element scorrevole

[gradasso]

In the case of the guide cylinder I think it is a question of tolerances and friction. the central cylinder does not impurities if there is practically no game. the central cylinder also does not suffer from the torque if the friction coefficient in the slide is reduced to the maximum, in fact the reactions to the present moments would be born from the central cylinder while they would not generate resistances to the shift because of the low friction. is the first thing that comes to mind by looking at the design

 

[Beth]

yes the central guide cylinder is h7 tolerance
and flows in bronze compass graphite h7

 

[gradasso]

h-h I think it's just pairing with very light game (at the limit also 0 games), as I thought, so I confirm what I said before. graphite as everyone knows is a lubricant suitable for strong pressure between surfaces, temperatures. . ..concluding, therefore, in my opinion, the forces are equal and must not balance any moment, that is, the pencil must balance 200 kgf

 

[Beth]

perfect, hello and thank you

 

[Beth]

hi to all, a question about this topic " sliding element "
for those who want to help me with a minimum explanation.
attached 2 cases where an element flows on action of a f force (hydraulic pencil) for a race of 20 mm .
I would like to stop the element mechanically with element b
correctly sized? (I thought of a plug) that does not break up just for a series
high of shots.
Let's forget the friction coefficients that will be very low.
I hope to be expressed correctly and have entered the data correctly

thanks in advance to all

 

[Stan9411]

Hello everyone, a question

there is no question ..

 

[meccanicamg]

there is no question ..

wants to know the diameter of the plug that resists to cut considering the impact of which we do not know how elastic/anaelastic.
However it has already been treated and a value can be estimated that it resists cutting forces in play statically and then oversized for dynamic action.
on the plane included born breaks the weight port in parallel and perpendicular to the sliding plane, then calculate the friction force and make the sum of the resulting. I'll stick it to the plug.

 

[Stan9411]

but do not send it in line, stop and then apply strength statically .. there will always be an impulsive dynamic effect that is function of the speed with which the object beats on the plug .. there are various theoretical models to obtain the impact force, but it depends on the geometry of the components.
In fact for crash testing simulations use non-linear dynamic fem solutors.. But I think I'm digging

 

[meccanicamg]

We could try to simplify, just to have an order of magnitude.p=5kg
\(p//=p•sen(\alpha)=1,7kg≈0,02kn\) then it turns out to be negligible.
♪
null inertia forces, acceleration nothing by application almost static force.
Attrite force \( fa=p\bot•fa•g=p•cos(\alpha)•g=0,05kn \) and also this is negligible.

There's only f=10kn.
material 1,1730 is nothing more than c45u to be tempered and invented.

then depends on how much it advances on the plug, the material of the plate and the plug.the calculation formulas are these:
all this with almost static application of force.

 

[meccanicamg]

incorrect courier by friction, the right formula is as follows:
\( fa=p\bot•fa•g=p•cos(\alpha)•g•fa=0,005kn\)
with fa=0,1... but still it is negligible.

 

[Beth]

infinitely thank you

 

[Beth]

Hello, everyone.
I hope someone will help me with my gaps, I hope the position in the discussion is right.
(b a dx by wedge c.
I would like to understand how to break the force of 10000n on the sides x and y of the wedge in a position of joke.
all components are steel.
If someone wants and patience to throw down a scheme with formulas, maybe it's the good time I learn something, I'm arguing about the previous explanations.

Thank you.

 

[meccanicamg]

usually if you use these formulas you can solve:
is a classic locking blade with wedge.
However someone will have elasticity and this is not contemplated in formulas that consider infinitely rigid bodies. However, the forces are so disproportionate....in a rigid manner in the first approximation.

 

[biz]

you can be useful This is what debate

 

[Beth]

usually if you use these formulas you can solve:
View attachment 72023is a classic locking blade with wedge.
However someone will have elasticity and this is not contemplated in formulas that consider infinitely rigid bodies. However, the forces are so disproportionate....in a rigid manner in the first approximation.

Thank you, I try to use the scheme

 

[Beth]

you can be useful This is what debate

I try to study