Question 6.3: The composite beam shown in Fig. 6-10a is formed of a wood b...
The composite beam shown in Fig. 6-10a is formed of a wood beam (4.0 in. × 6.0 in. actual dimensions) and a steel reinforcing plate (4.0 in. wide and 0.5 in. thick). The beam is subjected to a positive bending moment M = 60 k-in.
Using the transformed-section method, calculate the largest tensile and compressive stresses in the wood (material 1) and the maximum and minimum tensile stresses in the steel (material 2) if E_{1} = 1500 ksi and E_{2} = 30,000 ksi.
Note: This same beam was analyzed previously in Example 6-1 of Section 6.2.
Transformed section. We will transform the original beam into a beam of material 1, which means that the modular ratio is defined as
n=\frac{E_{2}}{E_{1}}=\frac{30,000_\ ksi}{1,500_\ ksi}=20The part of the beam made of wood (material 1) is not altered but the part made of steel (material 2) has its width multiplied by the modular ratio. Thus, the width of this part of the beam becomes
n(4 in.) = 20(4 in.) = 80 in.
in the transformed section (Fig. 6-10b).
Neutral axis. Because the transformed beam consists of only one material, the neutral axis passes through the centroid of the cross-sectional area. Therefore, with the top edge of the cross section serving as a reference line, and with the distance y_{i} measured positive downward, we can calculate the distance h_{1} to the centroid as follows:
=\frac{322.0_\ in.^{3}}{64.0_\ in.^{2}}=5.031_\ in.
Also, the distance h_{2} from the lower edge of the section to the centroid is
h_{2}=6.5_\ in.-h_{1}=1.469_\ in.Thus, the location of the neutral axis is determined.
Moment of inertia of the transformed section. Using the parallel-axis theorem (see Section 12.5 of Chapter 12), we can calculate the moment of inertia I_{T} of the entire cross-sectional area with respect to the neutral axis as follows:
=171.0_\ in.^{4}+60.3_\ in.^{4}=231.3_\ in.^{4}
Normal stresses in the wood (material 1). The stresses in the transformed beam (Fig. 6-10b) at the top of the cross section (A) and at the contact plane between the two parts (C) are the same as in the original beam (Fig. 6-10a). These stresses can be found from the flexure formula (Eq. 6-15), as follows:
\sigma_{1A}=-\frac{My}{I_{T}}=-\frac{(60_\ k-in.)(5.031_\ in.)}{}=-1310_\ psi\sigma_{1C}=-\frac{My}{I_{T}}=-\frac{(60_\ k-in.)(-0.969_\ in.)}{231.3_\ in.^{4}}=251_\ psi
These are the largest tensile and compressive stresses in the wood (material 1) in the original beam. The stress \sigma_{1A} is compressive and the stress \sigma_{1C} is tensile.
Normal stresses in the steel (material 2). The maximum and minimum stresses in the steel plate are found by multiplying the corresponding stresses in the transformed beam by the modular ratio n (Eq. 6-17). The maximum stress occurs at the lower edge of the cross section (B) and the minimum stress occurs at the contact plane (C):
\sigma_{2C}=-\frac{My}{I_{T}}n=-\frac{(60_\ k-in.)(-0.969_\ in.)}{231.3_\ in.^{4}}(20)=5030_\ psi
Both of these stresses are tensile.
Note that the stresses calculated by the transformed-section method agree with those found in Example 6-1 by direct application of the formulas for a composite beam.
