Question 6.1: A composite beam (Fig. 6-7) is constructed from a wood beam ......

Use a four-step problem-solving approach.
1. Conceptualize [hypothesize, sketch]: Use the general theory of flexure for composite beams.
2. Categorize [simplify, classify]:
Neutral axis: The first step in the analysis is to locate the neutral axis of the cross section. For that purpose, denote the distances from the neutral axis to the top and bottom of the beam as h_1\, and \, h_2, respectively. To obtain these distances, use Eq. (6-4).

E_{1}{\int_{1}}^{}y d A+E_{2}\int_{2}y d A=0\quad\quad  (6-4)

Evaluate the integrals in that equation by taking the first moments of areas 1 and 2 about the z axis, as
\int\limits_{1}ydA = \overset{—}{y_1}A_1 = (h_1  –  75 \,mm)(100\, mm × 150 \,mm) = (h_1  –  75\, mm)(15000 \,mm^2)
\int\limits_{2}ydA  =  \overset{—}{y_2}A_2 = -(156 \,mm – h_1)(100\, mm  ×  12 \,mm) = (h_1  –  75\, mm)(12000\, mm^2)
in which A_1\,  and  \,A_2 are the areas of parts 1 and 2 of the cross section, \overset{—}{y_1} \, and \, \overset{—}{y_2} are the y coordinates of the centroids of the respective areas, and h_1 has units of millimeters.
Substitute the preceding expressions into Eq. (6-4) to get the equation for locating the neutral axis as
E_1\int\limits_{1} ydA  +  E_2\int\limits_{2} ydA  =  0
or
(10.5 \,GPa)(h_1  –  75\, mm)(15000\, mm^2)  +  (210\, GPa)(h_1  –  75\, mm)(1200 \,mm^2)  =  0
Solve this equation to obtain the distance h_1 from the neutral axis to the top of the beam:
\quad\quad\quad h_1  =  124.8\, mm
Also, the distance h_2 from the neutral axis to the bottom of the beam is
\quad\quad\quad h_2  =  162\, mm  –  h_1  =  37.2\, mm
Thus, the position of the neutral axis is established.
Moments of inertia: The moments of inertia I_1 \,   and \,  I_2 \,of areas\, A_1\, and \,A_2 with respect to the neutral axis are found by using the parallel-axis theorem (see Section D.4 of Appendix D). Beginning with area 1 (Fig. 6-7),
I_1 = \frac{1}{12}(100 \,mm)(150\, mm)^3  +  (100 \,mm)(150\, mm)(h_1  –  75\, mm)^2  =  65.33  ×  10^6\, mm^4
Similarly, for area 2,
I_2  =  \frac{1}{12}(100\, mm)(12 \,mm)^3  +  (100 \,mm)(12\, mm)(h_2  –  6\, mm)^2  =  1.18 × 10^6 \,mm^4
To check these calculations, compute the moment of inertia I of the entire cross-sectional area about the z axis as
I = \frac{1}{3}(100\, mm)h^3_1  +  \frac{1}{3}(100 \,mm)h^3_2  =  (64.79  +  1.72)10^6 \,mm^4  =  66.51  ×  10^6 \,mm^4
which agrees with the sum of I_1 \,  and \, I_2.
3. Analyze [evaluate; select relevant equations, carry out mathematical solution]:
Normal stresses: The stresses in materials 1 and 2 are calculated from the flexure formulas for composite beams [Eqs. (6-7a and b)]. The largest compressive stress in material 1 occurs at the top of the beam (A) where y =  h_1 =  124.8 \,mm
Denoting this stress by σ_{1A} and using Eq. (6-7a) gives

\sigma_{x1}=-\frac{M y{E}_{1}}{E_{1}I_{1}+E_{2}I_{2}}\qquad\quad (6-7a)
σ_{1A}  =  -\frac{Mh_1E_1}{E_1I_1  +  E_2I_2}
\quad = – \frac{(6\, kN·m)(124.8 \,mm)(10.5\, GPa)}{(10.5\, GPa)(65.33 × 10^6 \,mm^4) + (210 \,GPa)(1.18  ×  10^6\, mm^4)}  =  -8.42 \,MPa
The largest tensile stress in material 1 occurs at the contact plane between the two materials (C) where y = -(h_2  –  12 \,mm)  = -25.2\, mm. Proceed as in the previous calculation to obtain
σ_{1C} = -\frac{(6 \,kN·m)(-25.2 \,mm)(10.5\, GPa)}{(10.5 \,GPa)(65.33 × 10^6\, mm^4)  +  (210 \,GPa)(1.18  ×  10^6 \,mm^4)} = 1.7 \,MPa
These are the largest compressive and tensile stresses in the wood.
The steel plate (material 2) is located below the neutral axis; therefore, it is entirely in tension. The maximum tensile stress occurs at the bottom of the beam (B) where y = -h_2 = -37.2\, mm. Hence, from Eq. (6-7b),

\sigma_{x2}=-\frac{M y{ E}_{2}}{E_{1}I_{2}+E_{2}I_{2}}\qquad\qquad(6.7b)
σ_{2B} = – \frac{M(-h_2)E_2}{E_1I_1  +  E_2I_2}
\quad = – \frac{(6 \,kN·m)(-37.2\, mm)(210 \,GPa)}{(10.5 \,GPa)(65.33  ×  10^6\, mm^4)  +  (210 \,GPa)(1.18  ×  10^6 \,mm^4)}  =  50.2 \,MPa
The minimum tensile stress in material 2 occurs at the contact plane (C) where y = -25.2  mm. Thus,
σ_{2C} = – \frac{(6\, kN·m)(-25.2\, mm)(210 \,GPa)}{(10.5 \,GPa)(65.33 × 10^6 \,mm^4) + (210 \,GPa)(1.18 ×  10^6\, mm^4)} = 34 \,MPa
These stresses are the maximum and minimum tensile stresses in the steel.
4. Finalize [conclude; examine answer—Does it make sense? Are units correct? How does it compare to similar problem solutions?]: The stress distribution over the cross section of the composite wood-steel beam is shown in Fig. 6-8.
Note: At the contact plane, the ratio of the stress in the steel to the stress in the wood is
σ_{2C}/σ_{1C} = 34 \,MPa/1.7\, MPa = 20
which is equal to the ratio E_2 /E_1 of the moduli of elasticity (as expected). Although the strains in the steel and wood are equal at the contact plane, the stresses are different because of the different moduli.