∗𝐹𝑠
𝜋∗𝑑𝑚∗𝑛𝑡∗𝑝
=
10990∗2
3.14∗12∗107∗3
= 1.81 𝑀𝑝𝑎
 • The bending stress at the root of the thread is
 M = 𝐹𝑠 ∗
𝑝
4
 R=
𝑝
4
 I =
𝜋∗𝑑𝑟∗𝑛𝑡 ∗(𝑝/2)3
12
 𝜎b =
𝑀∗𝑅
𝐼
=
6∗𝐹𝑠
𝜋∗𝑑𝑟∗𝑛𝑡∗𝑝
=
10990∗6
3.14∗10.5 ∗107∗3
= 6.23 𝑀𝑝𝑎
Figure(d):show the force acting on the thread to determine stresses on it.

20.  • The transverse shear stress at the center of the root of the thread
 𝜏 =
3∗𝐹𝑠
𝜋∗𝑑𝑟∗𝑛𝑡∗𝑝
=
10990∗3
3.14∗10.5 ∗107∗3
= 3.11 𝑀𝑝𝑎
 The root of the thread is a weakest point on the thread which there is three stresses act on it (axial
compressive stress , shear stress due to torque and bending stress)
 𝜎x = 𝜎b =
6∗𝐹𝑠
𝜋∗𝑑𝑟∗𝑛𝑡∗𝑝
= 6.23 Mpa
 𝜏𝑦𝑧 = 𝜏𝑇 =
𝑇𝑅∗𝑐
𝐽
=
19195.86∗5.25
1192.71
= 84.5 Mpa
 𝜎z= 𝜎B =
4∗𝐹𝑠
𝜋∗𝑑𝑟^2
= 127 Mpa
 𝜎′
=
(𝜎𝑥−𝜎𝑦)2+(𝜎𝑦−𝜎𝑧)2+(𝜎𝑧−𝜎𝑥)2+6(𝜏𝑧𝑥
2+𝜏𝑦𝑧
2+𝜏𝑥𝑦
2)
2
 𝜎′
=
(6.06)2+(−12.7)2+(12.7−6.06)2+6(8.452)
2
= 191.8 Mpa
 𝑛 =
𝑆𝑦
𝜎′ =
300
191.8
= 1.56
Figure(e):show the stresses act in the root of the thread

21. 4- Handle analysis :
 Bending stress in the handle due to the force acted by a person (Fh = 95.97 N)
 C = d/2 = 9.4/2 = 4.7 mm
 I =
𝜋∗𝑑4
64
=
𝜋∗9.44
64
= 383 mm4
 M = Fh * 200 = 95.97 * 200 = 19194 N.mm
 𝜎 =
𝑀∗𝐶
𝐼
=
19194∗4.7
383
= 235.5 𝑀𝑝𝑎
 N =
𝑆𝑦
𝜎
=
280
235.5
= 1.188
Figure12:show the force acting on the handle(Fh) by a person.

22.  We studied car jack and determined the safety of each member through the
appropriate calculations and practical considerations with reasonable assumptions .
From our force and stress analysis it was discovered that the most sensitive position
occur at the minimum height when theta = 70 so we made our analysis at that
position using most allowable force (F=4000N) .
 It worthy to mention that the jack was designed in minimal number of parts(Only
rivet joints are induced Removal of welding to avoid distortion) that simplified
assembly process and reduced its weight , but in the same time it can rise high load
by applying very small force (we need force =96 N to rise 4000 N ).