Kentledge sliding resistance: Eccentric force

[Martin.H]

I'm having problems working out when a kentledge block with an eccentric force will slide.

As per image below, say block has plan dimensions of A x B, a weight of W and the sliding force P is applied on the very edge of the block.

I've done a few circles in my thinking now and so far google hasn't yielded anything either!

Any input appreciated!

Martin.

 

[desertfox]

Hi
I might be missing something but won’t P= mu * W ?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Martin.H]

If P was applied through the centre of mass of the block, then I would agree. But as it's off to the side, there is a tendency to rotate, so I'm thinking some of the friction must be 'used up' resisting the eccentricity.

A bit like the way the stress block under an eccentrically loaded footing is non-uniform. But different in this case, I think, as the friction resistance is uniform over the base.

Martin.

 

[Martin.H]

Maybe the answer is to take moments about the far right corner, say:

P x A = W x (A/2) x mu



Martin.

 

[robyengIT]

"Maybe the answer is to take moments about the far right corner, say:

P x A = W x (A/2) x mu"

saying that you suppose there is a hinge in the upper-right corner

 

[desertfox]

Hi Martin.H

Well I take your point however if the force P is not applied through the centre of mass then either the block topples over or it rotates. So if it rotates then I think P= mu*W.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Martin.H]

"saying that you suppose there is a hinge in the upper-right corner"

No, I don't think so. But I'm possibly going mad so who knows...

My thinking:

I can take moments about anywhere and equilibrium, considering all forces (internal and external) present, must prevail. Same is true with taking moments on a beam.

There are only 2 forces in my example: P and W (the latter distributed evenly, in theory at least, over the base)

If P was applied in line with the centre of the block as opposed to eccentrically then taking moments about the edge:

P x (A/2) = (W x mu) x (A/2)

i.e. P = W x mu , which is correct and doesn't, I believe, assume a hinge at the arbitrary point I took my moment.





Martin.

 

[desertfox]

Hi Martin.H

Do you know what the sizes of concrete block are along with the value of P?


“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Martin.H]

"Well I take your point however if the force P is not applied through the centre of mass then either the block topples over or it rotates. So if it rotates then I think P= mu*W"

If I try and push a 500kg block directly, I wouldn't be able to slide it . If I bolted a 20m pole (horizontally) off the side of the block and pushed the end, I'm pretty sure I could make the block move (in some kind of rotational way). That says to me that the eccentricity of the load has significance.



Martin.

 

[HTURKAK]

The block will rotate due to the torque developing with T=P*A/2 and slide due to the force P..Of course if the force is great enough to rotate and slide the object..

 

[Martin.H]

HTURKAK,

Yes, you are correct, but the challenge is what is the rotational sliding resistance (Rs) of the block?

My solution so far is to say rotation sliding resistance Rs = 0.5 x W x A x mu, linear sliding resistance Ls = W x mu

and that, using your applied force notation:

T/Rs + P/Ls <1 (or < 0.66 to give me a FoS of 1.5)


I think my Rs is an underestimate as the extreme corners of the block have a greater lever arm from the rotation centroid. Maybe this is what the radius of gyration represents? Text book here i come!

Update: To work out the static friction resistance to rotation, it's as if I need the pure plastic torsional (polar?) modulus of a rectangle...anyone know how to work that out?

Martin.

 

[BAretired]

As soon as the block starts to rotate, the direction of P changes with respect to the block. Eccentricity increases, with a maximum dimension of half the length of the diagonal. The force required to continue rotation and translation diminishes until the diagonal is normal to P.

To calculate the force P required to initiate movement would involve finding the variable frictional stress on the underside of the block due to combined rotation and translation. I think it would be an iterative procedure, beginning with a guess at the location of the instantaneous center of rotation, not an inviting type of calculation.

BA

 

[desertfox]

Hi Martin.H

If the block starts to rotate it’s really just sliding on the floor therefore P= mu*W is valid in my book, whether the block goes straight or not it’s still sliding.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[jayrod12]

I see it something like this.

 

[BAretired]

@jayrod,

Are you aware that the sketch in the original post is a plan view?

Yes, I see what you are suggesting. That is beginning to make sense to me.
So movement would start as soon as the maximum stress exceeded μW/AB.

BA

 

[Martin.H]

That was my first thought, Jayrod, but we're talking about static friction. No movement means no strain, means no stress, so to my mind triangular distributions aren't valid.

Martin.

 

[BAretired]

That was my first thought, Jayrod, but we're talking about static friction. No movement means no strain, means no stress, so to my mind triangular distributions aren't valid.

Stress on the interface exists even if it is not enough to cause sliding. I think jayrod has a valid concept.

BA

 

[Martin.H]

@BAretired, my thought is the block either moves, it doesn't move. The block is rigid, so either all friction is overcome simultaneously (at the same 'stress'), or the block doesn't move.

The approach I think I'm going to adopt is basically what jayrod has shown, but with a 'plastic' (i.e. square) distribution of rotational stress.

Martin.

 

[Martin.H]

I still don't think that is fully correct though, as only one half of the summed stress diagram will be at the maximum 'sliding' stress (i.e. friction resistance), which means the other half still has some reserve friction capacity to be overcome in order for the block to move.

But it's conservative, so I think I'll take it!

Martin.

 

[rb1957]

"No movement means no strain, means no stress" ... why?

the question is what force P will over come the static friction and cause the block to move.
So what Jayrod posted is theoretically reasonable, the unknown is the tapering distribution reacting the torque.
A question is ... is it (the torque reaction) uniform over the block face, or proportional to the distance from the center ?

I guess the next question is does it move the instant static friction is overcome locally ? Could the torque reaction redistribute ??

another day in paradise, or is paradise one day closer ?