(a) Composite (b) Tronsformed timber
section section
FIGURE 29
Calculation Procedure:
- Compute the rigidity of the steel compared with that of the
timber
Let n = the relative rigidity of the steel and timber. Then n = EJE 1 = (30 x 106 )/(1.2 x IQ^6 )
= 25.
- Transform the composite beam to an equivalent
homogeneous beam
To accomplish this transformation, replace the steel with timber. Sketch the cross section
of the transformed beam as in Fig. 29b. Determine the sizes of the hypothetical elements
by retaining the dimensions normal to the axis of bending but multiplying the dimensions
parallel to this axis by n.
- Record the properties of each element of the
transformed section
Element I: A = 25(T)(^1 X 2 ) = 87.5 in^2 (564.55 cm^2 ); I 0 is negligible.
Element 2: A = 8(12) = 96 in^2 (619.4 cm^2 ); / 0 = !/ 2 (8)12^3 - 1152 in^4 (4.795 dm^4 ).
Element 3: Refer to the AISC Manual for the data; A = 25(2.85) = 71.25 in^2 (459.71 cm^2 );
/o = 25(0.98) = 25 in^4 (1040.6 cm^4 ); a = 0.55 in (13.97 mm); b = 2.09 in (53.09 mm).
- Locate the centroidal axis of the transformed section
Take static moments of the areas with respect to the centerline of the 8 x 12 in (203.2 x
304.8 mm) rectangle. Then j;m = [87.5(6.25) - 71.25(6.55)]/(87.5 + 96 + 71.25) = 0.31 in
(7.87 mm). The neutral axis of the composite section is at the same location as the cen-
troidal axis of the transformed section.
- Compute the moment of inertia of the transformed section
Apply the relation in step 3 of the previous calculation procedure. Then compute the dis-
tance c to the outermost fiber. Thus, I= 1152 + 25 + 87.5(6.25 - 0.31)^2 + 96(0.31)^2 +
71.25(6.55 + 0.31)^2 = 7626 in^4 (31.74 dm^4 ). Also, c = 0.31 + 6 + 2.09 = 8.40 in (213.36
mm).
- Determine which material limits the beam capacity
Assume that the steel is stressed to capacity, and compute the corresponding stress in the
channel CA. of element 3
