The Bending Moment Diagram BHEC (fig. 32), consists of the above parabola with its axis vertical and its apex A situated as described above, and drawn both ways till it meets the verticals through the ends of the load at H and E, and of the straight lines HB and EC. In the above equation W 2' to the parabola, the coefficient of x2 is and the principal equation to the curve is this is the principal equation to the curve on fig. 25, so that in the present case the parabola is the same as that for an uniform load of intensity w over the whole span. BH and CE are the same as if the whole load were at G, therefore they meet at R, a point on the vertical through G; if the load were concentrated at G, BRC would be the bending moment diagram. At any point as L, the ordinate to HR is greater than that to HA, because the former is the product of P into BL, while the latter is that same product minus the moment of the load area to the left of L; that is every point in HR is outside the curve. For such a point as Q, by substituting gQ for x in the equation to the parabola, its ordinate is again less than that of BH, so that BH is a tangent at H to the parabola, and EC is a tangent at E; T and V are the middle points of AU and AZ, therefore the half extent of load. TV = 1UZ = k COROLLARY. Since HAE is always the same parabola for the same value of w, it is evident that A will be higher when the parabola passes through B and C; that is, when the whole span is loaded. This is the same result as corollary 2, page 68, derived in this case by geometry. Graphical Solution.—With a scale of feet for horizontals,lay off B'C' (fig. 32) equal to the span, and K'F' equal to the extent of load in its proper position. From G the centre of the load, lay off GS towards Ò the centre of the span, and equal to the same fraction of GO, as the extent of load is of the span. Apply the parallel rollers to B'C'; place any parabolic segment against the rollers with its apex at any point A on the vertical through S, and draw the curve HAE between the verticals through K' and F; draw UAZ with the rollers; bisect AU in T, and AZ in V; produce TH to meet the vertical through B' in B, and VE to meet the vertical through C in C, and join BC. The accuracy of the drawing will be tested by observing that BC should be horizontal, and that the slopes should meet the vertical through G in one point R. A scale of say ft.-lbs. for verticals is constructed as follows: Place the segment with its apex on the vertical through the centre O, shift the rollers till the curved edge passes through B and C, and draw the curve BAC (not shown on fig. 32); then the height of the apex 4 should measure wl2, where w is the intensity in lbs. per foot, and l is the span in feet. Another method.-If the extent of the load be small, it will generally happen that BC will not be quite horizontal on account of the shortness of VE the line to be produced, and the following construction may be more satisfactory. Lay off the span BC, construct any triangle BRC with its apex R on the vertical through G; lay off Cf equal to half the extent of load, draw fT parallel to CR, and TV horizontal; TV is then equal to fC half the extent of load; make TA equal to TU, and DA is the maximum bending moment. This may be sufficient, but as many points on the parabolic arcs HA and EA as may be required can be plotted as on fig. 14. The scale for verticals is to be such that the height of R may measure upon it what the bending moment at G would be if the load were concentrated there. Graphical Solution for Particular Case.-Fig. 33. When the load extends to one end, say B, of the span, the first graphical solution given above may be made shorter and more direct thus:-Lay off BC equal to the span; from G lay off k GS towards O, so that GSGO; apply the rollers to C BC; place any parabolic segment against the rollers with its apex on the vertical through S, and shift the rollers till the curved edge passes through B; draw the curve BAE, and join EC. Construct a scale as described above. Cantilever uniformly loaded on a portion of its length.-Fig. 34. Let w lbs. per ft. run be the intensity of the load, and let kft. be its extent. The loaded part AD may be considered to be a cantilever of length k uniformly loaded, so that the bending moment diagram is the parabola AE, as in fig. 27, for the whole length loaded. Suppose now the whole load concentrated at B the centre of gravity of the load, then BEF would be the bending moment diagram, as in fig. 23; and for points between D and K, the moment is the same as for the actual distribution of the load, because for sections at such points, the whole load is to the left whether we consider it concentrated at B or spread over AD. It is evident that for sections between B and D, the concentrated load would be all to the left, while only part of the actual distributed load is so situated. EF is a tangent at E to AE, since B is the middle point of AD. The Bending Moment Diagram AEFK consists of the above parabola, with its axis vertical and its apex at the free end, drawn till it meets the vertical through the end of the load at E, and of EF the tangent at E to the curve. Graphical Solution.—With a scale of feet for horizontals, lay off AK (fig. 34) equal to the length, and from the free X Ac w=↓wk w lbs per ft. --C-ky- B OK end A lay off AD the extent of the load. Apply the parallel rollers to AK; place any parabolic segment against the rollers with its apex at the free end A, and draw the curve AEG to meet the verticals through the right end of the load and the fixed end, at E and G respectively; EG may only be dotted. Bisect AD in B; join BE with a dotted line, and produce it with a full line to meet the vertical through the fixed end at F, then KAEF is the bending moment diagram. Construct a scale of say foot MW(C-2) Mc-k= Wk Fig.34. F lbs. for verticals such that KF may measure Wc = k 2 where W total load in lbs. ; c = length, and k = extent of load, both in feet. Examples. 40. A beam 54 feet span is loaded uniformly for two thirds of its length from the left end with 10 cwt. per foot Find the position and magnitude of the maximum bending moment. See fig. 33. run. In this case c = 27 ft., and OG = 9 ft., measured to the left of 0. = From G lay off GS= 6 ft. GO, since the load extends over two thirds of the span, and the maximum moment occurs at S, that is at 3 ft. to the left of the centre. Suppose the whole load W = 360 cwt. is concentrated at G; then Taking a section at S, the portion of the span to the left is 24 ft., so that the load upon it is 240 cwt. acting downwards, and if supposed to be concentrated at its centre, its leverage about the section is 12 ft.; at the same time P acts upwards with a leverage of 24 ft., and 240 × 24-240 × 12 2880 ft.-cwts. max. 41. The left half of a beam 32 feet span is uniformly loaded with one ton per foot run. Find the position and magnitude of the maximum bending moment. Ans. The maximum occurs at the section four feet to the left of the centre, and its value is M 72 ft.-tons. = 42. A beam 50 ft. span is uniformly loaded from the right end for an extent of 10 feet, with two tons per foot run. Find the position and magnitude of the maximum bending moment. - 16 = Ans. The maximum occurs at the section 16 feet to the right of the centre, and its value is M 81 ft.-tons. 43. A beam 36 feet span is loaded uniformly from the middle point towards the left to an extent of 12 feet, with 2 tons per foot run. Find the position and magnitude of the maximum bending moment. See fig. 32. 6 ft., and GS OG In this case OG 2 feet, since the extent of load is one third of span; the maximum bending moment is at S, four feet to the left of the centre. Suppose the whole load W, 24 tons, concentrated at G, we have Taking a section at S, the extent of the load to the left is 8 ft., and is equivalent to 16 tons acting downwards with a leverage of 4 feet, while P acts upwards with a leverage of 14 feet; hence |