21.53 in
4
(896.15 cm
4
). Then Sx = 21.53/4.60
= 4.68 in
3
(76.69 cm
3
). This value agrees with that recorded in the AISI Manual.
Then M=f£x = 20,000(4.68)712 = 7800 ft-lb (10,576 N-m); w = 8M/L^2 = 8(7800)/13^2
= 369 Ib/lin ft (5.39 kN/m).
Establish the value of fy for deflection determination
Apply (10,320*^2 //c°-^5 )[l - 2580/0? V)] = (HID)(fc +fb)lfc -A', or (113.8//c°-^5 ) x (1 -
23.74//?^5 ) = 1.084(/c + 20,000)//c - 1.93. By successive approximation,^ = 13,300 lb/in^2
(91.7MPa).
Compute the corresponding effective flange width for
deflection determination
Thus, b = (10,320*//?^5 )[1 - 2580/(/c°^5 g/0] = (10,320 x 0.105/13,300°^5 )[1 - 2580/
(13,300^05 x 108.7)] = 7.462 in (189.5 mm).
Locate the centroidal axis of the cross section having this
effective width; check the value of fc
Thus h = 11.415 - 7.462 - 3.953 in (100.4 mm); ht = 0.415 in^2 (2.68 cm^2 ); A = 313 -
0.415 = 2.715 in^2 (17.52 cm^2 ); H= 8.670 in^3 (142.1 cm^3 );^ = 8.670/2.715 = 3.19 in (81.0
mm); yb = 4.81 in (122.2 mm);/c = (3.19/4.81)20,000 - 13,300 lb/in^2 (91.7 MPa), which
is satisfactory.
Compute the deflection
For the net section, /, = 26.8 + 3.13(3.19 - 2.77)^2 - 0.415(3.19 - 0.053)^2 = 23.3 in^4 (969.8
cm^4 ). This value agrees with that tabulated in the AISI Manual. The deflection is A =
5wL^4 /(3&4EIx) = 5(369)(13)^4 (12)^3 /[384(29.5)(10)^6 23.3] = 0.345 in (8.8 mm).
