Page 306 - Design of Reinforced Masonry Structures

5.26 CHAPTER FIVE

h (20 )( )
12
= = 35 .14 < 99
r . 683
h
For ≤ 99,
r
2
⎤
⎡ ( ) ⎥ ( ) 2
h
35 14
.
C = 1 − =− = 09 .337 (5.13 repeated)
1
⎢
P ⎣ 140 r ⎦ 140
Alternatively, C can be found directly from Table A.16 as follows:
P
)( )
h (20 12
= = 10 .16
t 23 .625
By interpolation from Table A.16, for h/t = 10.16
C = 0.939 – (0.939 – 0.926)(0.16) = 0.937
P
φP = φ(.80 )[.80 f A − A + f A C (5.12 repeated)
′
(
0
)
0
]
n m n st y st P
620 = 0.9(0.80)[0.80 (1.8)(558 – A ) + (60)(A )](0.937)
st
st
A = 1.97 in. 2
st
Try four No. 7 bars,
2
2
A = 2.41 in. > A st, reqd = 1.97 in. OK
st
2
2
(Alternatively, try eight No. 5 bars, A = 2.45 in. > A st, reqd = 1.97 in. OK)
st
Check longitudinal reinforcement compliance with the code.
.
ρ = A st = 241 = 0 0043.
A 558
n
r max = 0.04 r min = 0.0025 r provided = 0.0043
0.0025 < 0.0043 < 0.04 OK

The longitudinal reinforcement complies with the code requirements.
3
Lateral ties: Provide /8-in.-diameter lateral ties. Tie spacing should be the smallest of
1. Sixteen times the of longitudinal bar diameter = 16 ( /8) = 14 in. (governs).
7
3
2. Forty eight times the lateral tie diameter = 48( /8) = 18 in.
3. Least column cross-sectional dimension = 24 in.
Provide No. 3 lateral ties spaced at 14 in. on center.
Final design: Provide a nominal 24 × 24 in. CMU column with four No. 7 Grade
60 vertical bars with /8-in.-diameter ties spaced at 14 in. on center. Code requires top
3
and bottom ties to be within one-half of required spacing at top and bottom. Provide
the top and bottom tie at 7 in. See Fig. E5.6 for details.