fr = 2.54%f*m = 2.5%1350 lb/in2 = 91.8 lb/in2

Where:

Ig = (7.62)(23.62)3/12 = 8368 in4

Yt = 23.62/2 = 11.81 in

Since Mcr < Mmax; the lintel is cracked, therefore the effective moment of inertia, Ie, must be computed as

follows:

Ie = (Mcr/Mmax)3Ig + [1 - (Mcr/Mmax)3] Icr

Where:

Icr = The moment of inertia of the cracked section, in4

(b) Determine the lintel deflection, mt, as follows:

Since the line of action of the concentrated load is located off the lintel span and the vertical height of the

distribution triangle is only 2 feet, the effects of the concentrated load on the centerline deflection is negligible

and will be ignored.

)max = 0.30 in, as follows:

...Deflection O.K.

follows:

Where:

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