is expended in causing compression upon fk; and is the measure of the greatest compression that member can receive through fj. But fk is also liable to compression, or a tendency thereto, from the tension of fl when ƒ and g, are loaded, or g alone, and the other parts unloaded. This, however, in the former case, will never equal the weight at f, and in the latter, the compression will not exceed that just found, resulting from action of fj; as will be better understood here after. Now, as jr' represents the thrust of jk, if we take kv, on jk produced, equal to jr', - raise the vertical vx = ft', and join kx, the line kx represents the resultant of the forces kr and ft (representing thrust of jk and ƒk); and xy, drawn parallel with ke, represents the tension of ke This is the maximum stress of the diagonal ke. * XXX. For, when the left half of the truss has more load than the left hand abutment is required by the statical law to sustain, it is clear that a part of the weight on the left, is transferred from left to right past the centre through dl; that being the only member capable of effecting the transfer. It is also clear, tnat such transferred weight, together with the weight at 2, if any. is sustained by lk and ek, and causes pressure at i, equal to the weight sustained by lk and ek. Also, that this pressure at i, causes a horizontal thrust in ij, which is all transferred to lk (except when kg is in action), and gives a lifting power to lk, equal to 3 of that exerted by y (the vertical reach of the former being that of the latter), that is, equal to of the weight *The sides of the triangle kry, being parallel with the directions of the thrust of lk, the tension of ek, and the resultant kr, of the thrust of jk and ƒk, which 4 forces are in equilibrio at the point k. transferred by lk and ek together. But the lifting power of lk is further increased by of the weight sue. tained by fj, which increases the horizontal thrust of lk, the same as a like amount sustained by ; also, by the weight sustained by ek; this member having 5 times the vertical reach of lk. Now, as each one of these items results from the weight transferred through lk and ck, and is greater or less in proportion as the last named weight is greater or less (ƒ and g being unloaded), since all the conditions are the same, except as to amount of weight, it must follow that the greatest stress of ek, is when ƒ and g are unloaded, and all the other points b, c, &c., are fully loaded - unless it be when ƒ and g, or one of them, be wholly or partially loaded. But any weight at f, increases the thrust and lifting power of lk, through increased action of ÿj and fj both, while it diminishes the amount sustained by ek and lk, whence the action of ek, is diminished, inasmuch as it transfers to k, a less proportion of a less weight. = Again, weight applied at g, while ƒ is unloaded, relieves the tension of fj, and diminishes its lifting power represented by ft' and rx, and if of sufficient amount, relaxes fj, and brings tension upon gk; so that, when the weight at g equals w, or 7w", lk has a lifting power pressure at i, less what is due to the horizontal pull of gk, plus, amount due to horizontal pull of ek; while the weight bearing at k, equals 9" (being weight at e (= 7w") + 2w" through dl). Now if ek lifts as much as ky, lk must have as great a horizontal thrust as ij, and be capable of lifting16w" (= weight bearing at i), 54w"; which taken from 9w" bearing at k, leaves 3" sustained by ek. Then it remains to be seen whether gk sustains more than 33w", so as to reduce the horizontal thrust of lk below that of i With the truss fully loaded except at the point f, ÿ sustains vertically,16w", whence jk, having the same horizontal thrust exerts a depressive force=16w",=103w”, at j, leaving a balance of 53w", exerted by ÿj toward lifting the 7w" at g. Hence, only 13w" remains as the weight sustained by gk. Therefore, the horizontal pull of ek, is not less than that of gk, the horizontal thrust of lk, is not less than that of ÿ, and its lifting power, not less than 51⁄2", and ek does not lift more than 33w”, nor as much as when ƒ and g are without load, as determined by the process above explained. XXXI. To determine the greatest stress to which dl is liable, let the weights at e, ƒ and g be removed. f Then the pressure at i, due to the weights at b, c and d, equals 6w", that is, 1w" for weight at b, 2w" for that at c, and 3" for that at d. We therefore take jq" on ÿ produced, to represent the thrust of ÿj, produced by 6w"-draw q′′r" parallel with fj, and from q′′r" find ft" (of course less than ft), and having taken kv' on jk produced, equal to jr", raise the perpendicular v'x' = ft", and draw x'y' parallel with ek. Then, x'y' represents the tension of ek, from which we find ea", representing the vertical thrust of el at its maximum. Also ky' represents the thrust of kl; and, having taken ld' on kl produced, equal to ky', raise the vertical d'e', equal to ea," from e', draw e'f', parallel with dl, and meeting Im (produced, if necessary), in ƒ', and e'f' represents the tension of dl. We have a short way of verifying the correctness or otherwise of the last result, since we know that, in the state of the load here assumed, 6w", is transferred from the left to the right of the centre, necessarily through the tension of de, the only member capable of performing that office. Hence the tension of dl in this case, can be neither more nor less than what is due to a lifting power equal to 6w". Then, taking de' on dm, to represent 6w", and drawing the horizontal c'b', we have db' to represent the tension of dl, and, if e'f'db', the result is probably correct. We know, moreover, that 6w" is the greatest weight ever transferred past the centre of the truss, the left hand side having the greatest possible load, and the right hand side, the least possible. Therefore, db' represents the maximum stress for dl, which is equal to 6w"✅h2 + v2 XXXII. If the points b and c alone be loaded, we know that 3w" is transferred through dl, and there being no weight at d, this lifting force of dl, must be sustained by the thrust of dm. Having then, found f" representing thrust of ml, by a process similar to that by which we obtained f' in the preceding case, that is, commencing with jq", representing the thrust of ÿj under a weight equal to 3w", we take mg′ = lf",* on lm produced, raise the vertical g'i' equal to a line representing 3w", and draw "'"' parallel with cm, when we have to represent the tension of cm. XXXIII. Or, we may take ok' on ao produced, to represent the thrust of ao, due to the vertical pressure (= 11w') at a, resulting from.the weights at b and c,— draw the vertical k'', representing 7w", weight at b, and cutting on produced in m', and I'm' represents the lifting force exerted by bn; as is made obvious by forming the parallelogram l'n', upon the diagonal om' = *q'i' and j' shown in Diagram B, to avoid complication = = Take bo' l'm', and draw the horizontal o'p', then op represents the tension of bn, and om', the thrust of on. Take na,,= om', on on produced, draw a b, bp', parallel with bn, and, from b,, let fall b,c, representing the weight (=7 w") at c, and the part below nm, represents the list of em, whence we derive the tension of cm. The result should be the same as that obtained by the former operation. XXXIV. If the point b only, be loaded, we may take ok" to represents the thrust of ao resulting from a pressure of 6w" at a, let fall k"" cutting on in m", to represent the 7w" at b, and m"l" represents the vertical lift of bn. Make bo"m"!", and draw the horizontal o"p", and we have bp" representing the tension of bn. This is the maximum stress of bn, since bn, can only sustain the weight at b, less the excess of lifting power of ao over the depressing power of on, both having the same horizontal thrust; which excess is represented by k'm' and 'm", and is least when the weight bearing at a is least. But the bearing at a (and the lift of ao), can never be less than of the weight at b, and k′′m' etc., can never represent less than weight at b, or of the lift of ao, whence m'l" etc., can never represent more than weight at b; consequently bn can never sustain a weight greater than 5w" which is the amount represented by m"l" when b is fully loaded, and the remainder of the truss without load. XXXV. With regard to em, no simple and conclusive reason presents itself, why the result above obtained for the stress of that member when b and c alone are loaded, is the actual maximum. But, as the assumed conditior is precisely analogous, as far as the |