Page 210 - Design of Reinforced Masonry Structures

4.74 CHAPTER FOUR

whence
c = +3.01 in., −3.9 in.

Use c = 3.01 in.
a = 0.80c = 0.80(3.01) = 2.41 in.
′
.
(.
.
C = 0 80 f ab = 0 80 2 0 2 41 7 625) = 29 4 kipps
)
(.
(.)
.
m
m
⎛ d ′ ⎞ ⎛ 20 ⎞
.
ε ′ = ⎜ 1 − ⎟ ε = 1 − (. = 0 0. 00084 < ε = .
0 0025)
0 002
s ⎝ c ⎠ m ⎝ 301 ⎠ y
.
Hence, the compression reinforcement has not yielded, and stress in it is calculated
from Hooke’s law:
ε
f ′= ′E = (.0 00084 )(29 ,000 ) = 24 .36 ksi < f = 60 ksi
s s s y
⎛ ⎡ d′ ⎞ ⎤
−
C = A′ ⎜ ⎢ 1 − ⎟ ε E − 0 80 f ′ = 0 79)[224 36 0 80 2 0)] = 17 98 kips
.
.
( .
.
(
.
.
m ⎥
s s ⎝ c ⎠ m s
⎣ ⎦
T = A f = (0.79)(60) = 47.4 kips
s y
Check equilibrium using Eq. (4.111):
C + C – T = 0
s m
29.4 + 17.98 – 47.4 = −0.02 kip ≈ 0 OK
Calculate M from Eq. (4.116):
n
⎛ a ⎞
M = C m ⎝ d − ⎠ 2 + Cd d′)
−
(
n
s
⎛ 241 ⎞
.
= 29 4 20 − +17.998 20 2 0( − . )
.
⎝ 2 ⎠
= 876 21 k-in. = 73 02 k-ft
.
.
fM = (0.9)(73.02)= 75.72 k-ft
n
−
75 72 64 2 .
.
Percentage increase in moment strength = ( 100) ≈ 18%
64 2 .
4.13 LINTELS
4.13.1 General Description
A lintel is simply a transversely loaded beam placed over an opening in a wall to support
the loads above (Fig. 4.14). These loads might consist of distributed gravity loads (dead
load from the wall above plus the service loads from the floors and the roof). The load from
the wall might be due to the entire height of wall above (Fig. 4.15c) or that from the wall
portions contained between two or more openings at different levels in a wall (Fig. 4.15a