Page 298 - Design of Reinforced Masonry Structures

5.18 CHAPTER FIVE

From Eq. (5.11)
′
)
(
φP = φ(.080 )[.080 f A − A + f A C
]
n m n st y st P
= 0.9(0.80)[0.80 (1.5)(558 – 8) + (60)(8.0)](0.876)
= 719 kips
Check loads on column. Calculate factored loads.
Load Combinations:
1. U = 1.4D = 1.4(200) = 280 kips
2. U = 1.2D + 1.6L = 1.2(200) + 1.6(300) = 720 kips > 280 kips
P = 720 kips (governs)
u
fP = 719 kips ≈ 720 kips OK
n
Check longitudinal reinforcement compliance with the code.
.
ρ = A st = 80 . = 0 0143
A n 558 0 .
r max = 0.04 r min = 0.0025 r provided = 0.0143
0.0025 < 0.0143 < 0.04 OK
The longitudinal reinforcement complies with the code requirements.
fP = 719 kips ≈ P = 720 kips
u
n
The column can support the imposed service loads.

Example 5.3 Nominal strength of a square brick masonry column.
An 18 × 18 in. brick masonry column hav-
18'' ing an effective height of 20 ft is reinforced with
eight No. 8 Grade 60 bars (Fig. E5.3). It carries
a service dead load of 250 kips and a service
live load of 200 kips. ′ f = 2500 psi. Determine
m
fP for the column and check if the column can
n
support the imposed service loads.
18''
A = 8#8 Solution
s
Gr. 60
Given: An 18 × 18 in. brick column, h = 20 ft,
2
′ f = 2500 psi, A = 6.28 in. (eight No. 8 Grade
st
m
60 bars), D =250 kips, L = 200 kips.
Check dimensions and h/t ratio for code
compliance.
FIGURE E5.3
Nominal column width = 18 in. > 8 in. OK
Nominal column depth = 18 in. < 3(18) = 54 in. OK
h/t = 20(12)/18 = 13.33 < 30 OK
For an 18 × 18 in. brick column,
A = (18)(18) = 324 in. 2
n
2
A = 6.28 in. (eight No. 8 bars)
st