light description, should be well strengthened at the ends where they rest on the supports, to prevent the web doubling up under the vertical strain that it has to bear, which in most cases, even when the load is a central one, amounts to half that load plus half the weight of the girder itself. With a rolling load it may be a great deal more than that; in extreme cases it may amount to the whole load plus half the weight of the girder. Obviously the thin plate of the web, say or even 3, inch thick, Fig. 133, could not bear such a strain as that, without the assistance of vertical, angle, or irons; usually the latter would be used, and as they are strained as pillars, their sizes may be determined by the rules in (534), or by Table 82.

(404.) Table 76 gives the strength and stiffness for certain standard sizes of plate-iron girders whose proportions are given by Figs. 101 to 107. Thus, the girder Fig. 103, 2 feet deep, and say 30 feet long, will break by col. 5 with 1665 ÷ 30 = 55·5 tons in the centre:-the safe dead load may be taken at 55.5÷3 = 18.5 tons, with which the deflection by col. 6 and (674) would be 18.5 × 0000007606 × 27000 = 38 inch. With a load equally distributed all over the length, we should have had 18.5 x 2 = 37 tons, with which the central deflection would have been by col. 7 in the Table, 37 × 0000004754 × 27000 475 inch, &c.

TABLE 76.-Of the STRENGTH and STIFFNESS of WROUGHT PLATEIRON GIRDERS, 1 foot long.

(405.) "Tubular Beams of Plate-iron."-The strength of a tubular beam, like that of every other structure, is limited by the strength of the weakest part, this is usually the top plate, which is due to the tendency of thin plates of wrought iron to fail by Wrinkling with a strain much less than the crushing or tensile strength of the material. This, however, is not universally the case, but depends on the relative dimensions of the plates subjected to tension and compression respectively. For example, with No. 1 in Table 77, the top and bottom plates have equal areas, they have also the same total load to carry, occupying as they do the position of fulcrum and resistance to one another reciprocally. But while the bottom will bear 21.6 tons per square inch, namely, the mean tensile strength of plate iron by Table 1, the top will fail by wrinkling with 18.38 tons per square inch, as shown by col. (11): the top being the weakest will therefore govern the strength of the beam. The thickness of the bottom plate might in this case be reduced to × 18.38

21.6 = 6382, or about g inch, without at all affecting the strength of the beam:-both plates would then be strained in proportion to their strength, and they should fail simultaneously. In this case, in order to obtain equality of strength, the areas would be in the ratio of 6 to 5.

(406.) The rule for calculating the strength of tubular plateiron beams must therefore include the Wrinkling Strain as a fundamental datum; the ordinary rules in which that is neglected will not give correct results. Thus, taking as an example No. 1 in Table 77, whose section is given by Fig. 97, taking the value of Mr from col. 6 of Table 66 at 1.786 ton, Rule (330) gives (35.75 x 24) (34.25 × 22·5)

W =

T

35.75

=

x 1.78645 = 214

=

tons; but by col. 8, the experimental breaking weight 118 tons; hence 214 118 1·81, or an error of +81 per cent. (407.) A still more striking illustration would be given by the very thin tube No. 12 in Table 77, with which Rule (330) (243 × 15) (23.7523 x 14.752) gives W =

24

x 1.786 30

= 24 tons; but by col. 8, the experimental breaking weight

=

=

5.78 tons only; hence the calculated strength is 24 ÷ 5·78 4.16 times the experimental!

These illustrations show that the ordinary rules which may be correct for very thick plates which fail by crushing are not correct for thin plates which fail by Wrinkling.

(408.) The first step as a basis for correct calculation of the strength of Tubular Beams with thin plates is to find the Wrinkling strain by the Rule (308), which for beams is Ww=

twbw x 104, or in our case 24 × 104, or √03125 × 104, or 17677 × 104 18.38 tons per square inch, col. 11. Having thus found the wrinkling strain, and supposing it not to exceed the crushing strength, or 19 tons per square inch (201), we may take it as equivalent to ƒ in rule (514), and may then calculate the strength of the beam by that rule, which in our case becomes

W =

18-38 × 2 × (35.75 × 24)-(34.253 × 22.5}

3 x 540 x 35.75

=

= 122.5 tons, as in col. 9 of Table 77: experiment gave 118 tons as in col. 8; hence 122 5118 1.038, showing an error of+3.8 per cent. only, as in col. 10.

To vary the illustration we may take No. 11 in the same Table, whose section is given by Fig. 98. strain, the rule (308) becomes

W =

272

For the wrinkling 15.5 × 104, or

13.78 tons per square the rule (514) becomes

13.78 × 2 × (23.75 × 15·5) (23.206 × 14.956} {23-753

3 × 360 × 23.75

= 22.29 tons: experiment gave 23.33 tons, col. 8; hence 22·2923·33=956, giving 1.0·956044, or an error 4.4 per cent., as in col. 10.

of

(409.) In some cases the wrinkling strain comes out in excess of the absolute crushing strength of the material, or 19 tons per square inch (201), and of course it will not be realised, the metal failing by crushing; in that case we must take ƒ = 19 tons, whatever the wrinkling strain may be. An illustration of this is given by Nos. 7 and 8, in Table 77, whose section is shown by Fig. 99:-the rule (308) gives Ww = 75÷16.5

× 104 22.17 tons per square inch, as in col. (11), which being in excess of 19 tons must be rejected; and the value of f being taken as 19 tons, the rule (514) becomes

W =

19 × 2 × (243× 16·5) — (22-753 × 16}

3 x 360 × 24

= 58.2 tons :

experiment gave 54-82 tons; hence 58.254.82

=

1.062,

or an error of +6.2 per cent., col. 14. Obviously, if we had erroneously taken 22.17 tons, the error would have been greater, the breaking weight coming out 58.2 x 22.17 ÷ 19 = 67.91 tons, giving 67.9154.82 1.24, or an error of +24

per cent.

=

(410.) The cols. 9, 10, and 11 of Table 77 have been calculated throughout in this manner:-the general result shown by col. 10 is that the sum of all the errors is 58, and of all the errors is 83, giving on 16 experiments an average of (8358) 16 = 1.56, or 1.56 per cent. (959). The greatest + error = 21.8, and the greatest error = 24.9 per cent., thus showing nearly an equal range; and it should be observed that this range of error may possibly be due to the natural variation in the strength of the material:-thus, taking the simple tensile strength of plate-iron, where, of course, the case is not complicated by possible errors in the rules, &c., Table 149 shows a variation of +29 and 33 per cent. respectively. The range in the dimensions of the beams is worthy of note; the extreme sizes, or those of Nos. 2 and 16, are shown by Fig. 100, and it should be observed that the errors of these two sizes differ but little from one another, being 0.7 per cent. in the largest (No. 2), and 2.3 per cent. in the smallest (No. 16).

Extreme cases are crucial tests of the accuracy of any rules, and, as we have seen, the rules we have given bear that test satisfactorily. Other illustrations of the different methods of calculating tubular beams are given in (320), (412), (414).

(411.) It is shown in (405) that there should be equality in the resistances to tension by the bottom, and compression by the top of a tubular beam, and failing that, the weaker of the two will govern the strength of the beam. With most of the beams in Table 77 the top is the weaker, but in Nos. 7, 8, 9, and 10