Page 260 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.123

Let A = equivalent area of masonry required to carry the same force as the area of ten-
m
sion reinforcement. Then, for equilibrium in the horizontal direction, we must have
Force in equivalent area of masonry = Force in steel reinforcement
P = P (4.140)
m
s
where P = force in masonry in the equivalent area of masonry
m
P = force in steel reinforcement
s
Since P = A f , and P = A f , we have, from Eq. (4.140)
m m
m
s
s s
A f = A f
m m
s s
which yields
⎛ f ⎞
A = ⎜ ⎝ f ⎠ ⎟ A = nA (4.141)
s
m
s
s
m
in which (f /f ) = n has been substituted from Eq. (4.139).
s m
Equation (4.141) shows that in a reinforced masonry section with tension reinforcement
area equal to A , the equivalent masonry area can be considered as being equal to nA , which
s
s
is assumed concentrated at the centroid of reinforcing steel, A .
s
As explained heretofore, the fundamental premise in analyzing reinforced masonry
beams is that masonry below the neutral axis (i.e., in the tensile zone of the cross section)
is cracked (i.e., it is structurally absent) and the tensile force in the beam is carried by only
the reinforcement. Thus, for the purpose of locating the neutral axis, the area of masonry
below the neutral axis is omitted, and only the transformed area equal to nA is assumed
s
present with its centroid at the level of centroid of reinforcing steel.
Referring to Fig. 4.36, the neutral axis of the transformed section is located at some
distance kd (< d) from the extreme compression fibers so that the cross-sectional area in
compression is bkd (i.e., width times depth of masonry in compression). The distance kd
can be determined from statics (by equating the moments of compression area bkd and the
equivalent tension area nA , taken about the neutral axis):
s
⎛ kd ⎞
−
()(bkd ⎜ ⎟ = nA (d kd (4.142)
)
)
⎝ 2 ⎠ s
where d = depth of section from extreme compression fibers to the centroid of tension
reinforcement
kd = depth of neutral axis measured from extreme compression fibers
k = neutral axis factor (< 1.0)
b = width of beam cross section
For design purposes, we modify Eq. (4.142) by introducing r, defined as the ratio of ten-

sion reinforcement A to the cross-sectional area of masonry bd [i.e., r = A /bd, Eq. (4.38)]
s

s
so that A = r bd. Substituting for A in Eq. (4.142) and simplifying yields
s
s

b kd) + 2 n kd −ρ 2 2 n bkd =ρ 2 0
2
(
or
2
k + 2nrk − 2nr = 0 (4.143)

2
Equation (4.143) is a quadratic in k of the form: Ax + Bx + C = 0, and can be solved for k:
k = (nρ + 2 nρ − nρ (4.144)
2
)