Page 262 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.125

9" the beam (Fig. E4.32). Calculate the moments of inertia
of (a) gross section, (b) cracked section. Assume ′ f =
m
1500 psi. (Note: This data is taken from Example 4.12
which illustrated flexural calculations.)
Solution
a. For a nominal 8-in. wide CMU, actual width b = 7.63
in. The moment of inertia of the gross section is
20" [Eq. (4.145)]
24" 3 3
763 24)
I = bh = (. ) ( = 8790in. 4
g
12 12
b. Since the centroid of tension reinforcement is located
at 20 in. from the compression face of the beam, the
2
effective depth d = 20 in. A = 1.0 in. (one No. 9 bar).

s
.
ρ= A s = 10 =0 0066.
#4 bd (. ) (
763 20)
6
(
FIGURE E4.32 E 29 10 )
.
n = s = = 27 62
E m 700 1500( )
0

nρ = 27.62(0.0066) = 0.1823
ρ
ρ
k = ( n ) 2 +2 n − nρ
0
2
0
)
0
= (.1823 ) 2 + (.1823 ) −(.1823
= =0 4484
.
(Alternatively, from Table A.15, by interpolation, k = 0.4484)
Calculate I from Eq. (4.146):
cr
1
2
3
I = b (kd) + nA (d − kd)
s
cr
3
kd = 0.4484(20) = 8.97 in.
d − kd = 20 – 8.97 = 11.03 in.
3
76
I = (. )( .897 ) 3 +(27 .62 1 0 11 03 2
)( . )( . )
cr
3
=5196 in. 4
6

Example 4.33 Gross moment of inertia of a transformed section.
Calculate the gross moment of inertia of the transformed section for the rectangular
beam described in Example 4.32.
Solution
2
Given: CMU beam, b = 7.65 in., h = 24 in., d = 20 in., A = 1.0 in. , ′ f = 1500 psi.
s m
Assuming that there is no compression steel, the transformed area of steel reinforce-
ment in the uncracked beam cross section is calculated as follows:
6
(
n = E s = 29 10 ) = 27 62
.
(
E 700 1500)
m