CHAPTER VII. ON THE STRENGTH OF PILLARS. (136.) “ Theory of Pillars.”—The theory of the strength of pillars will be most easily understood by analogy with the transverse strength and stiffness of a beam of the same sizes and material. In Fig. 33 let A be a beam 10 feet long supported at each end and deflected } foot by a central load 2 = 20 lbs : now obviously, if that load is removed, the elasticity of the beam will cause it to react with a force of 20 lbs. in the direction of the arrow B, and the question becomes, what horizontal forces in the direction of the arrows C, C will counterbalance the vertical strain B. Let D, E, in Fig. 34 be two rods similar to A in Fig. 33 and resting on the plane f: for the purposes of illustration we may suppose them to be jointed by pins at a, b, c, d. With a vertical load of 200 lbs. at W, we have to find the horizontal strain at c,d ; by the “parallelogram of forces”: drawing the straight lines a c, c b, a d, db, we obtain a parallelogram, and, as is well known, the ratio of the length of the vertical diagonal a b, to that of the horizontal one c d, is also the ratio of the vertical strain to the horizontal strain. Thus, in our case, the vertical strain being 200, we divide a b into 200 equal parts and thus obtain a scale, which applied to cd, gives 20 lbs. as the horizontal strain on that line, that is to say, a Salter's balance between cd, showing a strain of 20 lbs., would support the vertical load W = 200 lbs. In Fig. 35, the two pillars are separated simply for the purpose of analysis, and we have two similar pillars F, G, &c., each loaded with 100 lbs., or half the former load. The strain shown by the Salter's balance at c d, acted in both directions, therefore substituting weights w, w, as in Fig. 35, we shall require 20 lbs. for each. We thus find that a beam, as in Fig. 33, 10 feet long, deflected foot by 20 lbs. will, when unloaded, react with 20 lbs., and cause a horizontal strain of 100 lbs. in the direction of the arrows C, C, and as shown at W, W in the figure. (137.) From this we obtain a general law connecting the vertical with the horizontal strains, and we find that if we take & bar of any given length resting on end-supports, and observe the central deflection with a certain transverse central load, we can calculate the equivalent load acting longitudinally and straining the same bar as a pillar by multiplying the transverse load by the length of the beam, and dividing the product by four times the deflection. Thus, in our case, 20 x 10 = (1 x 4) 100: hence we have the general rule : (138.) W = w x1 :(8 4). w = the transverse load in the centre of the same bar and in the same terms as W. 1 = the length of the pillar and beam, in inches. (139.) As an example of the application of the rule, we may take a pillar of Dantzic oak, say 1 inch square and 12 inches long. Mr. Hodgkinson gives as the result of his experiments the rule 24542 x $* • L' = W for pillars of that material with both ends flat. Here 24,542 lbs. is the theoretical breaking weight of a pillar 1 inch square and 1 foot long, as due by flexure, neglecting incipient crushing (163). Now, by Table 67, a beam of Dantzic oak 1 inch square, 1 foot between supports, loaded transversely with the safe or working load of 71 lbs., deflects .026 inch, which by our rule (138) is equivalent to 71 x 12 : (026 x 4) = 8192 lbs. longitudinally, straining the bar as a pillar, this strain being in the centre line, or the pillar having both ends pointed, and is that due to flexure only, as in Mr. Hodgkinson's rule. By (149) we obtain for the same pillar with both ends flat 8192 x 3 = 24576 lbs., which is almost exactly 24,542 lbs. as given by Mr. Hodgkinson. (140.) But it will be observed that we have taken the safe transverse load of 71 lbs., and the corresponding deflection, whereas Mr. Hodgkinson's rule gives the breaking weight. It is, however, a theoretical law with pillars, that a load which will produce the smallest deflection will equally produce a greater, sufficient to break the pillar by flexure. Thus, let us take a beam of any material whose elasticity is perfect, that is to say, one in which the deflections are simply proportional to the strains, and say 5 feet or 60 inches long, deflected 1 inch by 10 lbs. in the centre:—then by the rule (138) the equivalent load as a pillar will be 10 x 60+ (1 x 4) = 150 lbs. It would be the same with any other transverse load and corresponding deflection ; for instance, with 20 lbs. in the centre, the deflection would evidently be 2 inches, and W would be 20 x 60(2 x 4) = 150 lbs. as before. If we take an extremely small deflection, say toorth of an inch, the transverse load producing that deflection would evidently be Thoth of a pound, and the rule would give .01 X 60; (.001 x 4) = 150 lbs. as before. (141.) This fact conducts us to two remarkable laws :-/st. As the smallest possible deflection of this pillar requires a longitudinal strain of 150 lbs. to produce it, it follows that less than 150 lbs. would not produce any deflection whatever, but the pillar would be perfectly rigid and unyielding until that load was laid upon it. 2nd. That as 150 lbs. will with equal ease produce a deflection of To-ooth of an inch—or 1 inch -or any other amount, it follows that when 150 lbs. are laid on, the pillar will not only bend, but will go on increasing in flexure until it breaks. (142.) Such is the theory; Mr. Hodgkinson found, however, by experiments on various materials, that these laws do not hold good in practice, and that instead of a pillar showing no signs of bending until a certain load is laid on, and then suddenly bending and breaking, he found that there is no weight, however small, that does not produce a slight flexure, which increases progressively as the load is increased until the breaking-point is attained. (143.) Another remarkable result of Mr. Hodgkinson's experiments was, that the deflection of a pillar on the point of breaking by flexure is very much less than that of the same bar broken by a transverse strain. For instance, a pillar of Dantzic oak 14 inch square, and 5.04 feet long, broke with a deflection of .48 inch only. Calculating the ultimate deflection with a transverse load by the rules in (695) and taking the value of M from col. 2 of Table 67 at • 198, we obtain 5.048 x 198 = 1.75 = 2.9 inches, or about 6 times the ultimate deflection of the same bar as a pillar. A pillar of wrought iron, 10 feet long, and practically 3 x 1 inches, failed with a deflection of .6 inch only. Calculating the deflection under a transverse load for the “crippling strain only (373) we obtain 10 x .035 = 1 = 3.5 inches, or about six times the ultimate deflection as a pillar. But, in fact, the ultimate deflection of all pillars is very irregular and uncertain, for example, with two pillars 7feet long, 3 x 1 inches, although the breaking weights were nearly the same, one failing with 29,572 lbs., and the other with 29,666 lbs., the ultimate deflection was • 39 inch in one case, and .08 inch only in the other, the ratio for two precisely similar pillars being about 5 to 1. With cylindrical cast-iron pillars the same anomalies were found to prevail, the ultimate deflection being very small, and very irregular. (144.) With materials whose elasticity is imperfect (688) the ultimate deflection, or that with the breaking weight, is much greater in proportion to the load than the deflection with a small load such as would occur in practice, as is shown by col. 8 in Table 67, which, combined with the fact that the ultimate deflection of pillars is very small, seems to show that in calculating the strength of a pillar from the transverse strength and stiffness, a small load and corresponding deflection should be taken as a basis, rather than the ultimate deflection with the transverse breaking weight. The connection between the strength of pillars, and the transverse strength and deflection of the same materials, will be considered more at large in (296). (145.) “ Effect of Diameter and Length."-We may now search for the laws by which the diameter and length of pillars govern their strength. 1st for the length :-say we take the same beam as before (140), but of double length, namely, 10 feet, or 120 inches. It is shown in (659) that the deflection of a beam loaded transversely with a constant weight, is directly proportional to the cube of the length, or L':-in our case, the length being doubled, and 23 being = 8, we shall have 8 inches deflection, or eight times the deflection due to the length of 1 foot. Then, н by the rule in (138) the equivalent longitudinal strain as a pillar will be 10 x 120 + (8 x 4) = 37.5 lbs., which is 4th of the load borne by the pillar 5 feet long. Again, the deflection of a beam of half the length, or 30 inches, would, by the same reasoning, be inch, and the strength as a pillar 10 x 30 :(} x 4) = 600 lbs., which is four times the strength of the 5-foot pillar, and 16 times the strength of the 10-foot pillar. We thus find that the strength of pillars is inversely proportional to the square of the length :thus with lengths in the ratio 1, 2, 4, the strengths are in the ratio 1, \, . (146.) Searching now for the power of the diameter (or side of square pillars) :-say we take a beam 5 feet long, loaded as before with 10 lbs., &c., but 2 inches square. Then, by (659), the deflection with a constant transverse load is inversely as d x b, or in our case, 23 x 2 = 16, hence the deflection of the 1-inch beam, being 1 inch, that of the 2-inch beam will be teth of an inch, and the strength as a pillar 10 x 60 = 68 x 4) = 2400 lbs., which is 16 times 150 lbs., the strength of a l-inch square pillar of the same length. Again, with a beam of the same length, but 3 inches square, we have 33 x 3 = 81, and instead of 1 inch deflection, as with a beam 1 inch square, we have 31st part of an inch with the 3-inch beam, and the strength as a pillar becomes 10 x 60 = (1 x 4) or 10 x 60 x 81 ; 4= 12150 lbs., which is 12150 150 = 81 times the strength of the same pillar with a length of 1 foot. We thus find that the strength of pillars is directly proportional to the fourth power of the diameter, or side of square, for 1*, 24, and 34 1, 16, and 81, and this, as we have shown, is the ratio of the strengths of the pillars of those respective sizes. (147.) Combining these results we find that the strength of pillars is proportional to d' = L. By the same reasoning the strength of rectangular pillars will be proportional to d xb; L', in which d is the depth, or smaller dimension and b = the breadth or greater dimension of the pillar. These theoretical laws should be correct for all materials, but the experimental researches of Mr. Hodgkinson have shown that timber pillars alone follow those laws exactly. Thus, with cast-iron pillars, he found the strength to be proportional to |