Page 302 - Design of Reinforced Masonry Structures

5.22 CHAPTER FIVE

reinforcement), and C (the column stability factor) which is a function of both the column
P
size and height. As a first step, a column size is assumed.
Because masonry construction consists of units (concrete and clay) of standard sizes
(discussed in Chap. 2), a nominal column size is first guessed. This guessed size depends
on the type of masonry (clay or concrete) and the type of cross section (square, rectan-
gular, or circular, etc.). These sizes are usually, but not always, different for concrete
and clay masonry because of differences in their standard sizes (e.g., a nominal 8 ×
8 × 16 in. concrete masonry unit is much larger than a standard brick). In all cases of
column design, cross-sectional dimensions should comply with code requirements dis-
cussed earlier. Note that h/t for the column must not exceed 30. Once a size is assumed,
the next step is to determine C for the assumed cross section and the given column
P
height. With these two parameters known, the required area of reinforcing steel, A , is
s
easily calculated from Eq. (5.20). The required number of bars to satisfy A requirements
s
can be found from Table A.9. Of course, one must ensure that the area of selected longi-
tudinal reinforcement, A , conforms to the code’s limits on the minimum and maximum
st
percentage of longitudinal reinforcement as stated earlier. The number of bars selected
should be even (minimum four bars) to permit symmetrical arrangement of bars in the
column’s rectangular cross section, to be in conformity with the assumption that all bars
carry equal axial loads. Examples 5.6 to 5.8 illustrate the design procedure for axially
loaded columns.
Reinforcement bars and their grades were discussed in Chap. 3. Grade 60 reinforce-
ment bars are used more commonly than Grade 40 bars, which are generally available
in size Nos. 3 to 6 only; Grade 60 bars must be used when larger size bars are required.
Accordingly, it is a common practice to use Grade 60 bars for all column reinforcement. A
major advantage with using Grade 60 bars instead of Grade 40 bars is that a smaller num-
ber of Grade 60 bars can be used to provide the same strength as a larger number of Grade
40 bars. A large number of bars would create crowding in the limited space available for
placement of longitudinal reinforcement (cells of masonry units), particularly at location
of splices. Such a condition would be detrimental to grouting operation as it would prevent
free flow of grout.
Quite often, a column design problem narrows down to finding the area of longitudinal
reinforcement (A ) to carry loads for a known column cross section and height, and detail-
st
ing of lateral ties. See Example 5.6.

Example 5.5 Determination for area of longitudinal reinforcement for a
given column size.
Design a 24- × 24-in. CMU column to carry a service dead load of 200 kips and a
service live load of 300 kips. The effective height of the column is 28 ft. ′ f = 1500 psi,
m
f = 60 ksi.
y

Solution
Given: CMU column nominal 24 × 24 in., D = 200 kips, L = 300 kips, ′ f = 1500 psi.
m
f = 60 ksi.
y
Check dimensions and h/t ratio for code compliance.
Nominal column width = 24 > 8 in. OK

Nominal column depth = 24 in. < 3(24) = 72 in. OK
h/t = 28(12)/24 = 14 < 30 OK