Masonry Magazine August 1992 Page. 27
TABLE 3
Reinforcement
Pilaster and Pier Reinforcement
| Rebar # | Bar Diameter, in. | Bar Area, in.2 |
|---|---|---|
| 3 | 0.375 | 0.110 |
| 4 | 0.500 | 0.196 |
| 5 | 0.625 | 0.307 |
| 6 | 0.750 | 0.442 |
| 7 | 0.875 | 0.601 |
| 8 | 1.000 | 0.785 |
| 9 | 1.128 | 0.999 |
| 10 | 1.270 | 1.267 |
| 11 | 1.410 | 1.561 |
Joint Reinforcement¹
| Wire Size | Gage | Diameter, in. | Area, in.2 |
|---|---|---|---|
| W1.1 | 11 | 0.121 | 0.011 |
| W1.7 | 9 | 0.148 | 0.017 |
| W2.1 | 8 | 0.162 | 0.021 |
| W2.8 | 3/16 | 0.188 | 0.028 |
| W4.9 | 1/4 | 0.250 | 0.049 |
¹Area of steel listed is for one wire.
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Step 5
Check compressive stress in masonry and the tensile stress in steel.
Fь = 4/3(1/3f'm) = 1600 psi
fm = fa + fb = 81 psi + 744 psi
= 825 psi < 1600 psi OK
F, = 4/3(24 ksi) = 32 ksi > 29.3 ksi OK
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Step 6
Calculate shear stress.
V = 200 lb
fv = = 4.7 psi
bjd (12 in.)(0.934)(7.63 in./2)
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Step 7
Check shear stress.
Fv = 4/3 √f'm = 80 psi, 4/3(50 psi) = 67 psi maximum
Thus, Fv = 67 psi > 4.7 psi OK
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Step 8
Does not apply.
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Step 9
Neglect the soil pressure resistance to overturning. Calculate overturning and resisting moments, and the safety factor on overturning (Fig. 2). Note: Sliding must also be considered. Sliding is a function of the soil conditions, and is beyond the scope of this Technical Notes.
Mo = (W)(h)(moment arm about toe of footing)
= (20 psf)(10 ft)(5 ft + y)
Mr = [P + (footing unit weight)(xy)](x/2)
= [730 lb + (150 pcf)(xy)](x/2)
Try a 2 ft x 4 ft footing
Mo = 1400 ft-lb Mr = 3860 ft-lb
Mr
F.S.= = 2.8 > 2 OK
Mo
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fa = P = 730 lb = 81 psi
bkd (12 in.)(0.198)(7.63 in./2)
`
fs = M = 1000 ft-lb (12 in./ft)
Ajd (0.115 in.) (0.934)(7.63 in./2)
= 29.3 ksi
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Noise Barrier Wall Design Example #1
Hollow Brick Cantilever Wall
FIG. 3
5