Mnx = Mpx-(Mpx-MJ^-^}
\Ar-Ap/
The properties Sx and Zx of the cross section in Fig. 40 must now be calculated.
Sx = —, where c = - = ^-^- = 29 in (73.7 cm)
\^/ 2* Zi
The contributions of the two flanges and the web to the moment of inertia Ix are
RT^3
Elements ^-+AD^2
2 Flanges T 18 in x (I1n) + (lg in x 1 in)(2g 5 in)2l 2 = 2 9,244 in^4 (1,217,227 cm^4 )
Web 0.44inx^(56in)3 + Q = ^ ^ ^^ ^
35 647 in^4
Ix Sx = -^- 1230 in^3 (20,156 cm^3 )
To determine Zx, we calculate ^AD, where A is the cross-sectional area of each element
and D represents its distance from the centroidal x axis.
In calculating Zx, the upper and lower halves of the web are taken separately.
Elements AD
Flanges [(18 in x 1 in) x 28.5 in]2 = 1026 in^3 (16,813 cm^3 )
2 l/2 Webs [(28 in x 0.44 in) x 14 in]2 = 343 in^3 (5,620 cm^3 )
Zx 1369 in^3 (22,433 cm^3 )
Zx =1369 in^3 (22,433 cm^3 )
FIGURE 40