Page 187 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.51

From Eq. (4.13)

φM = φA f ⎛ d − a ⎞
n s y ⎝ ⎠ 2
.
= 09. (079 60 20. )( ) ⎛ − 194 ⎞
⎝ 2 ⎠
= = 811.8 k-in. = 67.65 k-ft
.
φM = 67 7 k-ft > M = 63..05 k-ft OK
n u
Check that reinforcement has yielded and e ≥ 1.5e , which should be verified from
y
s
Eq. (4.30). From Eq. (4.5)
.
c = a = 194 = 243 in.
.
080 080
.
.
.
c 243
.
.
= = 0 12 < 0 454 (4.30 repeated)
d 20
Therefore, steel has yielded and the assumption e ≥ 1.5e is satisfied.
s
y
φM = 67.7 k-ft
n
Check that M is not less than 1.3M (MSJC-08 Section. 3.3.4.2.2.2). Calculate M
cr
n
cr
from Eq. (4.53).
⎛ bh ⎞
2
M = f r ⎜ ⎟
cr
⎝ 6 ⎠
2
where h = the total depth of the beam. The modulus of rupture, f = 200 lb/in. for solid
r
grouted masonry in running bond laid in mortar S portland cement/lime or mortar cement
(MSJC-08 Table 3.1.8.2.1). The value of M remains the same as in Example 4.15, but
cr
calculations are repeated here for completeness.
( )
76
(. ) 24 2
3
M = (200 ) = 146 ,496 lb-in. = 12 . 21 k-ft
k
cr
6

1.3M = 1.3(12.21) = 15.88 k-ft
cr
φ M 67 7 .
M = n = = 75.22 k-ft > 1 3 . M = 15.88 k-ft OK
t
φ 09 .
n cr
Check that r ≤ 0.75r .
b
.
ρ = A s = 079 = 0 0052. < 0 75. ρ = 0 0175. (Table 4.5) OK
bd (. ) ( b
763 20)
Final design: Provide a nominal 8 × 24 in. CMU lintel with one No. 8 Grade 60
bar, and d = 20 in.