Eccentrically loaded bolted joint-Shear force. 1

[takio]

Hello

I managed to calculate resultant shear force acting on P1, which is caused by the far end force, 250lbs. (P1 has the highest shear force), thanks to the link below.
and it is 505 lbf.
but I have trouble finding shear force acting on the bolts, due to the force on the right most side, close to the bolts)
Would anyone please help me with this calculation?

Thank you

Eccentrically loaded bolted joint

 

[rb1957]

the four bolts react shear and moment, yes? (as you show lower right)

the direct shear is easy, yes?

how do you think four bolts might react moment ? hint, it's not as you show (with vertical and horizontal forces)
hint 2 ... think about a moment being a couple. the moment is about the center of the four bolts (as you show). if you replaced the four bolts with a circle (to make it easier to see), then you could easily draw a line from the center to a bolt (and it'd be a radius, yes?) ... can you see where this is going?

then you need vector addition.

you might want to research "bolt groups".

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

 

[WARose]

I did it myself (figuring the polar moment of inertia, etc). I don't know if I follow you sketch exactly....but I came out with the max. (resultant) shear being 480 lbs.

 

[takio]

Hello
Thank you for your comment again!

I just quickly calculated.
1)Moment caused by the load = 250*1
2)Assuming all the bolts are equally spaced in a circle. All the bolts are 3 inches away from the centroid.
3)Load that is tangent to the circle = 250 lbf in / 3 in = 166.67 lbf
4) X and Y direction of the tangential load.
(Fx^2+Fy^2)^(1/2)=166.67lbf
Fx and Fy are equal.
2*Fx^2=166.67 lbf
Fx=83.37 lbf

but, I found explanations on bolt group. I am looking into this.
Thank you so much!!

 

[WARose]

I found a good link (if you need a reference):

https://mechanicalc.com/reference/bolt-pattern-force-distribution

The number I spit out above assumed a (single) 250 lbs force cantilevered 21.5" to the c.g. of the (4) bolt group.

 

[le99]

@YOUNGTAKI,

The simplest way is to resolve the forces and get the equivalent forces about the bolt center. For the case shown,

∑Fy = 250 + 250 = 500 lbs
∑M = 250*(18.5 + 5.5/2) + 250*(5.5/2 - 1) = 5750 lb-in
Let single bolt shear = V, and assume the moment is equally distributed among all bolts in the group, thus
∑M = 4V*r
V = ∑M/4r = 5750/(4*3) = 479.2 lbs
Vy (max) = 479.2*0.707 + 500/4 = 463.8 lbs
Vy (min) = -479.2*0.707 + 500/4 = -213.8 lbs
Vx = +/- 479.2*0.707 = +/-338.8 lbs

You need to watch out for the direction of forces in the Y-dir though. Also, please check my numbers and let me know if any mistakes.

 

[ProgrammingPE]

The sketch doesn't make it completely clear what the dimensions are. This may not be correct, but I think you have bolts that are 5.5" apart in the horizontal direction (horizontal dimensions are 1" from the leftmost bolt to the load and 4.5" from the load to the rightmost bolt). The bolts are 2.5" apart in the vertical direction (calculated based on a 3" diagonal dimension from the centroid which is 2.75" horizontally and 1.25" vertically from the centroid).

I see a few issues with your quick calc:
[ul]
[li]The moment should be calculated from the centroid of the bolts, so 1.75", not 1"[/li]
[li]Since the horizontal and vertical dimensions are not the same, Fx and Fy are not equal[/li]
[li]The direct shear is not included (you may be adding it in later, though)[/li]
[/ul]

The links below are for two PDFs that show the bolt reactions and supporting calculations for each of your two loads. To determine the combined load, you will add the X components and Y components from each load together before determining their combined reactions.

Bolt Group Example - Load 1
Bolt Group Example - Load 2

I also made a video before that discusses How to Calculate Bolt Group Reactions which may help make things a little clearer.

Structural Engineering Software:

http://www.structuralcentral.com

Structural Engineering Videos:

http://www.youtube.com/structuralcentral

 

[takio]

Hello

I have recalculated as below. I hope this is right.

1) V=500 LBF
2)Total moment at centroid
Ma=250 LBF * 1.75 in
Mb=250 LBF * (18.5+2.75) in

Mt = 250 * (18.5+2.75+1.75) lbf in = 5750 lbf in
3) r = 3.02 in (bolt distance from centroid)
4) Primary shear load per bolt
F' = 500 lbf / 4 = 125 lbf
5) Secondary shear force
F" = Mr/4/r^2 = M/4/r = 5750/4/3.02 = 475.99 lbf
6) tan-1 (1.25/2.75)=24.44 deg
Resultant of F' and F"
Fc and Fd = (F'^2+F"^2+2*F'*F"*COS(24.44deg))^(1/2) = 592.05 lbf
Fa and Fb = (F'^2+F"^2+2*F'*F"*COS(155.56deg))^(1/2) = 365.86 lbf

 

[takio]

Hello Warose

For me, I got 502 lbf.
(250 lbs force cantilevered 21.5" to the c.g. of the (4) bolt group)
I just calculated with the same method above, but using 250 lbf

Thank you