greatest tension of the chain is at the top of the pier. The value of this tension is

The section of the chain must be such as to satisfy the relation

T = A . S . . . . . . (8),

T being the maximum tension, A the area of the section in inches, and S the coefficient of resistance of the iron to shearing.

The rods which sustain the platform or roadway in suspension bridges are subjected to forces of traction in the direction of their length. Each rod of suspension sustains a weight equal to the total weight of the roadway divided by the number of rods, together with the resultant weight on each rod arising from the load placed upon the platform or roadway.

Consequently the dimensions of the rods of suspension are determined by the formula of Chapter II., Part I.

5. Coefficients of the resistance of materials to shearing, in pounds avoirdupois per square inch. (Rankine's Applied Mechanics, p. 633.)

6. The stability of a chimney-piece arch.

Let ABCDE and FGHKL (Fig. 70) be the cross sections of two perpendicular walls supporting a chimney-piece arch, of which the cross section is the trapezoid CHKD, CH and KD being the parallel sides. The stability of an arch of this kind depends upon the strength of the abutments, determined in the following manner :—

E

D

H

K

L

d

B

k

Fig. 70.

P

Let a, b, be the lengths of the parallel lines DK, CH, and h the distance between them; H the height of each wall BC or GH;

the thickness of ED or KL to be determined; q lbs. the weight of a cubic foot of the material of the perpendicular walls; p lbs. the weight of a cubic foot of the arch. Then if P be the weight of

the arch for a unit of width (measured perpendicularly to the face DCHK),

Neglecting friction, the arch is kept at rest by the perpendicular reactions from the joints DC and HK, which (reactions), from the symmetry of the figure, are equal to one another. Denote each reaction by Q; and let a be the angle which the joint CD or HK makes with the horizon.

Now when the arch is about to overturn each wall round the extrados A or F, the pressure Q may be supposed to act at C or H, perpendicular to CD or HK, and, consequently, in a direction which makes an angle a with BC or GH. Wherefore the pressure Q at C or H, at this instant, can be resolved into two components, viz. Q sin a, which is horizontal, and Q cos a, which is vertical. Hence the moment (M) of Q about A is

and by the notation and the geometry of the figure

BC =

H, and AB ED + Cc = x + † (a − b).

Hence (1) becomes

MP. H. tan a - P{x + (a - b)}

In order to simplify this equation, put a b = 7; then, remembering that P = † (a + b) hp, we have

M = (a + b) hp H tan a –

· † (a + b) hp (x + 1⁄2 1) .

(3).

AE) is

Again: the area of AEDd (Dd being drawn parallel to

AE. ED, or (H+ h) x ;

and the distance of the centre of gravity of this area from AE is

ED, orx.

Hence the moment about A of the part of the abutment which corresponds to the face AEDd is

1⁄2 x2 (H+ h) q.

Similarly, the moment about A of the part of the abutment which corresponds to the face BCcd is

† 1 H (x + − 1) q.

Also, the area of D Cc is

Cc. Dc, or lh;

and the distance of the centre of gravity of this area from AE is

Ad+Cc, or x + f l.

Wherefore the moment of DCc about A is

† 1 h (x + f 7) q.

Consequently, if M' be the moment of the abutment ABCDE about the extrados A,

M' =

† x2 (H + h) q + § 1 H (x + ‡ l) q + † lh (x + † 1) q (4).

....

Now, in order that equilibrium may exist between the arch and its supports, the moments M and M' in (3) and (4) must be equal. Equating the values of M and M', we get for the determination of x the quadratic

(H + h) q . x2 + {l Hq + † 1 hq + 1 (a + b) hp } x

+ } l2 Hq + 1/2 l3hq + † (a + b) hpl − (a + b) hp H tan a = 0.