The thrust of the cylinder is 50 kg, how big of an oil pump is needed? How big is the cylinder? The

1. Design Input:

Vehicle model. wheelbase: 4250 + 1350mm;

load weight: 65t; Envelope mass: 5t; Curb weight:; Volume: 22m3

Lifting type: front top four-stage cylinder lifting form.

Second, the vehicle layout:

See Figure 1 layout type: the upper support of the cylinder is fixed on the front plate (see Figure 1).

After drawing 2, the inside length of the trunk is 6000mm, and the ground clearance of the rear plate after lifting 48 ° is 444mm.

Figure 2 III. Description of scheme calculation.

1. Analyze whether there is the possibility of backward rollover when the vehicle climbs (see Figure 3).

By knowing the maximum climbing degree of 35 with full load, the slope is calculated to be equal to. After drawing, it is known that the center of gravity of the goods on the slope with a slope is in front of the rear wheel and the ground support point, so the vehicle will not roll back when climbing the slope with full load.

Fig. 32, the selection of Liuqi front lifting four-stage cylinder 4TG-E185 4650, the cylinder parameters are: rated pressure is.

16MPa, the working volume is, the total stroke is 4650mm, the rod diameter of the cylinder at all levels is 185 mm, 160 mm, 135 mm, 110 mm, and the thrust of the cylinder is 43 t, 32t under the rated pressure of 16MPa

The force of the cylinder is shown in Figure 4, F is the thrust of the cylinder, and G is the total mass of the box after the weight plus the goods.

According to the moment balance, it can be concluded that the following formula must be met if the cargo is to be lifted smoothly:

f×b＞g×a

Figure 4, Table 1 (load 65t).

Theoretical thrust (t) f(t) b(mm) g(t) a(mm).

Table 2 (load 80t).

Theoretical thrust (t) f(t) b(mm) g(t) a(mm).

Therefore: F4 B4 g A4 is satisfied

3.System pressure calculations.

According to the required thrust of the cylinder and the cross-sectional area of the piston rod, the internal pressure of the cylinder can be obtained

In the case of load 65t:

In the case of load 80t:

4.CB-J2100 oil pump is selected, the oil pump parameters are: rated speed of 2300 rpm, rated pressure of 20MPa, driving power, hydraulic system volume efficiency is usually taken, check lifting time.

5.Calculation of the propeller shaft.

According to , the required torque for the rated pressure of the oil pump (20 MPa) can be determined:

n?The torque required for the M oil pump at 20MPa rated operation is 277N?m；

The power take-off is QH50, and the output rated torque is 500N?m；

Therefore, the torque of the transmission shaft must be greater than 278N?m, and must also be greater than 500N?m, so that the propeller shaft will not be destroyed.

Hydraulics for 50 kg of thrust? It's too extravagant, isn't it!! It is completely solved by driving the screw with a motor.

Summary. Hello, happy to answer your <>

What is a cylinder column?

The cylinder diameter is d (mm), the piston rod diameter is d (mm), and the working pressure is p (mp).

Then: rodless cavity thrust is f=

The rod cavity tension is f=

The unit of f above is n

According to the cylinder diameter 100, the rod diameter 40 is calculated, and the tensile force is f=

Bore 100 How many kg of cylinder thrust

Hello dear! I've seen your question and I'm trying to sort out the answer, and I'll get back to you in five minutes, so please wait a minute

Hello, happy to answer for you what is the cylinder column [Happy or Changsun]? Let the cylinder diameter be d (mm), the rod diameter of the chain plug of the undershirt is d (mm), and the working pressure is p (mp), then: the thrust of the rodless cavity is f = the tensile force of the rod cavity is f = the unit of the above f is n according to the cylinder diameter 100, the rod diameter is 40, and the tensile force is f =

Pneumatic kilogram force square centimeter base front meter (kgf cm2)) cylinder diameter 50mm (5cm) cylinder cross-sectional area = square centimeter) Therefore, the theoretical output under the force = kilogram force manuscript hall). Fight respectfully.

I hope the above is helpful to you If you are satisfied with me, please give a thumbs up Welcome to the consultation <> next time

Summary. Dear, glad to answer for you. The thrust of the cylinder is 300 kg, then the power required for the oil pump is kilowatts.

To calculate the required power of the oil pump, you must first know the flow rate of the oil pump. The flow rate of the oil pump depends on the thrust of the cylinder, that is, the amount of oil that pushes the cylinder per second. Assuming that the thrust of the cylinder is 300 kg and the amount of oil pushed per second is cubic meters, then the power required for the oil pump is:

Cubic meters per second 1000 kg cubic meters per second 300 kg 1000 kg = kilowatts.

The thrust of the cylinder is 300 kg, how much power is the oil pump?

Dear, glad to answer for you. The thrust of the cylinder is 300 kg, so the power required for the oil pump is kilowatts. To calculate the required power of the oil pump, you must first know the flow rate of the oil pump.

The flow rate of the oil pump depends on the thrust of the cylinder, that is, the amount of oil that pushes the cylinder per second. Assuming that the thrust of the oil cylinder is 300 kilograms, and the amount of oil pushed per second is cubic meters, the power required for the failure of the oil pump is: cubic meters per second 1000 kilograms cubic meters meters seconds 300 kilograms 1000 kilograms = kilowatts.

Knowledge expansion: The oil cylinder is a hydraulic cylinder, and the output force is proportional to the effective area of the piston and the pressure difference between its two sides. It can convert hydraulic energy into mechanical energy, and is suitable for places where heavy objects need to be supported for a long time, such as underwater, single acting, load retraction, nut bar self-locking, etc.

The hydraulic cylinder consists of a cylinder barrel, a cylinder head, a piston and a piston rod, a sealing device, a buffer device and an exhaust device. Among them, the buffer device and the discharge tank luck device depend on the specific application of the letter book.

The thrust of the cylinder is 300 kg, how much power is the oil pump?

Hello, dear, finch model! To solve this problem, we can first calculate the thrust of the cylinder according to the formula, and then calculate the required pump power according to the power of the pump. First of all, according to Newton's second law, the thrust f (unit:

Newton) is related to the piston area a (unit: square meters) and the piston movement speed v (unit: meters and seconds).

f=pa, where p is the pressure (unit: Newton square meters). Suppose the thrust of the cylinder is 300 kg, that is, 3000 Newtons, the piston area is 1 square meter, and the piston moving speed is 1 meter second.

Substituting these data into the formula, we get p=3000 1=3000 Newton square meters, that is, the pressure is 3000 Newtons square meters. Then, according to the power p (unit: watts) and the flow rate q (unit:

cubic meters and seconds) and pressure p (unit: Pascal) is: p=qp, where q is the flow rate of the pump.

Suppose the flow rate of the pump is 1 cubic meter per second and the pressure is 3000 Newton square meters. Substituting these data into the formula, we get p=1x3000=3000 watts, that is, the pump needs to reach 3000 watts. Therefore, in order to meet the thrust of the cylinder of 300 kg, it is necessary to use an oil pump with a power of at least 3000 watts.

In addition, in order to ensure that the oil pump can work stably and meet the demand, the flow rate and pressure regulation capacity of the oil pump also need to be considered. For example, the flow rate and pressure can be adjusted by adjusting the speed of the oil pump or by using a variable pump.

The most important parameters of a hydraulic pump are displacement and pressure, both of which you are not sure of, and it is difficult to choose. After determining the thrust, the next step is to determine the inner diameter of the cylinder, which is limited by the mechanical structure and cannot be too thick. But if the stroke is 1 meter, if it is too thin, it is necessary to check the stability of the pressure rod.

It is recommended to set the diameter of the piston at 60-80 mm and the piston rod at 40-50 mm. With the bore (piston diameter), the operating pressure of the system can be determined. Once the bore is determined, you also need to determine the speed you want to be in the cylinder, and the flow rate of the pump can be determined based on the speed, stroke and bore.

With the flow rate and pressure, the oil pump is ready.

Summary. Hello dear, happy to answer your <>

In general, assuming that the pressure of the cylinder is p and the piston area is a, the thrust f of the cylinder can be calculated by the following formula: f = p a;Wherein, the unit is N (Newton), P is the pressure of the cylinder, the unit is Pa (Pascal), and A is the piston area of the cylinder, and the unit is square meters. Since the diameter is 500, the radius is 250, and the following is obtained

a = r^2 = 250mm)^2 ≈ 196,350 mm^2；Substituting the piston area into the formula, if the pressure of the cylinder is assumed to be 10MPa, the thrust f is approximately: F = P A = 10MPa 196,350 mm 2 1, kn;Thus, the thrust of a cylinder with a diameter of 500 is 1,thousand newtons (kn).

What is the thrust of a cylinder with a diameter of 500.

Hello dear, happy to answer your <>

In general, assuming that the pressure of the cylinder is p and the area of the piston is a, the thrust f of the cylinder can be calculated by the following formula: f = p a;Among them, the unit is N (Newton), P is the pressure of the cylinder, the unit of attitude crack is Pa (Pascal), and A is the piston area of the cylinder, and the unit is square meters. Since the diameter is 500, the radius is 250, and the following is obtained

a = r^2 = 250mm)^2 ≈ 196,350 mm^2；Substituting the piston area into the formula, if the pressure of the cylinder is assumed to be 10MPa, the thrust f is approximately: F = P A = 10MPa 196,350 mm 2 1, kn;Thus, the thrust of a cylinder with a diameter of 500 is 1,kiloN (kn) per mu of land.

Dear, the following quarrel is related to the expansion, I hope it will help you <>

The thrust of the cylinder should not exceed its rated load, otherwise it may cause damage to the cylinder or other safety hazards. Therefore, it is necessary to be cautious when using the cylinder and follow the guidelines provided by the manufacturer.

Hello dear, it's a pleasure to answer <> for you

Generally speaking, assuming that the pressure start-up force of the cylinder is p and the piston area is a, then the thrust f of the laughing dust cylinder can be calculated by the following formula: f = p a;Wherein, the unit is N (Newton), P is the pressure of the cylinder, the unit is Pa (Pascal), and A is the piston area of the cylinder, and the unit is square meters. Since the diameter is 500, the radius is 250, and the following is obtained

a = r^2 = 250mm)^2 ≈ 196,350 mm^2；Substituting the piston area into the formula, if the pressure of the cylinder is assumed to be 10MPa, the thrust f is approximately: F = P A = 10MPa 196,350 mm 2 1, kn;Therefore, the thrust of a cylinder with a diameter of 500 is kilonewtons (kn).

Summary. Hello, I'm glad to answer for you, the five-ton cylinder is generally thrust. It doesn't matter the angle, the thrust of the cylinder is five tons.

If your load is tilted, you need to know that the product of the maximum length from the fulcrum of the cylinder to the load and the load is the final load. The tension of the cylinder needs to know the bore, rod diameter, and system pressure of the cylinder. to be able to calculate it.

Hello, I'm glad to give you a bright answer, the five-ton cylinder is generally thrust. It has nothing to do with the angle of the virtual key, the thrust of the cylinder is five tons. If your load is tilted, you need to know that the product of the maximum length from the fulcrum of the cylinder to the load difference and the load is the final load.

The tension of the cylinder needs to know the bore, rod diameter, and system pressure of the cylinder. to be able to calculate it.

The inner diameter of the cylinder is 50m'm。The column meridian is 28

It can be concluded that the thrust force of a 50-bore cylinder is 98 kgf and the pulling force is 82 kgf at a pressure of 5 kg, which is the kilogram force.

The system pressure is 5MPa

Yes. It's the cylinder.

Not a cylinder. Speak.

Can you wait.

Formula for calculating the time required for hydraulic cylinder stroke When the piston rod is extended, the time is (15 cylinder bore squared cylinder stroke) Flow rate When the piston rod is retracted, the time is [15 cylinder clear diameter squared - rod diameter squared answer fluid) oil book commotion cylinder stroke] flow rate cylinder bore unit m rod diameter unit m stroke unit m flow unit l min

Is it a rally? Yes.

f 5 tff 5000 kgf 5*10 4 N, p 15 MPa section Pa analysis: from p f s, the cross-sectional area of the oil cylinder is s f p 5*10 4 ) 10 (-3) square digging bend m If the diameter of the cylinder is d, then d 2 * root number ( s 2 * root number [ 10 (-3) ] m cm.