Structural Concrete - Hassoun

156 Chapter 4 Flexural Design of Reinforced Concrete Beams
maximum size of the aggregate), (ACI Code, Section 25.2.1).Vertical clear spacing between bars in
more than one layer shall not be less than 1 in. (25 mm), according to the ACI Code, Section 25.2.2.
Also for reinforcement of more than two layers, the upper layer reinforcement shall be placed
directly above the reinforcement of the lower layer.
The width of the section depends on the number, n, and diameter of bars used. Stirrups are
placed at intervals; their diameters and spacings depend on shear requirements, to be explained
later. At this stage, stirrups of 3 in. (10 mm) diameter can be assumed to calculate the width of
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the section. There is no need to adjust the width, b, if different diameters of stirrups are used. The
specified concrete cover for cast-in-place and precast concrete is given in the ACI Code, Section
20.6.1. Concrete cover for beams and girders is equal to 3 in. (38 mm), and that for slabs is equal
2
in. (20 mm), when concrete is not exposed to weather or in contact with the ground.
to 3 4
4.3.2 Minimum Width of Concrete Sections
The general equation for the minimum width of a concrete section can be written in the form
where
b min = nD +(n − 1)s + 2(stirrup diameter)+2(concrete cover)
n = number of bars
D = diameter of largest bar used
s = spacing between bars (equal to D or 1 in., whichever is larger)
(4.5a)
If the stirrup’s diameter is taken equal to 3 in. (10 mm) and concrete cover equals 3 in.
8 2
(38 mm), then
b min = nD +(n − 1)s + 3.75 in. (95mm)
(4.5b)
This equation, if applied to the concrete sections in Fig. 4.1, becomes
b 1 = 3D + 2S + 3.75in. (95 mm)
b 2 = 4D + 3S + 3.75in. (95 mm)
To clarify the use of Eq. 4.5, let the bars used in sections of Fig. 4.1 be no. 10 (32-mm) bars. Then
{ 5 × 1.27 + 3.75 = 10.10 in. (s = D) say, 11 in.
b 1 =
5 × 32 + 95 = 225 mm say, 250 mm
{ 7 × 1.27 + 3.75 = 12.64 in. (s = D) say, 13 in.
b 2 =
7 × 32 + 95 = 319 mm say, 320 mm
If the bars used are no. 6 (20 mm), the minimum widths become
{ 3 × 0.75 + 2 × 1 + 3.75 = 8.0in. (s = 1in.)
b 1 =
3 × 20 + 2 × 25 + 95 = 205 mm say, 210 mm
{ 4 × 0.75 + 3 × 1 + 3.75 = 9.75 in. say, 10 in.
b 2 =
4 × 20 + 3 × 25 + 95 = 250 mm
The width of the concrete section shall be increased to the nearest inch. Table 1 gives the
minimum beam width for different numbers of bars in the section.