Fig. 29.

"We consider an arch as a truss with a succession of straight pieces; we can calculate it graphically the same as any other truss, only we will find that the absence of central or inner members (struts and ties) will force the line of pressure, as a rule, faraway from the central axis. Thus, if in Fig. 29, we consider A B C D as a loaded half-arch, we know that it is held in place by three forces, viz.:

1. The load BCLM which acts through its centre of gravity as indicated by arrow No. 1.

2. A horizontal force No. 2 at the crown C D, which keeps the arch from spreading to the right.

3. A force at the base B A (indicated by the arrow No. 3), which keeps the arch from spreading at the base. Now we know the direction and amount of No. 1, and can easily find Nos. 2 and 3. In an arch lightly loaded, No. 2 is always assumed to act at two-thirds way down C D, that is at F (where CE = EF = FD = 1/3 CD). In an arch heavily loaded, No. 2 is always assumed to act one-third way down C D, that is at E; further the force No. 3 is always assumed to act through a point two-thirds way down B A, that is at II (where B G = GH = HA = 1/3 B A). The reason for these assumptions need not be gone into here. Therefore to find forces Nos. 2 and 3 proceed as follows: If the arch is heavily loaded, draw No. 2 horizontally through E (C E being equal to 1/3 C D), prolong No 2 till it intersects No. 1 at O, then draw O H (H A being equal to 1/3 B A), which gives the direction of the resistance No. 3. We now have the three forces acting on the arch concentrated at the point O, and can easily find the amounts of each by using the parallelogram of forces. Make O I vertical and (at any scale) equal to whole load (or No. 1), draw I K horizontally, till it intersects O H at K; then scale I K, and K O (at same scale as O I), -which will give the amount of the forces Nos. 2 and 3. The line of pressure of this arch A B C D, is Line of pressure therefore not through the central axis, but along not central. E O H (a curve drawn through E and H with the lines Nos. 2 and 3 as tangents is the real line of pressure).

Now let in Fig. 30, A B C E F D A represent a half-arch. We can examine A B C D same as before, and obtain I K = to force No. 2; K O = to resistance (and direction of same) at U, where U C = 1/3 C D;OI being equal to the load on D A. Now if we consider the whole arch from A to F, we proceed similarly. L G is the neutral axis of the whole load from A to F, and is equal to the whole load, at same scale as OI. That L G passes through D is accidental.

Fig. 30.

Make EM = 1/3 EF and draw L M; also G H horizontally till it intersects L M at H, then is G H the horizontal force or No. 2. We now have two different quantities for force No. 2, viz.: I K and G H, I K in this case being the larger. It is evident that if the whole half-arch is one homogeneous mass, that the greatest horizontal thrust of any one part, will be the horizontal thrust of the whole, we select therefore the larger force or IK as the amount of the horizontal thrust. Now make S P = to IK and P Q = to No. 1, or load on A D and P R = to L G or whole load on F A, at any scale, then draw Q S and R S. Now at O we have the three forces concentrated, which act on the part of arch A B C D, viz.: Load No. 1 (=P Q), horizontal force No 2(=SP) and resistance K O (= Q S). Now let No. 3 represent the vertical neutral axis of the part of whole load on F D, then prolong K O until it intersects No. 3 at T; then at T we have the three forces acting on the part of arch E C D F, viz.: The loud No. 3 (= Q R), the thrust from A B C D, viz.: O T (= S Q), and the resistance N T (= R S). To obtain N T draw through T a line parallel to R S, of course R S giving not only the direction, but also the amount of the resistance N T. The line of pressure of this arch therefore passes along P O, O T, T N. A curve drawn through points P, U and N - (that is, where the former lines intersect the joints A B, D C, F E) - and with lines P O, O T and T N as tangents is the real line of pressure. Of course the more parts we divide the arch into, the more points and tangents will we have, and the nearer will our line of pressure approach the real curve.

Now if this line of pressure would always pass through the exact centre or axis of the arch, the compression on each joint would of course be uniformly spread over the whole joint, and the amount of this compression on each square-inch of the joint would be equal to the amount of (line of) pressure at said joint, divided by the area of the cross-section of the arch in square inches, at the joint, but this rarely occurs, and as the position of the line of pressure varies from the central axis so will the strains on the cross section vary also.

Let the line A B in all the following figures represent the section of any joint of an arch (the thickness of arch being overlooked) C D the amount and actual position of line of pressure at said joint and the small arrows the stress or resistance of arch at the joint.

We see then that when C D is in the centre of A B, Fig. 31, the stress is uniform, that is the joint is, uniformly compressed, the amount of compression being equal to the average as above. As the line of pressure C D approaches one side, Fig. 32, the amount of compression on that side increases, while on the further side it decreases, until the line of pressure C D, reaches one-third way, Fig. 33 (that is, DB= 1/3 AB). Then we see there is no compresStress at intra-dos and extrados.

Fig. 31.

Sion at A, but at B the compression is equal to just double the average as it was in Fig. 31. Now, as C D passes beyond the central third of A B, Fig. 34, the compression at the nearer side increases still further, while the further side begins to be sub-jected to stress in the opposite direction or tension, this action increasing of course the further C D is moved from the central third. This means that the edge of arch section at B would be subject to very severe crushing, while the other edge (at A) would tend to separate or open. When the line C D passes on to the edge B, the nearer two-thirds of arch joint will be in compression, and the further third in tension. As the line passes out of joint, and further and further away from B, less and less of the joint is in compression, while more and more is in tension, until the line of pressure C D gets so far away from the joint finally, that one-half of the joint would be in tension, and the other half in compression.

Fig. 32.

Tension means that the joint is tending to open upwards, and as arches are manifestly more fit to resist crushing of the joints than opening, it becomes apparent why it is dangerous to have the line of pressure far from the central axis. Still, too severe crushing strains must be avoided also, and hence the desirability of trying to get the line of pressure into the inner third of arch ring, if possible.

But the fact of the line of pressure coming outside of the inner third of arch ring, or even entirely outside of the arch, does not necessarily mean that the arch is unstable; in these cases, however, we must calculate the exact strains on the extreme fibres of the joint at both the inner and outer edges of the arch (intrados and extrados), and see to it that these strains do not exceed the safe stress for the material. The formulae to be used, are: For the fibres at the edge nearest to the line of pressure v = p/a + 6. x.p/a.d. (44)

Fie. 33.

Fig. 34.

And for the fibres at the edge furthest from the line of pressure

Pressure On Joint

v = p/a - 6. x.p/a.d (45)

Where v = the stress in lbs., required to be exerted by the extreme edge fibres (at intrados and extrados).

Where x = the distance of line of pressure from centre of joint in inches.

Where a = the area of cross section of arch at the joint, in square inches.

Where p = the total amount of pressure at the joint in lbs.

Where d = the depth of arch ring at the joint in inches, measured from intrados to extrados.

When the result of the formula? (44) and (45) is a positive quantity the stress v should not exceed (c/f), that is the safe compressive stress of the material. When, however, the result of the formula (45) yields a negative quantity, the stress v should not exceed (c/f),.that is the safe tensile stress of the material, or mortar.

The whole subject of arches will be treated much more fully later on in the chapter on arches.

Fig. 35.