Use of buttress.

Use of tie-rods.

Example VIII. A semi-circular dome, circular in plan, is 40' inside diameter. The shell is 5' thick at the spring and 2' at the crown. The dome is of cut-stone. Will it stand?

"We draw a section (Fig. 114) of the half-dome and treat it exactly the same as any other arch. The only difference is in the assumption of the weights. Instead of assuming the arch 1' thick, we take with each voussoir its entire weight around one-half of the dome. Thus, in our case, we divide the section into six imaginary voussoirs. The. weight on each voussoir will act through its centre of gravity. Now weight No. I will be equal to the area of the top voussoir multiplied by the circumference of a semicircle with A B as radius. Similarly, No. II will lie equal to the area of the second voussoir multiplied by the circumference of a semi-circle with A C as radius; No. III will be equal to the area of voussoir 3, multiplied by the circumference of a semi-circle with A D as radius, etc. The vertical neutral axes Nos. I, II, III, etc., act, of course, through the centres of gravity of their respective voussoirs. Now the top voussoir measures 5' 6" by an average thickness of 2' 1" or 5 1/2.2 1/12 = 11,46 or, say, 11 1/2 square feet. As A B (the radius) measures 2' 9", the circumference of its semi-circle would be 8',6. Taking the weight of the stone at 160 pounds per cubic foot, we should then have the weight of No. I = 11,5. 8, 6. 160 = 15824 pounds, or, say, 8 tons.

 Similarly we should have: No. II = 12,4. 25,1. 160 = 49708 pounds, or, say, 25 tons. No. III = 14,2. 40. 160 = 90880 " " 45 " No. IV = 17,4. 52,7. 160 = 146716 " " 73 " No. V = 21,2. 62,8. 160 = 213017 " " 106 " No. VI = 28,7. 69,1. 160 = 317307 " " 158 "

Fig. 113.

We now make a b = 8 tons, b c = 25 tons, c d = 45 tons, d e = 73 tons, ef= 106 tons and f g = 158 tons. We find the horizontal pressures gl h1 g2 h1, etc. (in Fig. 113), same as before. In this case we find that the largest pressure is not the last one, but g5 h5 which measures 78 units; we therefore select the latter and (in Fig. 114) make a o = g5 h5 = 78 tons. Draw o b, o c, o d, etc., and construct the line a 1 i1, i2 i3 i4 i5 K6, same as before. In this case we cannot tell at a glance which is the most strained voussoir joint, for at joint H I the pressure is not very great, but the line is farthest from the centre of joint. Again, while joint J L has not so much pressure as the bottom joint M N, the line is farther from the centre. We must, therefore, examine all three joints.

Fig. 114.

We will take II I first. The width of joint is 2' 5" or 29". The pressure is a c, which scales 83 tons or 176000 pounds. The distance of K2 from the centre of joint P is 16". The area of the joint is, of course, the full area of the joint around one-half of the dome, or equal to II I multiplied by the circumference of a semi-circle with the distance of P from a g as radius. The latter is 10' 6", therefore area = 2 5/12. 33 = 80 square feet or 11520 square inches, therefore:

Stress at I = 176000/11520 + 6. 176000.16/11520.29 = + 65,9

Stress at H = 176000/11520 - 6. 176000.16/11520.29= - 35,3.

For joint J L we should have: the width of joint = 50". The pressure = of= 272 tons or 544000 pounds. The distance of K5 from the centre of joint is 8", while for the area we have 50. 66 2/3. 12 = 40000 square inches, therefore:

Stress at J = 544000/40000 + 6. 540000.8/40000.50 = + 26,6 pounds, and

Stress at L = 544000/40000 - 6. 540000.8/40000.50 = + 0,5 pounds.

For the bottom joint M N we should have the width of joint = 60", The pressure = o g = 428 tons or 856000 pounds. The distance of K6 from the centre of joint is 4", while for the area we have 60. 70 2/3 12 = 50880 square inches, therefore:

Stress at M = 856000/50880 + 6. 856000.4/50880.60 = +23,5 pounds, and

Stress at N =856000/50880 - 6.856000.4/50880.60 = +10 pounds.

The arch, therefore, would seem perfectly safe except at the joint II I, where there is a tendency of the joint to open at H Had we, however, remembered that this is an arch, lightly loaded, and started our line at the lower third of the crown joint, instead of at the upper third, the line would have been quite different and undoubtedly safe.

The dangerous point K2 in that case would be much nearer the centre of joint, while the other lines and joints would not vary enough to call for a new calculation, and we can safely pass the arch. One thing must be noted, however, in making the new figure, and that is that the horizontal pressures gl h1, g2h1, etc. (in Fig. 113), would have to be changed, too, and would become somewhat larger than before, as the line a1 6 would now drop to the level of a2. A trial will show that the largest would again be g3 h5, and would scale 80 units or tons, which should be used (in Fig. 114) in place of a o.

As regards abutments, there usually are none in a dome; it becomes necessary, therefore, to take up the horizontal thrust, either by metal bands around the entire dome, or by dovetailing the joints of each horizontal course. We must, of course, take the horizontal thrust existing at each joint. Thus, if we were considering the second joint II I the horizontal thrust would be g2 h1; or, if we were considering the fifth joint J L the horizontal thrust would be g5 h5. For the lower joint M N we might take its own horizontal thrust g6 h6, which is smaller than g5 h5, provided we take care of the joint J L separately; if not, we should take the larger thrust.

If we use a metal band its area manifestly should be strong enough to resist one-half the, thrust, as there will be a section in tension at each end of the semi-circle, or a = h/2. (t/f) (65)

Where a = area, in square inches, of metal bands around domes, at any joint.

Where h = the horizontal thrust at joint, in pounds.

Where (t/f) = the safe resistance of the metal to tension, per square inch.

In our case, then, for the two lower joints, if the bands are wrought-iron, we should have, as h = 80 tons = 160000 pounds.