Masonry Magazine August 1992 Page. 30
Design Example #3: Reinforced Brick Masonry
Pier and 4 Inch Panel Noise Barrier Wall
Type S portland cement/lime mortar, f'm 3500 psi,
Em 3.0 x 106 psi
Wall dimensions shown in Fig. 5, running bond, full bedding. W1.7 ladder type joint reinforcement in every third course, brick height is 2.25 in.
Joint reinforcement: fy 80 ksi, E, 29 × 106 psi, n = 9.7
Pier reinforcement: fy 60 ksi, E, = 29 × 106 psi, n = 9.7
Loads: wind 20 psf. seismic 16 psf
panel weight 40 psf. pier weight = 120 pcf
The maximum permissible span between piers will be used based on the given amount of joint reinforcement.
Step 1:
Based on acoustical considerations, the wall height shall be 10 ft minimum.
Step 2:
Critical load combinations result in the following design values:
a) For the panel span between piers (L is in feet)(per foot of panel):
Mpx (1/8)(W)(L)2 = (1/8)(20 psf)L2 = 2.5 L2
Vpx (0.5)(W) (L) (0.5) (20 psf)L = 10 L
b) For the panel span between caissons, assume the panel is supported along its entire base by the ground.
c) For an interior brick masonry pier:
M = (0.5)(2)(Vpx)(h)2 = (0.5)(2)(10L)(10 ft)2 = 1000 L
V= (2)(Vpx)(h) = (2)(10L)(10 ft) = 200 L
P=(pier weight)(b)(t)(h) = (120 pcf)(15.63 in.) (19.63 in.)(10 ft) = 2557 lb
Step 3:
Calculate the maximum permissible spacing of piers with the given joint reinforcement based on the design moment, Mpx. Assume the distance from the extreme compression face to the joint reinforcement is 3 inches.
A, 0.026 in.2 per foot of panel (Table 4).
pn = An (0.026 in.2)(9.7) / bd (12 in.)(3 in.) =0.007
k = √(pn)2 + 2pn - pn =(0.007)2 + 2(0.007) - (0.007) = 0.111
j = 1-k/3 = 0.963
Mpx = A,Fajd = (0.026 in.2)(4/3)(30 ksi)(0.963)(3 in./12 in./ft) = 249 ft-lb
Set the maximum allowable moment equal to the applied moment to calculate the maximum permissible pier spacing.
Mpx 249 ft-lb = 2.5L2
Lmax 9.97 ft, try a pier spacing of 9 ft
Mpx 2.5 (9 ft)2 = 203 ft-lb
Step 4:
Calculate the masonry compressive stress and the steel tensile stress.
f = 2Mpx 2(203 ft-lb) (12 in./ft) / jkbd² (0.963)(0.111)(12 in.) (3 in.)2 = 422 psi