Impossible Question: Maritime Freight

[Dave1128]

Hi,

I have a question that has been bugging me for a while. I was on a ship where the vessel was rolling from left to right. On board the ship was a sled with angled sides of ~ 60°. the mass of the sled looked to be about 50 Tonnes and the only thing stopping the sled from moving around were some chocks around the base which were shaped parallel to the sides of the sled i.e. ~60° (chamfered chock 40mm x 40mm). I would say the acceleration at the peak of each roll (if that's the right terminology) was about 1.5g i.e. as the ship begins to right itself the sled if forced into the chock at 1.5g. the gap between the sled an chock was about 2mm so insignificant.

My question is what forces were at work to stop the sled from just riding over the chocks and causing havoc. Discounting friction?

Fnorm = 50,00kg * 9.81 = 490,500N
Fgrav = Fnorm

Fapp =50,000 Kg * 1.5g = 735,759 N

Does the angle of the chock exert a force at 60° vertical? how would I calculate this? are there any other forces at play?


Regards

Dave

 

[AlephZero]

I would say the acceleration at the peak of each roll (if that's the right terminology) was about 1.5g i.e. as the ship begins to right itself the sled if forced into the chock at 1.5g.

I suspect your estimates for the acceleration are way to high, unless you and the rest of the crew were literally strapped into seats at the time. You would not have been able to move around (even using safety ropes etc) if the accelerations were as large as that.

For example if the roll period was 10 sec (that's fast, for a large ship) and the maximum angle was +/-30 degrees (that's large!) = pi / 6 radians, the maximum angular acceleration would be (2 pi x 0.1)^2 x pi/6 = about 0.2 radians/sec^2. If the deck was say 10m above the center of roll of the ship, that would only give a linear acceleration (sideways) of 2 m/s^2 or about 0.2g.

I think even 0.2g is probably a huge over-estimate of the actual acceleration.

 

[haruspex]

I suspect your estimates for the acceleration are way to high, unless you and the rest of the crew were literally strapped into seats at the time. You would not have been able to move around (even using safety ropes etc) if the accelerations were as large as that.

For example if the roll period was 10 sec (that's fast, for a large ship) and the maximum angle was +/-30 degrees (that's large!) = pi / 6 radians, the maximum angular acceleration would be (2 pi x 0.1)^2 x pi/6 = about 0.2 radians/sec^2. If the deck was say 10m above the center of roll of the ship, that would only give a linear acceleration (sideways) of 2 m/s^2 or about 0.2g.

I think even 0.2g is probably a huge over-estimate of the actual acceleration.

Yes, that's the force needed tangential to the roll to reverse its direction (assuming SHM), but the chock also needs to supply a force in that direction to help support the sled's static weight, i.e. g.cos(pi/3) = g/2.
The sled won't override the chock unless the total force on it acts at an angle below the normal to the chock (i.e. >30 deg to the vertical). We have 0.2g at 60 deg to vertical and g vertical, giving a resultant at atan(sqrt(3)/11) = 0.16 radians, about 9 degrees.

 

[Dave1128]

Cheers guys, sort of makes sense to my puny brain.

 

[Dave1128]

Hi,

Thought I understood this but i really don't. So i have attached a quick sketch to try and explain.

can someone explain what is happening in the sketch and how i would calculate the forces.

apologies if the answers submitted already explain this but i don't understand.

Note the sled and chock are both angled at 60 deg.cheersdave

 

[AlephZero]

The way to attack this is to find all the forces that must be acting on the sled to keep it stable. Then, see if those forces are physically possible given the geometry of the situation.

To keep it simple, assume there is no friction between the sled and the deck, and the base of the sled is not perfectly flat, so it only contacts the deck at its two ends and not all the way along its length.

The geometrical proportions of the sled (the ratio of the height of the CG above the deck and its length) affect the answer as well as the size of the sideways acceleration. Call the length L and the height of the CG position H. Call the weight of the sled W.

The forces on the sled are
1. Its weight W acting down at its center of mass.
2. A sideways force of 0.2 W acting to the right, also at its center of mass
3. The vertical reaction force V3 of the deck on the left hand corner (assume there is no horizontal force, because we are ignoring friction between the sled and the deck)
4. The vertical and horizontal compoents V4 and H4 of the force at the stop.

We can find V3 by taking moments about the stop.
V3 x L + 0.2W x H = W x L/2
We can then find the V4 by resolving vertically
V4 = W - V3
The horizontal force H4 is 0.2W (acting to the left).

Then check if V3 > 0, otherwise the left hand end of the sled will lift off the deck and it will topple over the stop (unlikely, unless it is very tall or the sideways acceleration is very big).

Finally, compare the direction of the resultant force at the stop with the 60 degree angle, to see whether the sled will try to slide up the stop or be wedged into the corner between the stop and the deck.

 

[haruspex]

The geometrical proportions of the sled (the ratio of the height of the CG above the deck and its length) affect the answer as well as the size of the sideways acceleration. Call the length L and the height of the CG position H. Call the weight of the sled W.

The forces on the sled are
1. Its weight W acting down at its center of mass.
2. A sideways force of 0.2 W acting to the right, also at its center of mass
3. The vertical reaction force V3 of the deck on the left hand corner (assume there is no horizontal force, because we are ignoring friction between the sled and the deck)
4. The vertical and horizontal compoents V4 and H4 of the force at the stop.

We can find V3 by taking moments about the stop.
V3 x L + 0.2W x H = W x L/2
We can then find the V4 by resolving vertically
V4 = W - V3
The horizontal force H4 is 0.2W (acting to the left).

Then check if V3 > 0, otherwise the left hand end of the sled will lift off the deck and it will topple over the stop (unlikely, unless it is very tall or the sideways acceleration is very big).

Finally, compare the direction of the resultant force at the stop with the 60 degree angle, to see whether the sled will try to slide up the stop or be wedged into the corner between the stop and the deck.

The question as posed stated "ride over the chock". You seem to be answering the slightly different question of "tip over the chock". Of course, that might be the question intended. (The dimensions of the chock were not initially provided.)

Not sure about some of your forces. Also confused about whether you're analysing the case of hitting the right-hand chock or the left-hand one.
The 0.2W is not horizontal. It is parallel to the deck, with the deck tipped at 30 degrees.
The reaction from the (smooth) deck is at right angles to the deck, not vertical.
If V3 is at the left corner of the sled when about to tip over a chock then presumably you're thinking of the left-hand chock. But then you take moments about (the base of?) the chock and find a nonzero contribution from V3. hence my confusion.
The net force from the chock (V4&H4) acts at right angles to the chock face, so at 30 degrees to the vertical, and through the corner of the chock.

If the chock height is h (measured up its face), and the normal forces from chock and deck are C, D respectively, the sway force from the rolling is X (=0.2W we reckon), and writing c, s for cos, sin of 30 degrees, I get:

Perp to deck: W.c = D + C.s
||'l to deck: X + W.s = C.c
Moments about chock/deck/weight contact point:
D.h = X.H + W(H.s - L.c/2)
Can eliminate C, D from these equations and obtain an expression for the minimum value of h to prevent tipping. (Could be negative.)

 

[AlephZero]

The question as posed stated "ride over the chock". You seem to be answering the slightly different question of "tip over the chock". Of course, that might be the question intended. (The dimensions of the chock were not initially provided.)

No, I'm answering both of those. If the weight on the left hand side goes to zero it will tip. At the right hand side, ignoring friction, there are two physical forces. One is a vertical from the deck, the other is normal to the face of the chock. The resultant of those forces can be at any angle between the two. If the diirection of the force required to hold the sled in place is outside of that angle, it will start to slide up the chock.

Also confused about whether you're analysing the case of hitting the right-hand chock or the left-hand one.

Hitting the right hand chock. I took the OP's diagram as assuming the relative motion of the sled was left to right.

The 0.2W is not horizontal. It is parallel to the deck, with the deck tipped at 30 degrees...

Sure, if the deck was tipped the details would be different but the method is the same. I was answering using the OP's diagram where the deck is not tipped..

 

[haruspex]

OK, I see what you're doing, but there are still a couple of issues.
I think it's clear from the original statement that we're concerned with the point where the deck is at maximum tilt. Indeed, the 0.2W horizontal force would not be present when the deck is level.
And you've ignored the height of the chock in the tipping calculation. The tipping is quite tricky because its nature will be a tip-and-slide. The corner of the sled will slide away from the chock, as the sled tips to stay in contact with the top of the chock.

 

[AlephZero]

I think it's clear from the original statement that we're concerned with the point where the deck is at maximum tilt.

It's easy to include that. Just change the direction the weight of the sled acts. That produces a smaller force normal to the deck, and a bigger sideways force.

Indeed, the 0.2W horizontal force would not be present when the deck is level.

You seem to be ssuming the rolling is SHM with constant amplitude. There is no reason why there shouldn't be sideways acceleration (equivalent to a horizontal force) when the deck is level.

And you've ignored the height of the chock in the tipping calculation.

We could both probably list lots more things we have ignored, but I think the point of the OP's quesition was to get some basic understanding of what is going on, not to see the results of a 100,000 DOF finite element model of including plastic deformation of the sled and/or chock during the impact, shifting of the contents of the sled because of the rolling motion, etc, etc ...

 

[haruspex]

It's easy to include that. Just change the direction the weight of the sled acts. That produces a smaller force normal to the deck, and a bigger sideways force.

Sure, but given the original question I'd say this was a major consideration. In fact, at 30 degrees it's more significant than the 0.2W.

You seem to be ssuming the rolling is SHM with constant amplitude. There is no reason why there shouldn't be sideways acceleration (equivalent to a horizontal force) when the deck is level.

Again, the original question concerned roll. OK, it needn't be SHM, but that does seem a reasonable basis for estimating it, giving 0.2 (and it wasn't me that estimated 0.2 ;-).
Besides, I shudder to imagine the sort of seas that would produce other sideways accelerations to compete with that at the extremes of the roll.

Btw, I was wrong when I said the chock dimensions were not originally supplied - they were.

 

[Dave1128]

Hi guys

thanks for taking the time to reply to my question.

So from the information you supplied I got the following.

Mass = 50,000kg = 490,500 N @ CoG 2897(l),1545mm(h) (sled is not a uniform shape)

The length of the base is 5350mm (I assumed that because we're taking moments about the stop l should be the distance V3 and V4)

∴ V3 * 5.35m + 98,100 * 1.545 = 490,500 * 2.675m

V3 = 216,920 N

V4 = 490,500 - 216,920 = 233,980N

OR if L in the equation should have been 2.897N

V3 = 400,594N

V4 = 89,905N

Have I made a mistake since V3 is a positive number?

So for the resultant of V4, H4

From the first attempt above V4 = 233,980 and H4 = 98,100

= 253,712N @ 67° (so does that mean it will slide over?)

Next attempt

V4 = 89,950 and H4 = 98,100

133096N @ 43° (again does this mean it won't slide over)


I will add in the 30° roll in later but first can you tell me if I have done the above correctly?


Regards

Dave