Page 412 - Design of Reinforced Masonry Structures

WALLS UNDER GRAVITY AND TRANSVERSE LOADS 6.63

First iteration:

5Mh 2 5 M( − M ) h 2
δ = cr + ser cr
ser
48EI 48EI cr
m
mg
−
×
(
12
.
.
= 518.. )(96 20 2 )( ) + ( 5 2882 1896))(20 12 ) 2
48 (1800 )(443 . ) 48 (1800 )(20 .54 )
3
= . 0 143 1 .6
+
= . 1 743 in.. > 000. h = 168 in. NG
.
Therefore, increase wall reinforcing so as to increase I which would reduce d . Try
ser
cr
2
No. 4 vertical bars @ 16 in. on center; A = 0.15 in. (Table A.23).
s
P + A f 1 81 + 0 15 60)
. (
.
A = u s y = = 018. in. 2
se
f y 60
n = 16.11 (Table A.14)
.
ρ = A se = 018 = 0 0039.
bd () ( .
12 3 81)
rn = (0.0039)(16.11) = 0.063
From Table A.15, nr = 0.063, k = 0.2975
c = kd = (0.2975)(3.81) ≈ 1.13 in. < 1.25 in. (shell thickness)
Therefore, the centroidal axis lies in the shell and the section can be treated as a rect-
angular section.
−
I = bc 3 + nA d c) 2
(
cr
3 se
12 1 13)
= () ( . 3 + 16 11 0.118 3 81 1 13)− . 2
(.)(
)( .
3
= 26 6 . in. 4
δ = 5Mh 2 + 5 M( ser − M ) h 2
cr
cr
ser
48EI 48EI
mg m cr
−
×
.
12
.
(
= 518.. )(96 20 2 )( ) + ( 5 2882 1896))(20 12 ) 2
48 (1800 )(443 . ) 48 (1800 )(26 . )
3
6
+
= . 0 143 1 .236
= . 1 379 inn. < 000. h = 168 in. OK
.