Section Modulus Reduction Due to Bolt Holes

[BadgerPE]

Good evening all,
I am checking a contractors sketch of a pipe trapeze made out of a single angle section with unequal legs. The long leg is oriented in the vertical direction and the short leg is oriented in the horizontal direction. 6 unequal pipe loads are hung from the short leg by unequal sized rods. My question is, how would I go about determining the reduction in the section modulus to effectively determine the bending capacity of the angle? I apologize in advance if there is a simple solution to this question, but I have been staring at it for a couple hours and could use a little push in the right direction. Thanks much!

PS Sorry for the crude sketch

 

[Ron]

Assuming the bolt holes are where the arrows are in your sketch, your reduction in Sx is minimal for bending. More of the bending resistance is in the long leg.

I would check the section for torsion.

 

[Lion06]

I agree with Ron on the torsion. The shear center of the angle is on the other side of the long leg, so the eccentricity is not insignificant.

 

[rb1957]

the angle will want to bend and twist. you could analyze for bending about the principal axes of the angle. how is the angle restrained ?

 

[BadgerPE]

The angle is restrained at each end by a 3/8" plate welded to the top of the short leg which is then fastened to the purlin above.

 

[rb1957]

welds in tension ? and the short leg is doing all the work (bending about it's weak axis)

could work, depending on the load. it'd be nicer to weld the angle to an angle piece, welding on the long leg, aligned to the load, weld in shear, IMHO

 

[miecz]

To answer you original question, you can get the strong axis section modulus of the angle by using the section of an angle with the top leg shortened by the diameter of your largest hole. Say your hole is 13/16". Then calculate the section modulus of an angle with a 2-3/16" horizontal leg. That solution doesn't work for other section properties, like for torsion.

Seeing that your load lines up with your resistance, torsion may not be such an issue. But that connection sounds like a problem.

Could you show a sketch of that connection?

 

[BadgerPE]

I used a 1/4" Fillet weld down both sides of the plate. Capacity of weld was determined to be 11.1 k/in using AISC Eqn J2-5.

 

[Lion06]

miecz-

Torsion would not be a problem only for the connections, right? The load is applied relatively far from the shear center of the angle (which would be off to the left of the vertical leg in the graphic), which would imply quite a bit of torsion for the actual angle design, no?

 

[miecz]

EIT

You're right, there would be torsion in the angle. But with this support, it seems as though the torsion in the angle would be highest near the support, where the bending would be lowest.

 

[Lion06]

True.

 

[miecz]

I'm still confused about how the connection works. I think of the beam shear as being carried by the vertical leg, but it seems like that leg isn't attached to anything. So I'm having trouble picturing how the shear travels from the vertical leg to the support. Seems like the weld closer to the vertical leg would take all the load. On the other hand, the torsion would tend to load the other weld. Hard to imagine without a sketch.

 

[BadgerPE]

Sorry. Missed that the sketch didn't properly upload in my last post. I am starting to see that there are some larger issues to account for here.

 

[BAretired]

The shear center of an angle is at the intersection of the middle surfaces of the two legs, so the eccentricity is a little less than 1.5" assuming the loads are applied at the center of the 3" horizontal leg.

If the two hanger attachments align with the line of applied loads, there is no net moment to be carried by the hanger.

The angle itself carries bending moment and torsional moment with maximum value at one or both supports.

BA

 

[miecz]

Seems like we all agree that, if the hangers line up with the center of the hanger plate, then the angle has maximum torsion at it's end, but the hanger plate sees no moment. Something's not adding up here...

 

[rb1957]

if the loads and reactions line up, then there is no nett torque (simple free body shows this).

but take a section thru the angle between the loads. the long (inloaded) leg of the angle is the stiffer loadpath for shear, so this'll be reacting the applied loads, and therefore the torque on the section. the torque in the angle will be highest near the supports.

there's no moment to the rest-of-the-world, but there is twist in the angle ...

 

[BAretired]

I agree with rb1957.

miecz,

What is not adding up? If the reaction is R, the torsion at the support is R*e where e is the eccentricity to the shear center of the angle. If the hanger is not connected to the vertical leg, the horizontal leg must be capable of carrying the reaction in localized bending.


BA

 

[miecz]

I'm beginning to think that the torsion at the end of the angle is zero, since the hanger plate sees no moment. In this case, any torsion in the angle would grow towards the mid-span. At this point, I'm wondering if there is any torsion anywhere in the angle. This all assumes, of course, that the pipe hangers give a vertical load only.

 

[BadgerPE]

I really appreciate all the help guys. I think I'm going to have a meeting with my supervisor to figure this one out though. I am really confused and I can't really seem to get my head around the design of this. Thanks again for all the help!

 

[BAretired]

miecz,

The torsion at the end of the angle is NOT zero. It is maximum. If the reaction is offset by 'e' and the loads are offset by 'e', the hanger sees no moment but the angle sees a torsional moment because its shear center is offset by 'e'.

It is no different than a loose lintel in a brick wall. The angle has torsion at each end but if the brick is centered on the support, the supporting brick sees no moment.

BA