This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.
Arch ribs may be classified in three ways: first, those which have fixed ends and no hinges; second, those which have a hinge or joint at each end; and third those which are hinged at both ends and in the center. The first class is by far the most common, and is the simplest and cheapest to construct; but, as will be developed later, it necessitates a very considerable allowance for temperature stresses which, under very unfavorable conditions, are even greater than the maximum stresses due to loading. The temperature stresses of a two-hinged arch are less severe, while those for a three-hinged arch may be neglected; but the construction of hinges in arch ribs adds considerably to the cost.
In the following demonstration, the arch rib is considered as a single line OCB (Fig. 228), which is assumed to have the properties of an arch rib - namely, the moment of inertia, modulus of elasticity of the material, and the consequent resisting moment. The curved line PQR represents the special equilibrium polygon corresponding to some one condition or loading. Although this line is drawn as a curved line, it is assumed to be a curve which is made up of a large number of correspondingly short lines, each of which corresponds to a section of an equilibrium polygon similar to those described under "Voussoir Arches." This equilibrium polygon is yet to be determined.
In Church's "Mechanics of Engineering," Chapter XI, is given the mathematical proof of three, general equations which apply to this problem. No demonstration will here be made of these three equations, which are as follows;
The practical meaning of the first of these equations may be described as follows (see Fig. 228): ds represents one of an even number of very short sections into which the length OCB of the arch rib has been divided. M represents the transverse moment acting on the arch rib at that section under the particular condition of loading which is being considered. E is the modulus of elasticity of the material, and I is the moment of inertia of the section. At some of the sections the moment is positive, and at some it is negative. The product of M and ds, divided by the product of E and I, is therefore sometimes positive and sometimes negative. According to this equation, the summation of these various products for each short section (ds) of the rib equals zero; or, in other words, the summation of the positive products will exactly equal numerically the summation of the negative products.
The other two parts of Equation 48 must be interpreted similarly, the only difference being that in each case the term Mds/EI - is multiplied by the corresponding value of y for one of the equations, and by x for the other. This group of three equations (48) has nothing to do with the form of the special equilibrium polygon PQR.
It may also be proved by analytical mechanics, that if the curve PQR represents the special equilibrium polygon corresponding to some system of loading, and z represents the vertical distance between the arch rib and the special equilibrium polygon at any section, then the moment M at that section a of the rib, equals Hz, in which H is a constant which may be determined from the force diagram. The curve PQR represents a typical special equilibrium polygon which crosses the arch rib at two points. These points of intersection indicate points of contraflexure, where the transverse moment changes its direction of rotation, and where it is therefore zero. When the special equilibrium polygon is above the rib curve, we call the moment positive; and when it is below, we call it negative. When it is positive, it means that there is tension in the lower part of the rib, and compression in the upper part. The conditions are, of course, the reverse of this when the curve is below the rib. We may therefore substitute Hz for the value of M in the group of Equations 48; and since H and E are both constant for all points, from the principle enuncia ted in Article 423, we may not only place them outside of the sign of summation, but may even drop them altogether, since the summation equals zero; and we may therefore transform Equations 48 to the following:
Whenever we are investigating the mechanics of an arch rib which has a constant moment of inertia, we may simplify Equations 49 by dropping out altogether the I of the denominators of those equations; but since arch ribs are usually made with deeper sections near the abutments, the I will be greater near the abutments. Calling the I at the center Ic, then I equals nIc, in which n is a variable. If we substitute this value of I in the denominators of Equations 49, then, since Ic is a constant quantity, it may be placed outside of the summation sign, and even dropped altogether, which practically means that we substitute n for I in Equations 49. We shall also substitute for z its value z" - z' (see Fig. 228), and shall rewrite Equations 49 as follows, by making the substitutions:
It will later be shown how we can draw a line (marked vm in Fig. 228) which will satisfy the following equations:
Since the arch rib (represented by the curve OCB) is assumed to be symmetrical about its center C, and since vm is horizontal, any position of vm which will satisfy the first of Equations 51 will also satisfy the second.
It is another principle of the science of summations, that if we have a series of terms whose summation equals zero, and also have another series of terms whose summation equals zero, but whose terms are made up of the difference of two terms, one of which corresponds in each case to the terms of the first summation, then we may say that the summation of the other corresponding terms is likewise zero. For example, the first one of Equations 50 consists of a series of terms which may be rewritten: z" /ds - z'/n ds
The first one of Equations 51 is the summation of a series of terms, each with the form z'/n ds In each of these summations the different terms corresponding to the variable values of z' exactly correspond. We may therefore say that the summation of a series z" of corresponding terms, each one of the form - ds, will exactly equal zero; and we may therefore write Equation 52 as given below. We may also combine the second part of Equation 50 with the second part of Equation 51 in a similar manner, and obtain
Equation 53 as given below It will be found more convenient to separate the third part of Equation 50 into two summations, one of which consists of a series of terms z"/n yds, and the other of a series of terms consisting of z'/n y ds; and since the difference of these summations equals zero, then the summations must equal each other, and we may therefore write Equation 54:
An infinite number of equilibrium polygons may be drawn which will satisfy Equation 52 and 53. An equilibrium polygon may be drawn by trial, and the values of the summations for each side of Equation 54 may be determined. But since the position of the line vm is definitely determined by Equation 51, then the value of the right side of Equation 54 is fixed, and we only need to alter the pole distance of the trial equilibrium polygon in the inverse ratio of the required change in z", and then Equation 54 will be satisfied. Since changing all the values of z" in the same ratio does not alter the satisfaction of Equations 52 and 53, the changing of the pole distance does not vitiate the previous work. The value for the true pole distance is thus obtained, by which the true curve PQR may be graphically drawn out. We may then determine the moment at any point, which is the product of Hz for any point of the curve, in which z is the vertical distance between the center line of the arch rib and the finally determined equilibrium polygon, and H is the pole distance corresponding to that polygon.
It will be shown later that the thrust and the shear for any point of the curve equal the projection onto the tangent and normal respectively of the proper ray of the force diagram. It should be noted that the above equations apply only to symmetrical arch ribs which have their abutments at the same level. Under such conditions, Equation 53 is always satisfied when Equation 52 is satisfied. In the very rare cases where arches have to be designed on a different basis, some of the simplifications given above cannot be utilized, and the work becomes far more complicated. The solution of these very rare cases will not be given here.