Also, if be the angle at the centre made by a vertical radius, and the radius of the springing line,

Since this angle is less than 60°, the greatest thrust may be considered to act at the springing line. Let N be the thrust at the extrados of the key, and y the lever arm of N with respect to an axis of moments passing through the intrados of the springing line; then taking moments about this axis,

Ny = (r + h)2 (r - h) sin2 0 sin'

The calculation of the respective terms of (1) will be best effected by logarithms.

(r + h)2 (v − h) sin2 0 sin2 + 0 = 3450·9 .

and therefore

N=1228 nearly.

We next proceed to find the thickness of pier from the formula (Art. 124)

А

Now

A = 1 × 54 × (9+ 2·5) − × 452 (2 0 – sin 2 0)

× 452 × 3269236

310-5165505 = 145 nearly.

m =

=

=

§ × 542 × (9 + 2·5) — 453 († ◊ sin 0 + } cos3 0 — })

Wherefore, remembering that N = 122·8, and 7 = 3 + 9 + 2.5 14.5, we get by substitution in (6), taking S = 1,

This is the thickness of pier for mere equilibrium, as the coefficient of stability has been taken equal to unity.

2. Find the thickness of pier for safety in the last example when the coefficient of stability is taken equal to 2.

Ans. 98 feet.

3. The span of a segmental arch is 15 feet, rise or height of segment 2 feet, thickness of arch, a foot and a half, thickness of horizontal surcharge above the highest part of the extrados, half a foot, and depth of pier below the springing line, eight feet and a half; find the thickness of pier for stability of the arch, the coefficient of stability being taken at 14, and the curve of pressure being supposed to pass through the vertical joint at the key, at a distance from the extrados of the key of one-third the thickness of the vertical joint, and through the base of the pier at a distance from the exterior edge of the pier of one-third the thickness of the pier. Ans. 4.74 feet.

Section 7.

THE ELLIPTICAL ARCH.

126. PROP. To find the horizontal thrust of an elliptical arch with a horizontal surcharge.

The thrust due to an elliptical arch can be re

duced to the thrust of a circular one in the

following manner :—

Denote the half-span

or semi-transverse axis OB of the elliptical intrados Bmb, by r; the rise or semi-conjugate axis Ob by r'; and the thickness ab at the key by h.

Let us compare this arch with a circular arch

Bmb of the same span and of a vertical thickness a'b' at the key such that this thickness shall be to the corresponding thickness ab of the elliptical arch, in the ratio of OB to Ob, or of r to r'.

ť.

Draw a vertical line Dť, meeting the section of the elliptical arch in m, n, the section of the circular arch in m', n', the tangent at the point a in t, and that at a' in t. Then by the property of the ellipse, the ordinate Dm of the ellipse is to the corresponding ordinate Dm' of the circle, in the constant ratio of r' to r, whatever may be the position of the vertical line Dť.

Hence, by the preceding proposition,

Oa: Od Dm Dm'.

From m and m' draw to Oa', the perpendiculars mx and m' x'; then because Dm = Ox and Dm' = Ox',

We see from this that the vertical distances of the points m and m' below the respective extrados a and a', are in the constant ratio of r' to r.

The same property evidently holds for any two corresponding points in the elliptical and circular arches, provided the corresponding vertical depths of the arches are in the constant ratio of r' to r.