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Figure 2.24 Schematic of beam for bending dynamics. Figure 2.25 Schematic of differential beam segment.

cross-sectional neutral axes associated with pure bending in and normal to the plane of the diagram in Fig. 2.24. For simplicity, however, we will only consider uncoupled bending in the x-y plane, thus excluding initially twisted beams from the development. The bending deflections are denoted by i>( \', 0 in the v direction. The x axis is presumed to be straight, thus excluding initially curved beams. We will continue to assume for now that the properties of the beam allow the x axis to be chosen so that bending and torsion are both structurally and inertially uncoupled. Finally, the transverse beam displacement, v, will be presumed small to permit a linearly elastic representation of the deformation.

A free-body diagram forthe differential beam segment shown in Fig. 2.25 includes the shear force, V, and the bending moment, M. Recall from our earlier discussion on torsion that an outward-directed normal on the positive x face is directed to the right, and an outward-directed normal is directed to the left on the negative x face. By this convention, V is the resultant of the transverse shear stresses in the positive v direction (upward in Fig. 2.25) on a positive x cross-section face and in the negative v direction on a negative x cross-section face. In other words, a positive shear force tends to displace the positive x face upward and the negative x face downward, as depicted in Fig. 2.25. The bending moment, M, is the moment of the longitudinal stresses about a line parallel to the z-axis (out of the paper in Fig. 2.25) at the intersection between the cross-sectional plane and the neutral surface. Thus, a positive bending moment tends to rotate the positive x face positively about the "-axis (in the right-handed sense) and the negative x face negatively about the "-axis. This will affect the boundary conditions, as noted below. The distributed loading (with units of force per unit length) is denoted by q. The equation of motion for transverse beam displacements can be obtained by setting the resultant force on the segment equal to its mass times its acceleration, which yields and leads to dV d2v

where m — p A is the mass per unit length, p is the material density, and A is the cross-sectional area. We must also consider the moment equation. We note here that the

cross-sectional rotational inertia (about an axis normal to the page) will be ignored because it has a small effect. Taking a counterclockwise moment as positive, we sum the moments about the point a to obtain

which, after we neglect the higher-order differentials (i.e., higher powers of dx), becomes

0 0