(a) Section for positive moment (b) Section for negative moment
FIGURE 58. Steel beam encased in concrete, (a) Section for positive moment; (b) section for
negative moment.
tive flange width =^1 X 4 L = !/ 4 (28.5)12 = 85.5 in (2172 mm); spacing of beams = 120 in
(3048 mm); I6t + 11 = 16(4.5) + 11 = 83 in (2108 mm); this governs. Transformed width
= 83/9 = 9.22 in (234.2 mm).
Assume that the neutral axis lies within the flange, and take static moments with re-
spect to this axis; or^1 / 2 (9.22y^2 ) - 11.77(10 -y) = O; y = 3.93 in (99.8 mm).
Compute the moment of inertia. Slab: C/3)9.22(3.93)^3 = 187 in^4 (7783.5 cm^4 ). Beam:
515.5 + 11.77 x (10-3.93)^2 = 949 in^4 (39,500.4 cm^4 ); I= 187 + 949 = 1136 in^4 (47,283.9
cm^4 ); Sc = 1136/3.93 = 289.1 in^3 (4737.5 cm^3 ); Sbs = 1136/14.07 = 80.7 in^3 (1322.4 cm^3 ).
- Transform the composite section in the region of negative
moment to an equivalent section of steel; compute the
section moduli
Referring to Fig. 58&, we see that the transformed width = 11/9 = 1.22 in (31.0 mm). Take
static moments with respect to the neutral axis. Or, 11.77(10 - y) -^1 X 2 (1.22^^2 ) = O; y =
7.26 in (184.4 mm). Compute the moment of inertia. Thus, slab: (^1 X 3 ) 1.22(7.26)^3 = 155.6
in^4 (6476.6 cm^4 ). Beam: 515.5 + 11.77(10 - 7.26)^2 = 603.9 in^4 (25,136.2 cm^4 );'/= 155.6
- 603.9 = 759.5 in
4
(31,612.8 cm
4
). Then Sc = 759.5/7.26 - 104.6 in
3
(1714.1 cm
3
); Sts =
759.5/10.74 - 70.7 in^3 (1158.6 cm^3 ).
- Compute the bending stresses at midspan
The loads carried by the noncomposite member are: slab, (4.5)150(10)/12 = 563 Ib/lin ft
(8.22 kN/m); stem, 11(15.5)150/144 = 178 Ib/lin ft (2.6 kN/m); steel, 40 Ib/lin ft (0.58
kN/m); total = 563 + 178 + 40 = 781 Ib/lin ft (11.4 kN/m). The load carried by the com-
posite member = 145(10) = 1450 Ib/lin ft (21.2 kN/m). Then Mn = (^1 Xs)781(28.5)^212 =
951,500 in-lb (107.5 kN-m); Mc = (Y 2 o)1450(28.5)^212 = 706,600 in-lb (79.8 kN-m);/c =
706,600/[289.1(9)] = 272 lb/in^2 (1875 kPa), which is acceptable. Also, fbs =
(951,500/64.4) + (706,600/80.7) = 23,530 lb/in
2
(162.2 MPa), which is acceptable.
- Compute the bending stresses at the support
Thus, Mc = 706,600(^20 Xi 2 ) = 1,177,700 in-lb (132.9 kN-m);/c = l,177,700/[ 104.6(9)] =
1251 lb/in
2
(8.62 MPa), which is satisfactory. Also,/, = 1,177,700/70.7 = 16,60.0 lb/in
2
(114.9 MPa), which is acceptable. The design is therefore satisfactory with respect to
flexure.
- Investigate the composite member with respect to horizontal
shear in the concrete at the section of contraflexure
Assume that this section lies at a distance of 0.2L from the support. The shear at this sec-
tion is Vc = 1450(0.3)(28.5) = 12,400 Ib (55.2 kN).
