Page 540 - Design of Reinforced Masonry Structures

7.102 CHAPTER SEVEN

17.5'' 6.5'' 24''

4'' e s3
e s2
e m = 0.0025 e s1

FIGURE E7.17C Strain distribution for Trial 3 iteration.

Bar 1:

f s ′= ε s1 E s = (0 00193. )(29 000, ) = 55 97. ksi (compression)
C = ′′= 044 (5597 ) = 2463. kips (compression)
.
.
A f
s
s s
Bar 2:
0 00093 29 000 =
f = ε E = (. ) ( , ) 26 97 ksi (tension)
.
s2 s2 s
Tensile force in Bar 2 = A f = 0.44(26.97) = 11.87 kips (tension)
s s2
Compression force in masonry:
C = 0.80 ′ fat = 0.80 (2.0)(14)(7.625) = 170.80 kips
m
Total compression force = C + C = 170.80 + 24.63 = 195.43 kips.
s
Total tensile force = 11.87 + 3.53(60) = 223.67 kips > (C + C )
s
Check equilibrium: ΣF = 0
y
C + C – T = 170.80 + 24.63 – 223.67 = -28.24 kips ≠ 0
s
Therefore, 14 in. < a < 18 in.

Trial 4: Assume a = 16 in.
c = a = 16 = 20in.
.
.
080 080
Calculate forces in masonry and reinforcing bars as before. The strain distribution
diagram is shown in Fig. E7.17D
⎛ 16 ⎞
0 0025 =
ε = ⎜ ⎟ (. ) 0 0020 < ε = 0 00207 ( f < f )
.
.
s1 ⎝ 20 ⎠ y s y
0 00205 =
ε = 4 (. ) 0 0005 < ε = 0 00207 ( f < f )
.
.
s2 y s y
20