π

0 == 1·5708, sin 0 = 1, cos 0 = 0. Hence, substituting 2

these values in the preceding equation, we get

= 100 × 150 × 08575 = 1286 lbs.,

the horizontal thrust on the joint at the springing line.

Obs. It will be seen by a comparison of this result with that in example 1, that the thrust at the reins is considerably more than at the springing line.

4. The radii of the intrados and extrados of a semicircular arch are respectively 10 and 12 feet; find the greatest horizontal thrust on the arch, the density of the masonry being the same as in Example 1, viz. 150 lbs. per cubic foot.

Ans. 2,003 lbs.

5. In the last example find the horizontal thrust on the springing line of the arch.

6. Find the horizontal thrust on a joint angle of 30° with the vertical, in Example 4.

Ans.

1,489 lbs.

which makes an Ans. 1,213 lbs.

7. If the height of pier of the arch in example 4 be 15 feet, find the thickness of the pier in order that it may support the arch with safety; a cubic foot of the arch and pier being taken at 150 lbs., and the coefficient of stability at 1.9.

Ans. 6'08 feet.

Section 3.

CIRCULAR ARCHES WITH SURCHARGES.

112. PROP. To find the conditions of equilibrium of a

circular arch with a horizontal surcharge.

The following method of investigation, with some slight modifications, is applicable to an arch of any form.

Let the vertical height CD of the circular arch ABCD be denoted by h, the height GD of the surcharge by h1, the radius of the intrados by r, the weight of a unit of width of the arch and its load between CD and a joint EF by P, the density of the arch by 8,, and that of the surcharge by 82.

Then, if N be the horizontal thrust at D, the highest point of the extrados, arising from the weight P, y the lever arm of N in reference to the extremity E of

H

Fig. 54.

G

D

C

the joint EF, and x the lever arm of P with respect to the same point E, one condition of equilibrium (Art. 108) is

Now Px may be considered as the algebraic sum of the moments of (DGHd + CDdc +ƒCcF+ƒFO - EOC), with respect to the point E.

Denoting, then, the angle at O by 0, the area DG Hd is

h1 (r + h) sin 0;

and the horizontal distance of the centre of gravity of this area from E is

(h) sin - h sin 0,

or (r – h) sin 0.

Wherefore, as the density of the surcharge is d2, the moment of the surcharge about E for a unit of width is

The following moments about E are deduced in a similar way:

Moment of CDdc = 12 h (h) 8, sin2 0,

ƒCcF = 1⁄2 (r2 – h2) d, {r − ( r + h) cos 0} sin2 0.

Again: the area of the triangle ƒFO is

(r+ h) sin 0 (r + h) cos 0,

or (r+ h)2 sin 20;

and as the horizontal distance of the centre of gravity of this triangle from E is

{(r+ h) - h} sin 0,

or (2rh) sin 0,

1 12

the moment of fFO about E = (r+ h)2 8, (2r – h) sin 20 sin 0.

Lastly, the area of the sector OEC is r2 0, and if z1 be the distance of the centre of gravity of the sector from O, the horizontal distance of the centre of gravity of the sector from E is

Wherefore the moment of the sector OEC about E is

Collecting all these moments with their proper signs, and equating the result to Px, we get, after obvious reductions, the equation,

Px

--

= 1 (r+h){(r− h) (h, d2 + hd, + rd,) — } §, (r + h) (r−2h) cos e sin'

In order then that rotation may not take place about E, the

Pa

value of the thrust N must be greater than equations (2) and (3).

given by the

y

N

the thrust N, u and v being the lever arms of P and N in reference to an axis through the extremity F of the joint EF, in this case acting at D.

The following moments about Fare readily obtained:

Moment of the surcharge DH = h (r + h) sin 0 d2 × 1⁄2 (r + h) sin 0 =h1 (r+ h)2 sin2 0 d;

Moment of CDdc= h (r + h) sin 0 d1 × 1 (r + h) sin

Moment of ƒ CcF= {r − (r + h) cos 0} (r+ h) sin 0 &1 × 1⁄2 (r+h) sin ✪

O

=

1⁄2 (r + h)2 sin2 0 dı {r − (r + h) cos 0} ;

Moment of fF 0 = † (r + h) sin 0 (r + h) cos 0 d, × } (r + h) sin 0

The last of these must be taken with a negative sign.
Collecting these moments, we get

Pu = † (r + h)3 {h1 d2 + h d1 + r di

(r + h) &, cos e sin2 0

v = Cƒ = OC — Of = r —

(r + h) cos 0

Hence another condition of the stability of the arch is that N must be less than the minimum value of P 2, determined by

the preceding equations.

The condition given in Art. 112 will, in general, be sufficient for determining the stability of an arch.

STRENGTH OF PIER.

114. PROP. To find the thickness of pier necessary to support an arch as in the preceding article with a horizontal surcharge.

N

R

K

H

F

d

I

A f

B

P

E

Fig. 55.

G

D

с

Let EF be the joint of the arch on which the thrust N is the greatest as in the preceding Articles. Then by the principles of the lever the weight P of the part CDEF of the arch and its load, and the thrust N, may be supposed to act at the intrados E when the joint EF is about to open outwards. Let a and b be the lever arms of P and N in reference to the exterior edge R of the pier.

Then if P be the weight of the part LBSR of the pier, and t its thickness; P2 the weight of the

part HFEBLK of the arch, pier, and surcharge, and c the lever arm of P, with respect to R, the condition of equilibrium is

Nb L Pa + P1t + P2c;

the weights P, P1, P2, being referred to a unit of width of the arch, surcharge and pier.

Let S be the coefficient of stability (Art. 112); then

Nb S = Pa + 1⁄2 P1t + P2c..

I

(1).

Let & be the density of the masonry (including the arch) below a horizontal line through the intrados C; &, the density of the surcharge above this line; the angle which the joint EF