fr = 2.5of*m = 2.5o1350 psi = 91.8 psi

Where:

Ig = (7.62)(7.62)3/12 = 281 in4

Yt = 7.62/2 = 3.81 in

Since Mcr > Mmax;the lintel is not cracked, therefore Ig is used in lieu of I e.

by modifying equation 8-4 as follows:

)max = 0.30 in, as follows;

...Deflection O.K.

Where:

W = The weight of the triangular shaped wall segment, lbs.

= (50psf)(4.0ft)(2.0ft)/2 = 200 lbs

Fv = Fv = 1.0of*m = (1.0)o1350 psi = 36.7 psi

Fv = 36.7 > fv. = 6.36 psi

...Shear O.K.

follows:

Where:

Abrg = The bearing area of the lintel, in2

= 8 in 7.62 in = 61 in2

Fbrg = 0.25(f'm)

= 0.25(1.350) psi = 338 psi

Fbrg =338 psi > fbrgmax =7.3 psi

...Bearing O.K.

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