where represents the load per inch of the length of the beam necessary to produce rupture. In the case of a rectangular beam, this equation becomes 2. To determine the form of the beam of greatest strength having a rectangular section of given breadth b, let y be taken to represent its depth PQ at a point P, and x its horizontal distance from the point A. Then " I= 11⁄2by3, c1 = {y; also P1P1 (equation 637) representing the moment of the resultant of the pressures AP about the centre of gra upon vity of PQ=μax-x2; therefore by equation (637) μax — μx2 = 1 Sby2; the equation to an ellipse, whose vertex is in A, and its centre at C. 3. To determine the beam of absolute maximum strength, let it be assumed, as in Art. 422., that the area of the section of the rib is exceedingly small as compared with the areas of the sections of the flanges; and let the area of the section of the lower or extended flange be n times that of the upper; then, as in Art. 422., I A1 { (n + 1)(d)2 + nd2 2) + 12ny2 с 6 2y+d1+(n+2)d2 also Pip1=μax-1px; whence, by equation (637), 4. If it be proposed to make the rib or plate uniting the two flanges everywhere of the same depth *, and so to vary the breadths of the flanges as to give to the beam a uniform strength at all points under these circumstances; representing by y the breadth of the upper flange at a horizontal distance x from the point of support, we shall obtain, as in Art. 423., Moreover, Pip1 =μаx-μx2=ux(2a-x); whence we obtain by substitution in equation (637), and reduction, x(2a—x) = (5d;) {(n+1) (d ̧2+nd,2) + 12no2} y (673); the equation to a parabola, whose axis is in the horizontal line bisecting the flange at right angles, its parameter represented by the coefficient of y in the preceding equation, and half the breadth of the flange in the middle determined by the formula 6ca2μ {(n + 1) (d ̧2 +nd22)+12nc2} Sd1 (674). The equation to the lower flange is determined by substi yda tuting for y, in equation (673), ; whence it follows that the breadth of the lower flange in the middle is equal to that of the upper multiplied by the fraction yd nd 425. A RECTANGULAR BEAM OF UNIFORM SECTION, AND UNIFORMLY LOADED THROUGHOUT ITS LENGTH, IS SUPPORTED BY TWO PROPS PLACED AT EQUAL DISTANCES * As in Mr. Hodgkinson's construction. where represents, as before, the load upon each inch of the length of the beam, b its breadth, c its depth, and a, the length of the portion AC. Again, it is evident that the point of rupture of the portion AB of the beam is at E. Now the value of P11 (equation 637) is, in respect to the portion AE of the beam, μ2a(a—a,) — —μ‚a2; 2a representing the whole length of the beam, μ, the load upon each inch of the length of the beam which would produce rupture at E, and therefore a the resistance of each prop in the state bordering upon rupture; I also = bc2. Whence, by equation (637), μ‚a(a—a,)— 1μ‚а2=μ‚a(¦a—a1)=} be2S; 426. THE BEST POSITIONS OF THE PROPS. If the load be imagined to be continually increased, it is evident that rupture will eventually take place at A or at E according as the limit represented by equation (675), or that represented by equation (676), is first attained, or according as μ, or μ is the less. Let be conceived to be the less, and let the prop A be μι moved nearer to the extremity C; a, being thus diminished, , will be increased, and μ, diminished. Now if, after this change in the position of the prop, μ, still remains less than Hit is evident that the beam will bear a greater load than it would before, and that when by continually increasing the load it is brought into the state bordering upon rupture at A it will not be in the state bordering upon rupture at E. The beam may therefore be strengthened yet further by moving the prop A towards C; and thus continually, so that the beam evidently becomes the strongest when the prop is moved into such a position that, may just equal μ This position is readily determined from equations (675) and (676) to be that in which 427. A RECTANGULAR BEAM OF UNIFORM SECTION AND UNIFORMLY LOADED IS SUPPORTED AT ITS EXTREMITIES, AND BY TWO PROPS SITUATED AT EQUAL DISTANCES FROM THEM TO DETERMINE THE CONDITIONS OF RUPTURE. mined by the equation ux, P, it being observed that, at the section of rupture, the neutral line is concave to the axis of z, and therefore the second differential coefficient (equation 543) negative. The value of P is that determined by equation (551); so that where a represents the distance AE, and na the distance AB. ds The curvature of the neutral line being everywhere exceedingly small, may be assumed 1. The expression for the radius of curvature in terms of the rectangular co-ordinates resolves itself therefore, in this case, into the second differential coefficient. Let P represent the intersection of the neutral line with the plane of rupture, and μ, the load per inch of the whole. length of the beam which would produce a rupture at P. Now the sum of the moments of the forces impressed on AP (other than the elastic forces on the section of rupture) is represented, in the state bordering upon rupture, by P11112; or, since P111, it is represented by P,2; 2μ1 whence it follows by equation (637) that the conditions of the rupture of the beam between A and B are determined by Eliminating the value of P, between equations (551) and (679), we obtain Sbc2 3a2 { 2 8n(2n-3) } Substituting this value of p, in equation (679), and re ducing Similarly, it appears by equation (547) that the point of greatest curvature between B and C is E; if the rupture of the beam take place first between these points, it will therefore take place in the middle. Let μ represent the load, per P P |