• A two-hinged arch has hinges only at the supports (Fig. 4.2a). Such an arch is statically indeterminate. Determination of the horizontal and vertical components of each reaction requires four equations, whereas the laws of equilibrium supply only three (Art. 4.1).
    Another equation can be written from knowledge of the elastic behavior of the arch. One procedure is to assume that one of the supports is on rollers. The arch then becomes statically determinate. Reactions VL and VR and horizontal movement of the support Sx can be computed for this condition with the laws of equilibrium (Fig. 4.2b). Next, with the support still on rollers, the horizontal force H required to return the movable support to its original position can be calculated (Fig. 4.2c). Finally, the reactions of the two-hinged arch of Fig. 4.2a are obtained by adding the first set of reactions to the second (Fig. 4.2d).
    The structural theory of Sec. 3 can be used to derive a formula for the horizontal component H of the reactions. For example, for the arch of Fig. 4.2a, x is the horizontal movement of the support due to loads on the arch. Application of virtual work gives

    where M = bending moment at any section due to loads on the arch
    y = vertical ordinate of section measured from immovable hinge

    I  moment of inertia of arch cross section
    A  cross-sectional area of arch at the section
    E  modulus of elasticity
    ds  differential length along arch axis
    dx  differential length along the horizontal
    N  normal thrust on the section due to loads

    Circular Two-Hinged Arch Example. A circular two-hinged arch of 175-ft radius with a rise of 29 ft must support a 10-kip load at the crown. The modulus of elasticity E is constant, as is I/A, which is taken as 40.0. The arch is divided into 12 equal segments, 6 on each symmetrical half. The elements of Eq. (4.8) are given in Table 4.1 for each arch half.
    Since the increment along the arch is as a constant, it will factor out of Eq. 4.8. In addition, the modulus of elasticity will cancel when factored. Thus, with A and I as constants, Eq. 4.8 may be simplified to

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