on center, 65 psf.

w = wDL +2lin + wwall

Where:

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

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.)

And;

(h*tan " + 0.5L - 0x*)

a=

(0.577W + 0.5L - x*)

=

=

[(0.577 2.0) + 0.5(12.67) - 7.0)]

=

0.49 ft

(c)

Determine the shear loading, V, as follows:

Where:

L = The design span length of the lintel, feet.

= 12.00 ft 0.67 ft = 12.67 ft

Where:

= 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.

Asb = (pe)(b)(d)

Where:

pe = 0.0027 (See table 5-9)

b = The lintel width = 7.62 inches

d = The effective depth of lintel = 20.62 inches

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