Question 5.14: A composite beam (Fig. 5-44) is constructed from a wood beam...
A composite beam (Fig. 5-44) is constructed from a wood beam (100 mm × 150 mm actual dimensions) and a steel reinforcing plate (100 mm wide and 12 mm thick). The wood and steel are securely fastened to act as a single beam. The beam is subjected to a positive bending moment M = 6 kN·m.
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 = 10.5 GPa and E_2 = 210 GPa.
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). 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=(h_1 – 75 mm)(100 mm \times 150 mm)= (h_1 – 75 mm)(15000 mm^2)
\int_{2}{ydA} =\bar{y} _2 A_2=-(156 mm – h_1)(100 mm \times 12 mm)= (h_1 – 75 mm)(1200 mm^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. (5-50) gives the equation for locating the neutral axis, as follows:
E_1\int_{1}{ydA} +E_2 \int_{2}{ydA} =0 ( 5-50)
or
(10.5 GPa)(h_1 – 75 mm)(15000 mm^2) + ( 210 GPa) (h_1 – 75 mm)(1200 mm²) = 0Solving this equation, we obtain the distance h_1 from the neutral axis to the top of the beam:
h_1 = 124.8 mm
Also, the distance h_2 from the neutral axis to the bottom of the beam is
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 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} (100 mm)(150 mm)^3 + (100 mm)(150 mm)(h_1 – 75 mm)^2= 65.33 \times 10^6 mm^4
Similarly, for area 2 we get
I_2=\frac{1}{12} (100 mm)(12 mm)^3 + (100 mm)(12 mm)(h_2 – 6 mm)^2= 1.18 \times 10^6 mm^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} (100 mm)h_1^3+ \frac{1}{3} (100 mm)h_2^3
= 10^6(64.79 + 172) mm^4 = 66.51 \times 10^6 mm^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).
\sigma _{x1}=-\frac{MyE_1}{E_1I_1+E_2I_2} \sigma _{x2}=-\frac{MyE_2}{E_1I_1+E_2I_2} (5-53a,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 \sigma_{1A} and using Eq. (5-53a), we get
\sigma _{1A}=-\frac{Mh_1E_1}{E_1I_1+E_2I_2}=-\frac{(6 kN\cdot m)(124.8 mm)(10.5 Gpa)}{(10.5 GPa)(65.33 \times 10^6 mm^4) + (210 GPa)(1.18 \times 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. Proceeding as in the previous calculation, we get
\sigma _{1C}=-\frac{(6 kN\cdot m)(-25.2 mm)(10.5 GPa)}{(10.5 GPa)(65.33 \times 10^6 mm^4) +(210 GPa)(1.18 \times 10^6 mm^4)}= 1.7 MPa
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 = 37.2 mm. Hence, from Eq. (5-53b) we get
\sigma _{2B}=-\frac{M(-h_2)E_2}{E_1I_1+E_2I_2}=-\frac{(6 kN\cdot m)(-37.2 mm)(210 Gpa)}{(10.5 GPa)(65.33 \times 10^6 mm^4) + (210 GPa)(1.18 \times 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,
\sigma _{2C}=-\frac{(6 kN\cdot m)(-25.2 mm)(210 GPa)}{(10.5 GPa)(65.33 \times 10^6 mm^4) + (210 GPa)(1.18 \times 10^6 mm^4)}= 34 MPa
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
\sigma_{2C} /\sigma_{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.