Load Derivation for Gantry Travel Obstruction 1

[bangerjoe]

Hi,

bit of an odd question. I have been asked to estimate forces on a crane gantry if the wheel is obstructed. I.e. breaks or snaggs on a rock.

i have the drive motor power (7.5 kW with stall torque of 142 N.m) but unsure if it is necessary to use.

surely if the gantry becomes obstructed, the driving wheel will keep driving with a force of V (horizontal reaction) x coeficient of friction for steel on steel i.e.0.25?

do i really need to consider the motor power.

thanks.

 

[BUGGAR]

You only need the motor power to determine the gantry speed if the speed is not already known. Assuming the obstruction stops the gantry, it is overpowering the motor and the motor is no longer moving the gantry and is out of the equation.
The trick is to determine the force in stopping the gantry. I use energy methods similar to a docking ship hitting the pier.
Let the discussion begin.

 

[jec67]

Took me a few days (including bruising up on dynamics):

Considering Impulse and momentum:

m(v1) + F(deltat)=m(v2) [EQN 1]
m = mass
F = Stooping Force
Delta t = time of contact between wheel and stop (obstruction)
v1 = velocity prior to impact
v2 = velocity after impact

The question is, what is delta t?

Introducing the Coefficient of Restitution (e)

e=(vb2-va2)/(va1-vb1)

For this case:

vb2,va2 = velocities of bodies b & a after collision
vb1, va1 = velocities of b & a before collision

Designating body a as the crane and body b as the stop (obstruction):

vb2=vb1 = 0
va1 = travel speed of the crane prior to collision
va2 = travel speed of crane after collision

Therefore, e=(-va2)/va1, ore e(van) = -va2

Sources list the coefficient of restitution for a crane sticking a steel stop as 0.8 to 1.0

Using e = 1.0 results in va2=-va1

The collision produces a velocity of the crane post-collision equal in magnitude, and in the opposite direction of the velocity before collision.

BUT, what is delta t???

I don't know

An upper bound would be:

Weight of crane(horizontal distance from front wheel to location of center of gravity)<= (vertical distance from wheel to center of gravity) H

H = stooping force

solving for H:

H<= (W(X)/y

Using this value of H in EQN 1, solve for delta t

Is this a reasonable value for delta t? You will need to evaluate

Another route is to solve EQN 1 for an assumed value of delta t. You will need to determine a"reasonable value for delta t" I don't have an answer there.

You should also check the stopping force (H) against other possible limiting values:

Force that will shear the axles
Force that will cause failure of the connections of the wheel to the frame
Other possibilities??

 

[Teguci]

Done this for bumpers and lanyards where the energy absorption is well determined. For this though, I'd think the approach would be something like:
initial kinetic energy = 1/2 m v^2 - horizontal
final elastic energy = 1/2 k x^2 where k is the force required to elastically deflect the gantry a unit of distance horizontally at the center of mass. Solve for x so that initial energy = final energy. Check that x does not exceed the elastic range of the gantry (it probably will but hopefully won't). If it exceeds the elastic range, you will need to find a place to dump energy inelastically (sacrificial eccentric brace, crumple zone, ...)

 

[jec67]

Teguci:

Excellent approach. Once the deflection (x) is determined, k(x) becomes the force (assuming we are still in the elastic range.

 

[BUGGAR]

And if it goes plastic, use E = Force (usually constant in plastic range) x Distance.