or by CP'; we obtain for the equation to the neutral line in respect to the part CA of the beam (Art. 360) Since, moreover, the forces impressed upon any portion CQ of the beam, terminating between A and E, are the elastic forces developed upon the transverse section at Q, the resistance pa of the support at A, and the load upon CQ, whose moment about Q is represented by CQ3, we have (equation 501), representing CQ by x, Representing the inclination to the horizon of the tangent to the neutral line at A by 6, dividing equation (532) by “, integrating it between the limits x and a,, and observing dy that at the latter limit =tan. ß, we have, in respect to the dx portion CA of the beam, EI/dy μ dx (—tan.ß)=ja-ja,' .... (534). Integrating equation (533) between the limits and a, and dy observing that at the latter limit =0, since the neutral line at E is parallel to the horizon, EI du dx ..(535); which equation having reference to the portion AE of the dy beam, it is evident that when x=a, de =tan. B. ΕΙ .. -tan.ß=ža(a—a‚)'—¡(a'—a‚,')=¦(a—a‚)(2aa—4aa ̧—à ̧”) عليكم Substituting, therefore, for tan. B in equation (534), and reducing, that equation becomes Integrating equation (535) between the limits a, and x, and equation (537) between the limits 0 and x, and representing the deflexion at C, and therefore the value of y at A, by D, EI y= x + {a(a—a,)' — ¦ a'} x . . . . . (538); -y=2+x*+ ..... the former of which equations determines the neutral line of the portion AE, and the latter that of the portion CA of the beam. Substituting a, for a in the latter, and observing that y then becomes D,; then substituting this value of D, in the former equation, and reducing, ΕΙ D1= μα 24EI {12a(a—a,)'—(ta”—a,')} .... Y = 1, x' — fa(x—a,)' + } a {3(a—a,)' — a'} x . . . . (540). por The latter equation being that to the neutral line of the tion AE of the beam, if we substitute a in it for æ, and represent the ordinate of the neutral line at E by y1, we shall obtain by reduction на 24EI {4(a,+2a)(a—a,)'—3a'} .... (541). If a,=0, or if the loading commence at the point A of the beam, the value of y, will be found to be that already deter mined for the deflexion in this case (equation 530). Now, representing the deflexion at È by D,, we have evidently D, D,-y1. 374. THE CONDITIONS OF THE DEFLEXION OF A BEAM LOADED UNIFORMLY THROUGHOUT ITS LENGTH, AND SUPPORTED AT ITS EXTREMITIES A AND D, AND AT TWO POINTS B AND C SITUATED AT EQUAL DISTANCES FROM THEM, AND IN THE SAME HORIZONTAL STRAIGHT LINE. Let AB=a,, AD=2a. Let A be taken as the origin of the co-ordinates; let the pressure upon that point be represented by P1, and the pressure upon B by P,; alsc the load upon each unit of the length of the beam by . If P be any point in the neutral line to the portion AB of the beam, whose co-ordinates are x and y, the pressures applied to AP, and in equilibrium, are the pressure P, at A, the load a supported by AP, and producing the same effect as though it were collected over the centre of that portion of the beam, and the elastic forces developed upon the transverse section of the beam at P; whence it follows (Art. 360.) by the principle of the equality of moments, taking P as the point from which the moments are measured, that Integrating this equation between the limits a,, and x, and representing the inclination to the horizon of the tangent to the neutral line at B by dy ΕΙ '—tan. ß‚) =¿μ(œ°—a‚”)—‡P,(œ°—a‚') . . . . (544). Integrating again between the limits 0 and x, EI(y-x tan. 6)—¿ μ(‡x —α ̧3x)—¿P1(x3-a ̧1x)... (545), Whence observing that when x=a,, y=0, 2 EI tan. B,a,'-P,a,' .... (546). Similarly observing, that if x and y be taken to represent the co-ordinates of a point Q in the beam between B and C, the pressures applied to AQ are the elastic forces upon the section at Q, the pressures P, and P, and the load x, we have d'y EI- 2 =μx2-P1x-P,(x—a,). .... (547). Integrating this equation between the limits a, and x, and observing that at the former limit the value of sented by tan. 6,, we have dy d. is repre Now it is evident that, since the props B and C are placed symmetrically, the lowest point of the beam, and therefore of the neutral line, is in the middle, between B and C; so 0, when x=a. Making this substitution in equation (548), dy that = da -EI tan. B, (a'—a,')—¿P, (a'—a,')—P, (a—a,)'.. (549). Since, moreover, the resistances at C and D are equal to those at B and A, and that the whole load upon the beam is sustained by these four resistances, we have P1+P1=μa. . . . . (550). Assuming a=na, and eliminating P1, P,, tan. B, between the equations (546), (549), and (550), we obtain Making x=0 in equation (544); and observing that the cor dy responding value of is represented by tan. B,, we have EI (tan. ß, —tan. ß,)=—¡μa,'+¿P,a,'. 2 Substituting for tan. B, and P, their values from equations (553) and (551), and reducing, BC of the beam, respectively, by D, and D,, and by x, the distance from A at which the deflexion D, is attained, we have, by equations (544) and (545), -EI tan. 6,= (x‚'—a‚”)—¿P‚(x,'—a,') The value of D, is determined by eliminating, between these equations, and substituting the values of P, and tan. 3, from equations (551) and (553). Integrating equation (548) between the limits a, and a, and observing that at the latter limit y=D,, we have EID, EI(a-a,) tan. ß, +¡μ {1(a*—a‚*)—a‚°(a—a,)} — 2 ¿P, {}(a'—a,')—a, (a—a,)} -¿P,(a—a,)'. Substituting in this equation for the values of tan. ß,, P1, P,, and reducing, we obtain pa (1—n)3 D1 = 48EI (3—2n) {n'— 2n3 —8n+6} . . ..... (556). Representing BC by 2a,, and observing that a, = AEAB-a-na-(1-n)a, 375. A BEAM, HAVING A UNIFORM LOAD, SUPPORTED AT EACH EXTREMITY, AND BY A SINGLE STRUT IN THE MIDDLE. If, in the preceding article, a, be assumed equal to a, or n=1, the two props B and C will coincide in the centre; and the pressure P, upon the single prop, resulting from their coincidence, will be represented by twice the corresponding value of P, in equation (552); we thus obtain tan. ẞ,=0. (558). |