Structural Concrete - Hassoun

3.18 Sections of Other Shapes 137
3.17 INVERTED L-SHAPED SECTIONS
In slab–beam girder floors, the end beam is called a spandrel beam. This type of floor has part of
the slab on one side of the beam and is cast monolithically with the beam. The section is unsymmetrical
under vertical loading (Fig. 3.36a). The loads on slab S 1 cause torsional moment uniformly
distributed on the spandrel beam B 1 . Design for torsion is explained later. The overhanging flange
width b − b w of a beam with the flange on one side only is limited by the ACI Code, Section 6.3.2.1,
to the smallest of the following:
1. b e = L/12.
2. b e = 6 t + b w .
3. b e = b.
If this is applied to the spandrel beam in Fig. 3.36b, then
1. b e = (20 × 12)/12 = 20 in. (controls).
2. b e = 6 × 6 +12 = 48 in.
3. b e = 3.5 × 12 +12 = 56 in.
Therefore, the effective flange width is b = 32 in., and the effective dimensions of the spandrel
beam are as shown in Fig. 3.36d.
3.18 SECTIONS OF OTHER SHAPES
Sometimes a section different from the previously defined sections is needed for special requirements
of structural members. For instance, sections such as those shown in Fig. 3.37 might be used
in the precast concrete industry. The analysis of such sections is similar to that of a rectangular
section, taking into consideration the area of the removed or added concrete. The next example
explains the analysis of such sections.
Example 3.13
The section shown in Fig. 3.38 represents a beam in a structure containing prefabricated elements. The
total width and total depth are limited to 14 and 21 in., respectively. Tension reinforcement used is four
no. 9 bars. Using f c ′ = 4ksi and f y = 60 ksi, determine the design moment strength of the section.
Solution
1. Determine the position of the neutral axis based on T = 4 × 60 = 240 K:
240 = 0.85f c ′ [2(4 × 5)+14(a − 4)]
where a is the depth of the equivalent compressive block needed to produce a total compressive
force of 240 K.
If 240 = (0.85 × 4) (40 +14 a −56), then a = 6.18 in. and c = a/0.85 = 7.28 in.
2. Calculate M n by taking moments of the two parts of the compressive forces (each by its arm),
about the tension steel:
= compressive force on the two small areas, 4 × 5in.
C ′ 1
C ′′
1
= 0.85 × 4(2 × 4 × 5) =136 K
= compressive force on area, 14 × 2.185