SOLICITUD ENTREGA EN COMODATO INMUEBLE A FUNDACIÓN INTEGRA

4.

for lever angles less than 45o from the horizontal plane, single arm toggle force is less than the cylinder force in extension; for lever angles greater than 45o from the horizontal plane toggle force is greater than the cylinder rod force in extension. as the cylinder nears the end of its stroke as it extends, the toggle force is many times the magnitude of the rod force.

in figure 3-5 the double levered toggle method, with the bottom arm anchored at a pivot point is shown. the cylinder must be free to pivot as the toggle moves. in order to input full thrust into the toggle when the toggle reaches the verticle over center position, the cylinder must be mounted so that its axis is perpendicular to the toggle when the toggle is at the over center point. if the cylinder is not perpendicular when the toggle is at the over center point, it will be acting at an angle to the load and its thrust into the toggle will be reduced (see outcome 3.1.4).

applying what we already know, we must understand that when using a double levered toggle the system must be kept in balance. a double lever toggle system will require twice the force to lift a load as a single lever system. thus the following equation can be used:

fig. 3-4: force exerted by a toggle.

Review 3.1.3.1: if a cylinder rod exerts 5,000 lb

of force extending, how much force would a single lever toggle exert when the 12 inch lever arm (h) is 1 inch (a) from the vertical axis (o)?

a. 8,000 lbs b. 45,312 lbs c. 59,790 lbs d. 60,210 lbs e. 62,312 lbs 09/07/07

3 -  • Hydraulic Specialist • Study Guide HS Manual # 401 - 07/10/0

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eq. 3-3a

toggle force = .5 x cylinder force x tan 0 or tf = .5 x cf x tan 0

a derivative of this equation is: cylinder force = 2 x toggle force x tan 0

it is also understood that the above equations listed under 3-3a can also be used for double lever Scissor applications as well

in the situations above, we are looking at a discreet position. Because it is a static condition, there is no difference between the double lever anchored and double lever scissors type system for force calculations.

fig. 3-5: a toggle mechanism with two equal length lever arms.

Review 3.1.3.2: a toggle mechanism with two 15

inch lever arms, such as the mechanism shown in figure 3-5, is closed by a hydraulic cylinder exerting 7,200 lbs of force. What would be the toggle force when the position of the cylinder rod eye is 1/2 inch from the vertical?

a. 107,880 lbs b. 107,935 lbs c. 215,750 lbs d. 215,870 lbs e. 216,000 lbs 09/07/07

Hydraulic Specialist • Study Guide • 3 -  HS Manual # 40 - 07/0/06

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Outcome 3.1.4: Compute the hydraulic pressure to support jib-boom

loads.

the jib boom is an example of a third class lever. that is, one end of the lever rotates about a stationary pivot, the cylinder acts against the lever somewhere between the two ends, and the free end of the lever is used to move the load.

Solving jib boom crane problems (such as the example illustrated in figure 3-6) for the force acting against the cylinder rod requires solving for the vertical load at the fulcrum. then, solve the right triangle for the force that acts against the cylinder on a line through the center of the rod.

for example, if the load at the end of the jib boom cylinder rod is 1500 lbs, the vertical load at the end of the cylinder rod can be determined from the balance of forces on the lever.

(eq. 2-4)

loadlbs x load distanceft = effortlbs x effort distanceft l x ld = e x ed 1,500 lbs x 6 feet = ef x 3 feet

ef = 3,000 lbs

the force acting against the end of the cylinder rod can be determined by solving the right triangle formed by the cylinder, the beam, and the wall using trigonometric functions. However, by inspection it can be seen that for a right triangle where the other two angles are 45°, the relative side lengths are 1, 1, and 2 which equals 1.414. therefore the force on the cylinder rod is 1.414 x 3,000 lb = 4,242 lb.

3 - 0 • Hydraulic Specialist • Study Guide HS Manual # 40 - 07/0/06

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in figure 3-6 the beam is horizontal, the support is vertical, and the clevis mounted cylinder is mounted at one of the standard angles: 45°, 30° or 60°. this is usually not the case. as soon as the cylinder rod moves the beam, the angles change. for example, in figure 3-7 the beam is shown 45° from the horizontal while the cylinder is mounted to a horizontal ground support at a 30° angle to the beam. moreover, as the cylinder rod extends and retracts the angles will constantly change. What will not change is the direction of the force to move the beam which always acts perpendicular (at 90°) to the beam.

the dotted lines shown on figure 3-7 identify right triangles that can be solved to calculate the magnitude of the forces that act perpendicular to the beam. to solve problems of this type:

1. use the vertical angle between the load and the boom to compute the load that acts perpendicular to the end of the beam.

2. calculate the mechanical advantage of the load at the end of the beam to the load that acts perpendicular to the beam at the cylinder rod end pin.

3. calculate the load that acts perpendicular to the beam at the rod end pin.

4. calculate the load supported by the cylinder and minimum pressure required to hold the load at the given angle.

Review 3.1.4.1: in the figure shown, the angle

between the cylinder rod and boom is 30° and ld = 2 x ed, what minimum theoretical pressure would there be in the cap end of a 3 inch bore cylinder?

a. 350 psi b. 500 psi c. 850 psi d. 1700 psi e. 1850 psi

Hydraulic Specialist • Study Guide • 3 -  HS Manual # 40 - 07/0/06

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fig. 3-7: perpendicular forces acting against a Beam.

Review 3.1.4.2: the hydraulic cylinder in figure 3-

7 has a bore diameter of 5 inches. What minimum pressure in the hydraulic cylinder would be required to hold a load of 10,000 lb at the angles shown?

a. 1,426 psi b. 1,440 psi c. 2,161 psi d. 2,882 psi e. 4,321 psi

3 -  • Hydraulic Specialist • Study Guide HS Manual # 40 - 07/0/06

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Task 3.2: Determine the motor characteristics required to move a

load.

Outcome 3.2.1: Solve formulas for torque, speed, and horsepower of

hydraulic motors.

low speed, high torque motors are used for winch and wheel motor applications. When a hydraulic motor is “geared down,” the speed of the output shaft is decreased as the torque is increased.

the gear drive in figure 3-8 uses a chain and sprocket reduction system between the hydraulic motor drive shaft and the skid steer loader drive wheels. the speed ratio can be computed in several ways: 1) as the quotient of the pitch diameter or number of teeth of the driven wheel shaft to the pitch diameter or number of teeth of the hydraulic motor drive shaft; 2) as the quotient of the number of teeth on the driven gear by the number of teeth on the drive gear; or 3) as the quotient of the diameter of the driven pulley by the diameter of the drive pulley. this relationship is expressed by the equation:

(eq. 3-4)

Speed ratio = output Shaft / input Shaft Sr = oS / iS

a close-up illustration of what is meant by pitch diameter for a sprocket is shown in figure 3-9. essentially, the pitch diameter is the distance between the chain pins on opposite sides of the sprocket measured on a line through the center of the sprocket shaft. the pitch diameter is not the same as chain pitch, which is the distance between the pins measured along the chain.

for example, if the pitch diameter of the hydraulic motor sprocket is 3 inches and the pitch diameter of the wheel sprocket is 9 inches, the speed ratio would be:

Speed ratio = output Shaft / input Shaft = 9 inches / 3 inches = 3:1 reduction

this means the hydraulic motor drive shaft turns three times every time the wheel shaft turns once. if a compound drive is used, that is a double or triple reduction through belts, chains or gears, the final speed ratio is the product of the individual ratios.

Hydraulic Specialist • Study Guide • 3 - 3 HS Manual # 40 - 07/0/06

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