D= Hodgkinson, we find that pillars of timber follow precisely the theoretical law (147), the strength of long pillars failing by flexure, being directly proportional to the fourth power of the diameter or side of square, and inversely as the square of the length :hence we have for cylindrical pillars the rules : (279.) F= Mp X D' ; L. (280.) V (F XL' = Mp). (281.) L Ý (Mp x D4 ; F). For square pillars these rules are modified and become : (282.) F = Mp X S': L'. (283.) S= V (F XL : Mp). (284.) L = y (Mp XS = F). And for rectangular pillars other than square, into:(285.) F = Mp x 13 xb L' (286.) » {Fx L? +(Mex6}. (287.) b = F x L = (Mp xt). (288.) L = 4 (Mp xt xb; F). S = the side of a square pillar in inches. pillar in inches. angular pillar in inches. F = the breaking weight by flexure in lbs., &c., dependent on Mo L = the length in feet. = Multiplier, whose value is given in lbs. and tons, by Table 34. (289.) Mr Hodgkinson made some experiments on square and rectangular pillars of Dantzic Oak and Red Deal, the results of Me which are given by Table 57 : col. 8 has been calculated by the rules. Thus, taking No. 5 as an example, the value of Mp for a pillar of Dantzic Oak is given by Table 34 at 27,000 lbs., S' or 134 = 9.379, and 2.522 6.35: then rule (279) becomes 27000 x 9.379 = 6.35 =39880 lbs., as in col. 11, this being due by flexure. This will require correction for incipient Crushing (163): by Table 32, C, or the crushing strength of Dantzic Oak = 6840 lbs. per square inch, and the area being 11 x 1} = 3:0625 square inches, Cp becomes 6840 x 3:0625 = 20950 lbs., as in col. 10, hence { Cp = 15724 lbs., and the rule (164) becomes 39880 x 20950 = (39880 + 15724) = 15020 lbs., the breaking weight, as in col. 8: experiment gave 14,305 lbs., hence 15020 • 14305 1:05, or + 5 per cent. error, as in col. 9. (290.) As an example of rectangular pillars, we may take No. 10:- the value of Mp for a pillar of Red Deal with both ends flat = 24000 lbs. by Table 34; then rule (285) becomes 24000 x 1.413 X 2.82 • 4.833", or 24000 x 2.803 X 2.82 •- 23.36 = 8121 lbs., the breaking weight by flexure as in col. 11. Correcting for incipient crushing: Table 32 gives 6167 lbs. for the value of C, or the specific crushing strength of Red Deal, and the area being 1.41 x 2.82 4 square inches, we obtain 6167 x 4 = 24668 lbs. for the value of Cp, hence { Cp 18500 lbs. Then the rule (164) becomes 8121 x 24668 ; (8121 + 18500) = 7525 lbs., as in col. 8: experiment gave 7681 lbs., hence 7525 • 7681 = •98, or 2 per cent. error. It should be stated that the actual dimensions of the rectangular pillars Nos. 10 and 11 were not given by Mr. Hodgkinson, but the areas were 4 square inches, and the ratios of the sides 2 to 1 in No. 10, and 3 to 1 in No. 11, and the dimensions in the Table were obtained from those data. (291.) Table 57 has been calculated in this way throughout : the sum of the minus errors in col. 9 = 44.3, and of the plus errors 39.6, which on the 11 experiments gives a mean of (44.3 – 39-6) -11 = -0.427, or less than } per cent. The maximum minus and plus errors = 15.5 and + 15.9 per cent. respectively, thus showing equality of range (959). With the exception of Dantzic Oak and Red Deal, we have no TABLE 57.-Of EXPERIMENTs on the STRENGTII of SQUARE and RECTANGULAR PILLARS of TIMBER. 1 Dantzic Oak 2 ends pointed 1.75 X 1.75 5.04 3,645 2,749 3,197 3,323 + 3:0 20,950 3,323 7,229 4,989 6,109 6,221 + 1.9 6,616 2 1 flat, 1 pointed 9 Red Deal 3.2 24,668 16,440 2:0 X 2:0 4.833 12,385 11,602 11,993 11,605, 1.41 x 2.82 7,681 7,525 10 2.0 8,121 experimental data for Timber pillars, and are compelled to resort to theory, in order to obtain the values of Mp for other kinds of Timber: this we have done in (303), and we have thus obtained most of the numbers in Table 34. This method is of course not so satisfactory as direct experiment, but it is shown in (139) that as applied to Timber, the theoretical and experimental results practically agree with one another. (292.) Table 58 gives the strength of square pillars of Red Deal calculated by the rule (282): we have selected the case of flat at one end and round at the other as approximating to ordinary conditions more nearly than any other. In most cases timber pillars are nominally flat at both ends, but this supposes that the surfaces between which the pillar is strained are perfectly parallel and unyielding, conditions which are seldom realised in practice: for example, when a suft-wood Bressummer is supported by a pillar, the effect of flexure in the latter is to compress the fibres of the former unequally, the soft wood yielding, so that the result is little if any better than it would have been with a round end. If we suppose that the foot of the pillar is well bedded on a large stone or cast-iron plate, and the upper end loaded by the Bressummer in the usual way, we should have in effect a pillar flat at one end and round at the other, being the conditions assumed in Table 58. (293.) Say we take the case of a pillar 7 inches square; then by col. 5 of Table 34, the value of Mp for a pillar of Riga Fir, with one flat and one round end = 7.14 tons, and with a length of say 12 feet, rule (282) becomes 7.14 x 74 = 122, or 7.14 X 2401 • 144 = 119 tons, the breaking weight by flexure. This requires correction for incipient crushing, being greater than {th Cp given by col. 4 of Table 58 (169). By col. 4 of Table 32, the specific resistance of Red Deal to crushing, or C = 2.75 tons per square inch, and as we have 72 49 square inches area, Cp becomes 2.75 x 49 135 tons, as in col. 2: Cp = 101 tons, col. 3: and 1 Cp = 33.8, col. 4. Then in our case, the rule (164) becomes Po = 119 x 135 • (119 +101) = 73 tons, as in Table 58. By the use of col. 4, we can easily determine when the correction for incipient crushing is necessary : thus, for a pillar 3 inches square 10, 12, and 14 feet long, the TABLE 58.-Of the STRENGTH of SQUARE PILLARS of breaking weights by flexure = 5.8, 4.0, and 2.9 tons respectively, wbich are all less than 6·18 tons or Cp by col. 4; the correction is therefore not required, as shown by the Table, but for the shorter lengths, 5, 6, 7, 8, and 9 feet, that correction is necessary (294.) Table 58 may be adopted for conditions of fixing other than that of flat at one end and round at the other, as in that Table. By (149) it is shown that the breaking weights by flexure are in the ratio 1, 2, 3, for the three cases—both ends pointed, -one flat, one pointed, -and both ends flat respectively. Thus a pillar 6 inches square, 16 feet long 36 tons by flexure from Table 58: then with both ends pointed we have 36 = 2 = 18 tons, which being less than 24.8 tons given by col. 4, correction for incipient crushing is not required (163). The same pillar with both ends flat 36 x 3 = 2 = 54 tons breaking weight by flexure, which being greater than 24.8 or 1 Cp by col. 4, correction for crushing will be necessary. Taking = |