Page 336 - Design of Reinforced Masonry Structures

5.56 CHAPTER FIVE

The negative root (−4.41 in.) is ignored as it is meaningless in this problem. With
c = 3.75 in., the compression reinforcement would be in tension because c = 3.75 in.
< d′ = 3.8 in. Therefore, c should be determined from the following equation [see Eq. (4.121),
for derivation of this equation]:
) − ′ A ε
(.064 ′ fb ) +c 2 ( ′ A ε E − A f c E d ′ = 0 (4.121 repeated)
m s m s s y s m s
2
The above equation is a quadratic of the form: Ax + Bx + C = 0, which can be solved
for x (= c)
−± B − 4 AC
2
B
x =
2 A
where x = c
A = 0 64. f b ′ = 0 64 2 0 15 625. ( . )( . ) = 20 kiips/in.
m
B = A′ε E − A f
sm s s y
= 1 2 0 0025 29 000.( . )( , ) −112 60)
.(
= 15 kips
C = ′ A ε E d ′ = 1 2 0 0025 29 0000 3 8)( . ) = 330 6 k-in.
,
.(
)
.
(
.
sm
s
Substituting in the above equation, we obtain
)(
15
(
c = −15 ± () 2 + 4 20 330 6 . )
(
220)
.
.
=+371in. − 446 iin.

c = 3.71 in. < d′ = 3.8 in. OK
a = 0.8c = 08 3711) = 2 97 in.
.(.
.
(
.
.
.
(
C = 080 ′ f ab = 080 20 297 15.6625) = 74 25 kips
)
.
.
(
)
m m
Because c < d ′, the strain in compression reinforcement is tensile. From Eq. (5.44)
( ) ( 38 )
.
′ d
ε
′ = c − ε m = 371 −1 0 0025(. ) = 6 06 10. ( −5 )) < ε = 0.00207
1
s
y
.

Therefore, the strain in the compression reinforcement is calculated from Hooke’s law as
f s ′= ′Eε s s = (.606 ×10 −5 )(29 ,000 ) = .176 ksi (tensile
)
C = ′′= 12 176. ( . ) = 211kips (tensile)
A f
.
s
s s
T = 72 kips (as before)
Equilibrium in the vertical direction requires that ΣF = 0. Thus,
y
C + C – T = 0
m
s
C + C – T = – 2.11 + 74.25 – 72.0 = 0.14 kip ≈ 0
m
s