Page 330 - Design of Reinforced Masonry Structures

5.50 CHAPTER FIVE

Point 1: Pure axial load, no bending moment
The axial load capacity can be determined from Eq. (5.12):
′
φ
.
)
φP = 080 080 f A − A + f A C
.
[
]
(
n m n st y st P
The value of C for a nominal 16 × 24 in. CMU column with an effective height of 24 ft
P
was determined as 0.793 in Example 5.9 (calculations not repeated here).
For a nominal 16 × 24 in. CMU column, A = (15.625)(23.625) = 369.14 in. 2
n
φ
[
.
φP = 080 080 f ′( A − A + f A C P
.
]
)
m
n
st
y
st
n
−
+
)
0
.
(
2
0
2
.
0
= 080 09))[ .80 ( . )(369 .14 2 .41 ) (60 )( .41 )]( .793)
= 417 .6kips
Point 2: Zero strain on tension side of column when a moment is applied
In this case, the neutral axis is assumed to be located on the tension face of the
column so that c = h = 23.625 in. (depth of column). Figure E5.11B shows the strain
distribution and forces acting on the column. The strain in the compression reinforce-
ment is calculated from the strain diagram as:
Fig. E5.11B Strain distribution and forces acting on the column when strain is zero
on the tension face of the column.
( ) (
ε
>
ε
′ = 1 − ′ d ε = 1 − 38 . ) (. = 0 0021 ε = 0 00207
.
.
0 0025)
s m y
.
c 23 625
15.625" e m = 0.0025 0.55 f ʹ m
3.8" eʹ s 3.8" s
C S
2
4#7
d = 19.825" C m
23.625" c = 23.625" s
2 M n
e T C S = A S F S
3.8"
Tension side
of column
FIGURE E5.11B
Therefore, the stress in compression reinforcement, ′ f = f = 60 ksi
y
s

Force in compression reinforcement is calculated as
C = ′( f ′ =) 1 . [60 − .80 ( . )] = 70 .08kips
2
0
2
A f ′− .0 80
0
s s s m
Because c = h = 23.625 in. > d = 19.825 in., the tensile reinforcement is in slight com-
pression. The strain in tensile reinforcement is given by
T ( ) ( 19 825 )
.
d
<
0 0025)
)
.
.
ε = 1 − ε = 1 − (. = 0 0004 < ε = 0 00207(compressive
c m 23 625 y
.
Therefore, stress in tensile reinforcement is calculated from Hooke’s law:
f = e E = (0.0004)(29,000) =11.6 ksi (compressive)
T
T s