Page:The Building News and Engineering Journal, Volume 22, 1872.djvu/146

 130 THE BUILDING NEWS. Fes. 16, 1872. ——— —-_—_ Orr ans multiplied by the secant of CB'B=EG'G. Each of the pillars FF’, GG’ supports half the weight lying between them. ‘The pressures on C’'B' and B’E’ are increased by that on E’G’. The tension on CBis increased by that on BE, We may pursue this process of removing the points of support till we form a truss of any desired length, loaded with any desired load, and the follow- ing general properties of the uniformly-loaded truss will become apparent : 1. Each tie supports the weight lying between its centre and the centre of the truss, and is exposed to a tensile strain equal to the weight supported, multi- plied by the secant of the angle between the tie and pillar. 2. Each pillar supports the weight lying between itself and the centre of the truss. 3. The pressure upon the upper chord at any point is equal to the sum of the tensions upon all the ties between that point and the pier, multiplied by the sine of the angle between the tie and pillar, or equal to the sum of the horizontal components of these tensions. 4, The tension upon the lower chord at any point is equal to the pressure upon the upper chord between the same two consecutive ties. A truss constructed as shown in Fig. 3 has all the bracing necessary to insure its stability with respect to a uniformly-distributed load. No rational pur- pose would be served by ties joining the points C'B, B'E, E'G, that is, by what are usually termed coun- ter ties. Anincrease of the load would bring no additional strain upon such ties, but would tend to loosen them if already screwed tight, The points connected by such ties tend to approach one another under the strain of a uniformly-distributed load. Let us now consider the case of an unequally-dis- tributed load, Suppose A (Fig. 4) to be one of the sup- ports and the portion of the truss next A to be loaded, the rest of the truss being without load. Let A repre- sent the pressure on the pier.* Any tie, as CB’, sus- tains a vertical pressure equal to A less the weight between A and ¢, the line cc’ being drawn through the middle of the panel BC. That is, the portion of the truss to the right of cc’ tends to separate itself from the portion of the left of it by a force equal to A less the weight Ac. It is evident that a tie joining BC’ would offer no resistance to this tendency. As we recede from A towards the right, it is evident that we shall come to a point g, where the weight between A and g is equal to A, thatis where the tendency of the right-hand portion to separate itself from the left-hand portion is null. Beyond gg’, as at hh’, the portion to the left of hh’ tends to separate itself from the portion to the right of it, and this tendency can only be met by ties, as GH’, sloping in the opposite direction. The point g is sometimes called the neutral point, since the shearing strain at that point is nothing. It may with equal pro- priety be called the breaking point, since the strain upon the upper and lower chord where it occurs is a maximum. Thus the compressive strain upon the upper chord at g’ is the sum of the horizontal components of the strains upon all the ties to the left of g'. The ties under strain to the right of g' sloping in a contrary direction, their horizontal components do not increase the compressive strain upon the upper chord, but servein their aggre- gate to balance the pressure coming from the left of g'. For the same reason the tension upon the lower chord has its maximum value at g'. It is, therefore, at this point, if anywhere, that rupture would take place under the assumed load. The breaking point shifts its position with every change in the distribution of the load. Thus, in an unloaded railroad bridge, the breaking point lies at the centre. A train enters the bridge, and the break- ing point leaves the centre and approaches the train, reaches its extreme position, returns, passes the centre, reaches its extreme position on that side, and as the head of the train leaves the bridge, the rear not having entered, it returns to the centre. It describes the same cycle of motions while the rear of the train is crossing the bridge. Since only the ties which descend towards the breaking point can be exposed to any strain, it fol- lows that between the extreme positions of the breaking point a truss must be provided with ties sloping in both directions; outside of these positions it requires only ties descending toward the centre. It is customary to speak of ties descending toward the centre as principal ties, and those descending from the centre as counter ties. I adopt this no- by three vertical forces 1, 1ts own weight acting down- wards. 2. The force arising from its connection with the portion of the right ec’, also acting downwards. 3. The pressure, A, acting upwards. As these forces are in equi- librium, the second is equal to the difference between the first and third, . menclature, although it implies a misunderstanding of the function of what is called the counter tie, as if it served to counteract or oppose the action of the principal tie. All ties are principal ties, or under strain, when the breaking point lies on the side towards which they descend, and counter ties, or out of use, in the contrary case. Tt is usual to calculate the strength of a truss with reference to a load of uniform intensity applied to the whole or a part of the length of the truss. Let A B (Fig. 5) represent a truss, b the breaking- point, c the centre, wy, the test load per linear foot. The maximum value of b c, or the greatest deviation of the breaking-point from the centre, oceurs when the test load extends from A to b. For, by the suppo- sition, the total weight between A and b is equal to the pressure at A. Now, any diminution of the loaded portion, as, for instance, by removing the load from 6 to b’ will diminish the weight between A and b by w, x bb’, but will diminish the pres- sure at A by w, x bb’ x ace eS That is, it will make the weight between A and b less than the pressure at A, and will consequently move the breaking-point nearer the centre. Again, by ex- tending the loaded portion to b”, we increase the pressure at A, without increasing the weight between A and b, and consequently move the breaking- point nearer the centre. Since, therefore, we cannot increase or diminish the length of the loaded part without diminishing the distance of the breaking-point from the centre, we conclude that bc represents the maximum yalue of that distance. This maximum yalue of b ¢ may be determined thus :— Let w represent weight of truss per linear foot. sy Wy ” 2, ape load ,, ” ” @ z span in feet. aaee - distance A b in feet. Then— a—x aw 2 (w + wy) & = — + wy r- z a w w a, orz?+2a—g¢g=— ”’ aby Dy w a whenceer= ta— (V1-+ wu, —1). 1 Ss w The negative value of x simply indicates that there is nothing in the conditions of the question to pre- vent the whole system lying to the left of A, as well as to the right. When the extraneous load is equal to the weight of the truss, foot for foot, we have « = a( V 2 — 1), or the greatest deviation of the breaking-point from the centre is 0°0858a, and the length of the portion requiring counter-ties 0°1716a, or about one-sixth of the span. When the load is twice the weight of the truss, foot for foot, we have « = }a(¥ 3 — 1) = 0°366a, or the maximum deviation of the break- ing-point from the centre is 0°134a. That is, the truss will need counter-ties a little more than one- When w, =0, t = = or the eae breaking-point is at the centre. We may here recapitulate these general properties of the truss under an equally-distributed load ;— 1. The breaking point is a point which includes between itself and the pier a weight equal to the pressure on the pier. fourth of its length. * 2. Its maximum deviation from the centre occurs when the portion of the truss between it and the nearest pier is uniformly loaded with the greatest allowable load. 5. To find its maximum deviation from the centre or least distance from the pier, we divide the maxi- mum load per linear foot by the weight of the truss per linear foot, add one to the quotient, and extract the square root of the sum, diminish the root by one, and multiply the result by the product of the span in feet, and the quotient obtained by dividing the weight of the truss per linear foot by the maximum load per linear foot. The result is the least distance of the breaking point from the pier. The above are true independently of the arrange- ment of ties and braces. The following are true for the form of truss under consideration—viz., inclined ties and upright pillars. Tie reaching from top of pillar to foot of succeeding pillar. or indeterminate when mw, equals o. Consider wy, how- ever, as very small with reference to w. In that case, wy 1 Ante is represented by 1 + 3 “1 with a degree of wo w accuracy which increases as ty diminishes. When, there- fore, w, becomes indefinitely small— c=a~0+3,2—-=- wy w. IR 4. Only that portion of the truss between the ex- treme positions of the breaking point requires counter ties. 5. Every pillar bears the weight of truss and load between itself and the breaking point. 6. Every tie descending towards the breaking point bears the weight of truss and load between the middle of the tie and the breaking point, and is ex- posed to a tensile strain equal to the weight sus- stained multiplied by the secant of the angle between tie and pillar. Ties descending from the breaking point bear nothing. 7. The compressive strain upon the upper chord at any point is equal to the algebraic sum of the hori- zontal components of the tensions upon all the ties between the point and either pier. 8. The tensile strain upon the lower chord is equal tothe compressive strain upon the upper chord between the same two consecutive acting ties, Let us now return to our original problem of a weight to be supported midway between two fixed points Aand B. Fig. 6 shows another mode in whick it might be accomplished. Erect the two inclined pillars AC' BC' and support the weight by a rod or tie C'C. To prevent any lateral strain upon the supports we must join CA CB by ties. Draw the two vertical lines aa’ bb’ through the middle points of the two pillars. Each pillar supports (at its middle point) half the weight included between these lines, and is under a compressive strain equal to the weight sustained multiplied by the secant of the angle BC’C. The tie CC’ supports the weight W and half the weight of the ties CA CB. Tbe tensile strain upon CA, CB, is equal to the compressive strain on the pillar multiplied by the sine of the angle CC' A or the horizontal component of that strain, which is also equal to the pressure of the pillars against one another at C’. Removing the points of support to D and E, add- ing the ties AA’, BB’, AD, BE, and the pillars A’D, B'E, A'C', B'C’, and the extraneous weights at A and B, we have the weight of the portion AA’ B'B supported precisely as the weight at C was supported in the former case. Proceeding in the same manner as in the case of upright pillars, we arrive at entirely analogous results for the form of truss shown at Fig. 7, which, omitting all intermediate steps by reason of their similarity to what precedes, may be stated thus :— 1. Only that portion of the truss between the extreme positions of the breaking point requires counter pillars, 2. Every tie bears the weight of the truss and load between itself and the breaking point. 3. Every pillar ascending towards the breaking point bears the weight of truss and load between centre of pillar and breaking point, and is exposed to a compressive strain equal to the weight sustained multiplied. by the secant of the angle between tie and pillar. Pillars ascending from the breaking point bear nothing. 4. The tensile strain upon the lower chord at any point is equal to the algebraic sum of the horizontal components of the compressive strains upon all the pillars between the point and either pier. 5. The compressive strain upon the upper chord is equal to the tensile strain upon the lower chord between the same two consecutive acting pillars. We have thus far considered the simple truss, in which the sloping member passes over only one panel. Let us now consider the system of multiple inter- sections, in which each sloping member embraces two or more panels. In doing so it will only be necessary to consider the system with upright pillars and slop- ing ties, since every principle demonstrable of this system has its corresponding truth in the system with inclined pillars and upright ties, arrived at by methods too similar to require repetition. It is evident that, for any given span, there is @ certain height of truss which will give the required strength with the least material; and also that for any given height of the simple truss there is a cer- tain width of panel more advantageous than any other. It is, no doubt, within the resources of analysis to determine, in a general manner, the best values of these quantities. Yet a more ready and, all things considered, a shorter method, would be to find them for any given span by trial. For a given span let h be the height and J the width of panel so determined. Let 1 Fig. 8, represent a truss of five panels, each equal in width to 21. Suppose it to restat Land M, and to sustain the equal weights W at each of the points F A BG. Let2 represent another truss with four panels of the width 2 1, and two of thewidthl. Let it rest at L and M, and sustain the five weights W, at HDCEK,. Thestrains upon the several members
 * The portion of truss to the left of ce (Fig.4) sacted upon
 * It might appear, at first sight, that x becomes infinite