Hydraulic Arm

Design And Fabrication of Hydraulic Arm

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Hydraulic Arm Hydraulic arm is basically the extended component of a machine

which can perform task’s and

operations that

are far from the reach of the main

machine body and require lots of small & delicate movements . As the name suggests “hydraulic arm” is the mechanical arm which operates with the help of hydraulic press that is provided from the main body of the machine. Both, the arm & machine body are equipped with hydraulic cylinders that delivers hydraulic press and make the arm move.

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Pascal's law says that when the pressure at any point in a static fluid in a closed system is changed, the change in pressure will disperse equally throughout the fluid. That is, the pressure at a point far away from the region of change will change by the same amount as a point nearby. A closed system can simply be an enclosed container, or it may be something more complex, such as two or more interconnected containers; the important thing is that no fluid can enter or leave the system. It is also important to note that, in physics, a fluid can be either a liquid or a gas. The law can be demonstrated by a number of simple experiments, and has important applications, such as in the hydraulic press. The principle was named after the French mathematician and philosopher Blaise Pascal who discovered it in the 1600s. It applies to static situations and not to dynamic conditions where other factors could affect pressure values. For example, it does not apply to fluids that are in motion or subject to changing temperatures.

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Demonstrations There are a variety of Pascal's law experiments that can be used to demonstrate the effect. Pascal himself showed that it worked by filling a barrel with water and inserting a long pipe into the top. When he poured water into the top of the pipe, the barrel burst. The weight of the water in the pipe caused an increase in pressure inside the barrel that pushed against the sides until they gave way. Perhaps the most basic way of demonstrating the law at home involves simply squeezing a balloon. In this example, the flexible walls of the container show how the increase in pressure caused by squeezing is dispersed throughout the balloon. The balloon will bulge uniformly in all directions, not just at the side opposite that which is squeezed. In another common demonstration, a bottle is filled to the top with water, and some matchstick heads are dropped into it so that they float. The neck of an inflated balloon is stretched over the bottle, and then lightly squeezed. The matchstick heads now sink some distance into the water. This is because the increase in pressure due to the squeezing of the balloon is transmitted down into the water, forcing some of it into the porous matchstick heads and 4

causing them to sink, due to the extra weight. When the pressure on the balloon is removed, the water pressure decreases, air pressure in the matchstick heads forces water out, and they float again. Applications Perhaps the best-known application of Pascal’s law is the hydraulic press, a device that converts a small force into a larger one. It generally consists of two connected chambers, each with a piston — a moveable barrier that can be pushed down or pulled up without allowing fluid to escape — and containing a fluid that cannot be compressed. One chamber-piston combination is larger than the other: this is the “output.” The idea is that a small force applied to the smaller, or “input,” piston will result in a larger output force. Pressing down on the input increases the pressure, and that increase will be the same against the larger output piston. Calculating the Output Force The output force is calculated by dividing the area of the output piston by the area of the input piston then multiplying the result by the input force. If the output piston has ten times the area of the

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input, the output force will be ten times the input force. For example, if the input force is 5 units, the input area is 2 units and the output area is 20 units, the output force will be 50 units. In this way, heavy objects can be lifted without the need to apply a large force. This does not mean that extra energy is appearing out of nowhere. The amount by which the output piston is raised will be less than the amount by which the input piston is pressed down, which evens things out. In the above example, if the input piston is pushed down 10 units, the output piston will be raised by 1 unit. The principle is similar to using a lever to lift a rock. Hydraulic mechanisms of many types, such as the braking systems on aircraft and some vehicles, rely on Pascal’s law.

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Principle behind hydraulic arm: Hydraulic arms works on the principle of hydraulic press which is governed by Pascal's law according to which “pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure ratio (initial difference) remains same.

where : ΔP :

Is the hydrostatic pressure or the difference in pressure at

two points within a fluid column, due to the weight of the fluid. ρ:

Is the fluid density.

g:

Is acceleration due to gravity.

Δh :

Is the height of fluid above the point of measurement, or the

difference in elevation between the two points within the fluid column. Pascal’s law can be written as : (P1)(V1) = (P2)(V2)

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According to this equation the ratio of pressure & volume remains same at both the ends of an unrestricted system like pipes, hydraulic cylinders etc.

In case of hydraulic press the force transmitted is proportional to the area of cross section of the end’s where force is applied.

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Pascal's Law

A demonstration of Pascal's Law Pascal's Law (from Blaise Pascal 1623 to 1662), comprises a set of principles formulated in 1648 and states that pressure applied to a confined fluid at any point is transmitted undiminished throughout the fluid in all directions and acts upon every part of the confining vessel at right angles to its interior surfaces and equally upon equal areas. This is the basic principle behind any hydraulic system - pressure applied anywhere to a body of fluid causes a force to be transmitted equally in all directions, with the force acting at right angles to any surface in contact with the fluid.

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Pressure is force divided by area. If you apply even a small force to a needle, the pressure at the tip is high because its area is small (look at the formula below). That is why it penetrates materials!

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The hydraulic jack Pictured in the animation above, it transmits force from one piston (small) to the other (larger). The force is transmitted by a liquid (oil). A liquid exerts the same pressure in all directions (as well as a gas) and as a result the two pistons will exert different forces because of the different surface areas they possess, according to the formula P=F/A.. The larger piston will travel a smaller distance but it will exert a larger force. The formula can be rearranged: P=F x A. Because the pressure (P) is the same in the whole system (which is the hydraulic jack), the force (F) will be larger where the area (A) of the piston is larger!

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This way it can lift a car, a lorry or anything... It is all about being able of building a large enough piston (engineering limitation). The force exerted by the large piston will be larger than the force applied to the little piston, although the oil pressure is the same everywhere in the circuit. If the area of the large piston is twice as big as the area of the little one, the force will be doubled. That is the point of this machine! The small piston moves a longer distance while the large piston moves a shorter distance, as you wuld expect from the principle of the conservation of energy. You exert a small force but for a large distance; on the other end of the jack a large force is exerted by only for a short distance.It is a trade off. It is similar to the gear system of a bicycle: when you use a soft gear, which allows you to pedal easily, you have to pedal more times in order to cover the same distance. In a modern hydraulic system, the force will be applied by a motor which pumps oil through the circuit. This is the principle of hydraulic machines, extensively used in the industry, especially in heavy machinery. 12

Car, train, lorry and aircraft control systems also make use of this technology.

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Examples of questions and problems about Force and Pressure 1) A needle will exert enormous pressure because the surface area (A) of the point where the force (F) is being applied is very small. That's why it penetrates skin and other materials.

2) If your mass is 50 kg , what is the pressure you exert on the ground when you are standing up, assuming that the surface area of your shoes is 0.05 m2. Answer: (remember that mass is not the same as weight) Assuming that g =10 m/s2 , F = 50kg x 10 m/s2 = 500 N (weight) A = 0.05 m2 P = F / A = 500 N / 0.05 m2 = 10000 N / m2

3) A jack is projected to lift a bus which has a mass of 4 000 kg. If the large piston (the one that moves up in the animation) has an 14

area of 1 m2, and the small piston on the other end of the circuit has an area of 0.05 m 2 , what is the minimum force that must be applied (to the little piston) in order to lift the bus? Answer: We want find out a force, which is given by : F = P x A . P is the same in the whole circuit , and to do this exercise we don't need to know its value so that we will simply call it P. Lets call the force on the small piston F s and the force on the large piston FL . Likewise, the area of the small piston will be called A s and the area of the large piston A L. Because the pressure (P) is the same in both pistons we can write: Fs = A s x P and FL = A Lx P We know the areas ( A s = 0.05 m2and A L= 1 m2), so that we can substitute them in the equations: Fs = 0.05 m2 x P FL = 1 m2 x P

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We have 2 equations and 2 unknowns. It is a system of equations. From now on you will use your maths skills. You could even pretend that you have x's and y's instead of F's , A's and P's , if it makes things easier to you. There are many ways to solve this system . We will do it by dividing the top equation by the bottom one:

Fs / FL = 0.05 / 1 That is all (Remember that P / P = 1) . So, Fs / FL = 0.05 Rearranging: Fs = 0.05 FL In order to lift the bus , the force on the large piston (F L) must be al least 40 000 N (because the mass of the bus is 4 000 kg). So, Fs = 0.05 x 40 000

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Fs = 2 000 N (Final answer) This example illustrates how a small force can be multiplied many times by using a hydraulic jack. This is the type of calculation that mechanical engineers do.

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The relationship between hydraulic pressure and force

Pressure is the force per unit area exerted on a surface, divided by the area of that surface. Pressure = Force/Surface Area Therefore hydraulic pressure is the force per unit area exerted by a fluid on the surface within the container. For the demonstration of the relationship between hydraulic pressure and force, Sphaera's interactive interactive instructor-led CBT products are simple yet effective. Buttons on screen provide the instructor with full control over the amount of force applied 18

whilst pressure and force are clearly indicated numerically and through use of colour. Hydraulic Brakes

A simple interactive hydraulic braking system. In a hydraulic braking system, the slave pistons have a larger area than the master cylinder. Force applied to the master cylinder is multiplied as the fluid exerts the resultant pressure on a greater surface area. To demonstrate the operation of a simple hydraulic braking system, Sphaera's interactive classroom training aid is simple yet effective. Buttons on screen provide the instructor with full control over the amount of force applied whilst pressure and force are clearly indicated through use of colour

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Simple Hydraulic System

A simple hydraulic system A simple hydraulic system such as that within a hydraulic jack consists of two different sized cylinders connected by a pipe. According to Pascal's Law, pressure exerted on the smaller piston is transmitted through the fluid to act on internal surface of the larger piston. Pressure is a property of the system (not the pistons) and is therefore experienced equally by each piston. Because each piston has a different surface area, the force exerted on each piston will be different, even though the pressure is the same. If the larger piston is twice the area of the smaller piston then the force on the larger piston will be twice as great. In order to create that extra force, the smaller piston has to be moved by twice the distance. It was this principle that was understood by Joseph Bramah when he patented the Bramah Press in 1795

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For the demonstration of how a simple hydraulic system works, Sphaera's interactive instructor-led CBT is simple yet effective. Buttons on screen provide the instructor with full control over the amount of force applied whilst pressure and force are clearly indicated numerically and through use of colour. Hydraulic Swash Plate Pump

A constant displacement hydraulic swash plate pump A constant displacement hydraulic pump uses a fixed swash plate to drive a set of pistons in and out as they revolve. At the top of their stroke, the pistons move over the inlet port and draw in fluid at low pressure. The fluid is then carried round and expelled through the outlet port at high pressure as the swash plate drives the piston into the cylinder. To demonstrate this principle, Sphaera's interactive hydraulic swash plate pump features 8 revolving cylinders that provide a 21

clear indication of hydraulic pressure and flow. Operation can be viewed at fast or slow speed.

A Frenchman

named

Pascal

discovered

that

a pressure

applied to any part of a confined fluid transmits to every other part with no loss. The pressure acts with equal force on all equal areas of the confining walls and perpendicular to the walls. Remember when you are talking about the hydraulic machine, you are talking about the way a liquid acts in a closed system of pipes and cylinders. The action of a liquid under such conditions is somewhat different from its behavior in open containers or in lakes, rivers, or oceans. You also should keep in mind that you cannot compress most liquids into a smaller space. Liquids don’t “give” the way air does when you apply pressure, nor do liquids expand when you removepressure.Punch a hole in a tube of toothpaste. If you pushdown at any point on the tube, the toothpaste comes outof the hole. Your force has transmitted from one placeto another through the toothpaste, which is a thick, liquidfluid. Figure 10-5 shows what would happen if youpunched four holes in the tube. If you were to press onthe tube at one point, the toothpaste would come out ofall four holes. You have illustrated a basic principle ofhydraulic

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machines. That is, a force applied on a liquidtransmits equally in every direction to all parts of thecontainer.We use this principle in the operation of four-wheelhydraulic automobile brakes. Figure 106 is a simplifieddrawing of this brake system. You push down on thebrake pedal and force the piston in the master cylinderagainst the fluid in that cylinder.

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Pascal's Principle Pressure is transmitted undiminished in an enclosed static fluid.

Any externally applied pressure is transmitted to all parts of the enclosed fluid, making possible a large multiplication of force (hydraulic press principle). The pressure at the bottom of the jug is equal to the externally applied pressure on the top of the fluid plus the static fluid pressure from the weight of the liquid.

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Hydraulic Press

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Automobile Hydraulic Lift A hydraulic lift for automobiles is an example of a force multiplied by hydraulic press, based on Pascal's principle. The fluid in the small cylinder must be moved much further than the distance the car is lifted.

For example, if the lift cylinder were 25 cm in diameter and the small cylinder were 1.25 cm in diameter, then the ratio of the areas is 400, so the hydraulic press arrangement gives a multiplication of 400 times the force. To lift a 6000 newton car, you would have to exert only 6000 N/400 = 15 N on the fluid in the small cylinder to lift the car. However, to lift the car 10 cm, you would have to move the oil 400 x 10cm = 40 meters. This is practical by pumping oil into this small cylinder with a small compressor.

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Basics of Hydraulics The question "What is hydraulics ?" can be answered in the following way. Hydraulics is the transmission and control of forces and motions through the medium of fluids. Short and simple. Hydraulic systems and equipment have wide-spread application throughout industry. For example: - machine tool manufacturing - press manufacturing - plant construction - vehicle manufacturing - aircraft manufacturing - shipbuilding - injection molding machines Prerequisites that hydraulics requires of the user and serviceman: - knowledge of the basic physical laws of hydrostatics and hydrodynamics - knowledge of the symbols of hydraulic control elements

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- knowledge of hydraulic circuit diagrams - knowledge of the maintenance of a hydraulic system Hydraulic to Electrical Analogy Hydraulics and electrics are analogous, because they both deal with flow, pressure and load. The components in each type of circuit perform similar functions and therefore can be related, a few examples are listed below:

Various forms of energy are converted to accomplish mechanical movement in the injection molding machine. Electrical energy is 28

converted to mechanical energy, which in turn is converted to hydraulic energy to operate and control the moving components of the machine. The hydraulic energy is converted to mechanical energy to achieve the final desired result, which may be "mold clamping pressure" or "material injection". The figure above summarizes the energy conversions for an injection molding machine. Click on the thumbnail for a larger view. Pascal's Law states that a pressure acting on a confined fluid is transmitted equally and undiminished in all directions. In the figure below, a 10 pound force acting on a 1 square inch area generates a pressure of 10 pounds per square inch (psi) throughout the container acting equally on all surfaces.

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Pascal's Law

This principle is important to remember, that the pressure in any portion of an hydraulic system is equal throughout that system. This statement is valid with the omission of the force of gravity, which would have to be added, according to the fluid level. Due to the pressures that hydraulic systems operate at, this smaller amount need not be considered e.g. a 32 foot head of water approximately equals 14.5 psi. (a 10 meter head of water approximately equals 1 bar.)

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Force Transmission in Hydraulics

Area and Force As the clamp piston is moved forward during the clamp close function, the pressure developed acts upon the clamping piston which has a certain size or area.

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A basic formula in hydraulics states that pressure multiplied by area to which that pressure is applied equals force. i.e. pressure x area = force p xA= F The formula can be manipulated to calculate any one of the three variables p, A or F, if any of the other two variables are known. As follows: p xA= F F/p=A 32

F / A= p

Pressure Hydraulic pressure is generated when a flowing fluid meets resistance which is generally related to the load that is being moved.

A force is applied via the lever to produce system pressure (p = F/A

or

F

=

p

x

A).

If more force is applied, the system pressure rises until the load moves, if the load remains constant the pressure will increase no further. The load can therefore be moved if the necessary pressure is generated. The speed at which the load moves will be dependent upon the volume of fluid which is fed to the load cylinder. For example, as the mold is opening or closing, the 33

pressure generated in the system represents the resistance of the toggle lever to movement. Adding to that resistance would be the weight (i.e. mass) of the mold and toggle lever and also the friction between the toggle lever bushings and the tie bars. When the two mold halves touch and the toggle begins to straighten out, the increasing pressure represents that which is required to stretch the tiebars in the generation of a particular clamp force. Similarly when injecting material into the mold the pressure generated in the injection system represents the resistance of the injection ram to movement. Adding to that resistance would be the mass of the injection ram and screw, the friction between all moving components and the resistance of the plastic melt as it is forced quickly into the mold cavity. Pressure Control In order to safeguard the system, pressure relief valves are installed. The valves serve to limit the amount of pressure that can develop in the hydraulic system since the various hydraulic components are expensive and they are subject to pressure limitations before failure occurs. One characteristic of fluid flow that is important to note here is that flow occurs always in the path of

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least resistance. Pressure would continue to rise in the circuit consistent with the load being moved. The pressure relief valve is always set to allow flow to travel through the relief valve well before pressure rises above safe levels and causes damage to the system and its components. In other words, the path of least resistance is employed here to safeguard the system after the other movements have taken place.

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Pressure

Override

An extremely important concept to understand about pressure relief valves is their pressure override characteristics. Pressure override is the difference between the pressure at which the relief valve just starts to crack open and the pressure at the full open position. For direct acting pressure relief valves this pressure differential can be as high as 30% and proportional pressure relief valves range from 10% - 20%.

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Pressure Intensification Another important concept to keep in mind is that of pressure intensification. This law of hydraulics is often forgotten when troubleshooting hydraulic circuits.

For example, if two pistons of different size are connected by a rod, the pressure existing on the smaller area will always be greater. This principle also applies to the cap side and the rod side of a normal double acting piston. If P1 = 1,000 psi and A1 = 10 square inches, then F1 = 10,000 pounds of force. If F1 = 10,000 pounds of force and if A2 = 5 square inches, then P2 = 2,000 psi.

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Speed in Hydraulics The speed of a hydraulic component can be calculated based on the formula below:

For example, given the conditions below the injection piston, therefore the screw, will move at 3.85 inches per second. However, this speed will not be possible if the pressure relief valve opens.

Hydrodynamics As well as understanding the concept of speed in hydraulics, it is 38

also important to have some insight into flow characteristics. For example, the drawing below shows that when oil is flowing through different diameter pipes an equal volume flows in an equal unit of time. If that is true and if the shaded quantity Q1 equals the shaded quantity Q2, then velocity V2 must be greater than velocity V1.

As the diameter of the pipe decreases, the flow rate will increase. Specifically, if the pipe diameter decreases by one half in the direction of oil flow, the cross sectional area will decrease by four times, and visa versa. Oil flow velocity through different pipe sizes can be calculated using the formula:

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The same gallons per minute will have to travel 4 times faster through the smaller pipe. Another important concept in hydrodynamics is how fluids flow based on certain critical flow speeds or as the result of meeting restrictions to flow such as bends in the pipe or system components.

One goal in the initial design of hydraulic power transmission systems is to encourage laminar flow as much as possible since an increase in turbulence will increase flow resistance and hydraulic losses as well. The diagram below illustrates the concept of turbulent flow

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Although turbulent flow is wasteful in most hydraulic applications, it is desirable to have turbulence in the oil flow as it travels through the heat exchanger for cooling purposes. If turbulence exists as the oil flows through the heat exchanger, more of the oil molecules come into contact with the heat exchanger cooling tubes and more efficient cooling is the result. Directional

Control

One of the main advantages of hydraulic based systems is that the oil flow direction is easily controlled. The drawing below shows a piston being extended, held stationary and then retracted, simply by changing the position of a directional valve. Even though the drawing is simple in nature, it still demonstrates the principle involved in directional control. In addition to simple directional control valves, we also employ proportional directional control

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valves on some machines to control the clamp opening and closing function

Working

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Syringe Hydraulic Arm

Hydraulics is used in many applications we see or use every day. The brakes in an automobile or the lift on the bucket of a tractor are two very common applications. We can experiment with simple hydraulics using plastic syringes for cylinders and small plastic tubing for the hydraulic hose. One definition I read for hydraulics was, “the movement of pressurized liquids through confined spaces”. Like working with gears, pulleys, or levers; a mechanical advantage can be realized by using different size cylinders on the

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end of the hydraulic connections. By trading distance moved with the amount of force the advantage is realized.

Moving the larger cylinder between 1 mark moved the smaller cylinder 2 marks.

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Hydraulic cylinders for lifting and tilting bucket are easily seen in this tractor my father constructed.

Testimonial to Popularity of the Project Sandy Gady 7th and 8th grade Math,Science, Design and Engineering teacher “I have done the hydraulic arms with the syringes, they are so cool it’s beyond description. I used to do them when I taught 6th grade in a small town. We had lots of retired people with saws, so they would cut up the wood we needed. I then enlisted several of them to come into the classroom and we taught kids how to use drills, screwdrivers and hammers to assemble the finished project. In the end, they had a hydraulic arm that moved in three directions. I’ve had contact with some of these students since, and many of them still have their hydraulic arms 20 years later. They still beam with pride when they talk about them. Cost of materials, about $5.00. Educational value, priceless”.

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Gathering the Parts Together

The tubing I used was the type used for aquarium airlines. Fuel line

for

model

glow

engines

would

work

but is more

expensive. There could be medical sources also.

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Syringes can be purchased where farm supplies are sold. I used the 12cc size as the pump and the 6cc size as the actuator cylinder.

I used one flat washer on the pivot point of the gripper part of the arm, not sure it is needed

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8/32 Machine screws were used to hold clamps on and for pivot points.

I used these copper plated ½” tube straps to fasten the actuators (syringes) to the wood.

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The upright part of the arm is a furring strip. This inexpensive wood is used when installing drywall.

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For the horizontal part of the arm, I used 5/8” square stock that comes in 36” lengths. Normally found next to the round dowels.

Two sizes of craft sticks are used in this project, one on top is the size of a tongue depressor and the bottom one is the size of a Popsicle stick.

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The base can be made from scrap boards also, best to pick out the better boards as some are warped. This board was labeled as 1” x 6” but true dimension was smaller.

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Making Up the Parts Normally I am making up kits for a class, so I will get all the parts fabricated first. That way few tools are needed to complete the project and it is safer for them not having to use saws or drills.

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Straighten the bend in the straps with a pliers.

The straps should look like this after 90 degree bends have been removed.

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Cut the furring strip to 8 ½” in length

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Cut the 5/8” square stock to 11” in length.

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Cut 1” from both ends of craft stick, the ends are used for building gripper parts and the center is used to connect syringe to pushrods.

Gripper pads are ¾” long cut from smaller craft stick material.

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A “right triangle” is cut from scrap wood that is at least ¼” thick. The long side is 1 ½” long and shorter side is ¾” long.

These are the wood parts needed before drilling the holes. The small triangle piece can be made from a scrap piece of ¼” wood.

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Construction

Drill holes for bearing pieces on the boom, I drilled both pieces together

Bearing pieces after drilling the two holes at once.

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Drilling the two holes on boom to mount the actuator (syringe). There is also a hole at the far right that is the pivot for this portion of the arm

Drilling hole at end for gripper assembly, this is drilled 90 degrees from the other holes.

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Hole for strap on the main support is close to the edge so be especially careful when drilling.

Larger holes on the ends are pivot point and smaller holes in the middle is where pushrod attaches.

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The pushrods for the gripper section can be made from paper clips that are straightened. Bend the curves out with your fingers and further straighten with a pliers.

One end of the pushrods will attach to the wood by making two 90 degree bends in the wire. Pictures shows the first bend, for this bending a needle nose pliers is needed.

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Make another bend a short distance from the previous bend in the opposite direction.

This is what the pushrod should look like, two will be needed. I was going to use music wire which is stiffer but it is also harder to bend.

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Plastic tubing to be used at a bearing for the wire pushrod can be cut from the tubing in a pen.

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Putting it Together - It is at this point that the students would start assembling the parts that had been fabricated.

Mounting the actuator to lift the arm. I find it easier to test the fit of all pieces before gluing the wood wedge.

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Actuator to lift the boom is securely strapped at the correct angle. I should have mounted this slightly higher.

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The straps need to be pulled over the syringe very tight so that it does not shift when pressure is applied.

Using two nuts tightened against each other should allow the nut to the inside to have a small gap from the wood and not come off. The two thin pieces of wood need to turn freely on the bolt. 66

To position the two craft sticks that will be part of arm bearing make sure that the arm will come down to where the actuator is almost pushed in.

Also make sure the arm can swing upwards without the binding at the hinge location.

67

There needs to be enough room inside the joint so the arm can move up and down

Syringe actuator in extended position should lift arm above horizontal position.

68

I built the two halves of the gripper using Ducocement so I used clothespin to hold parts together while glue dries when possible.

The pads are added next, no real way to clamp it but the Duco is sticky at the start and tends to hold pieces together.

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Note washer on the bolt at the pivot point should help the wood parts move easier. Do not tighten the head of the bolt down too tight, the parts must be free to move easily.

Plastic tubes attach one end of the pushrods to the syringe actuator.

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The wood piece with plastic tubes is glued to the end of the syringe actuator. Hot glue holds fairly well or you can use small wood screws to attach the wood to the end of the syringe.

Bring the gripper plates together, pull the actuator out to almost full extension, and insert the pushrod ends with the 90 degree bends into the small holes.

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Bend the wire 90 degrees downward into the plastic tubes.

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Carefully bend the wire back so the pushrods do not fall out. I was going to use music wire which would have been stiffer but the paper clip material is much easier to bend.

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Mark the center of the base and glue the upright portion of the arm to the base. I used hot glue for this and held it for a couple of minutes to make sure it was hard.

Cut off two lengths of the plastic tubing at least 12 inches long for each.

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Push the tubing on the syringe where the needle would normally attach. If the fit of the tubing is too loose, stretch the ends of the tubing while heating it will shrink the tubing at the ends. Too much heat is not good, just enough to shrink the tube slightly for a tighter fit on the syringe.

Actuator for gripper attached to boom. It must be placed at the correct spot so amount of travel is correct for the gripper.

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Attach one end of the other piece of tubing to the lower syringe.

Attach the larger 12 cc syringes to the opposite ends of the plastic tubing.

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Filling the Cylinders Part of the key to success with this project is getting as many of the

air

bubbles

out

of

the

lines

as

possible.

In

automobile brake lines this is known as “bleeding the brakes”. I am not

sure

I

have

the

best

procedure

for

this

but I will give some suggestions. Bring the actuator cylinder plunger down, pull the line off and pull in

some

water

into

the

line

with

the

larger

cylinder.

Put the line back on the small cylinder and push water into it. When it is full, turn the actuator cylinder upside down so the water does not drain out.

The next steps are to pull water into the large syringe and line, repeating this until the air bubbles are gone. No doubt it will take several strokes of pulling water through and pushing some of it out until the bubbles are gone.

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Small actuator cylinder with water to top of syringe end when positioned upside down.

I found that even after shrinking the ends of the tubing, some people were pulling the syringes off so I am taping the connections after the line is full of water.

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If glue is not holding the small piece of wood to the end the syringe, two small wood screws into the plastic should hold.

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We have designed the hydraulic arm that has four movements .

All the movements are hydraulic operated and uses 25 ML cylinders .

Except for the one grab & release movement which uses 10ml cylinder as it is a very small and delicate movement.

This hydraulic arm can grab a object and place it wherever you want inside it’s circle of reach.

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Raw Materials Used 1. Syringe’s (as hydraulic cylinders) 2. Water (as hydraulic fluid) 3. Plywood stick’s (for basic structure) 4. 2mm pipes (for hydraulic transmission) 5. Nuts & bolts 6. Insulation tapes 7. File’s ( for cutting the plywood) 8. Glue/ m-seal/fevi quick 9. Red & black tape 10.

Hard wood ( for lower frame)

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Fabrication Process 11.

In the fabrication of hydraulic arm the

following things were kept in mind: 12.

Each part should withstand the load this

hydraulic arm is meant to carry. 13.

The hydraulic cylinders must not leak

while operating the arm. 14.

All the moving parts move unrestricted

and frictional hindrance should be kept low. 15. 16.

Each part should be removable as to

make it repair friendly. 17.

The grab & release function should be

powerful enough to hold the desired load . 18.

The lower frame should not vibrate or

get misbalanced while arm is under operation.

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19.

The control cylinders should be so fixed

that they can be easily used without distorting the over all structure of the arm. 20.

The most important and unique

function that is the “rotational

movement”

should be performed with hydraulics not with motor’s or mechanically. 21.

All the cylinders should be fixed at an

optimal distance as to take full mechanical advantage.

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Basic lower frame under fabrication

STAGE 2 Fabrication of vertical support The vertical support is the main foundation for the arm as it holds the whole arm an raises it above the ground, and most importantly provides the rotational motion to the arm. The vertical support is mounted on the lower frame with a roller disc that is free to move rotationally only. It is tightened with the help of nuts and bolts on the roller disc and in between it a special place is made to mount the cylinder which will be used to raise the arm vertically.

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Stage 3 assembling of horizontal arm Two parallel plywood strips are cut equally and hand drilled

to fit into the place provided on the vertical

support. Another set of holes are drilled on them some distance apart to fix the moving piston from the cylinder to transmit motion. At the end of this horizontal arm another set of holes are drilled so that the final grab & release arm can be fixed. The dimensions of these arms must match each other so that the movement don’t get disturbed . Rotatory movement is the specialty of this hydraulic arm because in this arm we have used hydraulic power to operate the rotatory section. Even in the high tech hydraulic arms motors are used to deliver power to the rotatory section.

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Like

in

case

of

cranes

&

other

construction

and

manufacturing machines only motors are used to deliver motion to the rotatory part. But we have designed this arm to be operated fully with hydraulic power so we have used a roller disc on the base of vertical support and that roller disc is connected to the cylinder mounted on the frame base, when the pressure is applied the piston moves outwards and pushes the roller disc which then moves in a circular direction giving arm its rotatory motion.

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Advantages of using hydraulic power to operate the rotatory system:

1. Less energy & power consumption. 2. Easy to craft. Minimum mechanical constraints & moving parts .

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DISADVANTAGES : 1. Limited movements possible. 2. Can’t

work

efficiently

when

high

power

required. 3. Requires regular rectification. 4. Control unit & final touch 5. After all the fabrication is done the cylinders are filled with the working fluid in this case it’s water. 6. 7.

All the controlling cylinders are fixed on the basic frame . 8. After this the last decoration work is done to make the arm

presentable.

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Pascal's law Continuum mechanics

.

The effects of Pascal's law, as Pascal discovered in his 1646 barrel experiment Pascal's

law or

the principle

of

transmission

of

fluid-

pressure is a principle in fluid mechanics that states that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure

89

ratio (initial difference) remains the same. [1] The law was established by French mathematician Blaise Pascal.[2] Definition Pascal's principle is defined A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid This principle is stated mathematically as:

where is the hydrostatic pressure (given in pascals in the SI system), or the difference in pressure at two points within a fluid column, due to the weight of the fluid; ρ is the fluid density (in kilograms per cubic meter in the SI system); g is acceleration due to gravity (normally using the sea level acceleration due to Earth's gravity inmetres per second squared); is the height of fluid above the point of measurement, or the difference in elevation between the two points within the fluid column (in metres in SI). 90

The intuitive explanation of this formula is that the change in pressure between two elevations is due to the weight of the fluid between the elevations. Note that the variation with height does not depend on any additional pressures. Therefore Pascal's law can be interpreted as saying that any change in pressure applied at any given point of the fluid is transmitted undiminished throughout the fluid. [edit]Explanation If a U-tube is filled with water and pistons are placed at each end, pressure exerted against the left piston will be transmitted throughout the liquid and against the bottom of the right piston. (The pistons are simply "plugs" that can slide freely but snugly inside the tube.) The pressure that the left piston exerts against the water will be exactly equal to the pressure the water exerts against the right piston. Suppose the tube on the right side is made wider and a piston of a larger area is used; for example, the piston on the right has 50 times the area of the piston on the left. If a 10 kg load is 91

placed on the left piston, an additional pressure (nearly 1 N/cm2) due to the weight of the load is transmitted throughout the liquid and up against the larger

piston.

The

between force and pressure is

difference important:

the

additional pressure is exerted against every square centimeter of the larger piston. Since there is 50 times the area, 50 times as much force is exerted on the larger piston. Thus, the larger piston will support a 500 kg load - fifty times the load on the smaller piston. Forces can be multiplied using such a device. One newton input produces 50 newtons output. By further increasing the area of the larger piston (or reducing the area of the smaller piston), forces can be multiplied, in principle, by any amount. Pascal's principle underlies the operation of the hydraulic press. The hydraulic press does not violate energy conservation, because a decrease in distance moved compensates for the increase in force. When the small piston is moved downward 10 92

centimeters, the large piston will be raised only one-fiftieth of this, or 0.2 centimeters. The input force multiplied by the distance moved by the smaller piston is equal to the output force multiplied by the distance moved by the larger piston; this is one more example of a simple machine operating on the same principle as a mechanical lever. Pascal's principle applies to all fluids, whether gases or liquids. A typical application of Pascal's principle for gases and liquids is the automobile lift seen in many service stations (the hydraulic jack). Increased air pressure produced by an air compressor is transmitted through the air to the surface of oil in an underground reservoir. The oil, in turn, transmits the pressure to a piston, which lifts the automobile. The relatively low pressure that exerts the lifting force against the piston is about the same as the air pressure in automobile tires. Hydraulics is employed by modern devices ranging from very small to enormous. For example, there

93

are hydraulic pistons in almost all construction machines where heavy loads are involved. ]Applications  The siphon  The

underlying

principle

of

the hydraulic

jack and hydraulic press  Force amplification in the braking system of most motor vehicles.  Used in artesian wells, water towers, and dams.  Scuba divers must understand this principle. At a depth of 10 meters under water, pressure is twice the atmospheric pressure at sea level, and increases by about 100 kPa for each increase of 10 m depth.

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Hydraulic systems use a incompressible fluid, such as oil or water, to transmit forces from one location to another within the fluid. Most aircraft use hydraulics in the braking systems and landing gear. Pneumatic systems use compressible fluid, such as air, in their operation. Some aircraft utilize pneumatic systems for their brakes, landing gear and movement of flaps. Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container. A container, as shown below, contains a fluid. There is an increase in pressure as the length of the column of liquid increases, due to the increased mass of the fluid above. For example, in the figure below, P3 would be the highest value of the three pressure readings, because it has the highest level of fluid above it.

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If the above container had an increase in overall pressure, that same added pressure would affect each of the gauges (and the liquid throughout) the same. For example P1, P2, P3 were originally 1, 3, 5 units of pressure, and 5 units of pressure were added to the system, the new readings would be 6, 8, and 10. Applied to a more complex system below, such as a hydraulic car lift, Pascal's law allows forces to be multiplied. The cylinder on the left shows a cross-section area of 1 square inch, while the cylinder on the right shows a cross-section area of 10 square inches. The cylinder on the left has a weight (force) on 1 pound acting downward on the piston, which lowers the fluid 10 inches. As a result of this force, the piston on the right lifts a 10 pound weight a distance of 1 inch.

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The 1 pound load on the 1 square inch area causes an increase in pressure on the fluid in the system. This pressure is distributed equally throughout and acts on every square inch of the 10 square inch area of the large piston. As a result, the larger piston lifts up a 10 pound weight. The larger the cross-section area of the second piston, the larger the mechanical advantage, and the more weight it lifts.

The formulas that relate to this are shown below: P1 = P2 (since the pressures are equal throughout). Since pressure equals force per unit area, then it follows that F1/A1 = F2/A2

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It can be shown by substitution that the values shown above are correct, 1 pound / 1 square inches = 10 pounds / 10 square inches Because the volume of fluid pushed down on the left side equals the volume of fluid that is lifted up on the right side, the following formula is also true. V1 = V2 by substitution, A1 D1 = A2 D2  A = cross sectional area  D = the distance moved or A1/A2= D2/D1 This system can be thought of as a simple machine (lever), since force is multiplied.The mechanical advantage can be found by rearranging terms in the above equation to

98

99

Mechanical Advantage(IMA) = D1/D2 = A2/A1 For the sample problem above, the IMA would be 10:1 (10 inches/ 1 inch or 10 square inches / 1 square inch).

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REFERENCES

R . K. Bansal (Fluid mechanics & fluid machines, laxmi publications) D . S. Kumar (Fluid mechanics & fluid power engineering, s.kkataria& sons) V .P . Gupta (Hydraulic machines, cbs publishers) Research papers by the students of MIT (Massachusetts institute of technology) (available on internet)

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