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Abstract
This report is submitted by a group of students. It aims
to apply what learned
in class practically.It discusses a car jack screw. It gives an
introduction about the jack and its
advantages and safety standards. On the other hand, it contains
simple
calculation for the maximum force it can handle and factor of
safety it canprovide.
Introduction
A jackscrew is a type of jack that is operated by
turning a lead screw. It is
commonly used to lift heavy weights, such as the foundations of
houses, or
large vehicles.
Procedure and Results
ack bars calculations!"irst we need to calculate the
centroid for the cross section of the member
and to do that we divide the member into # areas as shown in
"ig.$.
"ig.$! The centroid for the cross section of the member
• %entroid %alculations!
X ’=a1 x1+a2 x 2+a3 x 3
a1+a2+a3 =
(2∗20) (1 )+(31∗2 ) (17.5 )+(2∗20)(34)(20∗2)+(31∗2)+(20∗2)
=17.5mm
$
http://en.wikipedia.org/wiki/Jack_(device)http://en.wikipedia.org/wiki/Leadscrewhttp://en.wikipedia.org/wiki/Leadscrewhttp://en.wikipedia.org/wiki/Jack_(device)
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Y ’=a 1 y 1+a 2 y 2+a 3 y 3
a 1+a 2+a3 =
(2∗20 ) (10 )+ (31∗2 ) (19 )+(2∗20)(10)(20∗2)+(31∗2)+(20∗2)
=13.93 mm
The &eutral Axis is $#.'# mm from the reference (ref
is the far side at the
bottom).After *nding the centroid, we need to *nd the moment of
inertia.
• +oment of Inertia %alculations!
I I$ - I - I# , I$ I#
I 1+ I 3=2∗
( 1
12
b h3+ A d2
)=2∗
( 1
12
∗2¿203+2∗20∗3.932
)¿2∗(1333.33+617.796 )=3904.252 mm4
I 2=( 112 b h3+ A d2)=(
112∗31¿23+31∗2∗6.072)
¿2305.05 mm4
Itot = I 1+ I 2+ I 3=3904.252+2305.05=6209.303
mm4
/ecause of the symmetry we have we will study on side of the
jack and we
will study the right side.
&ow we assume the critical point we will check it is #.0#1mm
away from the
&.A.
Then we will analysis the "max2. Then we will start the
&ormal stress
calculations.
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&ormal stress calculations!
σF =(
Fmax
2
)∗cosα
A = ( Fmax )∗
cos 40
2∗(2∗(2∗20)+(31∗2 ))=19.71 Mpa
( Fmax sinα 2 ) (120 )∗3.035σM =
M c
A =¿ ¿
6209.303=137.605 Mpa
3tot 3" - 3+ $14.#$1 +5a
• To *nd the shear we need to *nd the centroid for the
shaded area 6*g.#7!
89sh$4.1 mm
Y ’ s h=a 1 y 1+a 2 y 2+a3 y 3
a 1+a 2+a 3 =
(2∗3.035 ) (1.5175 )+(31∗2 ) (2.035
)+(2∗3.035)(1.5175)(3.035∗2)+(31∗2)+(3.035∗2)
=144.6
74.14
• :hear :tress!
τ =VQ
It
; 6Ash =.'?17 #@'.1' mm3
#
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τ =( Fmax2 )sin 40∗369.59
(6209.303)(4) =34.9159 Mpa
• sing +:: Theory!
Be will use +:: in our calculation to *nd the factor of safety
6n7 because it9s
more conservative than the other theories.
τmax=√(σ
2 )2
+(τ )2= Fmax√ (0.010775 )2+( .004783
)2=86.067 MPa
• "actor of safety calculation!
n :sy2 Cmax
n 60.1>=@[email protected]@4n .@4#
Be can see here that the factor of safety is bigger than $ and
that means the
design is acceptable.
:crew calculations!
Be have a single threaded power screw of coarse pitch series
M16X2, the
friction %oeDcient 6f7 of 0.$ and the maximum force 4.# E&
obtained from
the previous calculations.
+ajor diameter d $@ mm
The pitch 5 mm
Fead length mm
According to table 6?G$7 in the text book we used!
• The mean diameter!
=
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dm=d−0.649519 p → dm=16−0.649519 (2 )=14.7 mm
• The reHuired raising torHue!
T r= F dm
2 (l+! dm dm−!l )
T r=(7.3 ) (14.7 )
2 ( 2+ (0.1 ) (14.7 ) (14.7 )−(0.1 ) (2 )
)=7.723 "m
• The reHuired lowering torHue!
T #= F dm
2 (! dm−l dm+ !l )
T r=(7.3) (14.7 )
2 ( (0.1 ) (14.7 )−2 (14.7 )+(0.1 ) (2
) )=3.03 "m
ivet $ calculations!
• :hear !
V = F
F $ =
V
2 =0.5 F
%= F $
"ig.=! rivet $&Y =450 Mpa
&sy=0.577 &Y → & sy=265.42 Mpa
τ = F
$
A=
0.5 F
48
2
=9.95∗103 F
1
"9 J "9
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n=&sy
τ =2=
265.42
9.95∗103 F
F =13.34 '"
• apture !
A=(d∗t =2∗8∗2=34 ) m2
σ = F
A=
F
34 n=2=
460
0.0294 F =7.823 '"
• /earing !
σ o= F
Ab=
F
8∗2=
0.5 F
16=0.03125 F
n=2=460
0.03125 F
ivet calculations!
• :hear!V = F
+ " d '$.' "
F
$
=
V
2 =0.5 F
F = left ({M {r} rsub {B}} over {{{r} rsub {}}
! {2} " {{r} rsub {B}} ! {2}} ri#$t % = 0.
"ig.1! rivet
F $
K ") 0.1 " K 0.1 " 0
@
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&o shear stress
• upture
σ = M *
I 25
¿¿¿3¿15
¿¿¿32∗¿ I =¿
σ b+arin,=(91.92 F )∗(12.5)
2041.667 =0.5627 F MPa
n= &Y
σ b+arin,=2=
460
0.5627 F
• /earing !
σ b= F
t d= F
2∗5=0.5 F
60
n=2=460
0.0166 F
F =113.813 '"
The bearing force that can be used is "4.# E&
This rivet has the same force as the upper one but in the
inverse direction.
4
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Conclusion
we studies a real application on what studied theoretically in
class and faced
the problems and solved it, we learned that it9s not easy to
deal with the
outer world and enjoyed an experience in being engineer.
Be found that this jack, which has been studied, can handle a
maximum
force of 4.# E& with a factor of safety of .@.
:ome errors may occur due to the supposed terms, such as type of
material
6L$0#10 %OFM BOE7 and friction factor, or due to some measured
values,
such as the angle between the two bars of the jack.
?
