By V. I. Smirnov and A. J. Lohwater (Auth.)

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**Example text**

19). We use the convention that forces acting on the beam are positive if directed down wards. We isolate section N of the beam with abscissa x. Let y denote the displacement of the point on the neutral axis, and R the radius of curvature of the deformed axis. I t is shown in the theory of strength of ma terials that, with certain assumption regard ing the character of the deformation and the position of the beam relative to axes OX, OY, the equation of equilibrium is to be obtained as follows: we neglect the part of F I G .

The problem therefore amounts to investigating the bending of a supported beam under the action of continuously distributed loading f(x) = — kx. We start by calculating P0 and Ph the reactions of the supports. The total loading is Jfc|d| = fc/2 2 The reactions at the supports O and L due to the elementary loading fc£d| are, in accordance with the usual law of levers: kHl- ■f) At A k & At —- d | and —=— d£. £_±,, , , - , - , . x-o = °; (34) i/Ix-r = °- The general solution is: -h Z2*3 ( x* , . n n \ Constants Gx and C2 are found from conditions (34): C2 = 0; C1 = 7 60 -^-^, whence finally: V = 3 " ^ (3* 5 - 10/2 ^3 + 7/%).

Which corresponds to a maximum for \y\. 54 ORDINARY DIFFERENTIAL EQUATIONS [17 Maximum deflection thus occurs towards the end L and not at the centre, its value being: ^ — l/lx-if. 348- 360j57/ I8QJM-• 17. Lowering the order of a differential equation. We notice a number of particular cases in which the order of an equation can be lowered. 1. Let the function y and a certain number of consecutive derivatives of y:y',y", . . , y**""1*, be excluded from the equation, which has t h e form: &(xyyW,y(k+1\..