304. It fortunately happens that simple designs are best calculated for centres; for it would be very difficult to form anything like an accurate estimate of the strength of a complicated one. We will here show some approximate methods of determining the scantlings for the timbers of the designs already given; and add to one of them some examples, in numbers, which will tend further to explain the subject.

In the centre (Fig. 1, Plate XXXVII.) the stress may be considered, so far as it tends to strain the frame E D H; also the stress upon the pieces E H, H' E', when the whole load is upon them; and, lastly, the strain upon the posts G K, G' K'.

First. Let the pressure of the arch-stones between B and C be calculated, if the arch be circular, by Art. 297, and if it be elliptical, by Art. 294. Consider half this weight as collected at D, and acting in the direction D F, which will bo sufficiently accurate for our present purpose. Then, by attending to the rules in Arts. 14 and 17, Sect. L, the strains

* ' Transactions of the Society of Arts,' vol. xxxiii., p. 128 in the direction of each of the beams composing the frame E D H will be found; and the dimensions of the pieces that would resist them are to be determined by the rules for the stiffness of beams in Sect. II. or by the rule at the end of this Article.

Secondly. Compute the pressure of the arch between D and C, and consider it as acting at C in a vertical direction; then the strain on the beams E H, H' E, will be found by the rules above referred to.

Lastly. Let the whole pressure of the arch-stones between B and C, together with half the weight of the centre itself, bo considered as acting at the point E in a vertical direction, and find the dimensions of the supports K G, K' G', that would resist this pressure.

But in these calculations it must be observed, that if the length of any of the pieces in feet be less than about 1.25 times the breadth, or least dimension in inches, it will cripple at the joint rather than bend. Thus, if a piece be 8 inches in breadth, then its length must be 1.25 x 8, or 10 feet to yield by bending.

Therefore, when the length between the points where it is braced is less than in this proportion, instead of finding the scantlings by the rules for the stiffness of beams, they must be determined by the following rule.

Rule. - The pressure upon the beam in pounds divided by 1000 gives the area of the piece in inches, or that of the least abutting joint, if that joint should not be equal to the section of the piece.

As all long pieces in a centre may be rendered secure against bending by cross braces, or radial pieces notched on and bolted to them, this rule may almost always be applied for centres, instead of the rules in Sect. II.

305. In the centre (Fig. 2, Plate XXXVII.) the beams E F, F F', and F' E', constitute the chief support, and the arch being an ellipse, a considerable part of it will bear almost wholly upon the centre. But from what has been shown respecting the pressure of the arch-stones, it would appear that if we take the whole weight of the ring between D and C, and consider it to act in the direction H F at the joining F, it will be the greatest strain that can possibly occur at that point from the weight of the arch-stones. Produce the line H F to f, and make hf to represent the pressure, draw he parallel to the beam E F. Then, as hf represents the pressure of the arch between D and C, he will be the pressure in the direction of the beam F E; and ef, the pressure in the direction of the beam F F': and the beams must be of such scantlings as would sustain these pressures.

Let the weight of the arch from H to H' be estimated, and if two-thirds of this weight be considered to act at C in a vertical direction, it will be the greatest load that is likely to be laid at that point, and the dimensions for the parts of the truss FCF' must be found so as to sustain that pressure.

The frame E D F may be calculated to resist half the pressure of the arch-stones between B and H; that pressure being found by Art. 294.

The whole weight of the arch-stones from D to C, together with the weight of the centre itself, may be considered as acting in a vertical direction at E, and the supports at G E should be sufficient to sustain the action of this pressure.

To determine the scantling of that part of the rib which supports the weight between H and C, or D and H, etc, calculate the weight of that part of the arch which rests upon it, and consider it as a weight uniformly diffused over the length. The proper scantling will then be found by the rule in Art. 110. These bearings may be much shortened by lengthening the blocks against which the inclined beams of the truss abut.

Examples.

306. Example 1. - To determine the principal scantlings for the centre of a stone arch of 50 feet span according to the design shown by Fig. 1, Plate XXXVII.; a cubic foot of the stone weighing 130 lbs., the depth of the arch-stones being 3 feet, and the ribs 5 feet from middle to middle.

The arch is described with a radius of 26 feet; consequently, 27.5 feet is the radius of an arc passing through the middle of the depth of the arch-stones; and to find the length of this arch for one degree, multiply its radius by .01745329, which gives .48 for the length.

And 5x3x .48 = 7.2 feet, the solid content of one degree of the ring of arch-stones. But by Art. 297, W x 32.26 = 7.2 x 32.26 x 130 lbs. = 30,195 lbs. for the pressure of that part of the ring between B and C. Then suppose this pressure to act in the direction D F, and in order to render the operation more simple, call it 31,000 lbs. Draw df, in No. 2, Plate XXXVII., parallel to D F; set off df equal 31 parts, by any convenient scale; and draw e h parallel to the beam E H; also draw d e and d h parallel to the principal rafters of the frame EDO. Now when d h is measured by the same scale as df, it will measure 70 parts; and as both the rafters make the same angle with straining force, the strain on each will be 70,000 lbs. Let the abutting joint be equal to the section of the rafter, then 70000/1000

= 70 inches for the area of the section of each rafter, or nearly 8 1/2 inches square. The strain in the direction of the length of the tie-beam E H need not be calculated, because when it is sufficiently strong to resist the other strains to which it is exposed, its strength to resist tension will always be above what is necessary.

Our next operation is to calculate the weight or pressure of the arch-stones between B and C, which may be done by . common arithmetic. The weight of one degree of the ring is the same as before, that is, W = 7.2 x 130 = 936 lbs., and the space between B and C is 32 degrees; therefore n = 32. Hence cos. 1/2na - f sin. 1/2na = cos. 16° - fsin. 16°= .96126 - .27564 x .625 = .788985. And

W sin. n+1/2 a x.788985 = 936 x .284 x .788985 = sin. 1/2 a .00873

24022 lbs., the pressure of the part B C upon the centre.

By drawing lines parallel to the directions of the straining force and the beams, we find the pressure in the direction of the beam EH to be nearly 23,000 lbs; therefore, 2300/1000 =

23 inches, the area of the section of the beam; but it must be made a little larger than this, in order to have abutments for the other parts of the truss.

To find the dimensions of the supports required at K G, an approximate rule is given in Art. 304, and calculating according to this rule,

This force would produce a pressure of nearly 41,000 lbs. in the direction K G; hence 4100/1000 = 41 inches, the area of the support required at K.

Example 2. - To find the strain upon the principal supports of the centre, in Fig. 2 Plate XXXVII.; the depth of the arch-stones being 5 feet, the ribs 5 feet from middle to middle, and the stone 130 lbs. per cubic foot. The pressure of the arch-stones upon the centre, estimated by the methods detailed in Art. 295, will be about 130,000 lbs.; and measuring the proportions of the forces by the diagram, No. 3, Plate XXXVII., the pressure in the direction of the beam F F' will be 220,000 lbs., and the pressure in the direction F E will be 230,000 lbs.; consequently, the area of the horizontal beam should be 220 inches, and may consist of two beams 10 inches by 11 inches. The area of the inclined beam EF should be 230 inches, and may consist of two beams, 11 inches by 10 1/2 inches each.

Note. - It is observed by Prof. Rankine, "that if there were no friction between the arch-stones, the load upon the centre could be computed exactly. The friction between them, however, renders all formulae for that purpose uncertain." - Civil Engineering, p. 486.