This equation being of three dimensions in z, it follows that for certain values of y there are three possible values of %. The curve has therefore a point of contrary flexure, and is somewhat of the form shewn in the figure.

The LINE OF PRESSURE in a trapezoidal mass has been determined in my former paper. It is there shewn to be of three dimensions in ≈, and to have, like the line of resistance, a point of contrary flexure.

The POINTS OF RUPTURE being those where the line of resistance meets the Intrados or the Extrados of the mass may be determined by assuming in equation (7), y = a + z, tan a, and y = %, tan a; whence there is obtained

sec

{tan a

- a2 tan }

= 0......(8),

sec →ΣP sin
{tan a

- sec O tan a, ΣP cos

+ Pk cos p

--

tan a,} {tan a1 2 tan a2

2 tan a1 = tan → there is but one point of rupture in the

2 tan a, = tan → there is but one point of rupture in the

These single points of rupture are determined in the two cases by the equations

4. THE BUTTRESS.

If → be taken = 0 the trapezoidal mass will assume the position of the ordinary buttress, (Fig. 5), whose line of resistance is determined by the equation

~3 {tan'a, — tan3a,} + az2tana, +≈{ΣPsino + { a2 } +Σ ± Pk cos Þ
{tana, - tan a} + ax + ΣP cos

And its greatest possible height by the least root of the equation

The best dimensions of the buttress would seem to be those which bring the line of pressure to the center of the base. These may be determined by assuming in equation 12, y {a + % (tan a1 + tan a2)}, whence,

=

{tan'a, - tan'a,} +3ax {tana, +tana} -6% {2EP sin - (tan, +tana,) ΣPcos } Φ} - 6 {22 Pk cosaΣP cos }

=

0......(14).

To determine the line of pressure in the buttress, assuming → = 0 in equation (2), we have

Also assuming P, k,

y1 = a + C tan a1,

not to

be functions of C, and taking → = 0, y2 = C tan a,, we have, by equation (5),

2

Performing the integrations indicated in equation (15), substituting this last value of C, and reducing,

If απ = a2, (see Fig. 6), the mass may be taken to represent a pier or a wall of uniform thickness, and the equation (7) to its line of resistance will become

az2 {tan a1 — tan →} +≈{sec OΣP sin o +1a2} + sec → Σ ± Pk cos Þ
az + sec OΣP cos

Which is the equation to an hyperbola whose axis is inclined to the axis of at an angle represented by the formula

In the case in which = 0, or the intersections are horizontal, (Fig. 7), equation (17) gives for the equation to the line of pressure,

1

≈ΣP sin + =
↓ = acota.y = {} acota - ΣP costly+a

1

-

16

a3 cota − Σ + Pk cos Þ.....(21),

which is the equation to a parabola whose axis is vertical, whose para

The supposition a, == 0 gives the case of an upright Pier with horizontal intersections (Fig. 8), and the equation (18) to the line of resistance becomes

y =

≈ {EP sin + a2} + Σ + Ph cos p*
az+EP cos

(22),

the equation to a rectangular hyperbola, whose axis is by formula (19) inclined at an angle of 45° to the axis of %, whose asymptotes are therefore vertical, and the co-ordinates of whose center Care by formulæ (20),

CE being an asymptote to the hyperbola, it is clear that if AK be less than AD, that is, if

the line of resistance will not meet the extrados of the pier however great may be its height. But that if

22P sin be greater than a2,

it will somewhere cut the extrados; there is, therefore, in this case, a certain height of the pier beyond which, if it be continued, it will be overthrown.

This maximum height of the pier is determined by the equation

* This equation may be put under the following form, whence all the circumstances mentioned in the text are apparent,

6. THE STRAIGHT ARCH, OR PLATE BANDE.

Let → be now assumed = 7, the sections will then be vertical (Fig. 3),

2

the line of resistance will coincide with the line of pressure (Art. 2), and the equation common to both will be

In the case of a trapezoidal mass (Fig. 9), this formula gives by equation (7)

y =

- {tan atan a} -ax2+≈ΣP sin +E+ Pk cos
ΣΡ cos Φ

and the points of rupture are determined by the equation

~{tan a1 — tan ɑ2} +3azi +6{tan a1 ΣP cos -ΣP sin o}≈

12

If a1 = a2 (Fig. 10), the equation to the line of resistance and to the line of pressure becomes

y =

- az2 +≈ΣP sin + Σ + Pk cos

ΣΡ cos Φ

·(27).

The equation to a parabola whose axis is vertical, whose parameter and the co-ordinates of whose vertex are

2ΣP cos

is

a

The elements of this parabola are thus independent of the position of the mass.

Let us suppose the impressed forces P to be placed symmetrically (Fig. 11), then will the vertex of the parabola manifestly be situated midway between the extremities of the mass. Let the length of it

be 2b;