Question 6.3: The composite beam shown in Fig. 6-11a is formed of a wood b......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize: Use the transformed-section approach and compare results to those found in Example 6-1.
3. Analyze:
Transformed section: 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{210~\mathrm{GP}{a}}{10.5~\mathrm{GP}{a}}}=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(100 ~mm) = 20(100 ~ mm) = 2~m
in the transformed section (Fig. 6-11b).
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 the distance y_i measured positive downward, calculate the distance h_1 to the centroid as
{h}_{1}\,=\,\frac{\sum y_{i}A_{i}}{\sum A_{i}}\,= \frac{(75 ~mm)(100~ mm)(150 ~mm) + (156 ~mm)(2000 ~mm)(12~ mm)}{(100 ~mm)(150 ~mm) + (2000 ~mm)(12 ~mm)}
= \frac{4869 × 10^3 ~mm^3}{39 × 10^3~ mm^2} = 124.8~ mm
Also, the distance h_2 from the lower edge of the section to the centroid is
\quad\quad\quad h_2 = 162~ mm  –  h_1 = 37.2 ~mm
Thus, the location of the neutral axis is determined.
Moment of inertia of the transformed section: Using the parallel-axis theorem (see Section D.4 of Appendix D), calculate the moment of inertia I_T of the entire cross-sectional area with respect to the neutral axis as
I_T = \frac{1}{12}(100~ mm)(150 ~mm)^3 + (100~ mm)(150~ mm)(h_1  –  75~ mm)^2
\quad +\frac{1}{12}(2000 ~mm)(12~ mm)^3  +  (2000~ mm)(12 ~mm)(h_2  –  6 ~mm)^2
\quad = 65.3 × 10^6 ~mm^4 + 23.7 × 10^6 ~mm^4 = 89.0 × 10^6 ~mm^4
Normal stresses in the wood (material 1): The stresses in the transformed beam (Fig. 6-11b) 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. 6-11a). These stresses can be found from the flexure formula in Eq. (6-16) as

\quad σ_{x1} = – \frac{M_y}{I_T}\quad\quad (6-16)
\quad σ_{1A} = – \frac{M_y}{I_T} = – \frac{(6 × 10^6 ~N·mm)(124.8 ~mm)}{89 × 10^6~ mm^4} = -8.42 ~MPa

\quad σ_{1C} = – \frac{M_y}{I_T} = – \frac{(6 × 10^6 ~N·mm)(-25.2 ~mm)}{89 × 10^6 ~mm^4} = 1.7 ~MPa
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 in Eq. (6-18b). The maximum stress occurs at the lower edge of the cross section (B) and the minimum stress occurs at the contact plane (C):

\quad σ_{x2} = -\frac{M_y}{I_T}n \quad\quad (6-18b)
\quad σ_{2B} = – \frac{M_y}{I_T}n = – \frac{(6 × 10^6 ~N·mm)(-37.2 ~mm)}{89 × 10^6 ~mm^4}(20) = 50.2 ~MPa
\quad σ_{2C} = – \frac{M_y}{I_T}n = – \frac{(6 × 10^6 ~N·mm)(-25.2 ~mm)}{89 × 10^6 ~mm^4}(20) = 34 ~MPa
Both of these stresses are tensile.
4. Finalize: 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.
Balanced design: As a final evaluation of the wood-steel composite beam considered here and in Example 6-1, note that neither wood nor steel has reached typical allowable stress levels. Perhaps some redesigning of this beam would be of interest; consider only the steel plate here (you could also re-size the wood beam).
A balanced design is one in which wood and steel reach their allowable stress values at the same time under the design moment; this could be regarded as a more efficient design of this beam. First, holding the steel plate thickness at t_s = 12 ~mm, find the required width b_s of the steel plate, so the wood and steel reach allowable stress values simultaneously under design moment M_D. Then setting b_s = 100 ~mm, repeat the previous process but also find required plate thickness t_s to achieve the same objective. Assume that the allowable stress values for wood and steel are σ_{aw} = 12.7 ~MPa  ~ and ~ σ_{as} = 96 ~MPa, respectively.
Also assume that the wood beam dimensions are unchanged.
Using the transformed-section approach, write the expressions for the stresses at the top of the wood and bottom of the steel. Equate each to its allowable value as
\quad\quad σ_{aw} = \frac{-M_Dh_1}{I_T}\quad and \quad σ_{as} = \frac{-M_Dh_2n}{I_T}\quad\quad (a,b)
Next, solve each of Eqs. (a) and (b) for ratio M_D /I_T; then equate the two expressions to find the h_1/h_2 ratio for which allowable stress levels are reached in both materials:
\quad\quad \frac{h_1}{h_2}  =  n\frac{σ_{aw}}{σ_{as}} \quad\quad\quad\quad (c)
Expressions for h_1 ~  and  ~ h_2 can be obtained in terms of the transformed-section dimensions b, h, b_s, and t_s (Figs. 6-12a and b) by taking first moments about the z axis to get
\quad h_1 = \frac{h}{2} + \frac{(b_snt^2_s) + (b_shnt_s)}{(2bh) + (b_snt_s)} \quad and \quad h_2 = \frac{nb_st_s{\Bigg\lgroup}\frac{t_s}{2}{\Bigg\rgroup} + bh{\Bigg\lgroup}t_s + \frac{h}{2}{\Bigg\rgroup}}{(nb_st_s) + (bh)} \quad (d)
With some effort (and perhaps with computer assistance), rewrite Eq. (c) as
\quad\quad \frac{bh^2 + (2b_snht_s) + (b_snht^2_s)}{bh^2 + (2bht_s) + (b_snht^2_s)} = n \frac{σ_{aw}}{σ_{as}} \quad\quad\quad (e)
Collecting terms and solving for the required width of b_s for the steel plate (with thickness t_s unchanged) then substituting numerical values, width b_s (instead of the original width of 100  mm) is
b = 100 ~mm  ~h = 150 ~mm  ~t_s = 12 ~mm ~n = 20~ σ_{aw} = 12.7 ~MPa~ σ_{as} = 96 ~MPa
\quad\quad b_s = \frac{{\Bigg\lgroup}n\frac{σ_{aw}}{σ_{as}}{\Bigg\rgroup}(bh^2 + 2bht_s) – bh^2}{(2nht_s) + nt^2_s{\Bigg\lgroup}1 – n\frac{σ_{aw}}{σ_{as}}{\Bigg\rgroup}} = 69.2  ~mm \quad\quad (f)

\quad E_1 \int_{1}ydA  +  E_2\int_{2}ydA = 0 \quad\quad (6-12)
So for a 100  mm  ×  150  mm wood beam reinforced by a 69.2 mm × 12 mm steel plate (Fig. 6-12a) under any applied moment M that is less than or equal to M_D, the stress ratio σ_{1A}/σ_{2B} will be equal to σ_{aw} / σ_{as}. If M = M_D, then \sigma_{1A} = \sigma_{aw} and \sigma_{2B} = \sigma_{as}.
Alternatively, reformulate Eq. (e) to get a quadratic equation for the steel plate thickness t_s (with the original width b_s = 100 ~mm) to obtain
t^2_s\begin{bmatrix}nb_s{\Bigg\lgroup}1 – n\frac{σ_{aw}}{σ_{as}}{\Bigg\rgroup}\end{bmatrix} + t_s \begin{Bmatrix} 2h\begin{bmatrix}nb_s – b{\Bigg\lgroup}1 – n\frac{σ_{aw}}{σ_{as}}{\Bigg\rgroup}\end{bmatrix}\end{Bmatrix} + bh^2{\Bigg\lgroup}1 – n\frac{σ_{aw}}{σ_{as}}{\Bigg\rgroup} = 0\quad (g)
The solution of Eq. (g) results in a reduced steel plate thickness leading to a balanced design of the wood-steel composite beam as t_s = 7.46 ~mm. Once again, for b_s = 100 ~mm ~and  ~t_s = 7.46 ~mm, the stress ratio σ_{1A}/σ_{2B} = σ_{aw} /σ_{as} for applied moments M which are less than or equal to M_D.