Masonry Magazine August 1992 Page. 33
Step 4: Calculate the masonry compressive stress and the steel tensile stress.
fb = 2Mpx / jkbd2 = 2(188 ft-lb)(12 in./ft) / (0.976)(0.073)(12 in.)(4.9 in.)2 = 220 psi
f = Mpx / Ajd = (188 ft-lb)(12 in./ft) / (0.017 in.2)(0.976)(4.9 in.) = 27.7 ksi
Step 5: Check compressive stress in masonry and the tensile stress in steel.
Fb = 4/3(1/3fm) = 1333 psi > 220 psi OK
F, = 4/3(30 ksi) = 40 ksi > 27.7 ksi OK
Step 6: Calculate shear stress.
f = Vpx / bjd = 94 lb / (12 in.)(0.976)(4.9 in.) = 2 psi
Step 7: Check shear stress.
F, = 4/3 fm = 73 psi, 4/3(50 psi)= 67 psi maximum
Thus, F, = 67 psi > 2 psi OK
Step 8: Design the reinforced brick masonry pilaster.
Repeat steps 3 through 7.
Step 3a: Calculate required pilaster reinforcement.
As req'd = M / Fajd = (3375 ft-lb)(12 in./ft) / (4/3)(20 ksi)(0.9)(9 in.) = 0.188 in.2
Try 4 #3 bars, A, = 0.22 in.2 per pair of bars (Table 3).
Ignore compression steel.
pn = An / bd = (0.22 in.2)(10) / (11.63 in.)(9 in.) = 0.021
k =√(pn)2 + 2pn - pn =(0.021)2 + 2(0.021) - (0.021) = 0.185
j = 1-k/3 = 0.938
Step 4a: Calculate the masonry compressive stress and the steel tensile stress.
f = 2M / jkbd2 = 2(3375 ft-lb) (12 in./ft) / (0.938)(0.185)(11.63 in.)(9 in.)2 = 496 psi
f = P / bkd = 676 lb / (11.63 in.)(0.185)(9 in.) = 35 psi
f = M / Ajd = (3375 ft-lb)(12 in./ft) / (0.22 in.2)(0.938)(9 in.) = 21.8 ksi
Step 5a: Check compressive stress in masonry and the tensile stress in steel.
F= 4/3(1/3fm) = 1333 psi
fmfaf = 35 psi + 496 psi = 531 psi < 1333 psi OK
F= 4/3(20 ksi) = 27 ksi > 21.8 ksi OK
Step 6a: Calculate shear stress.
f = V / bjd = 1125 lb / (11.63 in.)(0.938)(9 in.) = 12 psi
Step 7a: Check shear stress.
F= 4/3m = 73 psi, 4/3(50 psi) = 67 psi maximum
Thus F, = 67 psi > 12 psi OK
With pilaster designed, check the induced vertical can- tilever deflection in the panel. When determining the flex- ural strength of the pilaster, the critical cross section does not include the panel. However, because the panel is built integral with the pilaster, the combined pilaster and panel cross section is used to determine if cracking will occur in the panel. Therefore, determine if the pilaster and panel cross section is cracked. Assume the modulus of rupture, fer. for hollow grouted brick masonry is equal to 4.0 fm = 219 psi (Ref. UBC 91). Refer to Fig. 6(b) for the location of the centroid of the combined cross section. Panel section properties are based on face shells only. ACI/ASCE/TMS code requires the maximum length of flange when determining stress distribution to be 6 times the thickness of the flange, i.e. 6 (5.63 in.) = 33.8 in. in each direction.
Ipilaster = (1/12)(b)(t)3 + (b)(t)(7.48 in.-5.81 in.)2 = (1/12)(11.63 in.) (11.63 in.)3 + (11.63 in.) (11.63 in.)(1.67 in.)2 = 1900 in.4
Ipanel = 2(1/12)(2)(33.8 in.) (1.25 in.)3 + (2)(33.8 in.) (tr) (4.15 in.)2 + (2)(33.8 in.) (ts) (4.15 in. - 0.63 in.)2 = 2(1/12)(67.5 in.)(1.25 in.)3 + (67.5 in.) (1.25 in.)(3.52 in.)2 + (67.5 in.)(1.25 in.) (0.85 in.)2 = 1130 in.4
I = Ipilaster + Ipanel = 3030 in.4
Mer = ferIg / (7.48 in.)(12 in./ft) = (219 psi)(3030 in.4) / (7.48 in.)(12 in./ft) = 7393 ft-lb
M = 3375 ft-lb < 7393 ft-lb. Thus, the pilaster and panel cross section is not cracked under service load. Check the flexural tensile stress in the panel based on uncracked sec- tion properties.
f = Mc / I = (3375 ft-lb)(4.15 in.)(12 in./ft) / 3030 in.4 = 55 psi