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3 x 594 x 12 the centre; by Rule (511) f becomes in that case

2 x 18 •594 x 36

or •594 x 18 = 10.7 tons per square inch. 2 But if we apply the same rule to extreme strains in wrought or cast iron bars, we obtain a fictitious value for f, for reasons given in (504), &c.; for example, Table 66 gives 4000 lbs. or 1.786 ton for the value of M, for the breaking-down load; hence 1.786 x 18 = 32.15 tons per square inch, the apparent value of f, which being 36 per cent. in excess of the real value, as shown in (506), the latter becomes 32.15 = 1.36 = 24 tons per square inch.

It will be observed that in these cases f = W x 18 simply (639), but that rule will apply to those cases only where the tensile and compressive strains are equal to one another for bars 1 inch square and 1 foot long, &c., but will not be correct for cast iron where those strains are very unequal. For example, a bar of cast iron 1 inch square and 1 foot long breaks with • 92 ton in the centre (335); hence f = .92 x 18 = 16:56 tons per square inch. But this is neither the true tensile nor crushing strength of cast iron, which, as found by direct experiment, is 7.142 and 43 tons respectively. In applying these rules to cast iron, f must be taken at the apparent value of 16:56 tons per square inch in calculating the breaking loads (504).

(521.) The meaning of this fictitious value of f is, that if the transverse strength = .92 ton, and the tensile and crushing strengths are equal to one another, then the value of both would be 16.56 tons per square inch; by Rule (496) we then obtain 16.56 Х

; = .92 16.56 + ~ 16.56 By taking 16.56 tons for the value of f, the rules in (510) coincide in their results with those in (323), with .92 for the value of M, in rectangular sections, and .92 = 1.7 = 5412 for circular and elliptical sections. Thus, for the hollow rectangular beam, Fig. 75, we found in (347) and by rule (330) the breaking weight = 3.558 tons : by Rule (514) we have

W =

W =

W =

16.56 x 2 x {4:04x 2.21) – (3.298 x 1.46}
3 x 72 x 4.04

= 3.558 tons also.

(522.) Fig. 201 gives the section of six elliptical beams experimented upon by Mr. E. Clark; the length was 6 feet, the maximum breaking weight = 3.595 tons; the minimum = 2.918 tons, and the mean of the whole 3.207 tons. By rule (518) we obtain 3.1416 x 16.56 x {2-338 x 1:165) – (1.955°* •79}

72 x 2.33 2.738 tons. By Rule (334) we have (4.66x 2.33)-(3.91 x 1.58)

x •5412 = 6 = 4.66

2.736 tons, or practically the same as the other; if we take • 92 : 1.5 = •6133 for the value of Mt, the ratio 1:5 to 1.0 being, as we have shown (361), more correct than 1.7 to 1.0 for the breaking weight, W comes out 3.099 tons; experiment gave 3.207 tons, hence 3.099 : 3.207 = .9664, giving an error of 1:0 – 9664 = .0336 or - 3.36 per cent.

For some reason the analytical method followed in (348) does not give correct results in this case ; the area of the section = 3.676 square inches, the maximum tensile strain at A 7.14 tons per square inch, and the mean at B 3:57 tons; hence (3.76 x 3.57 x H x 2.33 36) x 2 = 2.265 tons, showing an error of 30 per cent.

(523.) If we take for cast iron the working or safe tensile strain at krd of the breaking weight, f becomes 7.142 • 3 = 2.381 tons per square inch. Table 92 shows that with 2.355 tons the extension and compression are precisely equal to one another. In that case the neutral axis of a rectangular section will be in the centre and the strain at the top and bottom, or f will also be equal, namely, 2.355 tons per square inch. Under these conditions the rules in (510) are quite correct, as we have shown (617), and by Rule (511) we

2.355 x 2 x 18 obtain W =

or 2.355 ; 18 = .1314 ton in the 3 x 12

centre only. The breaking load by experiment = .92 ton, or •92 •1314 = 7 times the working load as thus found, so that to secure the equality of strains on which the rules in (510) are based, the transverse load must not exceed 4th of the breaking weight, the tensile strain being then frd and the compressive strain 1th of their respective breaking weights.

(524.) As the transverse load is increased, the neutral axis moves towards the edge under compression until the breaking weight is reached, when it becomes as in Fig. 168. When the bar is loaded to įrd of the breaking weight or •3067 ton, it is not correct to assume that the tensile and crushing strengths are also strained to 3rd of their respective ultimate resistances, but that, on the contrary, the tensile resistance is much more, and the crushing much less than frd of their ultimate values (356), (617).

(525.) In calculating the breaking weights on beams by the rules in (510) it is therefore necessary to take a fictitious value for f as we have done for cast iron (520); thus for glass, col. 4 of Table 79 gives W 262 lbs., therefore f must be taken at 262 x 18 4716 lbs., whereas, by cols. 1 and 2, the real values of T and C = 2560 and 30,150 lbs. respectively. Again, for Ash by the same Table, W = 681 lbs. by col. 4, therefore f must be taken at 681 x 18 12260 lbs., the real values of T and C being 16,576 and 9023 lbs. respectively.



(526.) “ Load on Roofs.—The load on roofs is, Ist, the weight of the roof itself, and 2nd, the vertical pressure due to the Wind. The weight of the materials of the roof is not simply proportional to the span or area, but increases more rapidly than either; it is composed 1st, of the weight of the principals or trusses; 2nd, of the purlins, &c.; and 3rd, of the slates or other covering. It is shown in (490) that the ratio of the weight of a beam or truss to its safe load rises with the dimensions, and that to such an extent that there is with every beam a length with which it would break with its own weight and could carry no extra load whatever. Then the ratio of the weight of the purlins, &c., to the area rises with the span also, because it is found expedient to increase the distance between the principals for large roofs; for instance, with roofs, say, 60 feet span, the pitch of the principals would be from 7 to 10 feet, but with very large roofs it might be 30 feet; with such a distance the strength of the cross-beams, purlins, &c., must be very great as compared with those for small roofs. The weight of the slates, &c., would be constant for all

spans, and may be taken at 10 lbs. per square foot.

Wind.—The horizontal force of the wind in the greatest hurricane in this country is 80 lbs. per square foot, but the vertical pressure, with which we have to deal, is a very uncertain one: it has usually been taken at 40 lbs. per square foot.

(527.) Say we take a roof 60 feet span with trusses 8 feet apart, each truss therefore carries 60 x 8 480 square feet of roof; now the weight of a truss for such a case would be about 22 cwt. or 2464 lbs., and is equivalent to a pressure of 2464 • 480 = 5 lbs. per square foot; the weight of the purlins may be taken at 5 lbs. and of the slates

= 10 lbs. per square foot; then allowing 40 lbs. for wind, we obtain a total pressure of 5 + 5 + 10 + 40

60 lbs. per square

foot. (528.) For very large roofs we may take as an example the St. Pancras Station roof, where the straining weight of the truss, 240 feet span,

= 35 tons or 78,400 lbs., the distance between the trusses was 30 feet, the area of roof borne by each truss, measured on plan, becomes 240 x 30 = 7200 square feet, equivalent to 78,400 = 7200 9.2, or say 10 lbs. per square

foot. The weight of the covering of slates, purlins, &c., with large roofs is very great, for reasons already given; thus at New Street, Charing Cannon Lime Street,

St. Pancras, Birmingham, Cross, Street, Liverpool,

[blocks in formation]

respectively, the estimated weight of the coverings, as given by Mr. W. H. Barlow, was :20 37 37 38

36 lbs. per square foot. The weight of the principals being :25 27 37

54 tons.


(529.) In the St. Pancras roof the estimated total pressure, exclusive of the strain due to the weight of the truss itself, was taken as 70 lbs. per square foot; the truss, as we have seen, = 10 lbs., hence the total = 80 lbs. per square foot, leaving 34 lbs. for the vertical pressure of the wind. We thus have, truss = 10, covering = 36, wind = 34, and the total = 80 lbs. per square foot.

For ordinary roofs, say 60 to 70 feet span, we may, for convenience of calculation, take the total pressure at 56 lbs. or 1 cwt., or oth ton per square foot, being 5 lbs. for weight of truss, 5 lbs. for purlins, &c., 10 lbs. for slates, and 36 lbs. for wind. Large roofs, however, should in all cases be subjected to special calculation.

(530.) “Strains on Roofs.—Let Fig. 179 be an outline of a truss of the form commonly adopted for wooden roofs of small span, in which, for the purposes of calculation, we have taken the strain as concentrated at certain points : for example, if the weight of that part of the rafter, purlins, slates, &c., between

10, then half of that weight or 5 is discharged at n, and 5 at p. Similarly the part between m, n gives 5 at m and 5 at p, &c., &c.: we thus obtain 10 at each point, except at the ends where we have 5 only, which last, being discharged direct on the supports, will have no effect in straining the truss.

In order to analyse the combined effect of these weights, we may take them separately. In Fig. 176, the weight A = 10, evidently gives 5 at B, and 5 at C: then by the parallelogram of forces, making a, b = 5 by a scale of equal parts, that force is resolved into two forces a, d 11.18, and a, c =

10, hence the strain on the rafter E 11:18, and that on the tie-rod D 10 by the same scale.

In Fig. 177, a weight of 10 at G gives 7.5 at H, and 2-5 at

n and

P =

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