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parallel, and upper or underside horizontal, whosebearing or distance between the supports is AD anddepth in the middle BC. Let the horizontal line ACequal i AD be the axis or assimptote of the logarith-mic curve BE, whose subtangent is equal to the mo-dulus of fracture of the given material. Draw FG atany point F in the horizontal line AC at right anglesto AC, cutting the curve in G. FG is the depthrequired at F. In like manner the depth of anypart of the other half of the beam, &c. may be ob-tained.*

BC = n. Subtangent = f

Log.FG = Log

FC —f x 2.302585 (Log. FG — Log. n.)

TO DETERMINE THE SECTION OF AN ABUTMENT WALL,

UPON THE SUPPOSITION THAT THE EXTREME POINT

OF ITS BASE IS FIXED AND IMMOVABLE.

Considered as a Strutt.

Fig. 7. Let GD be the vertical height of an abutmentwall, AD a horizontal line where it leaves the found-ation or base line, EF the base line of a weight W,and EFG the angle, EF makes with a vertical line.

Draw the horizontal line GEH. In GD continuedtake DT equal to FG. Upon GT as a diameter, de-scribe the circle GTL cutting AD in L. Make DKin AD equal to 2 DL and HG equal to DL. JoinHF and HK. K is a point in a parabola whose focal

Fig. 6. * By the approximation of'principles, a prismatick beam may be

assumed to contain within it an arch pendant or insistent, whoseheight at the key ;= n, versed sine — x, and semispan = y — halfthe distance between its supports, wherein the excess, that is thesegment EBH above or below the arch, as the case may be, mustbe considered part of the weight. See Table of Examples, twolast columns.