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Eccentric load cases.

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takio

Mechanical
Jul 9, 2019
37
Hello

I have a question on eccentric load cases as below.
Would anyone could help me with this?

Capture_km0x99.jpg

[link Loading of the connection by a force inclined to the contact surface not running through the connection gravity centre.][/url]
[link Derivation of Bolted Joints When Load Is Parallel to the Axis of Bolt - Design of Machine][/url]

[1] M=Fr*H - Fa*L : Why do we have to substract Fa*L? I thought theses two components has to be added together. ([1] & [2] : When calculating Moment, [1] is based on the point in the middle of the plate [2] is based on the pivot point. Is it because of Force direction? Can I assume the lever arm starts from the middle of a plate, if force direction is diagonal? (if the bolts are symmetrically installed on a plate)
[3] if W location is between the two bolts, does the equation "wt2=(W*L)/2/(l1^2+l2^2)xl2" still valid? . I actually think so.

Thank you
 
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How involved do you want it to get? For an over-constrained (by the number of bolts) bracket the actual bolt load will vary depending on the flatness and stiffness of the parts (as well as the relative stiffness of the bolts). A conservative approach would be to assume the mounting flange is infinitely stiff and the flange pivots about the end.
 
Q1
The vertical component of F, Fsin(a) would cause the bracket to turn ACW about the overturning point they've used. Whilst the horizontal component Fcos(a) that is greater (based on the perspective) is causing it to turn CW. Conservatively you could leave it on, though you'd be artificially increasing the moment.

Alternatively you could use M=F*H*cos(a), instead of resolving the resultant force into cartesian components.

Q2
Thats a bit more arbitary and open to judgement and interpretation. Usually a lot of people assume it will overturn about the centroid of the bolt group. With that assumption though the bolts on the right side of the first picture wouldn't actually see any load as the plate would be in bearing against the other surface.

For thick and stiff brackets you could also use the far edge which is what the second and third figures appear to be doing.

The moment wan't change, your choice of pivot for this just changes the load you expect in the bolts.

Q3
If the load W, is vertical and in the centre it won't cause overturning so you wouldn't use that equation. Instead it will produce prying where bracket bends. As the bracket bends it puts an additional tensile load in the bolts, usually around an extra quarter. If you want to look at that it would be worth while looking at a structural textbook or standard to calculate/treat it.

To minimise prying though you can reduce the edge distance and thicken the bracket so it's thickness is similar to the bolt diameter for steel usually.

If W acted upward it wouldn't as again the joint would be in bearing instead and should be zero.
 
1) Why do we have to substract Fa*L? ... cause the moment is CCW, whereas Fr*H is CW, so the resultant moment is the sum of the components, just one component is a -ve moment
(your positive moment is CW)

2) what the calc does is move the applied force to the bolt group center. research "bolt groups".

3) no. with W beyond the fasteners both reactions cause the same moment about W. With W between the fasteners the two reactions cause different moments about W ... one +ve the other -ve.

refresh your strength of materials texts. these are basic questions.

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

I would suggest doing a hand calc to reinforce your understanding (rather than relying on some you found on the internet).

another day in paradise, or is paradise one day closer ?
 
You'll notice that the attachment show a similar method of calculations on the shear stress of weldments under eccentric loading
 
Hello rb1957

I have calculated bolt group as below.

Capture111111_xsbivq.jpg
 
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