Page 182 - Design of Reinforced Masonry Structures

4.46 CHAPTER FOUR

From Eq. (4.13)
⎛ a ⎞
φM = φA f ⎜ d − ⎟
n s y ⎝ ⎠ 2
⎛ 333 ⎞ ⎞
.
= 09 10 60 20 − 2 ⎠ ⎟
.( . )(
⎜ )
⎝
.
= 990 1 lb-in . = 82 51 k-ft
.
φM = 82 51k-ft > M = 68 35k-ft OK
.
.
n u
Check the M / V d ratio.
u v
u
2 136 16)
u
V = wL = (. ) ( = 17 09kips
.
u
2 2
M (68 .35 )( )
12
u = = .24 > .10
Vd (.09 )(20 )
17
uv
Verify that reinforcement has yielded and e ≥ 1.5e . From Eq. (4.5):
s y
.
c = a = 333 = 416 in.
.
.
.
08 08
.
c 416
<
.
= = 0 208 0 538 (4.34 repeated)
.
d 20 0
.
Hence, steel has yielded and e ≥ 1.5e . Check that M is not less than 1.3M
s y n cr
(MSJC-08 Section. 3.3.4.2.2.2). 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
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)
( ) 2
24
( )
,
M = 200 9 = 172 800 lb-in. = 14 4 . k-ft
ct
6
1.3M = 1.3(14.4) = 18.72 k-ft
cr
φ
.
M = M n = 82 51 = 91 68 k-ft > 18 72 k-ft OK
.
.
φ 09 .
n
Check that r ≤ 0.75r .
b
.
ρ = A s = 10 = 0 0056. < 0 75. ρ = 0 0126. (Table 4.6) OK
bd () ( b
920)
Final design: Provide one No. 9 Grade 60 bar, with d = 20 in.
2
Commentary: We could have also used two No. 6 bars (A = 0.88 in. ) instead of one
s
2
No. 9 bar (A = 1.0 in. ) if necessary.
s