Calculation of Jack force
Publish: 2021-04-04 13:55:08
1. F=P*S=3X(10^7)*pi*(0.085^2)=680940.208N
2. (1)
if the atmospheric pressure is ignored: P = w / A2 = 5000 / (1000 / 10 ^ 6) = 5 * 10 ^ 6 Pascal = 5000kp
if the atmospheric pressure is not ignored and 100KP is added, the answer is 5100kp
(2)
when lifting heavy objects, the pressure of small plunger is greater than or equal to that of big plunger
the minimum pressure of small plunger is 5000kp * 100 / 10 ^ 6m ^ 2 = 500N
after 10:1 labor-saving lifting rod, Force F = 500N / 10m ^ 2 = 50N
so the minimum value of F is 50N
(3)
the simplest calculation is to use the function transformation:
force F work = the potential energy increased by the weight rising
suppose the weight rising h
then 5000n * H = 50N * 6 / 1000m = 0.3j
H = 0.06mm
PS: the value is so small, which is not consistent with the reality, and the rest of your data are very consistent with the reality,
so I estimate that your stroke s refers to the small plunger stroke
in this case:
use the total volume balance of liquid at both ends of the jack to list the equation
A1 * s = A2 * h
calculate: H = s * A1 / A2 = 0.6mm
if the atmospheric pressure is ignored: P = w / A2 = 5000 / (1000 / 10 ^ 6) = 5 * 10 ^ 6 Pascal = 5000kp
if the atmospheric pressure is not ignored and 100KP is added, the answer is 5100kp
(2)
when lifting heavy objects, the pressure of small plunger is greater than or equal to that of big plunger
the minimum pressure of small plunger is 5000kp * 100 / 10 ^ 6m ^ 2 = 500N
after 10:1 labor-saving lifting rod, Force F = 500N / 10m ^ 2 = 50N
so the minimum value of F is 50N
(3)
the simplest calculation is to use the function transformation:
force F work = the potential energy increased by the weight rising
suppose the weight rising h
then 5000n * H = 50N * 6 / 1000m = 0.3j
H = 0.06mm
PS: the value is so small, which is not consistent with the reality, and the rest of your data are very consistent with the reality,
so I estimate that your stroke s refers to the small plunger stroke
in this case:
use the total volume balance of liquid at both ends of the jack to list the equation
A1 * s = A2 * h
calculate: H = s * A1 / A2 = 0.6mm
3. Pressure times area equals force
20 * 10 ^ 6 * 25 ^ 2 * 3.14/4/10 ^ 6 = 9817n = 1T
20 * 10 ^ 6 * 25 ^ 2 * 3.14/4/10 ^ 6 = 9817n = 1T
4. Ha ha, originally won't, carefully pondered, a little thought, only for reference
suppose that the input force = F1, the output force of Jack = f
the final result is f = 2F1 * L * (d ^ 2 + H ^ 2) / (d * h)
General H & lt& lt; 50. The approximate result is f = (2L / h) * F1, which shows that it is a force doubling mechanism, that is, the distance type of saving labor and cost
I don't know whether the result is right or wrong, for reference only.
suppose that the input force = F1, the output force of Jack = f
the final result is f = 2F1 * L * (d ^ 2 + H ^ 2) / (d * h)
General H & lt& lt; 50. The approximate result is f = (2L / h) * F1, which shows that it is a force doubling mechanism, that is, the distance type of saving labor and cost
I don't know whether the result is right or wrong, for reference only.
5. At present, the general steel strand is 1860MPa secondary relaxation, and the design use is 0.75, that is, only
1860MPa secondary relaxation is used × 0.75=1395MPa The cross-sectional area of 15.2 diameter steel strand is 140m2. So the tension force P = 1395 × 140=195300N=195.3KN
selection of tensioning force: count the number of holes in your prestressed anchor cable, one hole is 195.3kn, and the number of holes multiplied by the number is the tensioning force
oil pump selection: select the oil pump that can exceed the tension. Look at the instruction manual of the oil pump and see how much the oil cylinder area of the oil pump is. Divide the tension force by the area of the cylinder to get the reading of the oil gauge.
1860MPa secondary relaxation is used × 0.75=1395MPa The cross-sectional area of 15.2 diameter steel strand is 140m2. So the tension force P = 1395 × 140=195300N=195.3KN
selection of tensioning force: count the number of holes in your prestressed anchor cable, one hole is 195.3kn, and the number of holes multiplied by the number is the tensioning force
oil pump selection: select the oil pump that can exceed the tension. Look at the instruction manual of the oil pump and see how much the oil cylinder area of the oil pump is. Divide the tension force by the area of the cylinder to get the reading of the oil gauge.
6. The relay room is installed to prevent the long-distance pipe jacking from moving. The relay room should not be pushed out. After the pipe jacking, remove the oil cylinder in the relay room and weld the expansion joint
7. Uneven force may cause rotation, or the weight may swing unsteadily.
8. Find 6155 company!! a playboy
9. The construction requirements show how many tons a steel strand can bear; For example; One can bear 20 tons; Then there are five channels in one channel; 20 times 5 is 200 when multiplied by the theoretical coefficient 2 (or 1.5); The number is how much Jack to use; Bigger jack is good for your oil pump;
10. If the pressure is the same, f = mg = 1 times the fourth power Newton of ten, but the pressure received by the two is different, the top cover should be smaller. According to the formula P = f divided by s, it can be calculated
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