The manner of laying arches has been de-scribed in the previous chapter, while in the first chapter was given the theory for calculating their strength; all that will be necessary, therefore, in this chapter will be a few practical examples. Before giving these, however, it will be of great assistance if we first explain the method of obtaining graphically the neutral axis of several surfaces, for which the arithmetical method has already been given (p. 7.). To find the vertical neutral axis of two plane surfaces A B E F and B C D E, proceed as follows: Find the centres of gravity G11 of the former surface and G, of the latter surface. Through these centres draw G11 H11 and G1 H1 vertically.

Draw a vertical line a c anywhere, and make ab at any scale equal to surface A B E F, and at same scale make b c = B C D E. From any point o, draw o a, ob and oc. From the point of intersection gl of o c with H1 G1 draw gl g11 parallel o b till it intersects H11 G11 at g11; from g11 draw g11 g parallel with o a till it intersects o c at g; through g draw g G vertically and this is the desired vertical neutral axis of the whole mass.

Where there are many parts the same method is used.

We will assume a segmental arch divided into five equal parts. Calling part A B C D = No. I; part D C E F = No. H, etc., and the vertical neutral axis of each part, No. I, No. H, etc.

Neutral axis found graphically.

Fig. 98.

Draw a f vertically and at any convenient scale, make

 a b = No. I or ABCD bc = No. H or D C E F, cd = No. III de = No. IV and ef = No. V

From any point o draw oa, ob, oc, od, oe and of.

From point of intersection 1 of vertical axis No. I and oa, draw 1 g parallel with o b till it intersects vertical axis No. II at g: then draw g h parallel with o c till it intersects vertical axis No. III at h; similarly draw h i parallel with o d to axis No. IV and i j parallel with o e to axis No. V, and, finally, draw j 5 parallel with of till it intersects a o (or its prolongation) at 5. A vertical axis through 5 is the vertical neutral axis of the whole mass. Prolong j i till it intersects o a at 4, i h till it intersects o a at 3, and h g till it intersects o a at 2.

A vertical axis through 4 will then be the vertical neutral axis of the mass of Nos. I, II, III and IV; a ver-tical axis through 3 will be the vertical neutral axis of the mass of Nos. I, II and III; and through 2 the vertical neutral axis of the mass of Nos. I and II. Of course the axis through 1 is the vertical neutral axis of No. I. We will now take a few practical examples.

Fig. 99.

Fig 100.

Example I.

In a solid brick wall an opening 3 feet wide is bricked over with an 8-inch arch. Is this strong enough?1

The thickness of the wall or arch, of course, does not matter, where the wall is solid, and we need only assume the wall and arch to be one foot thick. If the wall were thicker, the arch would be correspondingly thicker and stronger, so that in all cases where a load is evenly distributed over an arch we will consider both always as one foot thick. If a wall is hollow, or there are uneven loads, we can either take the full actual thickness of the arch, or we can proportion to one foot of thickness of the arch its proportionate share of the load.

In our example we assume everything as one foot thick. The load coming on the half-arch B J I L Fig. 100, will be enclosed by the lines A L and IA at 60° with the horizon. We divide the arch into, say, four equal voussoirs B C = CF = FG = GJ. (The manner of dividing might, of course, have been arbitrary as well as equal, had we preferred.) Draw the radiating lines through C, F and G, and from their upper points draw the vertical lines to D, E and H.

Now find the weight of each slice, remembering always to include the weight of the voussoir in each slice. We have, then, approximately,

Arch In solid wall.

Voussoirs arbitrary.

 No. I (A B C D) = 3' x 1' 2 x 112 pounds = 168 pounds No. II (D C F E) = 2 1/8' x 1' 2 x 112 " = 119 " No. III (E F G H) = 1 1/2' x 7 ' 16 x 112 " = 73 " No. IV (H G J I) = 7/8' x 7/16' x 112 " = 43 " Total = 403 "

1 For convenience, most of the figures are duplicated: the first, showing manner of obtaining the horizontal thrust; the second, the manner of obtaining the line of pressure.

Fig. 101.

As the arch is evidently heavily loaded at the centre, we assume the point a at one-third the height of B L from the top, or

L a = LB/3 =2 2/3" and draw the horizontal line a 4.

As previously explained, find the neutral axes:

Horizontal pressures.

 1 g1, of part No. I, 2 g2. of parts No. I plus No. II, 3 g3 of parts Nos. I, II and III, and 4 g4 of the whole half arch.

Now make at any scale 1 g1 = 168 units = No. I; similarly at same scale 2. g2 = 168 + 119 units = 287 = No I and No. II, and

3 g3= 287+ 73 units = 360 = No. I + No. II + No. III.

4 g4 = 360 + 43 units = 403 = weight of half arch and its load. Now make: C l1 = 1/3 C L1 = 2 2/3".

Similarly, F 12 = 1/3 F L2 = 2 2/3"; also, G l3 = 1/3 G L3 = 2 2/3",

And, J l4 - 1/3 J I = 2 2/3".

Through g1, g2, g3 and g4 draw horizontal lines, and draw the lines 1 l1 2 l2, 3 l3 and 4 l4 till they intersect the horizontal lines at h1, h1, h3 and h4; then will g1 h1 measured at same scale as 1 g1 represent the horizontal thrust of A B C D; g2 h1 the horizontal thrust of A B F E; g3 h3 the horizontal thrust of A B G H and g4 h4 the horizontal thrust of the half arch and its load. In this case it happens that the latter is the greatest, so that we select it as our horizontal pressure, and make (in Fig. 101) a o = g4h4= 620 pounds, at any convenient scale.

 Now (in Fig. 101) make a b = 168 pounds = No. I. bc = 119 " = No. II. c d = 73 " = No. III. dp. = 43 " = No. TV.

Draw pb, oc, od and oe.

Now begin at a, draw a 1 parallel with o a till it intersects axis No. I at 1; from 1 draw 1 i1, parallel with o b till it intersects axis No. II at i1; from i1, draw i1 i2 parallel with o c till it intersects axis No. Ill at i2; from i3 draw i2 i3 parallel with o d till it intersects axis No. IV at i3; from i3 draw i3 K4 parallel with o e. A curve through the points a, K1, K2, K3 and K4 (where the former lines intersect the voussoir joint lines) and tangent to the line a 1 i1 i2 i3 K4 would he the real curve of pressure. The amount of pressure on joint C L1 would be concentrated at K1 and would be equal to o b (measured at same scale as a b, etc.). The pressure on joint F L2 would be concentrated at K2 and be equal to o c. The pressure on joint G L3 would be concentrated at K3 and equal to o d. The pressure on the skew-back joint J I would be concentrated at K4 and be equal to o e. The latter joint evidently suffers the most, for not only has it got the greatest pressure to bear, but the curve of pressure is farther from the centre than at any other joint. We need calculate this joint only, therefore, for if it is safe, the others certainly are so, too. By scale we find that J K4 measures 2 1/2", or K4 is (x =) 1 1/2" from the centre of joint; we find further that o e scales 740 units, therefore (p ==) o e = 740 pounds.