Minimum Bolt Diameter & Distance for Bolt Shear Force, 200kN

[Ry122]

the shear stress in each bolt is to be limited to 250mpa. i need to determine the minimum diameter of the bolts and the distance from the end edge of the centre plate to the centre of the bolt holes. All plates are 6mm thick and the punching shear stress for the centre plate is not to exceed 200MPa.
P=200kN
http://users.on.net/~rohanlal/Untitled54.jpg

I know I have to work out the total area of the shear plane required to resist the shearing force so i can then determine the diameter of the bolts but I don't know how to determine the distance from the end edge of the centre plate to the centre of the bolt hole or how to find the shear force.

 

[PhanthomJay]

the shear stress in each bolt is to be limited to 250mpa. i need to determine the minimum diameter of the bolts and the distance from the end edge of the centre plate to the centre of the bolt holes. All plates are 6mm thick and the punching shear stress for the centre plate is not to exceed 200MPa.
P=200kN
http://users.on.net/~rohanlal/Untitled54.jpg

I know I have to work out the total area of the shear plane required to resist the shearing force so i can then determine the diameter of the bolts but I don't know how to determine the distance from the end edge of the centre plate to the centre of the bolt hole or how to find the shear force.

Each bolt is in double shear, such that the 4 bolts each see , in shear, their equal share of one-half the applied force. Calculate the required area and diameter of each bolt to take this force. For punching shear, note that the full applied force divides equally amongst each of the 4 bolts, in bearing, and that the far bolts at the edge have to cut through two shear planes for edge distance shear failure.

 

[Ry122]

what is edge distance shear failure, and how exactly is it calculated?

 

[Ry122]

For the bolt diameter, is this correct?
50+25+25=100
100/2=50
50/4=7.5
x = diameter
250=7.5/18x
Are the bolts said to be in double shear because each bolt is parallel with another bolt along the same axis as the direction of the forces?

 

[PhanthomJay]

For the bolt diameter, is this correct?
50+25+25=100
100/2=50
50/4=7.5
x = diameter
250=7.5/18x
Are the bolts said to be in double shear because each bolt is parallel with another bolt along the same axis as the direction of the forces?

It looks like you're trying to cram a semester's course in Strength of Materials into a one week period. I am unsure where your numbers are coming from. A bolt is in double shear when there are 2 shear planes in the one bolt to take the load in that bolt. If the double shear concept is confusing, forget it. Instead, take a free body diagram of the top plate. The applied load at the right is 100kN. So there must be 100kN shear total on 4 bolts, which is ____kN per bolt. That's the number to use to determine the bolt area based on the allowable shear stress in the bolt. Then I don't know wher your 18 is coming from; the area of a bolt with diameter d is (pi)d^2/4.
Edge distance is often a matter of code, but for calcualation purposes without Code, the shear area for the plate is the plate thickness times the distance from the center of the hole to the edge. But there are 2 such areas acting before the bolt can tear thru the plate.

 

[Ry122]

That's the number to use to determine the bolt area based on the allowable shear stress in the bolt. Then I don't know wher your 18 is coming from; the area of a bolt with diameter d is (pi)d^2/4.

Doesn't A = the area the force is applied to? Because the force is being applied to only the x-y plane of the bolt, and not the area of the whole bolt, wouldn't it be the diameter multiplied by the length of the bolt? The width of the plates are 6mm and there are three of them. 6x3=18

 

[PhanthomJay]

Doesn't A = the area the force is applied to? Because the force is being applied to only the x-y plane of the bolt, and not the area of the whole bolt, wouldn't it be the diameter multiplied by the length of the bolt? The width of the plates are 6mm and there are three of them. 6x3=18

What you are calculating, albeit incorrectly, is the bearing stress on the bolt, which acts perpendicular to the surface area of the bolt that is bearing against one of the plates (but it's not over the 18 mm depth...the bearing force in the middle plate is opposite to the bearing force on the other plates, which is smaller, and in any case, they don't add). The shear stress in the bolt, that you are trying to find, acts parallel to the cross section area of the bolt. Shear stresses always act parallel to the area in question, not perpendicular to it.

 

[Ry122]

The cross section of the bolt is collinear to the applied force so doesn't the force that is being applied have to be perpendicular to the surface that the shear strain is being generated parallel to, to generate a shear force?

 

[PhanthomJay]

The cross section of the bolt is collinear to the applied force so doesn't the force that is being applied have to be perpendicular to the surface that the shear strain is being generated parallel to, to generate a shear force?

No, you are misunderstanding the concept of shear. Shear strains and stresses are parallel to the the cross section. You might want to look back at your other question where instead of bolts, there was a glued surface holding the plates together. The forces and shear stresses in the glue act parallel to its area. Or you might want to consider a simply supported beam with a 10kN load applied at mid point. In addition to the bending tensile or compressive stresses caused by that load, which act perpendicular to the beam's cross section, there is also a shear force in the beam of 5kN, which acts parallel to the beams cross section, creating shear stresses in that section in the plane of the cross section.

 

[cameron16]

Hi I am doing a problem similar to this to do with the edge failure shear so I was just wondering if this is right for this problem.
Like you said doing a fbd of the top part you get a shear of 100kn so over the 4 bolts that's 25kn each. So
Max punching stress=shear per bolt/thickness of middle plane * the distance to the bolt
so
200=25x10³/6xd
d = 20.833mm

I have no idea whether this is right I was just trying to go by what you said in the past things.
Im probably way off track so any help would be great.

Thanks a lot

 

[PhanthomJay]

Hi I am doing a problem similar to this to do with the edge failure shear so I was just wondering if this is right for this problem.
Like you said doing a fbd of the top part you get a shear of 100kn so over the 4 bolts that's 25kn each. So
Max punching stress=shear per bolt/thickness of middle plane * the distance to the bolt
so
200=25x10³/6xd
d = 20.833mm

I have no idea whether this is right I was just trying to go by what you said in the past things.
Im probably way off track so any help would be great.

Thanks a lot

Hi, cameron, welcome to PF!
Your answer for the edge distance, d, is correct (rounded up). However, your method is not quite correct. What you really have is the 50kN bearing force from the edge bolt to the middle plate, that splits amongst 2 shear planes in the middle plate, to give the 25kN shear force per shear plane, and the corresponding correct answer for 'd'. But look now at the top plate (or bottom plate), and the bolt bearing force is 25kN, which splits amongst 2 shear planes in the top plate to give a 12.5kN shear force per plane, such that the edge distance from the near left bolt to the top plate edge need be only 1/2 as much (11mm) as the middle plate edge distance to the near bolt. Do you see the difference?