Structural Concrete - Hassoun

16.2 Maximum Moments in Continuous Beams 559
Figure 16.4 Example 16.1.
2. Loads on slabs:
Dead load = 5 × 150 + 80 = 142.5psf
12
Live load = 130 psf
Factored load (w u )=1.2(142.5)+1.6(130) =379 psf
Loads on beams: A typical interior beam ABC carries slab loads from both sides of the beam, with
a total slab width of 12 ft.
Factored load on beam = 12 × 379 + 1.2 ×(self − weight of beam web)
The depth of the beam can be estimated using the coefficients of minimum thickness
of beams shown in Table 1. For f y = 60 ksi, the minimum thickness of the first beam AB is
L/18.5 = (24 × 12)/18.5 = 15.6 in. Assume a total depth of 22 in. and a web depth of 22 − 5 = 17 in.
Therefore, the factored load on beam ABCD is
w u = 12 × 379 + 1.2
( 17 × 12
144
)
× 150 = 4804 lb∕ft
Use 4.8 K/ft.
3. Moments in beam ABC: Moment coefficients are shown in Fig. 9.8. The beam is continuous on
five spans and symmetrical about the centerline at D. Therefore, it is sufficient to design half of
the beam ABCD because the other half will have similar dimensions and reinforcement. Because
the spans AB and BC are not equal and the ratio 26 is less than 1.2, the ACI moment coefficients
24
can be applied to this beam. Moreover, the average of the adjacent clear span is used to calculate
the negative moments at the supports.
Moments at critical sections are calculated as follows (Fig. 16.4):
M u = coefficient × w u l 2 n