TM 5-809-3/NAVFAC DM-2.9/AFM 88-3, Chap. 3
Design dead load is "w". Unit weights of 8-inch wall are; solidly grouted, 92 psf; and grouted at 32 inches
on center, 65 psf.
w = wDL +2lin + wwall
wlin = the weight of the lintel, lbs/ft
= (92 psf)(2.00 ft) = 184 plf
wwall = (92 psf)(0.67 ft) + (65 psf)(1.33 ft)
= 62 plf + 86 plf = 148 plf
w = 100 + 184 + 148 = 432 plf
Design uniform live load is wLL.
wLL = 250 plf
Uniform distribution of the concentrated live load on the lintel, wp, is determined as follows: (See figure
8-1 for an explanation of the terms used in this distribution.)
(h*tan " + 0.5L - 0x*)
(0.577W + 0.5L - x*)
[(0.577 2.0) + 0.5(12.67) - 7.0)]
Determine the shear loading, V, as follows:
L = The design span length of the lintel, feet.
= 12.00 ft 0.67 ft = 12.67 ft
(d) Minimum lintel depth without shear reinforcement, dreqd, is determined as follows:
= 1.0\of*m = o1350 lb/in2 = 37 lb/in2
Fv = The allowable shear stress, psi.
b = the actual lintel width, in.
= 7.62 in
For a 24 inch deep lintel, the actual effective beam depth, dact, is 20.62 inches.
dact = 20.62 in > dreqd = 19.01 in
...24-Inch Lintel Depth O.K.
(a) Determine the maximum moment due to loading, Mmax, as follows:
(b) Determine the area of reinforcement, Asb, required to provide a balanced steel ratio, pe.
Asb = (pe)(b)(d)
pe = 0.0027 (See table 5-9)
b = The lintel width = 7.62 inches
d = The effective depth of lintel = 20.62 inches