Shear Stress Bolted Joint Cantilevered Beam

[Cowea]

Hi Guys,

I am trying to use a formula on the Roymech website which is very applicable to a problem I am facing. Located at

http://www.roymech.co.uk/Useful_Tables/Screws/Bolted_Joint.html

Scroll down until you get to 'Strength of bolt joints withstanding bending forces'. I have copied and pasted the formulae below but it has not imported the numerical font sizes correctly!!

"Each Bolt withstands a shear Force Fs = Fv / (Number of bolts)
The resulting shear bolt stress τ n = Fs /A
Note: Each bolt is assumed to withstand the same shear force.
If there are x bolts( numbered n = 1 to x). Then the tensile force withstood be each bolt is designated Fnt i.e F1t,F2t, F3t....Fxt
A selected bolt withstands a tensile force of Fnt = ( Fv. Rv + Fh. Rh) . Vn / (V12 + V22....Vx2 )
The resulting tensile bolt stress sn = Fnt /A "

I have got a bit stuck running through the Fnt part of the equation where it is asking for V1 squared + V2 squared etc .....Vx2. For each bolt that I run through this calculation I need to sum all of the heights to EVERY bolt? Just find that a bit strange.

I have a point load of approximately 600kg's (6KN) at the end of my 1 metre beam but the cantilever platform itself also weighs approx 90kg (1mtr long). I am using 4 bolts in my example not 6. Basically I need a bolt that will be up to the job with a decent safety factor, the load will be placed on the cantilever platform approximately 9 times per day. The height from the pivot (pivot being the bottom of my pad) to the first row of bolts (2 bolts) is approximately 25mm and the height to the second row of bolts (2 off) is approximately 125mm.

Secondly.....My cantilever platform mounting pad is mounting to the web of an I-beam (so this is a single shear calculation), the centre line of my mounting pad will mate to the centre line of the I-beam web therefore making the holes on my mounting pad offset from the centreline of the I-beam .... Am I correct in saying that I should potentially put the 4 total bolts on the same row (instead of having two rows of bolts) to lower the stress on the structural I beam ( so that all bolts are fixed through the I-beam centre line rather than offset)? Although I imagine if I did this this would have a negative effect on my cantilever platform & bolt stresses.

If anyone could also show a worked through example that would be great! I have gone years through my career without using my university knowledge on this, if you dont use it you lose it!

I have ran just the platform through the stress analysis software so at a later date when I get back in work I can show you pictures of the platform assembly.

Kind regards

 

[dhengr]

Cowea :
Why don’t you show us a sketch of what you have, with loads, dimensions, material sizes and thicknesses, etc.; and do this so things are generally in proportion. We can’t see it from here, and I’m not going to spend a lot of time wading through a bunch of Vhn’s and Vvn’s when I don’t know how they relate to your problem.

 

[desertfox]

Hi Cowea

I am very familiar with that calculation and yes you do need to square all the distances.
Basically that formula assumes that your fabrication is completely rigid and so when it sees the bending load the frame pivots about one edge of the frame, now the only thing stopping it rotating is the four bolts close to the pivoting end of the frame. Of those bolts the ones furthest away from the pivot point carry the most load and those nearest proportionally less however in order to calculate those forces you have to solve a statically indeterminate problem and in order to do this we assume a unknown quantity µ which represents a (force/unit length) and solve a value for µ based on moments from the pivot point of the frame, once a µ value is calculated you then multiply this value by the distance from the pivot arm to the furthest bolt centre to give maximum bolt load, note that the units of length cancel out after the multiplication leaving force on its own .
I've uploaded a file demonstrating this, it works out the bolt loads exactly the same as that on the Roymech site.

 

[desertfox]

Hi Again

The file I uploaded shows 2µa+2µb the 2 is because I have 4 bolts i.e. 2 bolts at distance "a" and 2 at distance "b".
So in your case the equation would look something like this:-

6000*1m + 90*8.81*0.5m = 2µa^2+ 2µb^2

6441.45 = 2µ(0.025m^2 + 0.125m^2)

µ= 198198.46 N/m

so maximum bolt load would be 198198.46N/m*0.125m = 24774.8N

I've assumed the platform is horizontal and the 6kN acts vertically down at 1m distance and similarly for the mass of the frame but at 500m distance.

 

[rb1957]

yes, you do square the distance. you are solving a bolt group. fastener shear is proportional to distance from the CG (picture the bolt group rotating around the CG) and moment reacted by each fastener = F*d = k*d^2; sum individual moments = applied moment.

Quando Omni Flunkus Moritati

 

[robyengIT]

If You can read a little bit of Italian, it can help a lot (I think)

Regards