Page 322 - Design of Reinforced Masonry Structures

5.42 CHAPTER FIVE

equal to its yield strength f , whereas the stress in compression steel is determined from
y
the strain distribution diagram assuming compressive strain in the face of the column
equal to e (0.0025 for concrete masonry and 0.0035 for clay masonry).
m
Referring to Fig. 5.12, the following values of forces acting on the column are
obtained (governing equations from Chap. 4 are presented here for completeness; refer
to Section 4.5 for the details of their derivations):
Compression force in masonry Cm:
′
C = 080. f ab (5.21, 4.6 repeated)
m m
where a = depth of the compression block in the column cross section.
Compression force in steel C :
s
A f
C = ′′ (5.22, 4.104 repeated)
s
s s
where ′ A = area of compression reinforcement
s
′ f = stress in compression reinforcement (≤ f ) as calculated from Hooke’s law:
s y
′ f = ′ ε E ≤ f (5.23, 4.105 repeated)
s s s y
( )
′ d
ε ′ = 1 − ε (5.24, 4.107 repeated)
s m
c
( )
f ′= ′Eε = 1 − ′ d ε E (5.25, 4.108 repeated)
s s s m s
c
s( )
C = A′ − d′ ε E (5.26, 4.109 repeated)
1
s
c m s
Equation (5.21) gives an overestimated value of the force in masonry in com-
pression because the area of compression reinforcement, ′ A , was not deducted from
s
the compression area of masonry, the actual area of masonry in compression being =
ab – ′ A . To compensate for this overestimation, the force in compression reinforcement
s
can be expressed as
s ( )
⎡
⎤
C = A′ 1 − d′ ε m E − 0 80 f ′ (5.27, 4.110 repeated)
.
⎣ ⎢
⎦ ⎥
s
m
s
c
Tension force in tension reinforcement T:
T = A f (5.28, 4.7 repeated)
s y
Equating sum of all horizontal forces to zero for equilibrium in the horizontal direc-
tion, we have,
C + C – T = 0 (5.29, 4.111 repeated)
s
m
Substitution of values of various parameters in Eq. (5.29) yields
⎡ ( ) ⎤
′ d
=
.
080 ′ fab + ′ A s ⎣ ⎢ 1− c ε m E s − 080 ′ −f m ⎦ ⎥ A f = 0 (5.30, 4.112 repeated)
.
s y
m
In Eq. (5.30), the value of a is unknown. By definition,
a = 0.80c (5.31, 4.5a repeated)