Construct the banding-moment diagram
Apply the ACI equation for maximum midspan moment. Refer to Fig. 19c: M 1 =
(Ys)wL'^2 = (^1 / 8 )3.3(22)^2 - 200 ft-kips (271.2 kN-m); M 2 = (Vi 6 )WZ'^2 = 100 ft-kips (135.6
kN-m); M 3 = 100 ft-kips (135.6 kN-m).
Determine upon what area the moment of inertia should
be based
Apply the criterion set forth in the ACI Code to determine whether the moment of inertia
is to be based on the transformed gross section or the transformed cracked section. At the
support/^ = 4.43(40,000)/[14(20.5)] = 617 > 500. Therefore, use the cracked section.
Determine the moment of inertia of the transformed cracked
section at the support
Refer to Fig. I9d: nAs = 10(4.43) = 44.3 in^2 (285.82 cm^2 ); (n - I)A 5 = 9(1.58) = 14.2 in^2
(91.62 cm^2 ). The static moment with respect to the neutral axis is Q = - Yi(My^2 ) +
44.3(20.5 -y)- I4.2(y - 2.5) = O; y = 8.16 in (207.264 mm). The moment of inertia with
respect to the neutral axis is I 1 = (^1 X 3 )14(8.16)^3 + 14.2(8.16 - 2.5)^2 + 44.3(20.5 - 8.16)^2 =
9737 in^4 (40.53 dm^4 ).
