Beams with a Double Flange. A αι C B b Let ABFE be a section of a beam having two flanges ABba and EFfe formed by angle-irons riveted to a vertical plate CD, the material being symmetrically distributed with respect to the vertical line CD. Let O be the neutral axis, and KVG a line passing through the centre V of the vertical line CD parallel to AB and Er. Put A= the area of the section of the material, A1=area ABFE or 2 area ABGK, A2=2 area mjsi, A.=2 area jgb G, a2=2 area mjqv, a3=2 area jufG, D1=GB=GF, D2= sj, D3=bG, d2=jq, d3=fG, X=OV, I, moment of in = K e E V G m D ig w F ertia about KG, I=moment of inertia about O, W= the breaking weight upon the centre of the beam, l=the distance of the supports, S, S=1 the resistance of the material per square at the edges EF and AB respectively. To find the Neutral Axis. Taking KG as the axis of moments, Taking the moments of inertia with respect to the line KG, I1=moment inertia ABFE-2 moment inertia space bgsiqf —¦}(A‚D; — A‚D;—A ̧D?—a„d¦—a ̧d}) . (8.) which expresses the breaking weight when S is given. Let D', D'2,..., A', A1, . . ., &c., be the corresponding dimensions of a beam in all respects similar, and let r be the ratio of the linear dimensions, then D'rD', &c., A'=r2A,, &c., A'2D'2=r3A,D2, &c., That is, THE BREAKING WEIGHTS IN SIMILAR BEAMS ARE TO EACH OTHER AS THE SQUARES OF THEIR LIKE LINEAR DIMENSIONS. The method of demonstration here used in establishing this important theorem may be applied to any other form of beam. When the sections of the beams are similar, but the distance between the supports any quantity l1, then we have Suppose W in equation (11.) to be determined by experiment, then we are at liberty to assume AdC ין. where d is the depth of the beam, and C a constant determined by the assumed relation. From equation (14.), W'=r2W That is, THE BREAKING WEIGHTS IN BEAMS ARE FOUND BY MULTIPLYING TOGETHER THE AREA OF THE SECTION, THE DEPTH, AND A CONSTANT DETERMINED FROM EXPERIMENT ON BEAMS OF THE PARTICULAR FORM, AND DIVIDING THIS PRODUCT BY THE DISTANCE BETWEEN THE SUPPORTS. The value of l' in this formula is not restricted to the condition of similarity. In experiment 12, D1=35, D2=1·375, D2=3·22, d2=1·375, d2=3·2, W=24380 +80=24460, 7=84, A1=4·5×7=31·5, A2=1·375×·28 × 2=7, A ̧=3·22+1·845 x 2 11.8818, a=1.375 × 3 × 2=75, a,=3·2 × 1·825 × 2=11.68, A=A1-A2-A-a2-a-32·5-25.01=6.48, In experiment 13, W=21715+80=21800 nearly, 21800 × 84 (3·5+0611)—15·5 tons, 4 x 46.7854 × 2240 21800 × 84(3·5—·0611)—15 tons. 4 x 46.7854 x 2240 The values of S and S1, as determined by these calculations, being less for the beam in experiment 13 than they are for the beam in experiment 12, it follows that the latter has a better distribution of the material than the former. And at the same time the difference of the value of these constants is so small as to lead us to infer that the form of the beam in experiment 12 approaches to that of maximum strength with a given quantity of material. The sectional areas of the top and bottom flanges are to each other as 28:30 or 14:15, which is very nearly a ratio of equality. Experiments by Thomas Loyd, Esq., Inspector of Machinery, to ascertain the effect of a tensile strain upon bars of wrought iron under varied conditions. Twenty pieces of 18 SC bar iron, each 10 feet long, were cut out of the middle of twenty rods of iron. These 10-feet lengths were cut into two parts of 5 feet each, and marked with the same letter. A, B, C, &c., were first broken so as to get the average breaking strain. A2, B2, C2, were subjected to the constant action of three-fourths of the breaking weight for five minutes. The load was then taken off, and they were afterwards broken. It will be seen that the breaking strain was about the same as before, thus proving that the previous strain had not weakened them. In the first columns of the experiments it will be observed that the force required to break the bars was 32.57 tons, with a mean stretch of 9 inches upon twenty bars. In the second |