Page 335 - Design of Reinforced Masonry Structures

COLUMNS 5.55

The strain in the tensile reinforcement is equal to yield strain, therefore
T = A f = 1.2(60) = 72 kips
s y
Equilibrium in the vertical direction requires that ΣF = 0. Thus,
y
C + C – T – P = 0 (5.39 repeated)
s
m
P = C + C – T = 54.46 + 217 – 72.0 = 199.5 kips
m
s
P = PC = (199.5)(0.793) = 158.2 kips
P
n
fP = 0.9(158.2) =142.4 kips
n
Taking moments of C , C , and T about the centroidal axis of the column, we obtain,
s
m
h
h
ha
−
Td −
+
−
d′ +
M = C s( ) C m( ) ( )
n
2 2 2 2
.
.
.
−
= 54 46 ( 23 625 − − 3 8 . ) + 217 ( 23 625 86 . ) ( . − 23 625 ))
.
+ 72 19 825
2 2 2 2
+
+
.
.
= 436 36 1630 21 576 9
.
.
= 2643 47k-in. = 220 3k--ft
.
.
.
φM = 0 9 2643 47) = 2379 1 k-in. = 198 3. k-inn.
.(
n
Point 12: Zero axial load: fP = 0
n
In the absence of an axial load, the column behaves as a doubly reinforced beam. The
location of neutral axis in this case is determined from Eq. (5.34):
f A
)c
(.064 ′ fb )c 2 + ( ′ A ε E s − A f − . 080 ′′ − ′ A εε Ed′ = 0 (5.34 repeated)
s y
s m
m
s m
s
s
m
Equation (5.34) is a quadratic of the form: Ax2 + Bx + C = 0, which can be solved for
x [= c in Eq. (5.34)]:
−± B − 4 AC
2
B
x =
2 A
where x = c
A = 0 64. f b ′ = 0 64 2 0 15 625. ( . )( . ) = 2 0 kkips/in.
m
B = A′ε E − A f − 080. f A ′′
sm s s y m s
= 12000.( . 2 25 29 000) − 1 2 60) − 0 80 2 0 1 2)
,
(
.
)(
.
.
.
(
)(
= 13 08 kipps
.
C = A′ε E d′ = 1 2 0 0025 29 000 3 8.( . )( , )(.) = 330..6 k-in.
sm s
Substituting in the above equation, we obtain
2
20c + 13.08c – 330.6 = 0
whence
.
)(
(
.
c = −13 08 ± ( 13 08) 2 + 4 20 330 6 . )
220)
(
=+375in..− 441. in.
.