Page 539 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.101

From observation,

e > e = 0.0025 > e = 0.00207
m
y
s3
Strains in bars 3 through 10 have strains greater than the yield strain. Therefore,
stress in each of those bars equals f (= 60 ksi). Calculate forces in bars 1 and 2 based
y
on strains in those bars.
Bar 1:
f s ′= ε s E s = (.0 00206 )(29 000 ) = 59 .74 ksi (compresssive)
,

C ′ = ′′=A f . 044 5974 ) = 2629 kips compression)
.
m
( .
(
s s
s
Bar 2:
f = ε E = ( 0 00017 29 000 = 4 93 ksi (tensilee)
.
,
)
)(
.
s2 s2 s
T = 0.44 (4.93) = 2.2 kips (tension)
Compression force in masonry:
C = 0.80 ′ fat = 0.80 (2.0)(18)(7.625) = 219.6 kips
m
2
For 8 No. 6 bars in tension, A = 3.53 in. .
s
T = 3.53(60) = 211.8 kips
Total force in tension, ΣT = 2.2 + 211.8 = 214 kips
Check equilibrium: ΣF = 0
y
C + C − T = 219.6 + 26.29 − 214 = 31.9 kips ≠ 0
s
Therefore, a should be smaller than 18 in. as assumed.
Trial 3: Assume a = 14 in.
c = a = 14 = 17 5.in.
080 080
.
.
Calculate forces in masonry and reinforcing bars (as in Trial 2) based on strain com-
patibility. The strain distribution is shown in Fig. E7.17C.
⎛ 13 5 ⎞
.
0 002007 ( f < f )
ε = ⎜ ⎝ 17 5 ⎠ ⎟ (. ) 0 00193 < ε = . s y
0 0025 = .
y
s1
.
65 ⎞
⎛ .
ε = ⎜ ⎝ 17 5 ⎠ ⎟ (. ) 0 00093 < ε = . s y
0 002007 ( f < f )
0 00205 = .
y
s2
.
From observation,
e > 0.0025 > e = 0.00207 ( f = f )
y
y
s3
s
Stresses in all remaining bars in tension would be equal to f = 60 ksi.
y