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]

If the block moves then there won’t be any stress because the block is sliding, if the block is sliding then all the friction is overcome, to me P=muW we will have equilibrium ie static,if P increases sliding occurs and it doesn’t matter whether it’s rotating or travelling in a linear line, all the friction as been overcome.

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

 

[Martin.H]

Yes, rb1957, you're right to call me out on the movement/stress/strain statement, it isn't quite what I meant!

I was trying to make the point that as the block is effectively rigid, no movement will occur until the base interface shear exceeds the friction resistance at all points on the block.

i.e. at the moment of slip, the shear stress at the interface is the same everywhere, hence my assertion the stress diagram should be square, not triangular.


Martin.

 

[Martin.H]

@desertfox, I'm fairly certain this is not the case.

Martin.

 

[rb1957]

see attached ...

I think the block will allow the static friction to be exceeded locally.

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

 

[dauwerda]

Here is my stab at it:

edit *Of course this doesn't account for the equilibrium of the straight horizontal force so this isn't correct either.

 

[Martin.H]

Rb1957 yes, I agree the friction can be overcome locally, but to my mind, the block can't move until friction is overcome globaly, as the block base is ridgid. I.e. if the block starts to move then all of the block/ground interface is slipping, hence the sum friction is W x my. My guess is the direction the friction is acting varies over the base (some vector of rotation and translation).

I feel dauwerdas approach gives a more accurate picture of pure rotational resistance, and dividing into to more smaller rectangles would give a more accurate picture of the true rotational resistance, but how that interacts with the forward force is, to my mind, complex:

In the 4 quadrant example, if P/4 is added to the centre of each quadrant (in the same direction as P), then the sum vector of P and F can be determined. BUT, if F and P are the same in each quadrant, then the sum of those vectors will be different for each quadrant, but the vector sum in each quadrant needs to exceed P/4 x mu If sliding is to ocurr (1 quadrant can't slip without the the others having to slip also). My head!

I've been pushing a block across my desk and analysing how it moves. It's not helping!

I also wonder if I'm overlapping static and dynamic friction concepts.

Martin.

 

[desertfox]

Hi Martin.H

imagine for a minute that the block is doweled in place but free to rotate about the dowel.
If we apply a force now to cause the block to rotate:ie a force at one edge of the block, what forces does P force have to overcome in order for it to rotate about its dose?


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

 

[desertfox]

Hi Martin.H

Well here is how they deal with disc rotational friction, maybe we need something similar for your case.

http://mechanicsmap.psu.edu/websites/6_friction/disc_friction/discfriction.html

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

 

[BAretired]

I've been pushing a block across my desk and analysing how it moves. It's not helping!

I have been doing that too, but I find the desk to be a bit too smooth.

When I represent the block with a 12" Engineer's scale, centered on a small stack of 8.5"x11" paper and apply P at the 6" mark, the scale moves more or less straight forward, sometimes straying a little from parallel, possibly because the coefficient of friction is variable along the length of block.

When I apply force P at 12", the 0" point of the scale comes back toward me, but the 6" point advances slightly, which means the block rotates and moves in the direction of P. When P is applied at the 0" end, the 6" point also advances a similar amount, indicating simultaneous rotation and translation.

BA

 

[BAretired]

When a block starts to move, the frictional coefficient changes from static friction to kinetic friction, which is not zero, but considerably less than static friction.

I agree the friction can be overcome locally, but to my mind, the block can't move until friction is overcome globaly, as the block base is ridgid.

I see no justification for that statement. I agree that the block is rigid, but if static friction is exceeded locally, equilibrium can no longer be sustained globally without exceeding static friction elsewhere, using conventional design.

Edit: By dividing the area into many small elements, the "method of instantaneous centers" such as is used in steel design for multi-bolt, eccentrically loaded connections would provide a higher value for P than conventional design, but locating the instantaneous center is an iterative process which may not be worth the effort.

BA

 

[Settingsun]

I'd have thought it's a polar moment calculation, similar to bolt groups.

 

[MIStructE_IRE]

I’ve designed countless hoardings where the contractor ultimately bolts the upright timber to the side of the kentledge block and rightly or wrongly, probably wrongly, never considered this.