Figure 4b indicates the assumed deformation of the upper angle, A being the point of
contraflexure in the vertical leg. Since the true stress distribution cannot be readily ascer-
tained, it is necessary to make simplifying assumptions. The following equations evolve
from a conservative analysis of the member: c = 0.6«; T 2 = T 1 (I + 3a/4b).
Study shows that use of an angle having two rows of bolts in the vertical leg would be
unsatisfactory because the bolts in the outer row would remain inactive until those in the
inner row yielded. If the two rows of bolts are required, the flange should be connected by
means of a tee rather than an angle.
The following notational system will be used with reference to the beam dimensions:
b = flange width; d = beam depth; tf = flange thickness; tf = web thickness.
Record the relevant properties of the W14 x 38; d = 14.12 in (359 mm); tf = 0.513 in
(13 mm). (Obtain these properties from a table of structural-shape data.)
- Establish the capacity of the shop rivets and field bolts used
in transmitting the moment
From the AISC Specification, the rivet capacity in single shear = 0.6013(15) = 9.02 kips
(40.1 kN); rivet capacity in bearing 0.875(0.513)(48.5) = 21.77 kips (96.8 kN); bolt ca-
pacity in tension = 0.6013(40) = 24.05 kips (106.9 kN).
- Determine the number of rivets required in each beam flange
Thus, T 1 = moment/d= 315/14.12 = 22.31 kips (99.7 kN); number of rivets = T 1 MvCt ca-
pacity in single shear = 22.31/9.02 = 2.5; use four rivets, the next highest even number.
- Assuming tentatively that one row of field bolts will suffice,
design the flange angle
Try an angle 8 x 4 x % hi (203 x 102 x 19 mm), 8 in (203 mm) long, having a standard
gage of 2V2 in (63.5 mm) in the vertical leg. Compute the maximum bending moment M
in this leg. Thus, c = 0.6(2.5 - 0.75) = 1.05 in (26.7 mm); M = T 1 C = 23.43 in-kips
(2.65 kN-m). Then apply the relation / = MIS to find the flexural stress. Or,
/= 23.43/[(V 6 )(S)(OJS)
2
] - 31.24 kips/in
2
(215.4 MPa).
Since the cross section is rectangular, the allowable stress is 27 kips/in^2 (186.1 MPa),
as given by the AISC Specification. (The justification for allowing a higher flexural stress
in a member of rectangular cross section as compared with a wide-flange member is pre-
sented in Sec. 1.)
Try a %-in (22-mm) angle, with c = 0.975 in (24.8 mm); M = 21.75 in-kips (2.46
kN-m);/= 21.75/(^1 X 6 )(S)(O-STS)^2 - 21.3 kips/in^2 (146.8 MPa). This is an acceptable stress.
5. Check the adequacy of the two field bolts in each angle
Thus, T 2 = 22.31[1 + 3 x 1.625/(4 x 1.5)] = 40.44 kips (179.9 kN); the capacity of two
bolts = 2(24.05) = 48.10 kips (213.9 kN). Hence the bolts are acceptable because their ca-
pacity exceeds the load.
6. Summarize the design
Use angles 8 x 4 x % in (203 x 102 x 19 mm), 8 in (203 mm) long. In each angle, use four
rivets for the beam connection and two bolts for the column connection. For transmitting
the shear, the standard web connection for a 14-in (356-mm) beam shown in the AISC
Manual is satisfactory
RIVETED MOMENT CONNECTION
A Wl8 x 60 beam frames to the flange of a column and transmits a shear of 40 kips
(177.9 kN) and a moment of 2500 in-kips (282.5 kN-m). Design the connection, using
%-in (22-mm) diameter rivets of A141 steel for both the shop and field connections.
