Page 166 - Design of Reinforced Masonry Structures

4.30 CHAPTER FOUR

From Eq. (4.13)

φM = φA f ⎛ d − a ⎞
n s y ⎝ ⎠ 2
⎛ 041 ⎞
.
= 09 011 60 20.( . )( ) −
⎝ 2 ⎠
= = 117 58 k-in . = 9 8 k-ft
.
.
Verify that reinforcement has yielded and e ≥ 1.5e From Eq. 4.5,
y
s
.
in
.
c = a = 041 = 051 .
.
.
08 08
c 051
.
.
= = 0 026 < 0 538 (Eq. 4.34)
.
d 20 0
.
Hence, steel has yielded and the assumption e ≥ 1.5e is valid.
s
y
φM = 9.8 k-ft
n
φ M 98
.
M = n = = 10 89. k-ft < 1 3. M = 16 64. k-ft
n φ 09 cr
.
The nominal strength of the beam (10.89 k-ft) is less than 1.3 times the cracking
moment of the beam (16.64 k-ft). Therefore, this beam does not meet the code
requirements for the nominal cracking moment strength of the beam.
b. Beam reinforced with one No. 4 Grade 60 bar.
2
b = 8 in., d = 20 in., A = 0.20 in. (one No. 4 bar) ′ f = 2500 psi, f = 60 ksi.
s m y
Assume that tension reinforcement has yielded and e ≥ 1.5e (to be verified
s
y
later). Calculate from Eq. (4.9) the depth a of the compression block.
Af
a = sy (4.9 repeated)
.
080 fb ′ m
a = (. 020 )(60 ) = . 075 in
.
( . )( .
. 080 25 800 )
From Eq. (4.13)
⎛ d − a ⎞
φM =
n φA f ⎝ ⎠ 2
s y
⎛ 075 ⎞
.
= 09 020 60 20 −
)
)(
.( .
⎝ 2 ⎠
.
= = 211 95 k-in. = 17 66 k-ft
.
Verify that reinforcement has yielded and e ≥ 1.5e . From Eq. (4.5),
s
y
.
c = a = 075 = 094 in
.
.
08 08
.
.
.
c 094
= = 0 047 < 0 538 (4.34 repeated)
.
.
.
d 20 0