Page 185 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.49

Verify that reinforcement has yielded and that e ≥ 1.5e from Eq. (4.30). From
s
y
Eq. (4.5)
.
.
c = a = 388 = 485 in.
.
.
080 080
.
c 485
<
.
.
= = 0 243 0 454 (Eq. 4.30)
d 20
Therefore, steel has yielded and e ≥ 1.5e .
s
y
Check that M is not less than 1.3 M (MSJC-08 Section 3.3.4.2.2.2). Calculate
n
cr
M from Eq. (4.53).
cr
⎛ 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
r
solid grouted masonry in running bond laid in mortar S portland cement/lime or
mortar cement (MSJC-08 Table 3.1.8.2.1)
( )
76
3
(. ) 24 2
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 64 2
.
t
M = n = = 71 3. 3 k-ft > 1 3. M = 15 88. k-ft OK
φ 09
n cr
.
Check that r ≤ 0.75r .
b
.
ρ = A s = 079 = 0 0052. < 0 75. ρ = 0 0088. (Table 4.5) OK
b
bd (. ) (
763 20)
φM = 64 2 k-ft
.
n
Final design: Provide a nominal 8 × 24 in. CMU lintel with one No. 8 Grade 60
bar, and d = 20 in.
Example 4.16 Solve Prob. 4.15 assuming that high strength masonry with ′ f = 4000 psi
m
would be used. All other data remain the same as in Example 4.15.

Solution
From Example 4.15, M = 63.05 k-ft. From Eq. (4.48)
u
M bd 2
u = (M , k-ft, b and d in inches)
φ k 12 000 u
,
n
From Eq. (4.45)
φ f ′
1
0
(
φk = [ ω − .625 ω)]
n m