Chap, XXIII.
CENTERING.
1407
into part* representing 576,000; this must be set off upon each of the lower struts, toexpress their sustaining force, which is shown by the dotted lines at, as; and as theexterior strut receives some assistance from the outer arched rib, to the extent of 518,400pounds, we must set off this quantity in continuation of these lines from t to e, which willform one side of the parallelogram ae ; the other side will be as. The parallelogram is to
\
Fig. 2349.
o
pitot’s system.
be completed by drawing $x and ex, equalin length and parallel to their opposingsides. A line drawn diagonally, as ax, willthen express the supporting strength of allthe pieces, supposing the pressure had beenin the direction of the diagonal, and as themaximum of the weight is in the perpen-dicular direction, and not obliquely, wemust let fall a perpendicular from the pointx, and let this line cut the base line ofthe arch; then the distance ac must beset off upon it at b, so that ac shall beequal to cb. From the point b, draw aright line parallel to a.r, of the samelength, which will cut the perpendicularat y ; join xb, and draw ay parallel to it;the parallelogram axby will be com-
pleted, and xcy is a vertical diagonal, the proportionate length of which will expressthe strength of the rib; this upon the scale will be found to be 2,850,000 pounds,making the centre equal in strength to more than four times that required for theload it has to support. All the timbers in the upper part of the centre may have theirstrength calculated in the same manner as we should proceed for a pair of principalswith a king-post, making due allowance for the extra support given to a centre by theouter arched rib, which does not exist in a roof. The force of each strut being taken at432,000 pounds, and that of the curve at 518,400, we must draw two lines from themiddle of the top of the king-post, viz. mv, parallel to the lower strut c, making its lengthequal to 432,000, and the second line, upon the upper strut ms, must also have a similarlength added to 518,000, or 950,000 pounds given to it. Complete the parallelogrammsrv, and draw the horizontal line rk, from the lowest point of the parallelogram, cuttingthe line mq in k, and make kq equal to km ; then mq will represent the weight that can becarried by the upper portion of the centre, which will be found equal to 1,160,000 pounds,greatly exceeding the strength required. The chief part of the load is sustained on theupper portion of the centre, therefore it requires that the struts should have had a greaterscantling than the lower, although there is abundant strength throughout. The strainof the stretching-piece A A is denoted by rk and mq, which is the actual load sustainedby the upper portion of the centre; one-sixth part or less of the strength of the cohesionof this stretching-piece is sufficient for the horizontal thrust to which it is exposed.