(^ 0 Pn = 768 kips for this case is also tabulated on p. 4-73 of the AISC LRFD Manual)
The 768-kip design strength is considerably more than the 238-kip (1640 Mpa) design
strength of a noncomposite W8 x 40 column under the same conditions.
Related Calculations. This procedure is the work of Abraham J. Rokach,
MSCE, Associate Director of Education, American Institute of Steel Construction. SI val-
ues were prepared by the handbook editor.
ANALYZING A CONCRETE SLAB
FOR COMPOSITE ACTION
A W18 x 40 interior beam is shown in Fig. 47. Steel is A36, beam span is 30 ft O in (9.14
m), and beam spacing 10 ft O in (3.04 m). The beams are to act compositely with a 5-in
(12.7-cm) normal-weight concrete slab;/,' = 5.0 ksi (41.3 kN). Determine: (a) The effec-
tive width of concrete slab for composite action; (b) Vh (the total horizontal shear force to
be transferred) for full composite action; (c) The number of 0.75-in (1.9-cm) diameter
shear studs required if Fu = 60 ksi (413.4 kN).
Calculation Procedure:
- Find the effective width of concrete slab for composite action
For an interior beam, the effective slab width on either side of the beam centerline is the
minimum of
-| - -^p = 3.75 ft = 45 in (114.3 cm)I -iMIU 5.00 ft (1.52m)The effective slab width is 2 x 45 in = 90 in (228.6 cm).
- Determine the total horizontal shear force for full
composite action
In positive moment regions, Vh for full composite action is the smaller of
0.85/c'^c = 0.85 x 5 ksi x (90 in x 5 in)= 1913 kips (8509 kN)