# RE: Steel Connections

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Steel Connections
• From: "Bill Allen" <Bill(--nospam--at)AllenDesigns.com>
• Date: Thu, 29 Jun 2000 06:09:37 -0700
```Hmmmm....

I guess I'm missing something. Instead of just blindly taking values from a
table, I worked through an example.

Ref: AISC ASD Manual, 9th ed., p. 4-52 "SINGLE-PLATE SHEAR CONNECTIONS TABLE
X"

First of all, in the nomenclature there is an item:

"n=number of bolts, in."

I don't think the "in." part should be there.

For the example of a 3/8" shear plate, (2)-3/4"Ø A325-N Bolts:

L=6"
n=2 (not in., just 2)
bolts pitch = 3 in.
aL=3 in. (distance from weld line to bolt line)

In the Design Procedure:

1. Calculate plate capacity in yielding, R-sub-o:
R-sub-o=0.4*36*L*t
R-sub-o=0.4*36*6*.375
R-sub-o=32.4k

2. Calculate fillet weld size, D (in sixteenths) to develop R-sub-o:
based on aL=3 in. -> a=3/6=0.50
based on aL=n x 1 in. = 2 in. -> a= 2/6=0.333
Larger of the two: a=0.50
C (from Table XIX and k =0) = 0.787
D=R-sub-o/(L x C)= 32.4 / (6 x 0.787) = 6.86 or 7/16" F.W.
<=0.75*t=0.75 x .375 = 0.281" = 5/16"

Based on Table X, 3/8" F.W. is required (should be 5/16" ?)

3. Calculate bolt group capacity, R-sub-b:
R-sub-b = C x r-sub-v
From Table XI based on l=3 in, C = 0.88
r-sub-v = 9.3k per bolt (Based on Table I-D, p. 4-5)
R-sub-b = 0.88 x 9.3 = 8.2k

Consistent with the verbage in the AISC manual, this procedure appears to be
conservative. Talbe XI places the load eccentrically on the bolts where
actually the load is applied directly to the bolts via the beam. In my
opinion, based on the FBD, the load should either be eccentric on the weld
or the bolts, not both. In this case, the shear tab is supporting the beam
and the load is applied directly onto the bolts. This appears to be quite a
penalty for a two bolt connection (44%). This is not so apparent in
connections where there are more bolts in a row:

2 bolts	44%
3 bolts	74%
4 bolts	83%
5 bolts	88%
6 bolts	91%
etc.

4. Calculate net shear fracture capacity of plate R-sub-ns:
R-sub-ns=0.3 x 58 x [L-n(d-sub-b + 1/16)] x t
R-sub-ns=0.3 x 58 x [6-2(0.75+1/16)] x 0.375
R-sub-ns=28.6k

R-sub-ns is greater than R-sub-b and does not govern.

Sooo....I would probably design based on Table I-D, consider shear block
stresses at coped flanges, etc. And, yes, Mr. Glaser, unless I hear
something that would compell me otherwise, I would say that a two bolt
connection consisting of (2)-3/4"Ø A325-N Bolts with the proper fillet weld
and shear tab thickness is good for 18.4k unless proven otherwise. As I
stated above, the problem I have is that I don't believe there is
eccentricity on the bolts, just the weld. I would be interested in knowing
what experience, if any, beyond flipping pages in the AISC manual you are
referring to.

Back to the original question. I would make a table and a detail with the
table as a schedule for my drawings grouping classes of WF beam connections.

For example, for W8s & W10s, I would use a 2 bolt connection, etc. with a
properly designed fillet weld and tab plate thickness and tablulate the
capacity. In the design, I would check the reactions to ensure the actual
reactions were less than the capacity of the detail and let it go if it is.
If the design reaction is greater than the capacity in my "standard detail",
I would have to decide to either create a special detail or to bump up the
beam size (depth) so that it would accommodate the capacity of my standard
detail (I normally lean towards the latter).

Regards,

Bill Allen, S.E. (CA #2607)
ALLEN DESIGNS
Laguna Niguel, CA
http://www.AllenDesigns.com

||-----Original Message-----
||From: Neil Glaser [mailto:NGlaser(--nospam--at)pkainc.com]
||Sent: Wednesday, June 28, 2000 4:51 AM
||To: seaint(--nospam--at)seaint.org
||Subject: Re: Steel Connections
||
||
||
||
||John Riley wrote:
||
||> "A two bolt connection, single shear, 3/4" ASTM A325N is
||good for (2)(9.3k) =
||> 18.6 kips." ..... "The two bolt connection would yield 91% of the
||> table-derived
||> requirement, and that's close enough for me."
||
||Are you saying that a 2 bolt shear tab connection is good for
||18.6 kips?
||According to my experience, and based on AISC Table X-A
||(ASD), a 2 bolt shear
||tab with 3/4" A325 bolts is only good for 8.2 kips.
||Therefore the connection
||only yields 40% of the table derived requirement for your
||example.  Is that
||still close enough for you?
||
||Bolt shear is not the only failure mode that you should be
||checked when
||designing a connection.  It is only one failure mode and is
||not the critical
||failure mechanism in this condition.
||
||
||

```