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Required Clamping force
- Thread starter r4c3r
- Start date
GregLocock
Automotive
What programs have you been using? What inputs do they need?
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Greg Locock
SIG
lease see FAQ731-376 for tips on how to make the best use of Eng-Tips.
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Greg Locock
SIG
lease see FAQ731-376 for tips on how to make the best use of Eng-Tips.Hi r4c3r
First calculate the component force acting parallel to the slope and trying to slide the cabinet across the floor, this will be a portion of the overall mass of the cabinet due to the angle of the slope.
Next using the simple rule of friction
mu=F/R
mu = friction coeff
F = force trying to slide the cabinet
R = normal reaction
From the formula after transposing you can workout what R needs to be to stop the cabinet sliding.
Finally you need a big enough bolt or bolts to resist the shearing force on them produced by the inclined cabinet.
desertfox
First calculate the component force acting parallel to the slope and trying to slide the cabinet across the floor, this will be a portion of the overall mass of the cabinet due to the angle of the slope.
Next using the simple rule of friction
mu=F/R
mu = friction coeff
F = force trying to slide the cabinet
R = normal reaction
From the formula after transposing you can workout what R needs to be to stop the cabinet sliding.
Finally you need a big enough bolt or bolts to resist the shearing force on them produced by the inclined cabinet.
desertfox
IMHO, there are two forces acting on the cabinet attachments ...
in-plane ... the component of the cabinet weight in the plane of the slope, conservatively, neglect friction which opposes this force
out-of-plane ... take moments about the lower edge of the cabinet, there could be tension loads on the upper attmts (stopping the cabinet from tipping over)
as posted previously ... draw a free body
in-plane ... the component of the cabinet weight in the plane of the slope, conservatively, neglect friction which opposes this force
out-of-plane ... take moments about the lower edge of the cabinet, there could be tension loads on the upper attmts (stopping the cabinet from tipping over)
as posted previously ... draw a free body
racookpe1978
Nuclear
Hmmmn.
If the cabinet is on a sloped floor, and you need to "clamp" it down (increase its relative "weight" down so the resultant force perpendicular to the sloped surface above that of gravity alone) so friction forces will prevent movement, then you really just need to increase the shear force parallel to the sloped surface.
Do, rather than clamping it down to the sloped surface, use a screw, bolt, lag bolt, or anchor bolt to hold the object down. It will be more positive, and the shear loads will be carried by the bolt/rod/screw/glue. Or increae the fiction force available with Velcro, glue, anti-skid plastics, etc.
If the cabinet is on a sloped floor, and you need to "clamp" it down (increase its relative "weight" down so the resultant force perpendicular to the sloped surface above that of gravity alone) so friction forces will prevent movement, then you really just need to increase the shear force parallel to the sloped surface.
Do, rather than clamping it down to the sloped surface, use a screw, bolt, lag bolt, or anchor bolt to hold the object down. It will be more positive, and the shear loads will be carried by the bolt/rod/screw/glue. Or increae the fiction force available with Velcro, glue, anti-skid plastics, etc.
MechEng2005
Mechanical
If it's a small angle, you could get a reasonable approximation just by putting weights in or on the cabinet until it doesn't slide anymore. You will need slightly more weight inside the cabinet than if you were bolting normal to the surface due to the direction of the applied force.
However, for a small angle, this may be easier than trying to determine the coefficient of static friction for the pair...
-- MechEng2005
However, for a small angle, this may be easier than trying to determine the coefficient of static friction for the pair...
-- MechEng2005
The clamping force must satisfy
mu*(W*sin@+Fc)>W*cos@
so
Fc>W*cos@/mu-W*sin@
The only trick is to get mu (static coeff friction empirically by changing the above inequality to an equation) Fc is then the force along the incline that just stops it from sliding)
Fc clamp force normal to the plane
mu static coefficient friction
W weight
@ angle of incline
By the way, why wouldn't you simply bolt the thing down so it can't move along the inclined plane or is this an academic exercise?
i guess we're all beating up on the OP 'cause we'll all thinking a similar solution ... attach the cabinet with fasteners so it can't move (so long as the fasteners are strong enough).
but maybe the OP is saying (w/o saying it) "i can't fasten the cabinet down". then indeed the problem is all about coefficient of friction ... between the cabinet and the sloping surface, between the clamp and the cabinet, and between teh clamp and the sloping surface.
the weight component normal to the surface is W*cos@, and the in-plane component W*sin@. the friction force of the cabinet (opposite to the weight component) is k*W*cos@. the resultant of these forces (W*sin@-k*W*cos@) is resisted by the friction force of the clamp (k2*P), and presumably you have n clamps.
of course a completely different approach is the free body solution, which will tell you the normal reaction for the clamps in order to react the tipping moment of the weight, and should be sufficient for friction as well.
but maybe the OP is saying (w/o saying it) "i can't fasten the cabinet down". then indeed the problem is all about coefficient of friction ... between the cabinet and the sloping surface, between the clamp and the cabinet, and between teh clamp and the sloping surface.
the weight component normal to the surface is W*cos@, and the in-plane component W*sin@. the friction force of the cabinet (opposite to the weight component) is k*W*cos@. the resultant of these forces (W*sin@-k*W*cos@) is resisted by the friction force of the clamp (k2*P), and presumably you have n clamps.
of course a completely different approach is the free body solution, which will tell you the normal reaction for the clamps in order to react the tipping moment of the weight, and should be sufficient for friction as well.
r4c3r,
A single 4-40UNC screw can easily exert 250lb clamping force. An appropriate, intelligent number of reasonably sized bolts will exert many orders of magnitude more force than that.
I claim that unless your cabinet is something huge...
[ol]
[li]The force pulling your cabinet down the incline is well below the weight of your cabinet.[/li]
[li]The tensile strength of four 3/8-16UNC bolts (or M10[ ]bolts) is way, way in excess of the weight of your cabinet.[/li]
[li]The flanges your bolts clamp onto is more of an issue than the bolts themselves.[/li]
[li]The cost of 3/8[ ]bolts is way, way below the cost of the engineering needed to verify that #4 screws will work and to make sure that their flanges work.[/li]
[/ol]
JHG
A single 4-40UNC screw can easily exert 250lb clamping force. An appropriate, intelligent number of reasonably sized bolts will exert many orders of magnitude more force than that.
I claim that unless your cabinet is something huge...
[ol]
[li]The force pulling your cabinet down the incline is well below the weight of your cabinet.[/li]
[li]The tensile strength of four 3/8-16UNC bolts (or M10[ ]bolts) is way, way in excess of the weight of your cabinet.[/li]
[li]The flanges your bolts clamp onto is more of an issue than the bolts themselves.[/li]
[li]The cost of 3/8[ ]bolts is way, way below the cost of the engineering needed to verify that #4 screws will work and to make sure that their flanges work.[/li]
[/ol]
![[thumbsup2]](/data/assets/smilies/thumbsup2.gif "[thumbsup2] [thumbsup2]")
> this is for SnTMan- Status
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