Masonry Magazine August 1992 Page. 29
TABLE 4
Area of Steel Per Vertical Foot of Wall (in.2)1
| Joint Reinforcement | | Brick Height |
|---|---|---|---|---|
| Size | Course | 2.25 in. | 2.75 in. | 3.63 in. |
| W1.1 | every | 0.050 | 0.041 | 0.033 |
| | alternate | 0.025 | 0.021 | 0.017 |
| | third | 0.017 | 0.014 | 0.011 |
| W1.7 | every | 0.077 | 0.064 | 0.051 |
| | alternate | 0.038 | 0.032 | 0.026 |
| | third | 0.026 | 0.021 | 0.017 |
| W2.1 | every | 0.095 | 0.079 | 0.063 |
| | alternate | 0.047 | 0.039 | 0.032 |
| | third | 0.032 | 0.026 | 0.021 |
| W2.8 | every | 0.126 | 0.110 | 0.084 |
| | alternate | 0.063 | 0.053 | 0.042 |
| | third | 0.042 | 0.035 | 0.028 |
| W4.9 | every | 0.221 | 0.184 | 0.147 |
| | alternate | 0.110 | 0.092 | 0.074 |
| | third | 0.074 | 0.061 | 0.049 |
1Area of steel listed is for one wire.
k = √(pn)2 + 2pn - pn
= √(0.0222)2 + 2(0.0222) - (0.0222) = 0.19
j = 1-k/3 = 0.936
Step 4: Calculate the masonry compressive stress and the steel tensile stress.
f=
2Mpx
2(703 ft-lb)(12 in./ft)
jkbd²
(0.936)(0.19)(12 in.)(3 in.)2
= 878 psi
f=
Mpx
(703 ft-lb)(12 in./ft)
Ajd
(0.077 in.2)(0.936)(3 in.)
= 39 ksi
Step 5: Check compressive stress in masonry and the tensile stress in steel.
F4/3(1/3f'm) = 1333 psi > 878 psi OK
F, 4/3(30 ksi) = 40 ksi > 39 ksi OK
Step 6: Calculate shear stress.
f=
Vpx
188 lb
bjd (12 in.) (0.936)(3 in.)
= 6 psi
Step 7: Check shear stress.
F = 4/3 fm=73 psi, 4/3(50 psi) = 67 psi maximum
Thus, F, 67 psi > 6 psi OK
Step 8: Assume the pier design will be governed by deflection limitations of the panel, not the required flexural strength of the pier.
Calculate the maximum vertical cantilever deflection the panel may undergo without exceeding the allowable tensile stress value for the panel found in Table 2. From Table 2:
F, 4/3(40 psi) = 53 psi
The maximum moment that may be applied to the panel within the allowable tensile stress is calculated as follows:
M =
Filg (53 psi)(1/12)(15 ft)(3.63 in.)3
C
(3.63 in./2)
= 1740 ft-lb
Calculate the maximum load per vertical foot, Wimax, which may be applied to the panel without exceeding Mmax in the panel. Two locations must be considered: 1) the bottom of the panel where the moment due to vertical cantilever deflection is a maximum and 2) the middle of the panel where the combination of Mpy and the moment due to vertical cantilever deflection is a maximum.
1)
Wmax=
2Mmax 2 (1740 ft-lb)
= 13.6 lb/ft
L2
(16 ft)2
2)
Wmax
8(Mmax-Mpy) = 8(1740 ft-lb-75 ft-lb)
L2
(16 ft)2
= 52 lb/ft
Therefore, wmax 13.6 lb/ft. Calculate the corresponding deflection of the panel.
Amax=
WmaxL4 (13.6 lb/ft)(16 ft)4(144 in.2/ft²)
81 Em
8[1/12(15 ft)(3.63 in.)3] (2.8 × 106psi)
= 0.096 in.
Based on Amax and the out-of-plane load on the pier, 2Vpx calculate the minimum required moment of inertia of the pier.
Ix req'd=
2VpL4
2(188 lb/ft)(16 ft)4(1728 in.3/ft³)
8EAmax
8 (29 x 106 psi)(0.096 in.)
= 1912 in.4
Try a W 24 x 76, I, 2100 in.4 > 1912 in.4 OK
Check the moment strength of the pier:
ФМ = 528 ft-kip > M 62.6 ft-kip OK
Check shear strength of the pier:
V = (0.9) (0.6) (fy)(Aweb)
= (0.9)0.6)(50 ksi) (21 in.) (0.44 in.) = 249 kips
V = 7.8 kips <249 kips OK
Step 9: Sliding and overturning resistance of the caisson
is a function of the lateral soil pressure and is beyond the scope of this Technical Notes.