Page 301 - Design of Reinforced Masonry Structures

COLUMNS 5.21

From Eq. (5.11)
′
φ
)
.
]
(
.
[
φP = 080 080 f A − A + f A C
n m n st y st P
= 0.80(0.9)[0.80(1.8)(369 – 6) + (60)(6.0)](0.856)]
= 544 kips
Check loads on column. Calculate factored loads.
Load Combinations:
1. U = 1.4D = 1.4(150) = 210 kips
2. U = 1.2D + 1.6L = 1.2(150) + 1.6(225) = 540 > 210 kips
P = 540 kips (governs)
u
fP = 544 > 540 kips OK
n
Check longitudinal reinforcement compliance with the code.
ρ = A st = 60 . = 0 0163
.
A n 369 0 .
r max = 0.04 r min = 0.0025 r provided = 0.0163
0.0025 < 0.0163 < 0.04 OK
The longitudinal reinforcement complies with the code requirements.
fP = 544 > P = 540 kips
n
u
The column can support the imposed service loads.

5.6 DESIGN PROCEDURE FOR REINFORCED
MASONRY COLUMNS

5.6.1 Determination of Longitudinal Steel for Given Column Sizes
The design procedure for an axially loaded column of a given height is simple. Initially, the
column dimensions are assumed, and the axial load reduction coefficient C is calculated
P
for the given h/r (or h/t) ratio, from which fP is determined. For designing a column, it
n
is required that
fP ≥ P (5.19)
u
n
The relationship fP = P can be used in Eq. (5.11), the resulting expression being
n
u
]
)
P = φ(.080 )[.080 f ′( A − A + f A C (5.20)
y
P
st
n
m
u
s
The general procedure for sizing a structural member (e.g., beams, columns, slabs)
requires assuming (guessing) a member size, and then checking its adequacy to carry
design loads and compliance with code requirements. The same procedure is used for sizing
a masonry column.
Equation (5.20) can be used to design a reinforced masonry column. It contains three
unknowns: A , which depends on the column size (unknown), A (the area of longitudinal
n
st