Question 5.14: A composite beam (Fig. 5-44) is constructed from a wood beam...

Neutral axis. The first step in the analysis is to locate the neutral axis of the cross section. For that purpose, let us 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, we use Eq. (5-50).

E_1\int_{1}{ydA}  +  E_2\int_{2}{ydA}=0

The integrals in that equation are evaluated by taking the first moments of areas 1 and 2 about the z axis, as follows:

\int_{1}{ydA}=\overline{y}_{1} A_1=(h_1  –  3  in.)(4  in. \times 6  in.)=(h_1  –  3  in.)(24  in.^2)

\int_{2}{ydA}=\overline{y}_{2} A_2=-(6.25  in.  –  h_1)(4  in. \times  0.5  in.)=(h_1  –  6.25  in.)(2  in.^2)

in which  A_1  and  A_2  are the areas of parts 1 and 2 of the cross section,  \overline{y}_{1}  and  \overline{y}_{2}  are the y coordinates of the centroids of the respective areas, and  h_1  has units of inches.

Substituting the preceding expressions into Eq. (5-50) gives the equation for locating the neutral axis, as follows:

E_1\int_{1}{ydA}  +  E_2\int_{2}{ydA}=0

or

(1500  ksi)(h_1  –  3  in.)(24  in.^2)  +  (30,000  ksi)(h_1  –  6.25  in.)(2  in.^2)=0

Solving this equation, we obtain the distance  h_1  from the neutral axis to the top of the beam:

h_1=5.031  in.

Also, the distance  h_2  from the neutral axis to the bottom of the beam is

h_2=6.5  in.  –  h_1=1.469  in.

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 can be found by using the parallel-axis theorem (see Section 10.5 of Chapter 10, available online). Beginning with area 1 (Fig. 5-44), we get

I_1=\frac{1}{12}(4  in.)(6  in.)^3  +  (4  in.)(6  in.)(h_1  –  3  in.)^2=171.0  in.^4

Similarly, for area 2 we get

I_2=\frac{1}{12}(4  in.)(0.5  in.)^3  +  (4  in.)(0.5  in.)(h_2  –  0.25  in.)^2=3.01  in.^4

To check these calculations, we can determine the moment of inertia I of the entire cross-sectional area about the z axis as follows:

I=\frac{1}{3}(4  in.)h^{3}_{1}  +  \frac{1}{3}(4  in.)h^{3}_{2}=169.8  +  4.2=174.0  in.^4

which agrees with the sum of  I_1  and  I_2.

Normal stresses. The stresses in materials 1 and 2 are calculated from the flexure formulas for composite beams (Eqs. 5-53a and b).

σ_{x1}=-\frac{MyE_1}{E_1I_1  +  E_2I_2}                 σ_{x2}=-\frac{MyE_2}{E_1I_1  +  E_2I_2}

The largest compressive stress in material 1 occurs at the top of the beam (A) where  y=h_1=5.031  in.  Denoting this stress by  σ_{1A}  and using Eq. (5-53a), we get

σ_{1A}=-\frac{Mh_1E_1}{E_1I_1   +   E_2I_2}

=-\frac{(60  k-in.)(5.031  in.)(1500  ksi)}{(1500  ksi)(171.0  in.^4) (30,000  ksi)(3.01  in.^4)}=-1310  psi

The largest tensile stress in material 1 occurs at the contact plane between the two materials (C) where  y=-(h_2  –  0.5  in.)=-0.969  in.  Proceeding as in the previous calculation, we get

σ_{1C}=-\frac{(60  k-in.)(-0.969  in.)(1500  ksi)}{(1500  ksi)(171.0  in.^4)  +  (30,000  ksi)(3.01  in.^4)}=251  psi

Thus, we have found the largest compressive and tensile stresses in the wood.

The steel plate (material 2) is located below the neutral axis, and therefore it is entirely in tension. The maximum tensile stress occurs at the bottom of the beam (B) where  y=-h_2=-1.469  in.  Hence, from Eq. (5-53b) we get

σ_{2B}=-\frac{M(-h_2)E_2}{E_1I_1  +  E_2I_2}

=-\frac{(60  k-in.)(-1.469  in.)(30,000  ksi)}{(1500  ksi)(171.0  in.^4)  +  (30,000  ksi)(3.01  in.^4)}=7620  psi

The minimum tensile stress in material 2 occurs at the contact plane (C) where  y=-0.969  in.  Thus,

σ_{2C}=-\frac{(60  k-in.)(-0.969  in.)(30,000  ksi)}{(1500  ksi)(171.0  in.^4)  +  (30,000  ksi)(3.01  in.^4)}=5030  psi

These stresses are the maximum and minimum tensile stresses in the steel.

Note: At the contact plane the ratio of the stress in the steel to the stress in the wood is

σ_{2C}/σ_{1C}=5030  psi/251  psi=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.