488

HISTORY OF ENGINEERING.

Book I.

case the weight of half the bridge will be 28*354 x 3295503, which is equal to 9344*0692062tons ; the weight of the whole bridge will therefore be 18688*1384124 tons. As the lawwhich secures a maximum of strength produces an equilibration throughout the wholearch, the horizontal pressure throughout is the same as against the abutment. In adjustingthe strength of the abutment to resist the horizontal pressure, the pressure must be sup-posed to act in the line from which the arch springs.

Mr. Playfair observed, the pressure on the abutments will be found by multiplying halfthe weight of the bridge by the number 2*41605, or by multiplying the weight of thewhole bridge by 1 *20802; the pressure on each abutment will therefore exceed the wholeweight of the bridge by about This immense pressure must be counteracted by abut-ments of great size, and of the most solid structure ; it would be worth while to comparethis pressure with the thrust against the abutments or piers of some well-known stonebridge, such as that of "Westminster or Blackfriars ; one would be enabled by that means toform some notion of the comparative strength required in the abutments. The rule abovelaid down gives the pressure in the direction perpendicular to the face of the abutment;of that pressure one part is directed downwards, and may be had by multiplying theforegoing number by *4139, and hence the perpendicular pressure on the abutment is foundnearly equal to the weight of half the bridge.

The other part, that which acts on the abutment horizontally, and which it is of mostconsequence to know, is found by multiplying the pressure formerly computed, namely,that which is perpendicular to the face of the abutment, by *91, which brings out the hori-zontal thrust against the abutment, equal to the whole weight of the bridge multiplied by1 *099, or nearly equal to the whole weight increased by one tenth part.

The horizontal thrust at the crown may be found by multiplying the weight of thebridge by 1*188 ; I would not, however, have it understood that this last number is accurate,as the investigation of it involves the knowledge of the centre of gravity of the bridge, theexact determination of which would require more data and more time than I am in pos-session of.

Mr. J. Robeson _I cannot conceive any physical division of this frame, which will

enable me to answer this question; I do not believe it possible to divide it into greatwedges, which will abut on each other like so many stones.

Dr. Milner observed, that the construction of the arch of this bridge required the in-genuous aud mutual communication of the theorists and the practitioners, for computingthe horizontal drift or shoot of the bridge, and its action against the abutments.

Dr. Charles Hutton .— By the equal sections mentioned in the question may beunderstood either vertical sections of equal weight, or those perpendicular to the curve, ofequal weight or of equal length ; and whichever of these is intended, their thrust orpressure in direction of the curve may be easily computed, if wanted for the purpose ofmaking experiments on the strength of the frames, to know whether they will bear thosepressures, or what degree of pressure they will bear without being crushed to pieces. Butas it is evident that the frames next the abutment will suffer the greatest pressure of any,I shall here give a computation of the actual pressure there, which may be sufficient, since,if the frames at the abutments are capable of sustaining that greatest pressure, we may safelyconclude that all the others, from thence to the vertex, will be more than capable of sus-taining the lesser loads or pressures to which they are subject, and this computation willanswer the latter and most essential part of the question, viz. On what part, and with whatforce, will the whole act on the abutments ? Now, from the nature of an arch, it appearsthat the whole pressure on the abutments will be chiefly on the lower parts of the impost,where the lower frame rests on it, and where we shall, therefore, in our computation,suppose it to act; and in the calculation, the whole weight of the half-arch A () must besupposed united in its centre of gravity N. Then, if a vertical line N M be drawn throughthe centre of gravity N, by computation, it is found that DM is nearly equal to 160 feet,and consequently ME is equal to HO feet; also if N O be perpendicular to the impost, or

n

Fig. 484.

in the direction of the arch at O E, we shall have this proportion, viz. as MN (60) is to theweight of half the arch (3250 tons), so is N O (152) to the pressure on the impost in thedirection of the arch at O, and so is M E (140) to the horizontal thrust or pressure in thedirection ME. This gives 8233 tons for the pressure on the impost at O in the direction