(Contributed by E. H. Hawkins)

Arches such as occur in smaller buildings generally have their thickness determined by rule, but for larger examples it is more satisfactory to determine their stability by graphic means.

The Theory Of Arches Vaults And Buttresses 226

Semicircular Arch (see Fig. 172). - Take, for example, a semicircular arch of 27 feet span and 1 foot 9 inches in depth, occurring in a wall 2 feet in thickness, and carrying masonry of that thickness up to a horizontal line XY, being such as might occur in a nave arcade. According to the old Gothic practice, the courses

The Theory Of Arches Vaults And Buttresses 227

Fig. 173.

The Theory of Arches, Vaults', and Buttresses 119 were built horizontally for some height. As will be seen presently, this is theoretically correct, and in this case the actual arch is shown springing from a skew-back at DC. An arch is kept in equilibrium by the action of three forces, the thrust at the crown, the weight of the arch and its load, and the reaction of the abutments.

If an arch is symmetrically loaded the thrust at the crown must be horizontal, assuming that there is a vertical joint there, and not a keystone. At what point the thrust acts in AB (the depth of the arch ring) cannot be positively ascertained, and the theory regarding arches is therefore in a somewhat uncertain state; but it is generally assumed to act somewhere within the centre third, this being necessary if stability is to be secured.


Mark off the centre third of arch at n and p. Take any number of equal parts on XY, and draw vertical lines downwards from the points obtained to meet the extrados at a, b, c, d, e, f, g, and h.

Calculate the weights of the 4-sided figures so obtained, and draw vertical lines where the line of action of each acts on the arch, i.e. centrally between a and b, b and c, c and d, etc.

Suppose the thrust at crown to act at the point n. Through n produce the horizontal thrust to meet the line of action of the first load between a and b, and from the point where they intersect draw the resultant parallel to kb, as shown on the dotted trial diagram to meet the next line of action between b and c; and so proceed till you reach the line of action of the last and greatest load between g and h. Produce the line obtained (which will be parallel to kh) backwards till it cuts a horizontal line drawn from n at W; then W will be the centre of gravity of all the loads. Join W to the point E where the load between h and g impinges upon the inner line of the central third of the arch ring, and from h draw a line hm parallel to WE, cutting ak at m. Join am, bm, cm, dm, em, fm, gm, and hm. Now, starting from n, draw the polygon wEW, with its lower sides parallel to am, bm, etc.

The lower sides will then form the line of pressure. If this line is wholly within the centre third, then the arch will be stable. If not, the arch must be thickened or its form altered until it is possible to draw a line wholly within by the above method.

In this instance, if the arch rose from the springing VZ, the line of pressure would run outside, and rotation would tend to take place round V. It becomes, therefore, theoretically necessary to form the Gothic tas de charge, making DC the springing.

N.B. - In practice sufficient backing would effectually prevent any rotation of V, so that a complete semicircular arch might be employed.

Pointed Arch

To determine the stability of a pointed arch of 29 feet span and 2 feet 3 inches in depth, in a wall 2 feet in thickness which terminates in a gable end (see Fig. 173). This example is similar in working to the last, but it is found on trial that the line of pressure, if started as before from n, passes quickly outside the centre third. A point q is therefore taken nearer p, when it will be found to work satisfactorily.

Pointed Arch 228

Fig. 174.

The resultant thrust at X is represented completely in magnitude, line of action, and direction by gm on the diagram.

It is required to find the buttress needed to counteract this thrust (Fig. 174), - that is, to provide sufficient abutment to keep the resultant within the middle third of the masonry throughout. The thrust T1, which is equal to the thrust at X obtained from Fig. 173, but drawn to a smaller scale for purposes of book illustration, combined with weight W1 (of the masonry above a horizontal line at A), produces a resultant R1.

Pointed Arch 229

Fig. 175.

Resultant R1 in its turn, combined with W2 (weight of masonry between horizontal lines at A and B), produces resultant R2; and the buttress shown keeps these lines within the centre third as required.

To Determine the Stresses in a Quadrapartite

Pointed Vault

The line of pressure in both diagonal and transverse ribs is found as shown in Fig. 172, with the exception that, as in this case the actual joints are shown, the line of pressure need only pass within the centre third where it crossed them.

The weights are taken to act on centre of extrados of each voussoir, and, being graphically proportional to the amount of severy that they respectively carry, the lines gt and pa then represent the thrusts in the transverse and diagonal ribs respectively, which thrusts may be taken as meeting at a point A on plan.

The thrusts in the two diagonals are equal, and meet at an angle of 90 degrees at a point A1 on elevation, vertically over A on plan, producing the resultant shown, with the same line of action and direction on elevation as the diagonal ribs themselves, as represented by the line A1X. The thrust of the transverse rib is represented by A1Y, and meets A1X at A1, a point vertically over A on plan, to produce the total resultant thrust A1Z.