0 37. In example No. 36, calculate M。 independently, by taking a section at the centre. MP.c-2x 5-2 × 10-2 × 15 =7×20-10 20 30 80 ft.-tons. 38. A beam 39 ft. span supports twelve loads, each 10 cwts., and placed symmetrically on the span at intervals of 3 ft. Find the maximum bending moment, and the height of the parabola which gives the bending moments; also, from the equation to the parabola find the bending moments M1-5, M4.5 and M ̧, all measured from the centre. Here W 120 cwts.; c 195 ft.; l 39 ft., and n = 12. = Max. bending momt., M1 = }. = it is the ordinate of the middle point of the chord joining the tops of y... and y7.5 Y7.5° 39. A beam 80 ft. span supports a load of 100 tons. Find the bending moments at the centre, and at twenty feet from the end of span; first, if the load be uniformly distributed; and, second, if it be distributed in equal amounts at intervals of two feet. X For uniform load, M1 =}W.l = } × 100 × 80 = 1000 ft.-tons. For distributed load, n = 39, W 100, 80, and c = 40; n+1 W. ×0 x 100 x 80 1026 ft.-tons. n In this case the bending moments exceed byth, that is by th, of themselves, the bending moments for the uniform load. Beam uniformly loaded and with a load at its centre.-Fig. 29. Let U be the amount of the uniform load, then the bend U ing moment at x due to it alone is (c2x2); let W be the 4c load at the centre, then the bending moment at x due to it (c-x); summing these, we have W alone is 2 the equation to the bending moment for positive values of x, that is, for the left half of the span. Putting y instead of Mx, we have a curve, the ordinates of which are the bending moments for the left half of span; this curve is a parabola with its axis vertical, and its apex above BC the span. To find the position of the apex A1, it is only necessary to find that value of x which makes y greatest; now y is greatest when (c-x) (c + x + 2 Wo) is a maximum; and since the sum of these two factors is constant, their product is greatest when they are equal; putting then we have 2 Wc as the value of x which makes y greatest; the negative sign denotes that A, lies to the right of O, so that OS, is to be laid W off towards the right and equal to c. The height of A, is the value of y when we substitute this value for x; that is, apex of the parabola for the uniform load alone. A,DC is the same parabola with its apex at the symmetrical point A, and the portion DC gives the bending moments for the right half of the span. Since the co-efficient 4c of 22 is the principal equation to the parabolas A‚DB and A,DC, each referred to its own apex as origin, is this is also the principal equation to the parabola BAC for the uniform load alone, so that all three parabolas are identical. We might suppose the diagram for the uniform load alone to consist of two parabolas lying on the top of each other; and that upon the addition of the load W at the centre, they both move upwards, while the one moves towards the right and the other towards the left. The Bending Moment Diagram is BDC. It can be shown that the tangent at D to the parabola A,DC cuts off CE equal to the height of A. ODEC is an approximate bending moment diagram made with straight lines; and it is safe, since the ordinate of any point on DE is greater than the ordinate for the corresponding point on DC. W Graphical Solution.—With a scale of feet for horizontals, lay off BC equal to the span, fig. 29. From the centre O lay off OS, and OS, to the right and left, each equal to, and draw verticals upwards from S1, O, and S. Apply the parallel rollers to the span BC; place any parabolic segment against the rollers, as in fig. 16, with its apex on the vertical through S; shift the rollers till the curved edge passes through B, the end of the span on the opposite side of the centre from S1, and draw the curve BDA. Again, place the segment with its apex on the vertical through S2; shift the rollers till the curved edge passes through the end C, and draw the curve CDA,; then BDC is the bending moment diagram. The scale, of say ft.-lbs., for verticals is to be constructed such that OD measures (U+2W)l, where U and W are in lbs. and l is in feet. The same scale may be constructed as follows:-Place the parabolic segment with its apex on the vertical through O; shift the rollers till the curved edge passes through B and C; draw the curve BA,C, and make a scale of ft.-lbs. for verticals, such that OA, measures upon it Ul, where U is in lbs. and 7 is in feet. COROLLARY.-For the same uniform load, although different loads be put at the centre, A,DB is always the same parabola; as the load at the centre increases, the apex A, moves from the centre, and the arc DB is a part of the wing of that parabola further from the apex. Now the wing of a parabola gets flatter as its distance from the apex increases; hence, if U be constant and W be increased, BD becomes flatter and flatter; and if W be very great compared to U, BD is sensibly a straight line. Cantilever uniformly loaded and with a load at its free end. -Fig. 30. As in the previous case, add the bending moments at the section distant x from O the fixed end, due to the loads separately; thus We consider the bending moments on a cantilever negative as compared with those on a beam, and so they will be represented by ordinates drawn down from the span instead of up as in the case of beams. If we further put y instead of MÅ for the ordinate at the point on the curve corresponding to any value of x, then y will be Mx only for values of x from 0 to c; 'and |