Page 332 - Design of Reinforced Masonry Structures

5.52 CHAPTER FIVE

15.625" e m = 0.0025 0.80 f ʹ m

3.8" eʹ s 3.8" s
C S 2
d = 19.825" C = 15.86" C m
23.625" s 2 M n
4#7
e T = 0 T = 0
3.8"
FIGURE E5.11C Strain distribution and forces acting on the column when strain is zero in the tension
reinforcement.

For equilibrium in the vertical direction, we must have ΣF = 0. Thus, noting that T = 0,
y
C + C – T – P = 0 (5.39 repeated)
s m
P = C + C – T = 68.4 + 396.5 + 0 = 464.9 kips
s m
P = PC = (464.9)(0.793) = 368.7 kips
P
n
fP = 0.9(368.7) = 331.8 kips
n
Taking moments of C , C , and T about the centroidal axis of the column, we obtain,
s
m
ha
h
M = C s( ) C m( )
−
−
d′ +
n
2 2 2
( 23 625 ) ( 23 625 15 86 )
.
.
.
= 68 4 . − 38. + 39665 . −
2 2 2
= 548 06 1539 41
+
.
.
= 2087 5 . k--in. = 174 k-ft
φM = 0 9 2087 5 . ) = 1878 8 . k-in = 1556 6. k-ft
.
.(
n
Points 4 to 6 and 7 to 10
Points 4 to 6 and 7 to 10 are arbitrarily selected at the following locations in the
column cross section:
Point c, in.
4 17.0
5 14.0
6 Balanced condition
7 10.0
8 8.0
9 6.5
10 5.0
11 4.0
12 Zero axial load