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TABLE 65.-TRANSVERSE STRENGTH of TIMBER, &c.-continued.

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TABLE 65.-TRANSVERSE STRENGTH of TIMBER, &c.-continued.

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TABLE 66.-Of the TRANSVERSE STRENGTH of BARS 1 inch square, 1 foot long value of Μ, for Breaking load.

Materials.

T

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74323-318 56712 532 6720 3.000

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540

2410 334

1500 408

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548 345

454

1013 331
2826 724 3230] 7
(4) (5) (6) (7)

1478 10

4

(8)

=

working load of 1900 lbs. in the centre:-we may take Mr 78 lbs. from col. 3 of Table 67, and the rule (325) becomes 1900 × 10) (78 × 3) = 9 inches, the depth required.

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Again: to find the breadth for a beam of Pitch-pine, 12 feet long, 10 inches deep, to break with 24,000 lbs. in the centre:taking MT from col. 5 of Table 66, at 577 lbs., rule (326) becomes (24000 x 12) (100 x 577) 5.16 inches, the ÷ breadth required, &c.

=

TABLE 67. Of the TRANSVERSE STRENGTH and STIFFNESS of BEAMS, 1 inch square and 1 foot long.

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of Breaking-down load.

"Limit of Elasticity": Wrought-iron, and Steel & the Breaking-down loads.
tof Breaking-down load.

"Old Rule."-The rule (329) shows that in hollow square

sections the strength is proportional to

D1

very commonly used by practical men is:

(337.)

D

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d.

The old rule

In this rule a hollow beam is considered as composed of two solid beams, one having the external, and the other the internal dimensions; then the strength of the smaller one subtracted from that of the larger was supposed to give the strength of the hollow one. But it is overlooked that the ultimate deflection is inversely proportional to the depth (694), and that the full strength will not be realised if that deflection is not permitted. For instance, in Fig. 64, the hollow beam A is supposed to be composed of the two solid beams B, Fig. 65, and C, Fig. 66, but the ultimate deflection of the two latter will be in the ratio 1 to 2. Now, when combined as in A, and when breaking with the deflection due to B, the deflection of C is half only of that due with its own breaking weight, therefore half only of its strength is to be subtracted from the full strength of B in order to find the real strength of A. By the old rule the strength of A would be 43 - 23, or 64 8 56, but allowing half of C only, we obtain 64 460 as the actual strength.

-

obtain the same result by the rule (329), namely

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We should

D' - d'
D

, or

= 60, as before,

1.07, or 7 per cent. in excess of the old

(338.) Again: in Fig. 67, the internal dimensions are of the internal; for the former we have 43= 64, and of the latter becomes 33 × 3 20 25; hence the strength of the hollow

=

beam is 64 20-25 43.75. The rule (329) gives

256

or

43

4

=

81

or

175
4

=

4'-3' 4

= 43.75 also. The old rule gives

31 37, the correct rule being 43·7537 = 1·18, or

18 per cent. in excess of the old one.

Again: in Fig. 68, the internal sizes are of the external; hence, instead of deducting 343 = 42.875, we have to deduct

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