FIGURE 28. Compound beam.Calculation Procedure:
- Obtain the properties of the elements
Using the AISC Manual, determine the following properties. For the Wl 6 x 45, d = 16.12
in (409.45 mm); A = 13.24 in^2 (85.424 cm^2 ); /= 583 in^4 (24,266 cm^4 ). For the WT6 x 20,
d = 5.97 in (151.63 mm); A = 5.89 in^2 (38.002 cm^2 ); 1=14 in^4 (582.7 cm^4 ); yi = 1.08 in
(27.43 mm); ^ 2 = 5.97 - 1.08 = 4.89 in (124.21 mm). - Locate the centroidal axis of the section
Locate the centroidal axis of the section with respect to the centerline of the W16 x 45,
and compute the distance c from the centroidal axis to the outermost fiber. Thus, ym =
5.89[(8.06 + 4.89)]/(5.89 + 13.24) = 3.99 in (101.346 mm). Then c = 8.06 + 3.99 = 12.05
in (306.07 mm). - Find the moment of inertia of the section with respect to its
centroidal axis
Use the relation / 0 + AW- for each member, and take the sum for the two members to find /
for the built-up beam. Thus, for the W16 x 45: k = 3.99 in (101.346 mm); 10 + AIf 583 +
13.24(3.99)^2 = 793 in^4 (33,007.1 cm^4 ). For the WT6 x 20: & = 8.06 - 3.99 + 4.89 = 8.96 in
(227.584 mm); I 0 + Ak^2 = 14 + 5.89(8.96)^2 = 487 in^4 (20,270.4 cm^4 ). Then / = 793 + 487
= 1280 in^4 (53,277.5 cm^4 ). - Apply the moment equation to find the flexural capacity
Use the relation M =fllc = 20,000(1280)/[ 12.05(12)] = 177,000 Ib-ft (240,012 N-m).
ANALYSIS OFA COMPOSITE BEAM
An 8 x 12 in (203.2 x 304.8 mm) timber beam (exact size) is reinforced by the addition
of a 7 x i/ 2 in (177.8 x 12.7 mm) steel plate at the top and a 7-in (177.8-mm) 9.8-lb
(43.59-N) steel channel at the bottom, as shown in Fig. 29a. The allowable bending
stresses are 22,000 lb/in^2 (151,690 kPa) for steel and 1200 lb/in^2 (8274 kPa) for timber.
The modulus of elasticity of the timber is 1.2 x io^6 lb/in^2 (8.274 x io^6 kPa). How does
the flexural strength of the reinforced beam compare with that of the original timber
beam?