Hydraulic jack (DOC)

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					Basic Mechanic Practice result


Hydraulic jack
1.Preview
      Hydraulic jacks are typically used for shop work, rather than as an emergency jack to be
carried with the vehicle. Use of jacks not designed for a specific vehicle requires more than the
usual care in selecting ground conditions, the jacking point on the vehicle, and to ensure stability
when the jack is extended.

      A hydraulic jack uses a fluid, which is incompressible, that is forced into a cylinder by a
pump plunger. Oil is used since it is self lubricating and stable. When the plunger pulls back, it
draws oil out of the reservoir through a suction check valve into the pump chamber. When the
plunger moves forward, it pushes the oil through a discharge check valve into the cylinder. The
suction valve ball is within the chamber and opens with each draw of the plunger. The discharge
valve ball is outside the chamber and opens when the oil is pushed into the cylinder. At this point
the suction ball within the chamber is forced shut and oil pressure builds in the cylinder.


2.Equipment and Tool




   Hydraulic jack                  Thickness measurement tool
       Tape                           Caliper




3. Sketch




4. Working procedure
   1. Measure the distance A – B
   2. Calibrate the Thickness measurement tool
   3. The direction of the prope grove thickness measurement tool should be parallel to the pipe
      axis
   4. Measure thickness and outside diameter of the pump piston
   5. Measure outside diameter of the load piston
   6. Measure stroke of the pump piston

5. Calculation
Data given :

F input = 30 lbf       1 complete stroke = 2 pump
Load distance = 0,36
ft                     1 stroke = 1 second
L=2m                   1 J/m = 1 N = 0,225 lbf          Pump / effort
η = 80 %               1 inch = 2,54 cm                 OD = 22 mm = 2,2
                                                        cm
                                                        tickness = 5,62 mm

                                                        Load
                                                        ID = 32 mm
Problem :
   1.   F rod pump piston ?
   2.   Hydraulic pressure ?
   3.   F load piston ?
   4.   No of cycle ?
   5.   Actual HP ?

Solution :




( Frod x 3 cm ) + 30 lbf x 197 cm = 0
        1. F rod




        2. P Hydraulic
       Effort




                     r2


                          mm2              -6
                                                m2
   
        Load
        ID = 32 mm
               r2


                            mm2 = 803, 84 . 10-6 m2
        Phydraulic        effort
                                     6
                                         N/m2
                        .106 N/m2
      3. Fload piston             load
                                  6
                    = 96,34. 10 N/m2 x 803,84 . 10-6m2

     Tinggi effort = 4 cm
     Tinggi load = 16 cm
     V1            = A1 x tload
                         2
                           x 16 cm
                            . 10-2cm2 x 16 cm
                                  . 10-2cm3
      V2          = A2 x teffort
                       2
                        x 4 cm
                             -2
                                  cm2 x 4 cm
                             . 10-2cm3
      4. No of cycle =
                    = 12861,44 . 10-2cm3 / 363,52 . 10-2cm3
                    = 35,38 X

      5. Actual HP
     Distance = 36 x 2 stroke
             = 72 times
     HP = lb.ft/s or Nm/s
            = 8755,5 N
     2 stroke = 1s
     HP    =

             =




     Actual hp = 93,89 hp x η
                = 93,89 hp x 80%
                 = 75,114209 hp


6. Analysis
   With assuming input force 30 lbf the pump piston force calculation can be done by using
    sigma moment equal to zero
    Oil path flow as bellow
          - Handle up : oil tank – check valve 1 open – check valve 2 close -- pump barrel
          - Handle down           : pump barrel – check valve 1 close – check valve 2 open –
              bleed valve close – load barrel

            So when handle down base on Pascal law we can calculate hydraulic pressure and force
            at load piston.

    Volume of piston load equal to volume of oil displacement x no of pump piston stroke
    Actual horsepower is horsepower theoretic x efficiency

7. Result & Conclusions
Result :

F input               = 30 lbf = 133.33 N

F road pump piston    = 8755.5 N

F load piston         = 77441.94 N

Hydraulic pressure    = 96.34 x 106 N/m2

Actual HP             = 75.114 Hp

Conclusion

Hydraulic jack is an affective device to transform small force to the bigger force with the minimal
power usage




Pipe flow
1.Preview
       Fluid flow can be either laminar or turbulent. The factor that determines which type of flow
    is present is the ratio of inertia forces to viscous forces within the fluid, expressed by the
    nondimensional Reynolds Number, where V and D are a fluid characteristic velocity and
    Internal diameter.
where:
      is the mean fluid velocity (SI units: m/s)
   μ is the dynamic viscosity of the fluid (Pa·s or N·s/m² or kg/m·s)
   ν is the kinematic viscosity (ν = μ / ρ) (m²/s)
    is the density of the fluid (kg/m³)
   Q is the volumetric flow rate (m³/s)
   A is the pipe cross-sectional area (m²).

    Typically, viscous stresses within a fluid tend to stabilize and organize the flow, whereas
    excessive fluid inertia tends to disrupt organized flow leading to chaotic turbulent behavior.
    Fluid flows are laminar for Reynolds Numbers up to 2000. Beyond a Reynolds Number of
    4000, the flow is completely turbulent. Between 2000 and 4000, the flow is in transition
    between laminar and turbulent, and it is possible to find subregions of both flow types within a
    given flow field.

2. Equipment and tool




  Thickness measurement                   Caliper                         Pipe line




3. Sketch
Do        : Outside pipe diameter
Di        : Inside pipe diameter
t         : Thickness

4. Working Procedure
     1. Calibrate the Thickness measurement tool
     2. The direction of the prope grove thickness measurement tool should be parallel to the pipe
        axis
     3. Measure thickness of the pipe at four position in each measurement point
     4. Measure outside diameter of the pipe at each measurement point

5. Calculation

Data given :



Re = 80
Ρ = 0,075 gr/cm3
µ = 2000cP = 20 P = 20 gr/cm.s
1 poise = 1 gr/ cm.s

Problem :

     1. Fluid velocity ?
     2. Flow rate ?

Measurement result :

                            Post 1   Post 2    Post
     No                      mm       mm      3 mm
     1       OD              115     115      115
     2       t on 0°         6,06    6,05     6,28
     3       t on 90°        6,06    6,05     6,60
     4       t on 180°       6,06    6,05     6,60
     5       t on 270°       6,06    6,15     6,13
     Thickness Average       6,06    6,75     6,40
  Average = 6,06 mm + 6,75 mm + 6,40 mm
                    3
             = 6,404 mm
Solution :

     Find D
      OD average = 115 mm




     Find V
      V




     Find Q
      Q=VxA
        = 209,15 cm/s x A
        = 209,15 cm/s x r2
        = 209,15 cm/s x 3,14 x 5,1 cm x 5,1 cm
        = 16754,89 cm3/s

6. Analysis
Reynolds number can be defined for a number of different situations where a fluid is in relative
motion to a surface, when Reynolds number of the fluid flow, fluid properties and inside diameter
of pipe are known fluid velocity can be determine by using Reynolds number equation




7. Result and Conclusions
Fluid velocity         = 209.15 cm/s
Flow rate              = 16754.89 cm3/s




Strange of Material
1.Preview
      During testing of a material sample, the stress–strain curve is a graphical representation of
the relationship between stress, derived from measuring the load applied on the sample, and strain,
derived from measuring the deformation of the sample, i.e. elongation, compression, or distortion.
The nature of the curve varies from material to material. The following diagrams illustrate the
stress–strain behaviour of typical materials in terms of the engineering stress and engineering strain
where the stress and strain are calculated based on the original dimensions of the sample and not the
instantaneous values.




Typical yield behavior for non-ferrous alloys.
1: True elastic limit
2: Proportionality limit
3: Elastic limit
4: Offset yield strength
Ductile materials




                           Fig 1. A stress–strain curve typical of structural steel
1. Ultimate Strength
2. Yield Strength
3. Rupture
4. Strain hardening region
5. Necking region.
A: Apparent stress (F/A0)
B: Actual stress (F/A)

      Steel generally exhibits a very linear stress–strain relationship up to a well defined yield point
(figure 1). The linear portion of the curve is the elastic region and the slope is the modulus of
elasticity or Young's Modulus. After the yield point, the curve typically decreases slightly because
of dislocations escaping from Cottrell atmospheres. As deformation continues, the stress increases
on account of strain hardening until it reaches the ultimate strength. Until this point, the cross-
sectional area decreases uniformly because of Poisson contractions. The actual rupture point is in
the same vertical line as the visual rupture point.

      However, beyond this point a neck forms where the local cross-sectional area decreases more
quickly than the rest of the sample resulting in an increase in the true stress. On an engineering
stress–strain curve this is seen as a decrease in the stress. Conversely, if the curve is plotted in terms
of true stress and true strain the stress will continue to rise until failure. Eventually the neck
becomes unstable and the specimen ruptures (fractures).

      Less ductile materials such as aluminum and medium to high carbon steels do not have a
well-defined yield point. For these materials the yield strength is typically determined by the "offset
yield method", by which a line is drawn parallel to the linear elastic portion of the curve and
intersecting the abscissa at some arbitrary value (most commonly 0.2%). The intersection of this
line and the stress–strain curve is reported as the yield point.1




Brittle materials




                                    Stress Strain Curve for Brittle materials
      Brittle materials such as concrete and carbon fiber do not have a yield point, and do not strain-
harden which means that the ultimate strength and breaking strength are the same. A most unusual
stress-strain curve is shown in the figure. Typical brittle materials like glass do not show any plastic
deformation but fail while the deformation is elastic. One of the characteristics of a brittle failure is
that the two broken parts can be reassembled to produce the same shape as the original component
as there will not be a neck formation like in the case of ductile materials. A typical stress strain
curve for a brittle material will be linear. Testing of several identical specimen, cast iron, or soil,
tensile strength is negligible compared to the compressive strength and it is assumed zero for many
engineering applications. Glass fibers have a tensile strength stronger than steel, but bulk glass
usually does not. This is because of the Stress Intensity Factor associated with defects in the
material. As the size of the sample gets larger, the size of defects also grows. In general, the tensile
strength of a rope is always less than sum of the tensile strength of its individual fibers.


2. Equipment and tool




   Thickness measurement               Tensile strength test                Caliper




3. Sketch
Do      : Outside diameter
Di      : Inside diameter
t       : Thickness
P       : Pressure
L       : length


4. Working Procedure
     1. Calibrate the Thickness measurement tool
     2. The direction of the prope grove thickness measurement tool should be parallel to the pipe
        axis
     3. Measure the thickness of pipe
     4. Measure the width of specimen

5. Calculation
Data Given:




Specimen width = 40 mm
Thickness = 40 mm = 0,04 m
Fu = 9280 kgf
FS = 4
1N = 12991,2 gf

Problem :
WP ? ( Mpa )

Solution :
      =










6. Analysis
           The ultimate stress of the material can be known from the tensile strength experiment by
dividing ultimate force (Fu) with cross section area of the specimen. Beside the dimensional
variable ultimate stress of the pipe material also needed to calculate burst pressure than the
maximum allowable pressure can be found by dividing burst pressure with safety factor.

7. Result & Conclusions
Result :

Burst pressure

Maximum safe operating fluid

Conclusion :

       The maximum safe operating fluid pressure in normal condition depend on material
         properties, dimensional factor and safety design factor.
       Safety factor design is covering all condition beyond of the design consideration




Chain Block
1. Preview
             Chain Block is a tackle which uses an endless chain rather than a rope, often operated
        from an overhead track to lift heavy weights especially in workshops. Also known as chain
        fall or chain hoist.
             A chain Block is operated by hand. An operator will pull down on one of the chain
        loops on one side of the chain. This will turn a pulley mechanism inside the chain hoist
        housing. When this pulley turns, it will lift up the end of the other chain which usually has a
        hook on the end. By pulling down on one chain, the manual hoist is actually able to increase
        the mechanical work that is being done. This is caused by the gear ratio inside the manual
        chain hoist.
         Inside the Chain Block housing are two gears. One is smaller than the other. Most Chain
    Blocks use a 20 cm and 25 cm gear. The two gears are attached, so when one moves, the
    other moves. The chain is looped over the smaller gear, and then hangs down in a loop (the
    loop you pull on). The chain then continues on over the larger gear and down to a point or
    another loop, depending on the type of hoist you are using. The operator pulls on the section
    of chain which is looped over the smaller gear.
         When someone pulls down the chain as explained in Section 1, the smaller gear will
    turn. However, since it is smaller in size, it turns faster than the larger gear. It makes more
    rotations than the larger gear in a cycle. Since the larger gear is turning slower, it creates
    more force, in effect transforming the "pull" on the chain into a larger force. This is a
    mechanical advantage. An operator can put less force on the smaller gear, but still lift large
    objects. That is because the larger gear transforms that force into a much larger one.




2. Equipment and tool




           Tape                   Chain and Block
3. Sketch




                     Zero reference




4. Working procedure
1. Define the zero reference
2. Measure the distance of h1( effort distance ) and h3
3. Pull down the effort chain as the h1 distance
4. Measure the load displacement h2
5. h2 – h3 is load distance


5. Calculation
Data Given :
Finput (effort) = 30 lbf
Max load = 500kg = 5000 N
             = 5000 N x 0,225
             = 1125 lbf
1 J/m        = 0,225 lbf

Problem :
η ? ( efficiency )

Solution :
      Efford distance = 585 cm


      Load1 distance = 115 cm
      Load2 distance = 107 cm
      Δ Load load distance = 8 cm

      Vr

            = 73,125 cm

      MA =

            = 37,5

      η

           = 51.3 %

 6. Analysis
 Chain and block efficiency can be calculate by dividing actual mechanical advented with velocity
 ratio




 7. Result and conclusions
 Result :

 Chain and block Efficiency       = 51.3 %

 Conclusions :

      The load work is 51.3 % of the effort work
      Load force : effort force = 37.5 : 1
      Load distance : effort distance = 1 : 73.13
     So chain and block is very good in mechanical advented but poor in velocity ratio

 Speed Measurement

1. Preview
               Rotational speed (sometimes called speed of revolution) indicates, for example, how
 fast a motor is running. Rotational speed is equivalent to angular speed, but with different units.
 Rotational speed tells how many complete rotations (i.e. revolutions or cycles) there are per time
 unit. It is therefore a cyclic frequency, measured in hertz (revolutions per second) in the SI System.
The units revolutions per minute (rpm or 1/min) are more common in everyday life. Angular speed,
however, tells the change in angle per time unit, which is measured in radians per second in the SI
system. Since there are 2π radians per cycle, or 360 degrees per cycle, we can convert angular speed
to rotational speed by:



and



where

           is rotational speed (cycles per second)
           is angular speed (radians per second)
            is angular speed (degrees per second)

             For example, a stepper motor might turn exactly one complete revolution each second.
Its angular speed is 360 degrees per second (360°/s), or 2π radians per second (2π rad/s), while the
rotational speed is 60 rpm.

            Rotational speed is not to be confused with tangential speed, despite some relation
between the two concepts. Imagine a rotating merry-go-round. No matter how close or far you stand
from the axis of rotation, your rotational speed will remain constant. However, your tangential
speed does not remain constant. If you stand two meters from the axis of rotation, your tangential
speed will be double the amount if you were standing only one meter from the axis of rotation.

TACHOMETER

             Laser light tachometer for measuring the rotational speed of selected rotating body
without making physical contact with that body. The laser techology art is used to provide beam of
light, visible even under extremely bright ambient light conditions, to be used to measure the
rotational speed of rotating device from grather distance thaen previously possible. A diverging
beam of laser light is colimated and transmitted to a rotating body the RPM of which is to be
measured. The rotating body modulateds the beam and reflects it back to the tachometer where it is
focused onto a photodetector, which converts it into a electrical signal representative of the speed of
the rotating body. Provisions are made for direct digital readout and for supplying the signal to
other devices.



2. Equipment and Tool
        Tachometer                       Turning machine

3. Sketch




4. Working procedure
   1.   Put on a peace of reflector on work piece holder of turning machine
   2.   Adjust the transmission of turning machine as the speed requested
   3.   Run the machine on rapid or slow mode
   4.   Shoot the reflector with the tachometer




5. Calculation
Measurement result

                                 Machine table        Tachometer
   No       n/min      S/R
                                 A         B          A         B
   1          4         R       1000        -       1120        -
    2         3        R        725       118       810       129
    3         2        R        460        75       520        82
    4         1        R        290        45       323        51
    5         1        S        145        24       162        26



6. Analysis
table above is showing that all of tachometer measurement results are difference with the machine
table, it could be caused by :

   1. Unstable hand position during measurement
   2. Tachometer position wasn’t perpendicular to the reflector area
   3. It should be available speed tolerance of the turning machine so we can determine whether
      our measurement result in or out of the machine tolerance
   4. Instability of motor speed
   5. it could be problems at turning machine gear box




Measurement assignments

 No                                   Picture                                        Nilai
1   0.00 mm




2   0,35 mm




3   0,50 mm
3a   3,00 mm




4    53,00 mm




4a   38,30 mm
5   38,30 mm

				
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