# Effective net area - steel member

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### #1 PEin2010

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Posted 26 September 2010 - 09:00 PM

How is the effective net area calculated? I have read the theory in AISC D3.2, 13th edition. I'm not sure how it applies to a double angle connection with gusset plate in between. There is a single row of bolts (3 in length).

As I understand it, Effective net area, An = Ag (gross area) - effective hole diameter (bolt dia. + 1/16 + 1/16)*member thickness + s^2/4*g

I'm not sure when to use the the s^2/4*g term. Also the spacing for the bolts is given.

Any help is appreciated. Thanks!

### #2 bootlegend

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Posted 27 September 2010 - 01:43 PM

The g is a gage spacing for staggered bolts, which will be zero for a single row of bolts.

Net area (An) is the gross area (Ag) minus the area of the holes with a 1/16" increase in diameter, or diameter of bolt + 1/8".

Effective net area takes into account the shear lag factor (U) described in AISC section D3.3.

Ae = An*U

### #3 IL-SE

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Posted 27 September 2010 - 05:15 PM

The gage and stagger don't apply to a single row of bolts. See the commentary starting on page 16-250 for an example of a staggard connection where it would apply. The commentary also notes why the extra 1/16" is added to the hole diameter in effective area calculations and shows how to determine the values for U.

### #4 PEin2010

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Posted 27 September 2010 - 05:30 PM

QUOTE (IL-SE @ Sep 27 2010, 10:15 AM) <{POST_SNAPBACK}>
The gage and stagger don't apply to a single row of bolts. See the commentary starting on page 16-250 for an example of a staggard connection where it would apply. The commentary also notes why the extra 1/16" is added to the hole diameter in effective area calculations and shows how to determine the values for U.

so if there is a single row of bolts, for a double angle having a gross area of say 5.71 in^2, the net area (An)= 5.71 (gross area) - 2 (since it's a double angle)*effective hole diameter (bolt dia. + 1/16 + 1/16)*member thickness. Is that correct?

And to get the effective net area, it will be An*U. I read the relevant section, but not quite sure how U is calculated. I think U should be equal to 1.0 for my example, but somehow I don't get the right answer.

Thanks.

### #5 bootlegend

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Posted 27 September 2010 - 09:25 PM

PEin2010,

Look at Table D3.1 Case 2. This is the situation that have if you have. Look at the commentary figure C-D3.2 on pg 16.1-251 (AISC 13th edition).

L is the length from first bolt to the last bolt, or (number of bolts - 1)*(typical bolt spacing).
Xbar is found in the section properties tables.

### #6 IL-SE

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Posted 27 September 2010 - 10:04 PM

To take U=1, both of the legs on each of the angles would have to be connected to a gusset plate. since only one leg of each angle is connected to the gusset plate, you will have shear lag. Basically shear lag accounts for the fact that the tension is distributed to the entire net area (An), but then has to get to one leg at the connection to transfer to the gusset plate. You can calculate U from case 2 (U=1-x/L) or you could also use case 8 in table D3.1 as you really have 2 single angles that each take 1/2 of the load. Then it would just be U=0.80 if you have 4 or more bolts or U=0.60 if you have 2 or 3 bolts. The code lets you take the larger of the two U values.

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