Internal vs external forces ...........Oh dear.

[chas10]

Hi all

I am trying to analyse the reactions on a complicated piece of equipment carrying external forces. The equipment is supported by hydraulic jacks which apply internal forces or pre-loads to fixed surfaces to generate frictional support capacity

I have simplified the problem (hopefully) to illustrated may problem see attachment.

The first diagram shows a vertical steel beam with cantilevered end which is gripped by four jacks which push against fixed surfaces.

The jacking force is calculated to generate the required vertical resistance capacity to carry a vertical load of 50 tonnes and hold the beam in position.This is based on a coefficient of friction of 0.15.

This arrangement gives horizontal reactions on the support surfaces equal to the jacking forces.

I now add a horizontal load to the end of the cantilever.

I am trying to decide if the horizontal reactions change.

My first thoughts were that they would..........but I don't think they do. Something to do with Internal and external forces not being additive. The jacks provide an internal force, whilst the loads are external forces.

I have the feeling that unless the externally applied force exceeds the internally applied jacking force then the reactions stay as they are. Not sure wwhat happens to the externaly aplied horizontal reactions.

Only problem is I can't see a way to prove it.

This is I think akin to the sort of thing that happens when bolts are pre-loaded with an internal tension. Any externally applied tension is not additive, unless it exceeds the preload value.

It's like this but only in compression .....I think.

Can anyone assist with an explanation or confirm my thoughts

Many thanks

Charlie.

 

[frv]

Sure they do.

Remember that under elastic response, you can consider each loading to be independent and additive. i.e., take the first diagram with all the internal forces and calculate the reactions, then draw a second diagram with only the horizontal force and get your horizontal reactions. Then simply add (vectorial addition, of course) these horizontal reactions to your original structure. The only issue I can see here is that on one of the pistons the horizontal reaction may be reduced, thus reducing your frictional resistance.

 

[chas10]

Hmmm.

I,ve done what you suggest............but..........by my reckoning the resulting forces don't balance when I check for equilibrium. i.e. left hand forces don't equal right hand forces. I'm 10 tonnes out of balance.

This is driving me nuts!

Charlie.

 

[badservo]

If you have two identical hydraulic cyliders facing eachother with equal pressure and ignore seal friction then any horizontal force will cause them to move until one of them reach end of travel. If (as in your drawing) the right cylinder is bottomed out on retraction due to the horizontal force and the left cylinder can still extend then the force on the top right pad will be cylinder force + horizontal force, the top left pad will be cylinder force only. if you vertical beam can tilt then the forces for the bottom cylinders will be opposite

 

[frv]

You are not transferring the internal 5 and 15 ton (sorry- American English spelling )forces to your pistons; these, in turn will transfer to your supports.

 

[chas10]

frv

There was a time when we spelt it the same....then we went metric....

If I've done it correctlly the top left reaction becomes 68.3 tons (when in rome) (83.3-15)

Bottom left reaction becomes 88.3 (83.3+5)

Top right reaction becomes 98.3 (83.3+15)

Bottom right reaction becomes 78.3 (88.3-5)

Now equating forces i.e. left to right forces must equal right to left forces

left to right 68.3 +88.3 + 10 =166.6 Tons

right to left 98.3 + 78.3 = 176.6 Tons

10 tons adrift.........!Something is wrong here......may be it's me!

Badservo.

This is getting a bit heavy but.............. I recon the 10 ton additional external force cannot over come the 83.3 ton internal force being applied by the jack. Therefore in my opinion the jack won't bottom out. if you see what I mean.

Still not convinced

Chas

 

[frv]

You forgot the 10 tons you're actually applying

 

[chas10]

I think I have included it

See........ it's in the top line.....the little devil.

left to right 68.3 + 88.3 + 10 = 166.6
Right to left 98.3 + 78.3 =176

don't know what else I can do.

Chas

 

[frv]

Ok.. I believe you are forgetting to transfer the forces to the supports..

look at the 15 ton force at the top. This force is in a piston, so the force is constant within this piston. That means that the piston also applies this 15 ton load at the right, meaning your reaction at the right of that support would be 98.3 tons (to the left), right?

If you do a free body diagram around the the top "joint" you would have the following:

a 10 ton shear in the continuous member at the top (to the right)

an 83.3 ton reaction at the left support (to the right)

a 98.3 ton reaction on the right support to the left

and (and I suspect this is what you're missing) a 5 ton shear to the right at the bottom of this "joint"


This five ton shear is then resolved in your reaction at the bottom "joint"


Think of it.. your original structure satisfies static equilibrium; your secondary structure also satisfies static equilibrium. Then the summation of both your structures MUST satisfy static equilibrium..

 

[frv]

Chas10..

I've just reread your penultimate response and I believe I've figured out where this went astray...

The top left reaction doesn't change.. you now have shear forces in your members to provide equilibrium around the joint (see my previous post).. same applies to the bottom right..

If you draw a FBD around only the vertical member and the pistons at both the top and bottom of the member where the pistons grip it, I believe you'll see what I'm talking (is it proper to use the term "talking" when we are not actually engaging in conversation? anyway..)about..

 

[hokie66]

Frv is correct. When you add the 15 tonne force to the right at the top, you don't take it off at the left. Same reasoning at the bottom.

Chas10, don't let him talk you into spelling it "ton". Tonne and ton are different words with different meanings.

 

[frv]

hehehe..

are you referring to the standard ton of 2000 lb vs. the "imperial" ton of 2000 and some?

By the way.. I didn't talk to him into it.. in the US we would refer to it as ton vs. something else (maybe British ton?).. I really don't know, as we typically use "kips"..

 

[hokie66]

A tonne is 1000 kilograms. And the unit of force normally used metrically by structural engineers is kilonewtons (kN).

 

[chas10]

Many thanks all

Now I can see it ...............yes the top left and bottom right jack forces do not reduce, and the top right reaction increases by 15 tonnes with the bottom left reaction increasing by 5 tonnes..........This all then stays in equilibrium.( I'll pass on the tonnes/tons debate at this stage)

This is good news for me, because in the piece of equipment I am analysing the forces on the support structure can be controlled.

The jack forces are monitored and can be adjusted so when the reaction goes up on a jack the force can be reduced using the hydraulic control system back to the force required to provide the required vertical reaction from frction, whilst the other jacks do not loose compression forces.

By the way ..........this is someone else's concept which I have inherited.

There are some safety back ups and it's all a bit complicated to explain but I think now that I understand that the external loads do not reduce the reactions I am fine.

So I think I've got it now......and it helps a lot......Many thanks.......phew!

Kips well I know we have Kippers which some people enjoy but I think they just stink the house out.He He.

1/32" is a smaller dimension than a millimeter.

Many thanks all

Charlie

 

[apsix]

short ton = 907 kg (approx)
(long) ton = 1016 kg (approx)
metric ton (tonne) = 1000 kg