An arch may be defined as a plane-curved bar or rib supported and loaded in a way that makes it act in direct compression. For example, the structure in Fig. 1a is a parabolic symmetrical arch that is loaded with a distributed load that varies linearly over the span L of the arch. It is fixed at end A and supported by an immovable hinge at end B. The horizontal distance between the end supports of the arch is the span of the arch, and the line AB joining its points of supports is the springing line.


Fig. 1(a) Parabolic arch of variable thickness loaded as shown. (b) Parabolic arch loaded with redundant force XA, redundant moment MA, and applied distributed loading. (c) Free-body diagram of a segment of the arch.

For the arch in Fig. 1a, the span and springing lines of the arch are the same because points A and B are at the same level with respect to the y axis. The highest point C of the arch is the crown, and an arch may be a symmetrical arch or an unsymmetrical arch. If, for example, one end of the arch is lower than the other, then the arch is unsymmetrical. Various types of arches are shown in Figs. 2 and 3.


Fig. 2 Circular arches: (a) Fixed at one end and hinged at the other. (b) Fixed at both ends. (c) Three-hinge arch. (d) Two-hinge arch with supports at different elevation. (e) Two-hinge arch.

It is assumed here that the plane of curvature of the arch rib is also a plane of symmetry for each cross section of the arch, and the externally applied loads are assumed to act only in this plane. On this basis, we have a two-dimensional problem and the deformation of the arch will take place in the plane of symmetry. The maximum vertical distance from the springing line to the arch axis, denoted as H in Fig. 1a, is the rise of the arch.

Parabolic Arch of Variable Thickness

We consider the linearly elastic variable-thickness parabolic arch in Fig. 1a that is loaded by a distributed load of maximum intensity w0 and varying linearly over the span of the arch. At any arc length s from support A, the moment of inertia Is is assumed to vary as follows:

clip_image006 (1)

where Ic is the moment of inertia at the crown C of the arch, and

clip_image008 (2)


Fig. 3 (a) Two-hinge parabolic arch. (b) Two-hinge elliptical arch. (c) Two-hinge hollow arch. (d) Two-hinge with horizontal tie. (e) Twin circular arch

Since the parabolic arch in Fig. 1a is statically indeterminate to the second degree, the horizontal force XA and bending moment MA at support A are taken as the redundants. On this basis, the arch is reduced to one that is hinged at end B, supported by roller at end A, and loaded as shown in Fig. 1b.

By considering a segment A0 of the arch as shown in Fig. 1c, the normal force Ns, shear force Vs, and bending moment Ms at the end 0 of the segment may be determined by using the three static equilibrium equations. For example, by taking moments about the end 0 of the segment and assuming counterclockwise moments as positive, we have

clip_image012 (3)

Note that in Fig. 1c is the vertical reaction at end A in Fig. 1b, and it may be determined from this figure by using statics. By solving Eq. (3), we find

clip_image014 (4)

By setting equal to zero the sum of the forces in the vertical direction, Fig. 1c, and assuming upward forces as positive, we have

clip_image016 (5)

By setting equal to zero the sum of the forces in the horizontal direction of the segment, we find

clip_image018 (8.113)

The simultaneous solution of Eqs. (5) and (6) for Vs and Ns yields

clip_image020 (7)

clip_image022clip_image024 (8)

From Example 1.3 of Section 1.5, an equation analogous to Eq. (10) may be written for the complementary strain energy U of the parabolic arch. This equation is

clip_image026 (9)

where S is the arc length of the parabolic arch, A is its cross-sectional area at any coordinate s, E is the modulus of elasticity, G is the shear modulus, and K is the shear factor. The shear factor K may be determined as shown in Example 1.3. For rectangular cross sections K is 1.2.

When the rise of the arch is large compared to its thickness, say a ratio of 10 or larger, then the complementary strain energy due to Ns and Vs would be small compared to the one produced by Ms and it can be neglected. On this basis, Eq. (8.116) yields

clip_image028 (10)

The integrations in Eq. (9), or Eq. (10), may be simplified by using the expression

clip_image030 (11)

Thus, by substituting Eqs. (4) and (11) into Eq. (10), we find

clip_image032 (12)

Additional Methods

The values of the redundant force XA and redundant moment MA may be obtained from the minimizing conditions

clip_image034 (13)

clip_image036 (14)

Application of Eqs. (13) and (14) yields

clip_image038 (15)

clip_image040 (16)

By considering the geometry of the arch in Fig. 1a, we find

clip_image042 (17)

By substituting Eq. (17) into Eqs. (15) and (16) and performing the required integrations, we obtain the following two equations, which are in terms of the redundants XA and MA:

clip_image044 (18)

clip_image046 (19)

Simultaneous solution of Eqs. (18) and (19) yields

clip_image048 (20)

clip_image050 (21)

With known XA and MA, the values of Ms, Ns, and Vs may be determined from Eqs. (4), (7), and (8), respectively. They are as follows:

clip_image052 (22)

clip_image054 (23)

clip_image056 (24)

Semicircular Arch with Hinged Ends

We consider now the uniform semicircular arch in Fig. 4a that is hinged at the support points A and B and loaded by a uniformly distributed load w as shown. The moment of inertia I is uniform throughout the arch. Since the arch is statically indeterminate to the first degree, the horizontal reaction XA at the end A is taken as the redundant. On this basis, we have an arch that is hinged at end B, supported by roller at end A, and loaded by the reactive force XA and applied load w as shown in Fig. 4b.

By considering the free-body diagram of the arch segment AC in Fig. 4c and applying the three static equilibrium equations, the expressions for the normal force

Fig. 3 (a) Semicircular arch hinged at the end supports. (b) Semicircular arch loaded with redundant force XA and applied distributed load w. (c) Free-body diagram of a segment of the arch.

Additional Methods

Ns, shear force Vs, and bending moment Ms, may be determined. For example, by setting equal to the sum of the moments about point C of the segment, we find

clip_image060 (25)

The static equilibrium equations in the horizontal and vertical directions of the arch segment yield

clip_image062 (26)

clip_image064 (27)

Simultaneous solution of Eqs. (26) and (27) yields

clip_image066 (28)

clip_image068 (29)

By considering only the complementary strain energy due to bending and using Eq. (25), we find


The redundant reaction XA may be determined by using the equation

clip_image034[1] (30)

Equation (30) yields

clip_image072 (31)

By integrating Eq. (31) and solving for XA, we find

XA=0.4099wr (31a)

By substituting Eq. (31a) into Eqs. (25), (28), and (8.136), we find

clip_image074 (32)

clip_image076 (33)

clip_image078 (34)


  1. I have successfully followed through your worked example in one figure 1a on Indeterminate Arches up to equations (6) and (7). Continuing from this point seem to be a challenge as I my figures for equations (18) and (19) differ from yours. This invariably means that I could not complete the exercise.
    I find your work very useful and shall glad if you could let me have a detailed solution to this example. I like to observe that there is a factor of (L2) missing in the denominator of the load term in equation (4).
    With Regards Anji-Sob Erekosima.

  2. Can you please help me with an indeterminate arch problem. A semi circular arch, pinned at both ends (A and B = (redundant to 1 degree), having an applied moment acting clockwise at support A? To find the bending moment function?

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