Page 171 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.35

TABLE 4.6 Values of r b , 0.75 r b , and r max for Practical Combinations of ′ f and f y for Clay
m
Masonry (e = 0.0035).
mu
f y = 60,000 psi f y = 40,000 psi

r b r max r b r max
Eq. Eq. 0.75 Eq. Eq. 0.75
f ′ m f ′ m
psi (4.69) 0.75r b (4.76) r /r max (4.69) 0.75r b (4.76) r /r max psi
b
b
1500 0.0101 0.0075 0.0085 0.889 0.0172 0.0129 0.0151 0.856 1500
1800 0.0121 0.0091 0.0102 0.889 0.0207 0.0155 0.0181 0.856 1800
2000 0.0134 0.0101 0.0113 0.889 0.0230 0.0172 0.0201 0.856 2000
2500 0.0168 0.0126 0.0141 0.889 0.0287 0.0215 0.0251 0.856 2500
3000 0.0201 0.0151 0.0170 0.889 0.0344 0.0258 0.0302 0.856 3000
3500 0.0235 0.0176 0.0198 0.889 0.0402 0.0301 0.0352 0.856 3500
4000 0.0268 0.0201 0.0226 0.889 0.0459 0.0344 0.0402 0.856 4000
4500 0.0302 0.0226 0.0254 0.889 0.0516 0.0387 0.0453 0.856 4500
5000 0.0335 0.0251 0.0283 0.889 0.0574 0.0430 0.0503 0.856 5000
5500 0.0369 0.0277 0.0311 0.889 0.0631 0.0473 0.0553 0.856 5500
6000 0.0402 0.0302 0.0339 0.889 0.0689 0.0516 0.0603 0.856 6000

An approximate value of the minimum area of reinforcement required for masonry
beams can be calculated by equating the expression for 1.3 times the nominal cracking
moment (1.3M ) to the nominal strength of a beam as follows.
cr
⎛ bh ⎞
2
M = f r ⎜ ⎟ (4.53 repeated)
cr
⎝ 6 ⎠
Assuming h = 1.2d, the nominal cracking moment can be expressed as
⎛ b(. 12 d ⎞
2
)
M = f r ⎜ ⎝ 6 ⎟ ⎠ = .024 fbd 2
cr
r
. 13 M = ( . )13 ((.0 24 fbd 2 ) = . 0 312 fbd 2
cr r r
For a reinforced masonry beam, the nominal strength can be expressed from Eq. (4.12):
⎛ a⎞
A f d −
M =
n s y ⎜ ⎝ ⎟ (4.12 repeated)
⎠ 2
Assuming (d – a/2) ≈ 0.95d, Eq. (4.12) can be expressed as
M = 0.95A f d
n
s y
Equating M to 1.3M yields
cr
n
0.95A f = 0.312f bd 2
r
s y
whence
A ⎛ f ⎞
s = 0 3284. ⎜ r ⎟
bd ⎝ f ⎠
y