The cross section properties in the AISC table may be applied. The moment of inertia
refers to an equivalent section obtained by transforming the concrete to steel. Refer to
Sec. 1. Thus yte = 5 + 18.12 + 1 - 16.50 = 7.62 in (194 mm); Ste = Ilytc = 5242/7.62 =
687.9 in^3 (11,272.7 cm^3 ). From the ACI Code J 0 = 1350 lb/in^2 (9.31 MPa) and n = 9.
Then/c = M
c
l(nStc) = 7,502,000/[9(687.9)] = 1210 lb/in
2
(8.34 MPa), which is satisfacto-
ry-
- Record the relevant properties of the 1/1/18 x 55
Thus, A = 16.19 in
2
(104.5 cm
2
); d = 18.12 in (460 mm); /= 890 in
4
(37,044.6 cm
4
); S =
98.2 in^3 (1609 cm^3 ); flange thickness = 0.630 in (16.0 mm). - Compute the section moduli of the composite section where
the cover plate is absent
To locate the neutral axis, take static moments with respect to the center of the steel.
Thus, transformed flange width = 87.53/9 = 9.726 in (247.0 mm). Further,
Element A, in^2 (cm^2 ) y, in (mm) Ay, in^3 (cm^3 ) Ay^2 , in^4 (cm^4 ) I 0 , in^4 (cm^4 )
WlSx 55 16.19(104.5) O (O) O (O) O (O) 890(37,044.6)
Slab 48.63(313.7) 11.56(294) 562.2(9,212.8) 6,499(270,509) 101 (4,203,9)
Total 64.82 (418.2) 562.2 (9,212.8) 6,499 (270,509) 991 (41,248.5)
Then y = 562.2/64.82 = 8.67 in (220 mm); /= 6499 + 991 - 64.82(8.67)^2 = 2618 in^4
(108,969.4 cm^4 ); yb = 9.06 + 8.67 = 17.73 in (450 mm); ytc = 9.06 + 5 - 8.67 = 5.39 in
(136.9 mm); Sb = 2618/17.73 = 147.7 in^3 (2420 cm^3 ); Stc = 2618/5.39 = 485.7 in^3 (7959
cm^3 ).
- Verify the value of Sb
Apply the value of the K factor in the AISC table. This factor is defined by K^2 = 1 - Sb
without plate/Si, with plate. The Sb value without the plate = 317.5(1 - 0.732) = 148 in^3
(2425 cm^3 ), which is satisfactory. - Establish the theoretical length of the cover plate
In Fig. 60, let C denote the section at which the cover plate becomes superfluous with re-
spect to flexure. Then, for the composite section, w = 0.773 + 0.330 + 2.200 = 3.303
kips/lin ft (48.2 kN/m); P=IO kips (44.5 kN); Mm = 7502 in-kips (847.6 kN-m); R 0 =
64.45 kips (286.7 kN). The allowable values of M 0 are, for concrete, Mc =
485.7(9)1.35/12 = 491.8 ft-kips (666.9 kN-m) and, for steel, Mc = 147.7(24)712 = 295.4
ft-kips (400.6 kN-m), which governs. Then RJC - l/2\vx^2 = 295A; x = 5.30 ft (1.62 m). The
theoretical length = 36 - 2(5.30) = 25.40 ft (7.74 m).
For the noncomposite section, investigate the stresses at the section C previously lo-
cated. Thus: w = 0.773 kips/lin ft (11.3 kN/m); P=IQ kips (44.5 kN); R 0 = 18.91 kips
(84.1 kN); Mc = 18.91(5.30) - '/2(0.773) x 5.30^2 = 89.4 ft-kips (121.2 kN-m);/ 6 = -
89.4(12)798.2 = 10.9 kips/in^2 (75.1 MPa), which is satisfactory. - Determine the axial force F in the cover plate at its end by
computing the mean bending stress
Thus/mean = Mymean/I= 295.4(12)916.50) - 0.50)/5242 = 10.82 kips/in
2
(75.6 MPa); F =
AfmQan = 9(10.82) = 97.4 kips (433.2 kN). Alternatively, calculate Fby applying the factor
120/7 recorded in the AISC table. Thus, F = UQMII = 0.33(295.4) = 97.5 kips (433.7
kN).