Question 6.1: A composite beam (Fig. 6-7) is constructed from a wood beam ...
A composite beam (Fig. 6-7) is constructed from 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 wood and steel are securely fastened to act as a single beam. The beam is subjected to a positive bending moment M = 60 k-in.
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.
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. (6-3). 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} = \bar{y}_{1}A_{1} = \left(h_{1}-3_\ in.\right)\left(4_\ in.\times 6_\ in.\right) = \left(h_{1}-3_\ in.\right)(24_\ in.^{2})\int_{2}{ydA} = \bar{y}_{2}A_{2} = -\left(6.25_\ in.-h_{1}\right)\left(4_\ in.\times 0.5_\ in.\right) = \left(h_{1}-6.25_\ in.\right)(2_\ in.^{2})
in which A_{1} and A_{2} are the areas of parts 1 and 2 of the cross section, \bar{y}_{1} and \bar{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. (6-3) gives the equation for locating the neutral axis, as follows:
E_{1}\int_{1}{ydA}+E_{2}\int_{2}{ydA} = 0 (6-3)
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 12.5 of Chapter 12). Beginning with area 1 (Fig. 6-7), we get
Similarly, for area 2 we get
I_{2} = \frac{1}{12}\left(4_\ in.\right)\left(0.5_\ in.\right)^{3}+\left(4_\ in.\right)\left(0.5_\ in.\right)\left(h_{2}-0.25_\ in. \right)^{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}\left(4_\ in.\right)h^{3}_{1}+\frac{1}{3}\left(4_\ in.\right)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. 6-6a and b). 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 \sigma_{1A} and using Eq. (6-6a), we get
\sigma_{x1} = -\frac{MyE_{1}}{E_{1}I_{1}+E_{2}I_{2}} \sigma_{x2} = -\frac{MyE_{2}}{E_{1}I_{1}+E_{2}I_{2}} (6-6a,b)
\sigma_{1A} = -\frac{Mh_{1}E_{1}}{E_{1}I_{1}+E_{2}I_{2}}= -\frac{\left(60_\ k-in.\right)\left(5.031_\ in.\right)\left(1500_\ ksi\right)}{} = -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
\sigma_{1C} = -\frac{\left(60_\ k-in.\right)\left(-0.969_\ in.\right)\left(1500_\ ksi\right)}{\left(1500_\ ksi\right)\left(171.0_\ in.^{4}\right)+\left(30,000_\ ksi\right)\left(3.01_\ in.^{4}\right)} = 251_\ psiThus, 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. (6-6b) we get
= -\frac{\left(60_\ k-in.\right)\left(-1.469_\ in.\right)\left(30,000_\ ksi\right)}{\left(1500_\ ksi\right)\left(171.0_\ in.^{4}\right)+\left(30,000_\ ksi\right)\left(3.01_\ in.^{4}\right)} = 7620_\ psi
The minimum tensile stress in material 2 occurs at the contact plane (C) where y = -0.969 in. Thus,
\sigma_{2C} = -\frac{\left(60_\ k-in.\right)\left(-0.969_\ in.\right)\left(30,000_\ ksi\right)}{\left(1500_\ ksi\right)\left(171.0_\ in.^{4}\right)+\left(30,000_\ ksi\right)\left(3.01_\ in.^{4}\right)} = 5030_\ psiThese 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
\frac{σ_{2C}}{σ_{1C}}=\frac{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.