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W = 5000 lbs , and L = 24 inches, that T, or the descent of the weight = 1.206 inch, then by Rule (707),

5000 x 242 x 60 x 2 5000 x 576 x 60 x 2 M =

3.1416 x 24 x 1.206 3.1416 x 16 x 1.206 = 5.701,000, the value of Me in that case.

(712.) “ Ratio of Stiffness of Round and Square Bars.”--Comparing rules (707) and (709) we can find a general ratio for the torsional stiffness of round and square bars : thus, for 4 inches diameter, R = 2, and by Rule (707) we obtain 2

·03979. For a 4-inch square bar, Rule (709) 3.1416 x 2

6 6 gives

= .02344; hence we have •03979 = .02344

256 1.7 to 1:0 = the ratio of the stiffness of square and round


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Wrought Iron and


606 5.41 Steel

Lancewood Cast Iron

324 | 2.90 Larch Alder

5.53 049|| Lime-tree Asb ..

6.91 062| Oak, English Apple-tree

6.95 .062 Oak, Dantzic Beech

7.23 • 065 Pear-tree Birch

5.88 •052 Pine, Memel Boxwood..

10.00 .090 Pine, American Brazil-wood

12.50 •112 Plane-tree Chestnut (Horse).. 7.56 067|| Sycamore Deal..

3.82 034| Teak Elm ..

4.60 041 Walnut.. Fir, Scotch



9.00 081 8.60 077 6.46 .058 6.23 056 6.81 061 5.62 ·050 6.18 055 5:11 046 5.02 045 6.00 .054 7.80 070 9.30 083 6.72 060

bars. It is shown in (552) that the ratio of torsional strength of square and round bars is theoretically 1.2 to 1.0; but by experiment 1.6 to 1.0.

Practical Rules.”—The theoretical Rules may be put in

T =

more convenient form for practical use on the large scale; they
then become :-
(713.) For Circular sections :-

L'xl x W
T =

Do X CT (714.) For Square sections:

L xl x W

Si X Cm X 1.7 (715.) For Rectangular sections :

L'xlx (d+b) W
T =

d x 18 x CT X 3.4 In which D = diameter in circular sections in inches: S = side of square in inches : d and b = depth and breadth of rectangular sections in inches: 1 = length of bar twisted in feet: L = leverage in feet with which the weight W, in lbs. or cwts., &c., acts in twisting the bar: T = the descent of the weight in inches, due to the twisting of the bar: and Cr = a Constant from experiment, the value of which is given by Table 113 as reduced from the results obtained by Mr. Bevan.

To find the value of C, from experiment, the Rules become: (716.) For Circular sections:

L xlx W

D' x T (717.) For Square sections :

L' Xlx W

Si XT X1.7 (718.) For Rectangular sections :

L'xlx (do +62) W

dx 13 x T x 3.4 (719.) We may now give some illustrations of the application of the Rules: say we take the bar considered before (711), in which D = 4; 1 5 feet; L = 2 feet; and W = 5000 lbs.




T =

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Taking the value of Cr at 324 from Table 113, Rule (713)

24 x 5 x 5000 4 x 5 x 5000 becomes, T =

= 1.206 inch, 4' x 324

256 x 324 the descent of the lever and weight due to the twist. For a square bar of the same dimensions, Rule (714) becomes,

4 x 5 x 5000 T=

= .709 inch. We should obtain the 256 x 324 x 1.7 same result if we applied to this bar the rule for rectangular sections, for obviously a square bar may be regarded as a rectangular one with equal sides : then Rule (715) gives 4 X 5 X (16 + 16) X 5000

= .709 inch, as before. 64 x 64 x 324 x 3.4 Again : say we have a Deal Plank 3 x 11 inches, 12 feet long, twisted by 2 cwt., with a lever 3 feet long. Then taking Cfrom Table 113 at 034, Rule (715) becomes

3 x 12 x (11? + 3) x 2 T

= 6.76 inches, 118 x 38 x .034 x 3.4 the descent of the lever and weight.

(720.) Table 113 shows that there are very great differences in the Torsional stiffness of Materials: thus wrought iron has no less than 606 ; 3.82 158 times the stiffness of Deal. This is the more remarkable because by (571) and Table 84, the Torsional strength of say Yellow Pine and Wrought iron are in the ratio 10580 = 328 = 32.25 to 1.0 only. Again : the Transverse stiffness of say Memel Fir and wrought iron are by col. 4 of Table 105, in the ratio :0002223 = .00001565

13.5 to 1.0 only; whereas, as we have seen, the Torsional stiffness is 158 to 1.0. This shows that wood is not adapted for torsional strains, at least for cases where stiffness is required.

The Table 113 shows, also, that the torsional stiffness of Wrought iron and Steel are equal to one another, although, as shown in (571), the torsional strengths are in the ratio of 2 to 3. The stiffness of wrought and cast iron is about 2 to 1.



(721.) "General Principles.”—The Modulus of Elasticity is the tensile force that would stretch a bar to double its primitive length, and is usually expressed in pounds per square inch of area of the bar. Say we had a bar 10 inches long of some very elastic material, which stretches 1 inch by 20 lbs. per square inch; then obviously, to stretch it 10 inches, and thereby double the original length, we require 200 lbs. per square inch, which is, therefore, the Modulus of Elasticity for that material. It is here assumed that the elasticity is perfect, or that the extensions would be exactly proportional to the strain throughout, which would not be strictly true with any known material:moreover, it is assumed that the bar would bear stretching to double its original length without rupture, which is true with very few materials. The expression "Modulus of Elasticity" must be regarded as a conventional one, adopted for convenience of calculation, rather than as a statement of fact; thus, as applied to the bar we have just considered, it means that within certain limits and with moderate strains, the bar would stretch both of its length for each pound per square inch, tensile strain.

(722.) The Modulus of Elasticity may also be found from the shortening which a bar experiences by a Compressive strain, and is in that case the weight in pounds per square inch capable, theoretically, of reducing the length of the bar to nothing. Thus taking the same bar as before, say that a compressive strain of 20 lbs. per square inch on the bar 10 inches long was found by experiment to shorten it one inch, or to reduce its length to 9 inches. Then evidently we require 200 lbs. per square

inch to shorten it 10 inches, or to reduce its length to nothing, which again is, of course, not a statement of fact.

(723.) The Modulus of Elasticity may also be found from the deflection of a beam strained transversely by a known weight. It is well known that, with a beam supported at the two ends

and loaded in the centre, the weight generates a tensile strain on the lower fibres of the beam, which stretch and become longer than before. Similarly, a compressive strain is generated at the upper part of the section, which thereby becomes shorter than before. The combined effect of both is that the beam originally straight becomes curved or deflects, and we may then find the Modulus of Elasticity by the rule :

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In which E = the Modulus of Elasticity in pounds per square

b = breadth of rectangular bar, in inches.
d = depth
1 = length between supports, in inches.
d = deflection, in inches.

w = weight in pounds producing that deflection. This rule assumes that the material resists extension and compression with equal energy, or that the Modulus of Extension and Compression are equal, and without the knowledge that they are so, the result may be taken as giving a mean of the two.

(725.) When we proceed to find the Modulus of Elasticity experimentally, by the extension, compression, and deflection of any material, we find departures from the simple laws which we have so far assumed, which complicate the question very considerably.

1st. The Modulus, calculated from deflection, does not agree exactly with those found by direct extension or compression. 2nd. So far as observations on wrought and cast iron enable us to judge, bodies yield more to compressive than to equivalent tensile strains, and as a result of this, the “Modulus of Extension” is greater than the “Modulus of Compression." 3rd. The elasticity of all bodies is more or less imperfect, as manifested by the extensions, compressions, &c., increasing in a more rapid ratio than the strains, the result of which is that the Modulus is not constant, but is progressively reduced as the

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