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5.9 Composite Beams

Beams constructed of two or more materials having different moduli of elasticity are referred to as composite beams. Examples include multilayer beams made by bonding together multiple sheets, sandwich beams consisting of high-strength material faces separated by a relatively thick layer of low-strength material such as plastic foam, and reinforced concrete beams. The assumptions of the technical theory for a homogeneous beam (Section 5.6) are valid for a beam composed of more than one material.

5.9.1 Transformed Section Method

To analyze composite beams, we will use the common transformed-section method. In this technique, the cross sections of several materials are transformed into an equivalent cross section of one material on which the resisting forces and the neutral axis are the same as on the original section. The usual flexure formula is then applied to the new section. To illustrate this method, we will use a frequently encountered example: a beam with a symmetrical cross section built of two different materials (Fig. 5.15a).

FIGURE 5.15

Figure 5.15. Beam composed of two materials: (a) composite cross section; (b) strain distribution; (d) transformed cross section.

The cross sections of the beam remain plane during bending. Hence, the condition of geometric compatibility of deformation is satisfied. It follows that the normal strain ɛx varies linearly with the distance y from the neutral axis of the section; that is, ɛx = ky (Figs. 5.15a and b). The location of the neutral axis is yet to be determined. Both materials composing the beam are assumed to obey Hooke′s law, and their moduli of elasticity are designated as E1 and E2. Then, the stress—strain relation gives

This result is sketched in Fig. 5.13c for the assumption that E2 > E1. We introduce the notation

where n is called the modular ratio. Note that n >1 in Eq. (5.46). However, this choice is arbitrary; the technique applies as well for n > 1.

5.9.2 Equation of Neutral Axis

Referring to the cross section (Figs. 5.15a and c), the equilibrium equations ΣFx = 0 and ΣMz = 0 lead to

where A1 and A2 denote the cross-sectional areas for materials 1 and 2, respectively. Substituting σx1, σx2 and n, as given by Eqs. (5.45) and (5.46), into Eq. (a) results in

Using the top of the section as a reference (Fig. 5.15a), from Eq. (5.47) with

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or, setting

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we have

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This expression yields an alternative form of Eq. (5.47):

Equations (5.47) and (5.47') can be used to locate the neutral axis for a beam of two materials. These equations show that the transformed section will have the same neutral axis as the original beam, provided the width of area 2 is changed by a factor n and area 1 remains the same (Fig. 5.15d). Clearly, this widening must be effected in a direction parallel to the neutral axis, since the distance equ272_509a.jpg to the centroid of area 2 remains unchanged. The new section constructed in this way represents the cross section of a beam made of a homogeneous material with a modulus of elasticity E1 and with a neutral axis that passes through its centroid, as shown in Fig. 5.15d.

5.9.3 Stresses in the Transformed Beam

Similarly, condition (b) together with Eqs. (5.45) and (5.46) leads to

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or

where I1 and I2 are the moments of inertia about the neutral axis of the cross-sectional areas 1 and 2, respectively. Note that

is the moment of inertia of the entire transformed area about the neutral axis. From Eq. (5.48), we have

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The flexure formulas for a composite beam are obtained by introducing this relation into Eqs. (5.45):

where σx1 and σx2 are the stresses in materials 1 and 2, respectively. Note that when E1 = E2 = E, Eqs. (5.50) reduce to the flexure formula for a beam of homogeneous material, as expected.

5.9.4 Composite Beams of Multi Materials

The preceding discussion may be extended to include composite beams consisting of more than two materials. It is readily shown that for m different materials, Eqs. (5.47′), (5.49), and (5.50) take the forms

where i = 2, 3,..., m denotes the ith material.

The use of the formulas developed in this section is demonstrated in the solutions of the numerical problems in Examples 5.7 and 5.8.

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