The width of joint is, of course, 8", and the area = 8 x 12 = 96 square inches; therefore, from Formula (44)

Stress at edge J = 740/96 + 6. 740.1 1/2/96.8 = +16,26 and from (45)

Stress at edge I = 740/96 - 6. 740.1 1/2/96.8 = -0,85

Or the edge J would be subject to the slight compression of 16 1/4 pounds, and edge at I to a tension of a little less than one pound per square inch. The arch, therefore, is more than safe.

Stress at extrados and intrados.

Example II.

A four-inch rowlock brick arch is built between two iron beams, of five feet span, the radius of arch being five feet. The arch is loaded at the rate of 150 pounds per square foot. Is it safe?

In this case we will divide the top of arch A D into five equal sec-ions and assume that each section carries 75 pounds - (which, of course, is not quite correct). We find the horizontal pressures (Fig. 102) g1h1,g2h1, etc., as before, and find

Brick floor- , arch.

again g5 h5 the largest and equal to 575 pounds. We now make (Fig. 103) at any convenient scale, o a = g5 h5 = 575 pounds, and a b =

Arches 100182

Fig. 102.

bc = cd = de = ef=. 75 pounds and draw o a, o b, o c, etc. We now find the broken line a 1 i1 i2 i3 i4 K5 where:

a 1 is parallel with o a

i1 i2 " "

o c

i3 i3 " "

o e

1 i, is parallel with ob

i2 i3 " "

o d

i4K5 " "

of

In this case again evidently the greatest stress is on the skew-back joint C D, for it not only has the greatest pressure o f, but the curve of pressure passes farther from the centre than at any other joint. We find that C K5 scales 1 1/4 inches, therefore the distance of K5 from the centre is (x =) 3/4". We scale o f and find it scales 690 units, or (p )= 690 pounds. The joint is 4" wide and its area = 4 x 12 = 48 square inches.

Arches 100183

Fig. 103.

From Formulas (44) and (45) we have then:

Stress at C = 690/48 + 6. 690.3/4/48.4 = +30,6 pounds and

Stress at D = 690/48 - 6. 690.3/4/48.4 = - 1,8 pounds.

The arch, therefore, is perfectly safe.

Example III. Two iron beams, five feet apart, same as before, but filled with a straight 7" hollow fire-clay arch. The load per foot to be assumed at 140 pounds. Is the arch safe?

Of the 350 pounds on the half arch we will assume 80 pounds to come on each of the blocks and 30

Fireproof floor-arch.

pounds on the skew-back. We then (in Fig. 104) find, as before the horizontal pressures, g1 h1, g2 h1, etc. Again we find the largest pressure to be g5 h5, and as it scales 2040 units, we make (in Fig. 105) at any convenient scale and place o a = g5h5 = 2040 pounds. We also make ab = bc = cd = de = 80 units and ef= 30 units.

Arches 100184

Fig. 104.

Draw oa, ob, oc, etc. Drawing the lines parallel thereto, beginning at a we get the line a 1 i1, i2 i3i4 K5, same as before Imagining a joint at C D this would evidently be the joint with greatest stress, for the same reasons mentioned before. We find C K5 scales 2 5/8", and as C D scales 7 1/4' the point K5 is distant from the centre of joint.

Arches 100185

Fig. 105.

(x = )3 5/8 - 2 5/8 = l" as of scales 2100 units or pounds, and the joint is 7 1/4' deep with area = 74. 12 = 87 square inches, we have:

Stress at C = 2100/87 + 6. 2100.1/87.7 1/4 = + 44,14 pounds and

Stress at D = 2100/87 - 6. 2100.1/87.7 1/4 = + 4,14 pounds.

The arch, therefore, would seem perfectly safe. But the blocks are not solid; let us assume a section through the skew-back joint C D to be as per Fig. 106. We should have in Formulae (44) and (45) x, p, and the depth of joint same as before, but for the area we should use a = 3.1 1/2.12 = 54 square inches, or only the area of solid parts of block. Therefore we should have:

Fig. 106.

Arches 100186

Stress at C = 2100/54 + 6.2100.1/54.7 1/4= + 71 pounds, and Stress at D = 2100/54 - 6. 2100.1/54.7 1/4 = + 6,71 pounds.

There need, therefore, be no doubt about the safety of the arch.

Example IV.

Over a 20-inch brick arch of 8 feet clear span is a centre pier 16' wide, carrying some two tons weight. On each side of pier is a window opening 2 1/2 feet wide, and beyond, piers similarly loaded. Is the arch safe?

We divide the half arch into five equal voussoirs.

The amounts and neutral axes of the different voussoirs, and loads coming over each, are indicated in circles and by arrows; thus, on the top voussoir E B (Fig. 107) we have a load of 2100 pounds, another of G2 pounds, and the weight of voussoir or 228 pounds. The neutral axis of the three is the vertical through G, (Fig. 108). Again on voussoir E F (Fig. 107) we

Arch In front wall concentrated loads.

have the load 174 pounds, and weight of voussoir 228 pounds; the vertical neutral axis of the two being through G11 (Fig. 108). Similarly we get the neutral axes G111, GlV and Gv (Fig. 108) for each of the other voussoirs. Now remembering that 1 g, (Fig. 107) is the neutral axis of and equal to the voussoir B E and its load; 2 g2 the neutral axis of and equal to the sum of the voussoirs B E and E F and their loads, etc., we find the horizontal thrusts g1 h1 g2 h1, g3 h3, etc. The last g5 h5 is again the largest, and we find it scales 7850 units or pounds.

Arches 100187

Fig. 107.

The arch being heavily loaded we selected a at one-third from the top of A B. We now make (Fig. 108) a o = 7850 pounds or units at any scale; and at same scale make ab = 2390 pounds; b c = 402 pounds; c d = 432 pounds; d e = 2956 pounds, and e f= 1868 pounds. Draw o b, o c, o d, etc. Now draw as before a 1 parallel with o a to axis G1; also 1 i1 parallel with o b to axis G11; i1, i2 parallel with o c to G111, etc. We then again have the points a, K1 K2, K3, K4 and K5 of the curve of pressure. As K5 is the point farthest from the centre of arch-ring and at the same time sustains the greatest pressure (of) we need examine but the joint C D; for if this is safe so are the others. We insert, then, in Formulae (44) and (45) for p = of= 11250 pounds, and as K5 C measures 6 1/2", x = 10" - 6 1/2" = 3 1/2"; also as the joint is 20" wide, a = 12. 20 = 240 square inches.