Page 331 - Design of Reinforced Masonry Structures
P. 331
COLUMNS 5.51
The force in the tensile reinforcement is
T = A f = 1.2(11.6) = 13.92 kips (compressive)
st T
The force in masonry in compression is given by Eq. (5.21):
′
C = 0 80. f ab = 0 80 2 0. ( . )( a 15 625)( . ) = 25 akips
m m
Substituting a = 0.80c = 0.80(23.625) = 18.9 in. in the above expression, we obtain
C = 25 (18.9) = 472.5 kips
m
For vertical equilibrium of forces, we must have ΣF = 0. Thus,
y
C + C + T – P = 0
s
m
70.08 + 472.5 + 13.92 – P = 0
whence
P = 556.5 kips
Multiplying the above value of P by C = 0.793 (for slenderness), we have
P
P = PC = (556.5)(0.793) = 441.3 kips
n
P
fP = 0.9(441.3) = 397.2 kips
n
Taking moments of C , C , and T about the centroidal axis of the column, we obtain,
s
m
h
ha
h
−
d′ +
−
M = C s( ) C m( ) ( )
−
Td −
n
2 2 2 2
.
.
−
.
= 70 08 ( 23 625 − − 3 8 . ) + 472 5 . ( 23 625 18 9 . ) −13 92 19 825. ( . − 23..625 )
2 2 2 2
.5 1116 −
= 561 + .3 111 .5
= 1566 .3k-in. = 130..5k-ft
φM = 0 .(1566 . ) 1409 .7 k-in. = 117 .5 k--ft
3 =
9
n
Point 3: Assume that the neutral axis passes through the tension reinforcement.
When the neutral axis passes through the tension reinforcement, the strain in this
reinforcement is zero. Therefore, T = 0.
c = d = 23.625 – 3.8 = 19.825 in.
a = 0.8c = 0.8(19.825) = 15.86 in.
′
.
C = 0 80. f ab = 0 80 2 0 15 86 15 625. ( . )( . )( . ) = 396 5kips
m
m
The strain distribution would be as shown in Fig. E5.11C. The strain in compression
reinforcement is given by
( ) (
.
0 0025)
ε ′ = 1 − ′ d ε m = 1 − 38 ) (. = 0 00202. < < ε y = 0 00207.
s
h 19 825
.
ε
Therefore, f s ′= ′E s = (.0 00202 )(29 ,000 ) = 58 .58ksi
s
=
.
0820
A f ′− .080
. [5858
C = ′( s f ′ =) 12 . − . ( . )] 6844kips
m
s
s