Question 5.16: The composite beam shown in Fig. 5-47a is formed of a wood b...

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}=20

The 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. 5-47b).

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:

h_1=\frac{\sum{y_iA_i}}{\sum{A_i}} =\frac{(3  in.)(4  in.)(6  in.)  +  (6.25  in.)(80  in.)(0.5  in.)}{(4  in.)(6  in.)  +  (80  in.)(0.5  in.)}

=\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 10.5 of Chapter 10, available online), we can calculate the moment of inertia  I_T  of the entire cross-sectional area with respect to the neutral axis as follows:

I_T=\frac{1}{12}(4  in.)(6  in.)^3  +  (4  in.)(6  in.)(h_1  –  3  in.)^2

+ \frac{1}{12}(80  in.)(0.5  in.)^3  +  (80  in.)(0.5  in.)(h_2  –  0.25  in.)^2

=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. 5-47b) 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. 5-47a). These stresses can be found from the flexure formula (Eq. 5-62), as follows:

σ_{x1}=-\frac{My}{I_T}                     (5-62)

σ_{1A}=-\frac{My}{I_T}=-\frac{(60  k-in.)(5.031  in.)}{231.3  in.^4}=-1310  psi

σ_{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 σ_{1A}  is compressive and the stress σ_{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. 5-64).

σ_{x2}=-\frac{My}{I_T}n

The maximum stress occurs at the lower edge of the cross section (B) and the minimum stress occurs at the contact plane (C):

σ_{2B}=-\frac{My}{I_T}n=-\frac{(60  k-in.)(-1.469  in.)}{231.3  in.^4}(20)=7620  psi

σ_{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 5-14 by direct application of the formulas for a composite beam.