Page 311 - Design of Reinforced Masonry Structures

COLUMNS 5.31

However, a minimum area of reinforcement must be provided to comply with the code
requirements.
A st, min = 0.0025 A = 0.0025 (558) = 1.4 in. 2
n
2
Provide four No. 6 Grade 60 bars, A = 1.77 in. . Calculate fP from Eq. (5.12) with
st
n
2
A = 1.77 in. .
st
′
(
]
)
φP = φ(.080 )[.080 f A − A + f A C
n m n st y st P
= 0.9(0.80)[0.80 (1.8)(558 – 1.77) + (60)(1.77)](0.937)
= 612 kips > P = 400 kips
u
b. Moment capacity:
The column has an excess axial load capacity of 612−400 = 212 kips. Therefore, it can
resist some bending moment. Because the column square, it is symmetrical about both
axes so it would resist the same moment about either axis. As a result of bending, two
No. 6 bars along one face of the column would be in compression, and the other two bars
in tension. Figure E5.8B shows the strains and forces in the column due to bending.
e m = 0.0025
23.625'' 0.80 f’ m

e′ s 3'' a
C s 2
a = C m
4#6 C
3''
#3@12 in. 23.625'' NA d – a
3'' 2

s
e ≥ e y T = A s f y
3'' 3''
FIGURE E5.8B Strains and forces in the column due to bending.

Compression force in masonry [Eq. (4.6)]:

′
.
C = 0 80. f ab = 0 80 1 8. ( . )( a 23 625)( . ) = 34 02 akips
m
m
(Note: Theoretically, the area of compression reinforcement should be subtracted
from the masonry area in compression. However, this precision is ignored here as it
would not affect the final result appreciably. This procedure is discussed in the next
section).
By definition, a = 0.8c [Eq. (4.5a)]; therefore,
C = 34.02a = 34.02(0.8c) = 27.216c kips
m