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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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posted: | 1/27/2012 |

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