Page 245 - Design of Reinforced Masonry Structures

4.108 CHAPTER FOUR

From Fig. 4.29, find coefficient for T, resultant of tensile stresses, and multiply
by wl (for continuous span beams), where w = load on wall per linear foot.
Midspan:
T = 0.12(wl) = (0.12) (5000) (30) = 18,000 lb
Support:
T = (0.23) (5000) (30) = 34,500 lb

Calculate the area of tension reinforcement:
,
Midspan: A = T = 18 000 = 075 in.
2
.
s
,
F s 24 000
2
2
Provide two No. 6 bars, A = 0.88 in. > 0.75 in. OK
s
Assume that reinforcing bar is located near the bottom of beam,
d = (15) (12) – 4 = 176 in.
Support:
,
A = T = 34 500 = 144 in.
2
.
s
F s 24 000
,
2
2
Provide two No. 8 bars, A = 1.57 in. > 1.44 in. OK
s
Locate the bars near the top of the beam.
c. Calculate the shear stress and required shear reinforcement. Assume that critical
section is located at 0.15 l from the face of support, where l clear span between
n
n
supports.
0.15l = 0.15 (30 − 3) = 4.05 ft
n
At the critical section, the maximum shear (assuming three continuous spans) is
⎡ L ⎤ ⎡ 33 ⎤
405 15 .) = 54 750,
V = w ⎢ ⎣ 2 − (. + ⎥ ⎦ = ( 5000) ⎢ ⎢ ⎣ 2 − (. + ⎥ ⎦ lb
4 05 1 5 .)
The shear stress at the critical section is
,
v = V = 54 750 = 26 8 . lb/in.
2
bd (. )(
11 625 176)

2
.
(
.
Allowable shear stress = 1 1. f m ′ = 1 1 2000) = 49 2 lb/in. (without shear
reinforcement)
Therefore, no special shear reinforcement is required.
2
Minimum wall reinforcement = 0.002(11.625)(12) = 0.28 in. /ft *
*
( Total in both direction; no more than / 3 in one direction)
2
Provide No. 5 bars @ 24 in. o.c. in horizontal direction, A = 0.15 in. 2
s
Provide No. 5 bars at 24 in. o.c. in vertical direction, A = 0.15 in. 2
s
Total reinforcement,
2
2
A = 0.15 + 0.15 = 0.30 in. > 0.28 in. OK
s