102

CALCULATIONS RELATIVE TO THE

stances, the tube would carry (in the absence of a disturbing cause) theload ad infinitum.

After all the parts of the tube under strain were brought to bear,the experiment was resumed on the 10th instant, when the load wasprogressively increased from 77,527 lbs. to 126,138 lbs. or 56-3 tons,when it broke, by tearing the bottom asunder through the solid plates.

Now, if we compare the above results with the first experiment, weshall find, that by the simple addition of 5 cwt. more material, the bear-ing powers of the tube were raised from 3511° 56) tons; and even withthis latter weight, the cellular top is yet too strong for the bottom, andI have no doubt, if the next experiment comes up to my expectation,that the previous experiments, as respects the ratio of 5 to 3, of the topto the bottom, will be fully confirmed. Assuming however the presentto be the maximum, we should then have for the large tube the follow-ing results:—

Taking the area of the bottom of the model tube at 12 - 8 inches, thebreaking weight 56-3 tons, depth 54 inches, and the distance betweenthe supports 75 feet; we have (neglecting the weight of the tube), byMr. Hodgkinson’s formula, supposing the tube to be cast-iron,

26 x 12-8. x 54

900

= 19-95,

or say 20 tons. Now, the same area for the bottom, when composed ofplate iron, carried 56 - 3 tons. Hence56-3 x 2620

= 73-19

as the new constant for computing the strength of tubes of the same con-struction as the model. Taking therefore the new constant, we have for atube six times the size in every respect, and six times the thickness ofplates, or 450 feet span, 27 feet deep, and 460-8 inches area at the bottom,

73x 460-8 x 3245400

=2018 tons,

the breaking weight, exclusive of the weight of the tube itself.

Let us now compare these results with Professor Airy’s theory, that

the strengths are as the squares, and the weights as the cubes; then we

have half the weight of the tube added to the breaking weight, and this

sum multiplied by the square of 6, or

/5-15 \

( "2“ +56-S) x36 = 21!9