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cutting the axis in F, then HFTBAH is the sectionof the abutment wall required.

The direction of the weight, considered to be thetangent of the extrados of an arch.

The direction EW is resolvable into the extradosof any arch EA, the tangent of which is at right anglesto any line EF drawn from F to E any point in HGwhose radius of curvature at the vertex A does notexceed the modulus of fracture of the material whenthere is no incumbent weight, and when there is,

does not exceed c = ——- t , see page 43. of this tract,1 +

n

putting GF — n — AV, and when the abutting jointEF does not exceed HF.

Let the extrados EA be a circular arc, and thevertical line CA the radius = EFC. From any pointa in the extrados draw ac at right angles to the tan-gent at a, and let fall from a the vertical line ab equalGF, and draw the horizontal line be, cutting ac in c.Through VcF draw the intrados to the extradosAaE.

An abutment wall considered as an arch.

Fig. 9. Take GF = the modulus of horizontal thrust = thefocal distance of a parabola in GD the height of theabutment wall = the absciss GD of the parabolaGJBK. Draw the horizontal line GH of any length,and bisect it in E and join EF. From EH letfall the similar triangle EHB, EH and GF homo-logous sides, B is a point in the parabola: in likemanner, find any other point K. Through GBK drawthe parabola GBK, cutting the horizontal line DK inK. Continue WE at right angles to EF the direc-tion of a weight whose base is equal to EF till ittouch the parabola.