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by col. 8. Table 67 gives, in cols. 2 and 4, the values of C, and Cg for many materials, these being in fact the deflections of a bar 1 inch square and 1 foot long with the Safe and Breaking Loads respectively: in that particular case, Cs and Cg are identical with ag and B.
It will be observed that the values of Cg and Cg for the two Standard cases, are not simply proportional to the load, or “Factor of Safety." In Table 67, col. 7 gives the Factor of Safety, and col. 8 the ratios of the deflections with Safe and Breaking loads respectively: thus with Cast iron the ratio of the loads is 3 to 1 by col. 7, but the ratio of the corresponding deflections is 4 to 1 by col. 8: again with Ash, the ratio of the Loads 5 to 1, but the ratio of the deflections 10.7 to 1.0.
Thus, a beam of Ash, say 15 feet long, 7 inches deep, 3 inches wide, would give for the breaking weight by Rule (324), the Value of M, for safe load being 136 lbs. by col. 3 of Table 67, W = 72 x 3 x 136 = 15 1333 lbs., the deflection with which
153 x 1333 x .00026 by rule (659) is 8 =
= 1.13 inch de
73 x 3 flection with safe load. Now applying Rule (695) and taking Cg from col. 4 of Table 67 at 0354, we obtain dg 152 x .0354 • 7 = 1.13 inch deflection, as before: but with the breaking weight 88 = 15% 375 • 7 = 12 inches deflection, &c.
dg X Again, with a bar of Cast iron 3 inches deep, 4 inches wide, and 11 feet long, taking Mr for safe load = 688 lbs. from col. 10 of Table 64, Rule (324) gives W 32 x 4 x 688
2251 lbs. Safe load, with which Rule (659) gives 11 x 2251 x .00002886 8 =
= .8 inch deflection. By Rule 39 x 4 (695) we obtain much more easily dg 11x .02 = 3 = .8 inch also. We have here taken the deflection per lb. = ·00002886 from col. 6 of Table 64, or col. 4 of Table 105. The deflection of the same bar with the Breaking weight becomes og 112 x .0785 = 3 3.17 inches.
But these Rules need not be restricted to ds and 8B, but will apply equally well to any standard fraction of the breaking weight, so long as the great principle is maintained, that the beams compared shall be loaded always in some given and constant proportion to their strength. Thus for Wrought iron
and Steel, we found (374) (376) it convenient to take the “ limit of Elasticity” and the “Working Safe Load” as data : then putting og and og for the deflections with those strains, we have for Wrought iron, the Rules :(697.)
de = L x .035 = d.
og = L x .0235 ; d. For Steel these Rules become :(698.)
de L' x .0802 ; d.
Og = L' X.0481 ; d. Thus, with a wrought-iron bar 2 inches deep, 4 inches wide, and 12 feet long, taking from col. 5 of Table 66, MI 1500 lbs. for the safe working load, we have by Rule (324) W 22 x 4 x 1500 - 12 = 2000 lbs., with which the deflection by Rule (659)
128 x 2000 x .00001565 becomes =
= 1.69 inch. By Rule
28 x 4 (697) we have og = 122 x .0235 ; 2 = 1.69 inch also.
Again: with a Steel bar 1} inch deep, 5 inches wide, and 10 feet long, the load by which the bar will be strained to the “ limit of elasticity
» will be W 112 x 5 x 5600 = 10 6300 lbs., with which the deflection by Rule (659) becomes 108 x 6300 x .00001433
= 5.35 inches. By Rule (698) 11% x 5 we obtain og 102 x .0802 = 1.5 = 5.35 inches also.
(699.) When the load is not in the centre, and when the beam is not supported at both ends as is assumed in the ordinary rules, the case is complicated, but by combining the data given in (431) and (667) the matter may be simplified.
Thus, for example, if the beam of ash in (696) had been built into walls at both ends, the safe central load would have been 1:50, or 50 per cent. greater, but the deflection with that increased load would have been 1.0, or precisely as before.
Again : if the bar of cast iron in (696) had been fixed at one end, and the safe load had been equally distributed, that load would be 0.5, or half only of its former amount, but the deflection with that reduced load would be 6 times greater than before, and becomes • 8 x 6 4.8 inches, &c.
(700.) Again: say we require an Oak Bressummer to carry the front of a house, the estimated distributed load being 19 tons and the span 12 feet, the ends being built into the walls in the usual way. Say we try 12 inches square, then by col. 3 of Table 67, MT 102 lbs. for Safe load : then Rule (324) becomes W 122 x 12 x 102 = 12 14688 lbs., or 6.56 tons for the ordinary case of a beam merely supported at ends and loaded in the centre; but by (431) in our case, 6.56 x 3 = 19.68 tons safe distributed load, or very nearly the actual load. Then, for the deflection, Rule (695) gives 8 = 122 x .04;
0.48 inches for the deflection with safe load in centre, which by (699) becomes •48 x 1.25 = 0·6 inch deflection, when the load is distributed and the ends built in, as in our case. We have taken the value of Cs = .04 from col. 4 of Table 67.
(701.) Although, as stated in (695), the value of Cg should strictly be adapted to the special section of the beam, we may with moderate accuracy apply the value for simple rectangular beams to ordinary girders, at least where the beam is parallel. But when the flanges are bellied, as in Fig. 131, the deflections are about 40 per cent. greater than with parallel beams, as shown by Mr. Hodgkinson's experiments. For ordinary parallel girders of cast iron loaded to of the Breaking Weight, we have the Rule:(702.)
og = L X.02 - d. For cast-iron girders reduced in section progressively towards the ends, or having bellied flanges, as in Fig. 131, the Rule becomes :
og = L X.027 = d. Thus, for the large girders in Table 68 and Fig. 79, the length being 16 feet and the depth 14 inches, with bellied flanges, we obtain og = 162 x .027 14 = 0.4937 inch, the deflection with central safe load. The mean breaking weight was 38.3 tons, or 38:3 = 3 = 12.8 tons Safe load, the nearest load to which in Table 68 is 14 tons, with which the mean deflection = .0525 inch; therefore .0525 x 12.8 : 14 0•48 inch with safe load of 12.8 tons, &c., or nearly as calculated by the Rule.
ON TORSIONAL ELASTICITY.
(704.) “ Methods of Estimating."-There are two ways of measuring torsional elasticity. 1st, by the angle of torsion or the number of degrees of twist, that is to say, if the whole circle = 360°, the amount of torsion would be expressed by the number of degrees of that circle produced in a long bar by a torsional strain. The other and more convenient method is to express the torsional elasticity by the descent of the end of the lever by which the twisting weight is applied ; this method is applicable to those cases only where the angle of torsion is small, but as this is always the fact in practice, this is no objection.
(705). It is necessary to observe that in this method of estimating torsional elasticity, the descent of the straining weight is proportional to the square of the length of the lever ; thus in Fig. 173, we have a constant weight = 1, acting with leverages in the ratio 1, 2, 3, it will therefore strain the bar D in those same ratios, and the descent of the lever measured at E will be 1, 2, 3 also, but measured at the points of application of the weight, as in our case, we obtain 1, 4, 9 for the ratios of the descent of the weight as in the figure, or in the ratios of the
square of the length of lever for 1, 2, 3, = 1, 4, and 9
“Laws of Torsional Elasticity.”—The fundamental laws as determined by mathematicians may be expressed by the Rules :
For Circular Sections :
L XL x W x 2 (706.)
3.1416 X R4 x Me
W x L xlx 2 (707.)
3.1416 x R x T
For Rectangular Sections,
L' x1 x (a + b) x 3 (710.)
d x 13 x Me (711.) M. =
W x L x 3 x (d2 + b2) x 1
d x 58 x T
In which R = radius of circular sections in inches : S side of square: d and b = depth and breadth of rectangular sections: L = leverage in inches by which the weight W in pounds acts in twisting the bar: 1 = the length of bar twisted in inches: Me = Multiplier for torsional elasticity derived from experiment: and T = the elastic torsion measured by the descent of the weight in inches.
By Mr. Bevan's experiments, the mean value of Me for cast iron = 5,709,600 lbs. : wrought iron and steel were nearly equal to each other; a mean from eight experiments on wrought iron and three on steel gave Me = 10,674,540 lbs.
To determine the value of M,, say we find by experiment with a 4-inch round bar of cast iron 60 inches long, with