## 2

along the depth of a cross section of the beam vary linearly; i.e., a plane section before loading remains plane after loading. Based on this observation, the stresses at various points in a beam may be calculated if the stress-strain diagram for the beam material is known. From these stresses, the resulting internal forces at a cross section may be obtained.

Figure 3.23a shows the symmetrical cross section of the beam shown in Fig. 3.22. The strain varies linearly along the beam depth (Fig. 3.23b). The strain at the top of the section is compressive and decreases with depth, becoming zero at a certain distance below the top. The plane where the strain is zero is called the neutral axis. Below the neutral axis, tensile strains act, increasing in magnitude downward. With use of the stress-strain relationship of the material (e.g., see Fig. 3.11), the cross-sectional stresses may be computed from the strains (Fig. 3.23c).

FIGURE 3.22 Uniformly loaded, simply supported beam.

FIGURE 3.22 Uniformly loaded, simply supported beam. FIGURE 3.23 (a) Symmetrical section of a beam develops (b) linear strain distribution and (c) nonlinear stress distribution.

If the entire beam is in equilibrium, then all its sections also must be in equilibrium. With no external horizontal forces applied to the beam, the net internal horizontal forces any section must sum to zero:

J Cb J Cb where dA = differential unit of cross-sectional area located at a distance y from the neutral axis b( y) = width of beam at distance y from the neutral axis f(y) = normal stress at a distance y from the neutral axis Cb = distance from neutral axis to beam bottom Ct = distance from neutral axis to beam top

The moment M at this section due to internal forces may be computed from the stresses f(y):

The moment M is usually considered positive when bending causes the bottom of the beam to be in tension and the top in compression. To satisfy equilibrium requirements, M must be equal in magnitude but opposite in direction to the moment at the section due to the loading.

### 3.16.1 Bending in the Elastic Range

If the stress-strain diagram is linear, the stresses would be linearly distributed along the depth of the beam corresponding to the linear distribution of strains:

where f, = stress at top of beam y = distance from the neutral axis

Substitution of Eq. (3.58) into Eq. (3.56) yields f - yb(y) dy = - i* yb(y) dy = 0 (3.59)

Equation (3.59) provides a relationship that can be used to locate the neutral axis of the section. For the section shown in Fig. 3.23, Eq. (3.59) indicates that the neutral axis coincides with the centroidal axis.

Cb C, C,Cb C, where }CC'Jb(y)y2 dy = I = moment of inertia of the cross section about the neutral axis. The factor I/c, is the section modulus St for the top surface.

Substitution of f,/ct from Eq. (3.58) into Eq. (3.60) gives the relation between moment and stress at any distance y from the neutral axis: ## Renewable Energy 101

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