Abbildungen der Seite
PDF
EPUB

the diagram, except the last, making 70w",=10w, and require material = (1022 + 10v) мm.

The tension verticals ob and jg, sustain the weights (w) acting at band g when the truss is fully loaded, which is their maximum stress, and they require material for the two, = 2CM.

The thrust uprights el and fk, receive and sustain the maximum sustained by dl and ek respectively, which are the measures of their respective stresses of compression, being 6w" for el, and 10w" for fk, and the same for dm aud cn, making an aggregate weight of 44w, whence, their representative for material is 44UM.

XLII. With regard to stress of the lower chord, the tension of ac equals the horizontal thrust of ao, and of course is greatest when ao sustains the greatest weight; which is manifestly under the full load of the truss. The teusion of cd equals the horizontal thrust of ao (through ac), and the horizontal pull of oc; and must be greatest when the combined action of ao and oc is greatest. Now, although the weight borne by oc is greater by w" when the point b is unloaded, than under the full load, on the other hand, the weight on ao is less by 6w", so that the combined action of the two members, must be greatest when the truss is fully loaded; since no other change can increase the action of either.

Again, de sustains the horizontal action of ao, oc and nd, when the truss is under a full load; dl and em being inactive in that case, since the tendency to action is the same in each, whence neither can act. Therefore nd sustains simply the weight (w) at d; oc sustains the two weights at c and d,= 2w, while ao sustains the same, with the addition of the weight at b, making 6w

sustained by the three members contributing to the tension of de. Now while the maximum weights sustained by oc and nd, exceed by only 4w," what they sustain under a full load, neither can be brought under its maximum stress, without removal of load from b in one case, and from both b and c in the other, thereby diminishing the weight on ao, by 6w" in one case, and by 11w" in the other. Hence the stress of de is greatest under a full load of the truss, and as already seen, equals the horizontal action of weight equal to 6w upon oblique members of vertical reach equal to v, and horizontal reach equal to h. The maximum stress of de, then, equals 6w, and that of cd equals the same, less the horizontal pull of dn, due to the weight (w) at d, and is therefore equal to 5w-.

h

h

We have for the lower chord, then, one section, de (with length equal to h), exposed to stress...... = 6w

Two sections, cd and ef, with stress....

[blocks in formation]
[ocr errors]

= 5w_h

[ocr errors]
[ocr errors][subsumed][merged small]

whence, adding, multiplying by h, and changing w to M, we have to represent required material, 284 M.

XLIII. It is scarcely necessary to state that the oblique members ao, and oc, exert at o in the direction from o to n, the same force that they exert in the opposite direction upon cd. In fact, the thrust of on, and the tension of cd, simply act and react upon one another through the media of ao, oc and ac, whence the compression of on, must be just equal to the tension of cd; and furthermore, the thrust of nm is the indirect counteraction of the tension of de; and, as the two forces are in opposite and parallel directions, they must be equal, being in equilibrio. Also, mlk must sustain

the same compression as nm, throughout, since the diagonals meeting the chord at m and l, are inactive under the full load of the truss. Of course, kj is liable to the same maximum action as on.

From what precedes, it cannot fail to be obvious that the maximum action of all parts of the upper chord occurs at the same time with that of the lower chord, namely, under the full load of the truss. We have then, two sections liable to a compression of 5w, and three, liable to 61; whence the representative of required material, is 28 M. We may now sum up 2872 the material for the truss, required to support the assumed movable load through all the changes liable to take place, as follows:

[blocks in formation]

= 48M

= 59м

For Arch Truss, Comp. Mat., [XXVII]...............................

66

66 Tension do.,

XLIV. A trapezoidal truss without verticals (except at one panel width from the ends), is represented in Fig. 13. The members ob and oc, as also jg and fj, are supposed to be so formed and connected as to act by tension only, and the other diagonals, so as to be capable of acting either by thrust or tension.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

If a weight (w), be applied at b, it will cause a bearing of lw" at i, and 6w" at a, the same as in the case of Fig. 12. Now, this weight at b, might (if bn were removed, and oc were capable of withstanding compression), be suspended entirely by ob, and supported by ao and oc, in proportion to the bearing produced by it at a and i respectively. But as bn is able to act by tension, and oc unable to act by thrust, the 6w" bearing at a, acts through bo and oa, while the lw" bearing at i, must first act by tension upon bn; secondly, by thrust upon nd, since that is the only member meeting bn at n, capable of sustaining weight. Hence, the action of the weight is transferred to d, and through dl to l, thence to f, and so on through fj, and ji, to the abutment at ; acting alternately by tension and thrust, upon six oblique members, producing the same amount of action (=w"), upon each.

Another weight (w) at c, must cause pressure at i, equal to 2", and at a, equal to 5w". The action therefore, must be divided between cm and oc, in the proportion of 2 to 5, producing alternately tension and thrust, through the points m, e, k, g, j to i; on the right, and through co and oa, to a, on the left.

Thus far, the weights have acted upon independent systems of oblique members (except as to king braces), neither weight acting upon any member acted on by the other. But when a weight (w) is imposed at d, it must act upon the same members acted on by the weight at b. The 3w" to be transferred to i, must act by tension upon dl, in concert with the 1w" of the weight at b. But the pressure at a, must be increased by 4w", which can be transferred from d, only through tension of dn, and dn being previously occupied in carrying 1" from b, by compression, it follows, since the same member can not be under compression and extension at the same time, that the greater force must preponderate, dn being brought under tension due to the difference of 4w" tending to act by tension, and lw", tending to act by thrust, the action of dn, being changed from thrust under 1w", to tension under 3w". All the weight at b, then, is sustained by bo, together with 3w" from weight at d, making 10w", which, with 5w" from weight at c, makes 15w" pressure due at a, from weights at b, c and d.

=

In the mean time, the weight at d, having obstructed the passage of 1w" from b to the right hand abutment, has been obliged to make compensation, by sending 4w" of its own gravitating force, nstead of 3w" owed by it to the bearing at i

Similar changes of action take place when another weight (w) is applied at e; which tends to throw 3w"

« ZurückWeiter »