# Instantaneous impact force

1. Aug 23, 2011

### confused slug

I am trying to determine the maximum (instantaneous) impact force experienced by an object when it is struck by a hydraulic ram. I know both the pressure in the cylinder side of the ram and the level of deceleration (-g) experienced by the end of the rod on the ram when it stikes the object. Ingoring friction etc is it as simple as caluclating the effective mass (pressure x area of piston) and multiplying by the g?

2. Aug 23, 2011

### Unrest

Not really.

F=ma still applies to the stopping ram. m would be the mass of the piston/rod/etc. F would be the sum of all the forces on the ram:
- Impact force (what you want)
- Applied force (pressure x area), opposite direction.
- Friction (~0).

So something like:

(impact force) - (pressure x area) = (mass of piston) * (measured acceleration)

Convert g/etc. to appropriate units of course.

3. Aug 26, 2011

### confused slug

Hello Unrest

Pressure = 110 Bar (11 N/mm2)
Surface Area Pistion = 12250 mm2
Mass of Piston = 7 kg
G = 0.85

Would i be correct in calculating the impact force to be
(11 x 12250) + (7 x 0.85) = 134756 N

as 13.7 tonnes seems a bit high

4. Aug 26, 2011

### afreiden

.85 g's of peak acceleration during the impact?
What's the time duration of the impact if you don't mind my asking?

Also, the weight of your hydraulic ram (and all relevant connecting parts) is only 7kg, yet its "surface area" (i.e. cross-sectional area over which the piston's pressure(s) acts on the ram) is around 20 sq. in. ?

Your equation looks correct, assuming the piston is pushing during the collision. If it's an impulsive collision, then you need to find the impulse imparted to the thing you're hitting, not just the peak force..

5. Aug 26, 2011

### confused slug

The duration of the impact is approx 0.2 seconds.
The mass of the piston and the rod is 7 kg (only part moving)
The nominal bore diameter is 125mm with 300mm stroke

It is impacting a fixed object towards the very end of its stroke. The oil flow is automatically reversed when the oil flow rate to the cylinder side falls below a set rate. The rod side is then supplied with high pressure (210bar @ 70ltr / min) to retract the rod.

6. Aug 26, 2011

### afreiden

I guess I would be most interested to know the ram's velocity at impact and whether the net hydraulic pressure on the ram is really a constant 110 bar throughout he entire collision. When does the 210 bar deceleration pressure come into effect? After the collision is finished?

Regardless, it sounds like you do have all of the data that you need.

As stated by "Unrest," and unless there is other important information that you haven't mentioned, you simply solve for F_unknown(t) from the equation:

F_hydraulics(t) +/- F_unknown(t)=mass_ram*a(t)

Obviously you need the acceleration and hydraulic pressure(s) as a function of time, and time synchronized, which it sounds like you have.

I don't understand what you mean by "fixed object" in your last response.

7. Aug 27, 2011

### Unrest

I don't see any conversion factor for eliminating "g". Make sure all your units are consistent. Carry them through the calculations and make sure you end up with a unit of force (might not be newtons).

8. Aug 27, 2011

### confused slug

Would the true mass of the piston at the moment of impact be its mass due to its material (7 kg) plus the effective mass due to the pressure acting on the end of the piston due to the oil pressure (11 N/mm2 x 12250mm2) / 9.81 = 13736 kg giving a total mass of 13743 kg.

Pressure = 110 Bar (11 N/mm2)
Surface Area Pistion = 12250 mm2
Mass of Piston = 7 kg
G = 0.85

(((11 x 12250)/9.81) + 7) x 0.85 = 11682 N

rather than

(11 x 12250) + (7 x 0.85) = 134756 N

9. Aug 27, 2011

### Unrest

Don't use that effective mass, it makes a simple thing needlessly complicated. Mass is just mass. Only take shortcuts if you understand what they're for.

G isn't dimensionless. 1g = 9.81m/s2 or 9.81 N/kg

To check if the result is too high, calculate the static force that cylinder applies (pressure x area). The impact force should be greater than the static force.

10. Aug 27, 2011

### Unrest

That's quite significant. Are you sure it's not reversing at the time you took that maximum acceleration reading? As Alfrieden said, make sure you use the actual pressure at the same time as the acceleration.

However it looks like the momentum is negligible compared to the static pressure, so you can probably ignore the acceleration entirely.