and of a reinforced concrete arch culvert of 20′ o" span are given in Fig. 18. These culverts were designed by Mr. C. F. Graff for the Great Northern Railway, and are strong enough to carry from 20 to 40 ft. of railroad embankment. These culverts are made heavier than is necessary for ordinary highway culverts.

Plain Concrete Culverts.—Plans for an 8-ft. plain concrete culvert, as designed by Edwin Thacher for the highways in Porto Rico, are shown in Fig. 19.

Reinforced Concrete Arch Culvert.-Details of a reinforced conrete arch culvert of 19 ft. span, as designed by the Michigan State Highway Department, are given in Fig. 20. Standards have also been prepared for arch culverts with straight end walls (0° with axis of stream). Details and data are shown in the cut.

Relative Costs of Small Culverts.-The relative costs of small culverts, as calculated by A. R. Hirst, highway engineer, Wisconsin Geological and Natural History Survey, are given in Table X.

CHAPTER XXIII.

DESIGN OF CONCRETE ARCH BRIDGES.

Introduction.-An arch is a beam or framework in which the reactions are not vertical for vertical loads. Arches are divided, according to the number of hinges, into three-hinged arches, two-hinged arches, one-hinged arches and arches without hinges or continuous arches. Solid two-hinged and continuous arches constructed of masonry or concrete, only, will be considered in this chapter. For the analysis of a two-hinged arch with spandrel bracing, see the author's "The Design of Steel Mill Buildings," Chapter XIV.

DEFINITIONS.-The following definitions will be of assistance in discussing arches. Skewback. The inclined surface upon which the arch rests. to the stone or brick arch.

Abutment.-A skewback and the masonry which supports it.

Soffit.-The under or concave side of an arch.

Back. The upper or convex side of an arch.

The term applies more properly

Springing Line or Spring. The line in which the soffit meets the abutment. The inner edge of skewback.

Intrados.-The line of intersection of the soffit with a vertical plane parallel to the roadway. Extrados.-The line of intersection of the back with a vertical plane parallel to the roadway. Span. The horizontal distance between springing lines, measured parallel to center line of roadway.

Rise. The vertical distance of the intrados above a line joining the springing lines.

Crown. The highest part of the arch ring.

Haunch. The portion of the arch ring between the crown and the springing line.

Spandrel.-The space between the back of the arch and the roadway.

Two-hinged masonry or concrete arches are rarely used, but the theory of this type will be deduced as preliminary to the theory of the arch with fixed abutments.

=

STRESSES IN A TWO-HINGED ARCH. Reactions.-The vertical reactions of the two-hinged arch in (a), Fig. 1, are the same as for a simple beam having the same loads and span. The horizontal reactions will be H H, and will be equal to the pole distance of the force polygon in (b) that is used to draw the true equilibrium polygon. The value of H depends upon the elasticity of the arch and is not statically determinate.

Having calculated the vertical reactions V1 and V, by means of moments in the usual manner, and the horizontal components H, as will be described presently, the equilibrium polygon in (a) may be drawn using the force polygon in (b) Fig. 1. The requirements being that the equilibrium polygon must pass through the hinges, and that the force polygon must have a pole distance equal H.

The bending moment at any point in the arch will then be H⚫t, where H is the pole distance of the equilibrium polygon, and t is the intercept from the point at which the moment is to be determined to the string P3P4. The shear on the section of the arch m-n will be Ss, while the direct axial stress will be P, as shown in (b) Fig. 1.

Calculation of Horizontal Reaction, H.-Now for equilibrium in the two-hinged arch in Fig. 2, the span must remain constant. This relation will be expressed in the form of an equation of condition.

In Fig. 2 assume that the arch ring is rigid except the length ds which bends under the action of some external loading. Now the point C will move to E if the arch be not constrained, the horizontal deformation being CD = dx, and the vertical deformation being ED = dy. The angle CBE = do.

Now in a beam, as in Fig. 3, the stresses at any point in the beam will vary as the distance from the neutral axis, and from similar triangles we have

and

R: ds: c: A

R.Ac.ds

(2)

Now if S is the fiber stress on the extreme fiber, and E is the modulus of elasticity, we have AS: ds: E, and

Also

But from the common theory of flexure we have M.c = S.I, and substituting in (4)

(3)

(4)

(5)

(6)

(7)

(8)

Now if bending takes place over the entire length of the span of the arch the total horizontal deformation will be

=

(10)

M'Hy, where

Equation (ro) will be sufficient to determine the pole distance in Fig. 1. Now in equation (10) the value of M at any point in the arch will be M M' is the bending moment as calculated in a simple beam, H is the horizontal component of the reaction and y is the ordinate of the point in the arch as in Fig. 2. Inserting the value of M in equation (10), it becomes

Graphic Solution.-Now in Fig. 4 let polygon AEB be a random polygon drawn with an assumed pole distance H'; polygon ADB is the true equilibrium polygon drawn with the true pole distance H; and ACB is the linear arch.

But DF is not yet known. However, we have the relation that the ordinates to the two equilibrium polygons are inversely proportional to the pole distances; and

where r equals the ratio of the assumed pole distance to the true pole distance. Then

Now in equation (13) if E∙I is a constant, the arch may be divided into seginents of equal length; or if E. I is not a constant the arch may be divided into segments for which ds/E-I is a constant, and we may write

r = Zy2/ZEF.y

(14) Graphic Interpretation of Equations.-Referring to Fig. 4, it will be seen that the numerator in (14) is the summation of the products of the ordinates to the arch taken at the centers of the

segments into which the arch ring is divided; while the denominator is the summation of the products of the ordinates to the random equilibrium polygon taken at the centers of gravity of the segments into which the arch ring is divided, and the distances of the segment from the line AB. Graphic Solution of the Stresses in a Two-hinged Arch.-Divide the given arch ring into a number of segments, varying from 10 to 20 parts, in which ds/E-I equals a constant. Assume that the external loads act through the centers of gravity of the segments. Lay off the loads and construct a force polygon with a pole distance H', and draw the equilibrium polygon so that it will pass through the hinges A and B of the arch. Now to use equation (13), scale off the ordinate y of the center of each segment of the arch ring, and assume that these ordinates are horizontal loads acting at the points in the arch ring of which they are the ordinates; lay off these ordinates as horizontal loads and with the assumed pole distance H" draw an equilibrium polygon with the force polygon thus constructed. In like manner assume that the ordinates to