Page 413 - Design of Reinforced Masonry Structures

6.64 CHAPTER SIX

Calculate moment due to new d u1

M = 28.82 + 1.81(1.379) = 31.32 k-in
u2
(
δ = 5 Mh 2 + 5 M ser − M ) h 2
cr
cr
se2
48EI 48EI
mg m cr
−
×
12
3
(
.
= 518.. )(96 20 2 )( ) + 5 (.132 1896))(20 12 ) 2
48 (1800 )(443 . ) 48 (1800 )(26 6
. )
3
= . 0 143 1 .549
+
= . 1 692 inn. ≈ 000. h = 168 in. OK
.
Calculate moment due to new d u2
M = 28.82 + 1.81(1.692) = 31.88 k-in.
u2
The new value of moment, 31.88 k-in., is very close to the previous value of 31.32 k-in.;
hence further iteration is not necessary. Use M = 31.88 k-in. Calculate the nominal
u
moment, M , from Eq. (6.17):
n
⎛ a ⎞
M = ( A f + P ⎜ ) ⎝ d − ⎟ ⎠ 2
s y
n
u
where
P + A f 1 81 + 0 15 60)
.
(
.
a = u s y = = . 0 56 in. < 1.25 in. (shell thickness)
(
)
.
(
.
.
008 fb ′ 080 20 12) )
m
Hence, the compression block lies in the shell, and the section can be treated as a
rectangular section.
⎛ a ⎞
M = ( A f + P ⎜ ) d − ⎟
n s y u ⎝ ⎠ 2
⎛ 056 ⎞
.
−
)] 3
1
= [.81 + ( . (60 ⎜ .81− ⎟
15
0
⎝ 2 ⎠
= 38 16 k-in.
.

fM = 0.9 (38.16) = 34.34 k-in. > M = 31.88 k-in. OK
u
n
Therefore, No. 4 vertical bars spaced at 16 in. on center horizontally are adequate for
resisting the seismic lateral load.
MSJC-08 (Section 1.17.3.2.6) requires a minimum reinforcement area (sum of verti-
cal and horizontal reinforcement) of 0.002bt of the gross sectional area of the wall for
structures in SDC D, with reinforcement no less than 0.0007bt of gross sectional area
in any one direction.
A s, min = 0.002bt = 0.002(12)(7.625) = 0.183 in. 2
0.0007bt = 0.0007(12)(7.625) = 0.064 in. 2