Page 207 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.71

Compression force in steel C :
s
A f
C = ′′ (4.104)
s s s
where ′ A = area of compression reinforcement
s
′ f = stress in compression reinforcement and calculated from Hooke’s law:
s
ε
f s ′= ′ E s ≤ f y (4.105)
s
where ′ ε = strain in compression reinforcement
s
E = modulus of elasticity of reinforcing steel = 29,000 ksi
s
Note that stress in compression reinforcement, ′ f , is limited to a maximum of yield
s
strength of reinforcing steel, f . The strain in compression reinforcement is obtained from
y
similar triangles of strain distribution diagram (Fig. 4.13) in which the neutral axis is
located at a distance c from the compression face of the beam.
′ ε s = cd =− ′ d
− ′
ε c 1 c (4.106)
m
whence
⎛ d ′ ⎞
ε ′ = ⎜ 1 − c ⎟ ε (4.107)
⎠
⎝
m
s
Substituting Eq. (4.107) in Eq. (4.105), we obtain
⎛ d ′ ⎞
ε
f ′= ′E = ⎜ 1 − ⎟ ε E (4.108)
s s s ⎝ c ⎠ m s
Substituting Eq. (4.108) in Eq. (4.104), we obtain the force in compression steel:
⎛ d ⎞ '
C = A′ 1 − c ⎠ ε m E (4.109)
⎝
s
s
s
Equation (4.6) gives an overestimated value of the force in compression area of masonry
because the area of compression reinforcement, ′ A , was not deducted from the compres-
s
sion area of masonry (actual area of masonry in compression = ab − ′ A ). To compensate
s
for this overestimation, the force in compression reinforcement can be expressed as
C = A′ ⎜ ⎢ ⎛ ⎡ 1 − d′ ⎞ ⎟ ε E − 0 80. f ′ m ⎥ ⎤ (4.110)
⎣ ⎦
s s ⎝ c ⎠ m s

Tension force in tension reinforcement T:
T = A f (4.7 repeated)
s y
Equating sum of all horizontal forces to zero for equilibrium in the horizontal direction,
we have,
C + C – T = 0 (4.111)
m s
Substitution of values of various parameters in Eq. (4.111) yields
⎛ ⎡ d ′ ⎞ ⎤
f
.
080 ′ fab + ′ ⎜ ⎢ 1− ⎟ ε E − 080 ′ − AAf = 0 (4.112)
⎥
A
.
⎣ c ⎦
m s ⎝ ⎠ m s m sy