As before stated (96), loads which are in reality concentrated at the apices may be considered as uniformly distributed if they be such as a uniformly distributed load would bring on the apices in question.

Place the diagram of the moments of rupture (1) for the case of loading, support, &c. (7 to 33) immediately above (or under), and drawn to the same horizontal scale as the skeleton elevation (53, 59) of the girder. (See fig. 41, in which the load is supposed to be uniformly distributed (96, 19) along the upper boom.) Draw verticals right through the diagram at each of the loaded apices, whether on one or both booms (as E E' and G G' fig. 41.) If the lines to or from which the ordinates in the diagram are supposed to be taken be curved, draw chords to that curve at the intersections of the verticals just drawn with that curve (as in fig. 41.) A polygon will be thus formed, and the strain on any bay of either a loaded or unloaded boom will be given (97) by the ordinate to that polygon taken at the opposite apex. (See also 106.)

For the diagonals the shearing forces may be taken directly from the diagram for the case (34 to 52) drawn to a large scale (35), subject to the modifications given under "Web" in the cases above.

106. NOTE. If all the lines in the diagram for the moments of rupture, to or from which the ordinates are directed to be taken, be straight, then the strain in any bay of either boom, whether loaded or unloaded, may be found from the ordinate taken at the opposite apex, by dividing the value of that ordinate by the depth of the girder (unless 97 be complied with, when the latter operation must, of course, be omitted).

a central bay, and if of more than two systems of triangles then with a complete system of triangulation in the half span (as ABCDEFG), but any number of intermediate systems loaded uniformly on one boom for an equal distance on either side of the mid span (see 110): may be applied also to cases of continuous girders (30-33).

BOOMS.-Place the digram (5, 6, 97) for the moments of rupture (1) for the case above (or below) the outline sketch of the girder (59) as in fig. 42.

Draw chords and tangents (as shown in fig.) at points (C'E' G') corresponding with the divisions made by the "primary triangles." Two polygons will thus be formed, one by (or including*) the

* When (21) is applied the load being centrally situated, the chords and tangents will of course extend only along the parabola C E D (fig. 15).

*

chords, another by (or including *) the tangents. For the strain in any bay of the unloaded boom take the ordinate to the circumscribed polygon, at the centre of the bay; and for any bay (as E H) of the loaded boom take the ordinate to the inscribed polygon at the centre of the bay. If the ordinates represent the moments of rupture (4, 5, 97), divide this value by the depth (98) of the girder.

WEB.-Divide the load per unit of length (34) by the number of systems of triangles (= the number of bays in the base of one of the primary tri

angles) for a new unit of weight, w,; thus = w1 or =w'1. Then for

w
n

n

the strain on any pair (95) of lattice bars, -Multiply the shearing force (8 H in the formulæ, or the ordinate in the diagram-34, 35,-calculated or constructed with the new unit w1, or w',) developed at their apex by the secant of the angle they make with a vertical, or increase that shearing force in the proportion of the inclined length of the bar to its vertical depth.

108. Any Lattice Semi-girder loaded on One Boom only.

Booмs.-If the load be concentrated, see 110. diagram for the moments of rupture (4, 5) immediately above (or below) the outline sketch of the girder (59).

Trace out the system of triangulation which terminates fairly at the extremity of the girder (see the thick lines in the fig.) To the curve of moments (4, 5) draw chords and tangents at points (E', D') corresponding to the divisions made by the triangles just traced out (D, E, fig. 43). If necessary for the construction, the diagram may be continued into the abutment. Two polygons will thus be formed, when, whatever be the number of systems of triangulation,For the strain in any bay (84-87) of the unloaded boom take the ordinate,

If distributed, place the

Fig. 43.

at the centre of that bay, to the circumscribed polygon; and for any bay of the loaded boom, to the inscribed polygon. If the ordinates represent the moments of rupture (4, 5, 97), divide their value by the depth of the girder (98).

WEB.-If the load be concentrated, see (110). If distributed, divide the load per unit of length (w) by the number of systems of triangles (=the number of bays in the base of a primary triangle (4 in fig. 43) for a new unit of load -=w1

w

n

w1). Then for the strain on any pair (95) of braces forming

an apex on the unloaded boom,-Multiply the shearing force (S Hx in the formulæ, or the ordinate in the diagram for the case, calculated or constructed with the new unit w-34, 35), developed at their apex, by the secant of the angle which they (each) make with a vertical, or increase the shearing force (as above) in the proportion of the inclined length of the brace to its vertical depth.

* See note, p. 34.

109. Lattice-Girder, if whole, and of more than two systems of triangles, then not without a central apex on one or other of the booms (see fig, 44), loaded uniformly and equally on both the Booms for an equal Distance on either side of the Mid-span, may be applied to continuous girders (30-33).

ВOOMS.-Place the diagram (4, 5) of the moments of rupture (1) immediately

Fig. 44.

above (or below) the outline sketch (53, 59) of the girder (as in the fig., where the load is supposed to extend over the whole length of the girder).

Where the line to or from which the ordinates are taken is curved, draw chords to it at points (E' F'G'C') corresponding to the several apices (on both booms) of the system of triangulation which has an apex at the mid span (E F G C in fig.) For the strain in any bay (84-87) of either boom, take the ordinate to the polygon, thus formed, at the centre of that bay. If the ordinates represent the moments of rupture (4, 5, 97), divide their value by the depth of the girder.

*

WEB.-Divide the load per unit of length (w) by the number of systems of triangulation (= the number of bays in the base of one of the primary

triangles,—3 in fig. 44) for a new unit of load (

n

= w

02).

Then for the

strain in any lattice bar, or pair of bars. Multiply the shearing force (SH in the formulæ, or the ordinate in the diagram for the case-34, 35), developed at its or their mid-length, by the secant of the angle it or they (each) make with a vertical, or increase the shearing force (as above) in the proportion of the inclined length of brace to the vertical depth.

109A. If there be only Two Systems of Triangulation. Then for the strain in any bay (84-87) of either boom. Take the arithmetical mean between the moments of rupture at the two ends of the bay (p and q), and divide it by the depth of the girder.

The chords in the diagram will, however, give the same result.

110. Concentrated Loads on Lattice Girders. If a lattice girder be subjected to the action of a concentrated load at any apex, as at D (fig. 42), the system of triangles upon one apex of which the load is situated (ABCDEFG, &c.) should be considered as constituting the sole web of the girder-that is, as far as the concentrated load is concerned, for there is also the distributed weight of the girder (61). The strains from the concentrated load might be calculated as if the girder were a warren, and as if the intermediate bracing did not exist. Indeed, to introduce other systems of triangles for a stationary concentrated load would be an error in design, unless the boom upon which the load is placed be made sufficiently rigid to distribute part of the pressure upon adjacent intermediate apices, in which case the strains would be very indefinite.

If it be an application of case (21) the lines AF and G B (fig. 15) must be considered to form part of the polygon.

There will be no pairs (95), unless the number of systems of triangles be even.

GIRDERS WITH CURVED OR OBLIQUE BOOMS.

*

111. Any Curved or Oblique whole or Semi-girder with a Singletriangular Web, loaded on One Boom. (This will include many roof principals, bow-string girders, bent cranes, &c.)

BOOMS.-If alternate braces be vertical, then for the strain in any bay of either boom; if not, then for the strain in any bay of the unloaded boom only, -Divide the moment of rupture (Mx in the formulæ, or the ordinate in the diagram (4, 5), for the case 7 to 33) at the opposite apex by the length of the perpendicular let fall from that apex on to the bay. For the strains in the loaded bays, when all the braces are inclined, the best way is by the diagram as follows. Place the diagram of the moments of rupture, immediately above (or below) the outline sketch (53, 59) of the girder. If the lines in the diagram to or from which the ordinates are directed to be taken be curved, draw chords at points corresponding to the position of the apices in the loaded boom. Then for the strain in any bay of the loaded boom, divide the value of the ordinate to the just-formed polygon, taken at the opposite apex, by the length of a perpendicular let fall from that apex on to the bay (121). (See 154.)

WEB. For any brace, first determine the shearing force (34), acting at the middle of the bay on the loaded boom, which forms part of the same triangle with the brace in question. Next, find the vertical components (80, 81) of the strains in those bays which are opposite sides of a quadrilateral figure, whose diagonal is the brace in question. If either of these bays be part of a boom in tension, and sloped as a strut (91), or part of a boom in compression and sloped as a tie (91), add the vertical component of the strain in it (already found) to the shearing force first obtained. Again, if either of the two same bays be part of a boom in compression and sloped as a strut, or part of a boom in tension and sloped as a tie, subtract its vertical component (already found) from the result of the last operation (the addition, if any). The total resulting quantity must then be multiplied by the secant of the angle the brace makes with a vertical, or increased in the proportion of the inclined length of the brace to the vertical distance between its ends. If the sign of the result be negative (—), it shows that the nature of the strain on the brace is opposite to that which its position would have indicated according to the general rules (91). If the brace be horizontal, the shearing force must be disregarded, and the horizontal components of the bays substituted for their vertical components in the process detailed above. The strain in a horizontal bay can have no vertical component.

112. Any Curved or Oblique whole or Semi-girder with a single. triangular Web, loaded equally on Both Booms.

BOOMS. For the strain in any bay of either boom,-Divide the moment of rupture (Mx in the formulæ, or the ordinate in the diagram (4, 5) for the case 7 to 33), taken at the opposite apex, by the length of a perpendicular let fall from that apex on to the bay. (See 154.)

WEB. As in the last case, excepting that the shearing force must be taken at the mid-length of a line joining the centres of the bays, which are two opposite sides of a quadrilateral whose diagonal is the brace.

* See foot note (*), p. 31.

113. For Continuous Girders, and Girders fixed at the Ends, the foregoing methods may be used in connection with cases 24 to 33 and 46 to 52, provided that the conditions stated in the latter and below (113A) are complied with.

113A. Wherever there is a negative moment of rupture at (or for safety in practice near) the abutments, the girder should be anchored down at its extremities. (See foot notes, pp. 13 and 15.)

114. Fixing the Points of Inflexion of Continuous Girders.-The points of inflexion (23) may be practically fixed at any part of a continuous girder subject to a moving load, by severing either boom at that part; if the upper boom, the parts thus severed must be prevented from coming in contact. The structure is thus resolved into a series of independent girders, the strains upon which can then be most readily calculated.

Fig. 45.

B

EXAMPLE.-In the accompanying fig. (45), by severing (or really removing) the bays of the upper boom opposite C and D, AC and D B become virtually semigirders, each having to sustain in addition to the distributed load upon its length, half the total load on CD, suspended at the extremity (11, 39). CD is simply an independent girder supported at both ends.

115. The points of inflection may be considered as fixed in those con

tinuous girders, and girders fixed (or tied back) at the ends, whose depths vary as (or nearly as) the moments of rupture. The strains in these also may be calculated as if the several divisions were independent girders.

If the tension members running down from the towers (as in fig. 46) be made to act simply as suspension chains, the strains

on them may be obtained from 155 or 118.

CALCULATION BY THE COMPOSITION AND RESOLUTION OF FORCES. (APPLICABLE TO ALL CASES OF OPEN-WEBBED INDEPENDENT GIRDERS.)

116. The following principles should be applied to the calculation of the strains on the various members of an open girder (discontinuous) by (1st) finding the reaction of the supports from any weight in the girder, and (2nd) tracing this reactionary pressure throughout the various parts.