Page 327 - Design of Reinforced Masonry Structures

COLUMNS 5.47

The strain in compression reinforcement ( ′ ε ) can be calculated similarly from similar
s
triangles of strain distribution diagram:
′ ε s = cd ( )
− ′
′ d
−
ε c = 1 c
m
whence
( )
ε ′ = 1 − ′ d ε (compressive (5.42)
)
s m
c
Case 2: Zero strain in tension reinforcement, c = d (Point 3)
Figure 5.14 shows the column cross section, and the strain distribution and force diagrams cor-
responding to zero strain in the tension reinforcement. The strain in the compression reinforce-
ment is obtained from the strain distribution diagram for which Eq. (5.42) is valid. Thus,
( )
′ d
ε ′ = 1 − ε (compressive ) (5.42 repeated)
s m
c
b
e m = 0.0025
eʹ s dʹ
a/2
C s
a C m
h C = d

dʹ e T = 0 T = 0

FIGURE 5.14 Strain distribution and force diagrams for the case of zero strain in tension reinforce-
ment of the column.
Case 3: Neutral axis location such that d′ < c < d (strain in compression reinforcement
is compressive, Points 4 to 6, 7 to 11).
Figure 5.15 shows the column cross section, and the strain distribution and force dia-
grams corresponding to neutral axis positioned anywhere between the compression and
tension reinforcements. The strain in tension reinforcement is calculated from similar
triangles of the strain distribution diagram:
−
ε dc ( )
d
T = = −1
ε m c c

b ε m = 0.0025 0.8 fʹ
m
dʹ
εʹ s a/2
C s
c a C m
d
h
NA
dʹ ε T T = A S F S
FIGURE 5.15 Strain distribution and force diagrams for the case of neutral axis positioned
anywhere between the compression and tension reinforcements.