THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

102 5. Vibrating BarsFigure 5.7. An element of the bar showing shear forces and bending moments.the right end of the element. To sustain static equilibrium in the bent bar, the shearforces and torsions acting on the element must counterbalance each other so thatthere is no net turning momentum. As shown in Figure 5.7, taking the left end ofthe element as the reference pivot point we obtainM(x) − M(x + dx) = F y (x + dx) (5.27)The terms M(x + x) and F y (x + x) are now expanded in a Taylor’s seriesabout point x, with the result that Equation (5.27) becomesF y =− ∂ M∂x = κ2 E Â ∂3 y(5.28)∂x 3In Equation (5.28) the second-order and higher terms in dx have been discarded.In undergoing transverse vibrations the bar is in dynamic rather than static equilibrium.This requires that the right-hand side of Equation (5.27) must equal the rateof increase of the angular momentum of the segment. But as long as the displacementand the slope of the bar remain small, the variations in angular momentumcan be disregarded, and Equation (5.28) should serve as a good approximationof the correlation between the displacement y and the acting force F y . The netupward force dF y in element dx is given bydF y = F y (x) − F y (x + x) =− ∂ F y∂x =−κ2 ÊA ∂4 y∂x dx 4The element undergoes an upward acceleration under the impetus of the force, andthe equation of motion for the mass of the element, ρ Âdx, may now be written as:ρ Âdx ∂2 y∂t 2=−κ2 E Â ∂4 y∂x 4 dxSetting c = (E/ρ) 1/2 , as in the case of longitudinal waves, the last equation changesto∂ 2 y∂t 2=−κ2 c 2 ∂4 y∂x 4 (5.29)