the relation Mu =fybd^2 /4, or Mp = 36(0.5)(24)^2 /4 = 2592 in-kips (292.9 kN-m). This is sat-
isfactory.
- Compare the ultimate direct force at section b-b
with the allowable value
Thus, Pn = 93.7 + 43.6 = 137.3 kips (610.7 kN); or Pn = 206.0 - 68.7 = 137.3 kips (610.7
kN); e = 9 - 2 = 7 in (177.8 mm). By Eq. 2, x = 9 + 1 - [(9 + 7)^2 - 7 x 18]°-^5 = 4.6 in
(116.8 mm). By Eq. 1, /\allow = 36,000(0.5)(18 - 9.2) = 158.4 kips (704.6 kN). This is
satisfactory.
On horizontal sections above a-a., the forces in the web members have not been com-
pletely transferred to the gusset plate, but the eccentricities are greater than those at a-a.
Therefore, the calculations in step 5 should be repeated with reference to one or two sec-
tions above a-a before any conclusion concerning the adequacy of the plate is drawn.
DESIGN OFA SEMIRIGID CONNECTION
A W14 x 38 beam is to be connected to the flange of a column by a semirigid connection
that transmits a shear of 25 kips (111.2 kN) and a moment of 315 in-kips (35.6 kN-m).
Design the connection for the moment, using A141 shop rivets and A3 2 5 field bolts of
7s-in (22.2-mm) diameter.
Calculation Procedure:
- Record the relevant properties of the W14 x 38
A semirigid connection is one that offers only partial restraint against rotation. For a rela-
tively small moment, a connection of the type shown in Fig. 4a will be adequate. In de-
signing this type of connection, it is assumed for simplicity that the moment is resisted
entirely by the flanges; and the force in each flange is found by dividing the moment by
the beam depth.
(Q) Semirigid connectionFIGURE 4. (a) Semirigid connection; (b) deformation of flange angle.(b) Deformation of flange angleGage