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Bolt Calculation VDI 2230(2)

6060842 (Mechanical) 
14 Feb 08 15:22 
Hi,
I am interested in developing some simple software for our company to aid in the selection of bolts, ideally using the VDI 2230 and interested in anyones ideas for what to include. Basically I design simple LBrackets with the back part fixed to walls and horizontal part of the bracket carrying the load. Mostly the load is concentric to the fixings. So, the fixings be loaded axially and in shear and possibly a bending force. I have seen the one sheet VDI 2230 approximation for quick selection of fasteners and intend using this as a starting point
So, as a starting example , say I have this simple LBracket fixed to a wall with 4 bolts on a 100mm square pattern with the load concentric. The load is 100kg with its centre of gravity 100mm away from the way and 200mm lower than the centre of the bolt pattern. How would the calculation procede to select correct fasteners using the VDI 2230 method
Any help appreciated 

CoryPad (Materials) 
14 Feb 08 16:21 
Your force is not concentric, it is eccentric. Perhaps you meant axial, in contrast to transverse. VDI 2230 is used to analyze the situation at the actual fastener location. So, first you need to transform your global forces (applied forces, reactions, etc.) into a local condition at the fastener. This is done with simple statics, finiteelement calculations, etc. After you have local conditions, then you input your joint geometry and materials. Then, you determine the minimum required preload: friction grip to resist transverse forces, or sealing against a medium, or to prevent opening along the joint interface(s). Then, calculate stresses, surface pressure, fatigue, thread engagement, etc. I recommend purchasing the standard and one of the available commercial software packages. They are very helpful for this type of work. Regards,
Cory
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dimjim (Mechanical) 
15 Feb 08 9:05 
Like Cory, I do not understand how the horizontal forces resolve to the vertical bolt pattern giving you an eccentric loading of the bolt pattern axis. Are you saying that the forces on the horizontal part are off center and creating a torque force on the bolts? 

rb1957 (Aerospace) 
15 Feb 08 9:50 
like the others, if the load is away from the wall, it's creting a moment about the bolt group center, trying to pry the top of the angle off the wall. 

6060842 (Mechanical) 
15 Feb 08 12:04 
Okay sorry I will try to describe a bit better
Assume a flat plate 300mm x 300mm x 3mm Aluminium rigidly mounted to a wall(not interested in the fixings in this example, just assume it is bonded). Now, assume a mass of 100kg mounted to this plate by 4 bolts on a 200 x 200 pattern (equal about the centreline). Also assume that the centre of gravity of this mass is 200mm from the plate surface and in the theoretical centrepoint of the 4 hole pattern.
What i am trying to ask is what is the procedure for sizing these 4 bolts (ie Tensile \ Shear \ Bending moment calculation). Just looking for the steps involved with equations if possible
Thanks 

rb1957 (Aerospace) 
15 Feb 08 12:34 
yeah, i think that's pretty much what we had in mind. your post could have meant you were mounting a picture (to continue the wall reference), pretty much in the plane of the wall, so the bolts would only be loaded in shear (by the weight). but your supporting something away from the plane of the wall ... weight = W (lbs or N), distance d (" or mm). the moment acting on the bolts is W*d. the bolts react this moment as a couple, if the bolts a pitched p (" of mm) apart and if there are 4 bolts (2 pairs) then the tension load on the bolt is (W*d)/(p*2). this is a simplification, you could say that the angle is "heeling" at it's bottom edge (this'll give you a wider moment base, and slightly smaller loads). these bolts are also supporting the weight in shear, W/4. now you have to combine these loads together ... Rt = applied tension/allowable tension Rs = applied shear/allowable shear a simple formula is .. Rt^2+Rs^2 = 1/RF^2 RF = reserve factor = 1 + Margin of Safety now greg'll post that you can't solve the problem 'cause it's redundant (sorry mate, needless dig !! ... 

6060842 (Mechanical) 
15 Feb 08 13:10 
Thanks rb1957
I was just looking at the NASA fasteners design manual which describes combining loads as you replied
So in my example
Tension Load = ((100 * 9.81) * 200))/200 * 2 = 490.5 Shear Load = 100 * 9.81 / 4) = 245.25 

6060842 (Mechanical) 
16 Feb 08 8:11 
Thank Desertfox Guys just one last thing. I have looked at all of the tensions and shear in all directions(see attached PDF), But 1) How so I apply the combined shear and tension (Rt^2+Rs^2 = 1/RF^2) and see if M6 Grade 8.8 bolts would be acceptable? 2) What if I had to calclate for combined forces (ie x,y & z simultaneously) Many thanks for the help Tom 

rb1957 (Aerospace) 
16 Feb 08 8:27 
in your example y is tension, x and z are shear; combine x and z forces as vectors (shear = sqrt(Fx^2+Fz^2)
all this is in basic (very basic) texts 

Hi 6060842 I might have missed something here but if your 100kg mass is acting vertically then you cannot have tension acting in the x direction. The forces that I see acting are shear in the z direction and tension in the y direction due to the mass centre being 200mm from the plate. Also the tension in each bolt is not equal, the top two bolts carry more tension than the bottom two bolts because the brackets pivots about the bottom edge of the bracket due to the gravitional force of the offset mass. I have uploaded a file to show you how I calculate the tension in the bolts. Finally I would use a Mohr circle to find the resultant tensile stress in the bolts. Regards desertfox 

Tmoose (Mechanical) 
16 Feb 08 15:46 
Everybody Starts with the basic methods described previously by others. The thickness and other geometry of the bracket can change the bolt loading significantly, and in ways that basic static analyses miss completely, and are not always "intuitive." For instance, if a gusset or rib is involved, bolts further than a few diameters from the rib rely on the basic brackets thickness, and may be subjected to very little of the axial load. Multi bolt patterns on thin uniform brackets ( http://www.activerobots.com/products/motorsandwheels/easyroller/bracket750.jpg) rarely distribute loads in a simple geometrical way. And the structure the bracket bolts to can make contributions just as important. IN a structural joint The "Proper" fasteners allow installation torque that provides clamping force greater than operating loads. This is what hapens when bolts are asked to act as drive pins or dowels. http://www.atsb.gov.au/publications/investigation_reports/2002/AAIR/images/aair200205705_001.jpgActually similar bad things happen to dowels that are subjected to alternating loads. 

Hi Tmoose
At this stage the poster is only trying to calculate the external loads on the bolt, after that he can select a bolt preload that should be higher than external loads.
regards
desertfox 

6060842 (Mechanical) 
16 Feb 08 16:25 
Desertfox
Many thanks for the excellent reply. The tension in x was because i was looking individually if the forces were acting in all 3 planes seperatley  I think!!
Where does the 6.037 come from in your calculation?
I'd be interested in the Mohrs circle, i will look it up, unless you have some references
Also you were correct I would have looked at preload next
Many Thanks
Tom


6060842 (Mechanical) 
17 Feb 08 8:51 
Desertfox,
Sorry i didn't read it properly i see that u = moment/2*(L1^2 + L2^2)
Excellent, thank you
Tom 

Hi 6060842
Your welcome. Are you okay with the Mohr circle?.
Regards
desertfox 

6060842 (Mechanical) 
18 Feb 08 8:05 
Desrtfox
"Are you okay with the Mohr circle?"
Not really but I have taken too much of your time already. I'll try to work out how it applys for my example
Thanks
Tom 



