Matrix and Displacement Methods

Abstract

This chapter is devoted to numerical analysis of stability for different types of arches. Among them are two approaches – Smirnov’s matrix method and classical methods of structural analysis. A procedure for constructing the stability equation for different approaches is discussed.

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This chapter is devoted to numerical analysis of stability for different types of arches. Among them are two approaches – Smirnov’s matrix method and classical methods of structural analysis. A procedure for constructing the stability equation for different approaches is discussed.

1 General

Determination of critical loads on the arches can be achieved by the precise or approximate methods. By “precise methods” we will assume that this is done by integration of differential equation of the arch, or by some other method under which the initial design diagram of the arch does not have to be modified to accommodate a numerical procedure. Precise methods for finding critical loads have been discussed in Chap. 4.

The only way to determine critical loads of an arch with variable stiffness is by approximate methods. The terms “Approximate methods” means replacing the arch by set of the members, followed by precise methods of analyzing the modified design diagram. Here the engineer is faced with two important issues. The first is how to approximate the arch, and the second is which method of analysis to choose for analysis of the modified design diagram. In the general case, one must choose such an approximation of the design diagram, and such a method for analysis that will simplify the numerical procedures without compromising the numerical accuracy. There exists a variety of approaches and their variations. Among them are Smirnov’s matrix method, classical methods of structural analysis, and of course, finite element method.

Smirnov’s matrix method [Smi47], [Smi84] considers the arch as a series of curvilinear segments, each of which coincides with the corresponding portion of the arch.

No modification of the design diagram is performed under this method. Therefore, this method should be treated as an exact method in matrix form. Smirnov’s method is based on the classic concepts of the fictitious (conjugate) beam, elastic loads, and utilizes the tools of the matrix algebra.

Classical methods of structural analysis (the Force method, Displacement method and Mixed method in canonical form) [Kar10] are precise, but an approximate result is the consequence of a change in the design diagram of the arch by its approximate (modified) diagram.

Finite element method is implemented in modern computer software and allows the user to obtain the value of critical loads with a high accuracy. Nowadays this is an effective method for stability analysis of arches with peculiarities (piecewise-linear stiffness, nonlinearities, the need to account for secondary effects, etc.).

2 Smirnov Matrix Method

This method allows us to numerically determine the critical loads on the arches. The shape of the arch and the law under which the moment of inertia of the cross section of the arch changes along the axis line are unspecified. At the heart of the method lies a discretization of the system in association with elastic load method (see Sect. 1.6).

In the case of a circular arch, we can use two fundamental differential equations. The first equation is constructed with respect to radial displacements (Boussinesq’s equation)

$$ \frac{{{{\text{d}}^2}\upsilon }}{{{\hbox{d}}{s^2}}} + \frac{\upsilon }{{{R^2}}} = - \frac{M}{{EI}}. $$

The second equation is constructed with respect to bending moments in the curvilinear bar

$$ \frac{{{{\text{d}}^2}M}}{{{\hbox{d}}{s^2}}} + \frac{M}{{{R^2}}} = - q. $$

These equations are similar, so computation of radial displacements $\upsilon$ may be replaced by the computation of fictitious bending moments caused by the load $q = M/{{EI}}$; this load is replaced by elastic loads W [Smi84].

2.1 Matrix Form for Elastic Loads

Ignoring the axial forces, elastic loads, according to (1.20), become

$$ {W_n} = \frac{{{s_n}}}{{6{{E}}{{{I}}_n}}}\left( {{M_{n - 1}} + 2{M_n}} \right) + \frac{{{s_{n + 1}}}}{{6{{E}}{{{I}}_{n + 1}}}}\left( {2{M_n} + {M_{n + 1}}} \right), $$

(5.1)

where s _n is a length of an structural element between joints (n − 1) and n (Fig. 1.15).

In the stability problems (as opposed to the strength problems), ordinates of the bending moment diagram M and even the general profile of design diagram are not known in advance. Therefore, in the stability problems the bending moment diagram is approximated either by the straight line segments or a set of quadratic polynomials each of which passes through the three neighboring points (Sect. 1.6, Fig. 1.15, dotted line). Each of these approximations allows us to represent the elastic load at joint n in the form [Smi47], [Smi84]

$$ {W_n} = \frac{{{S_0}}}{{6{{E}}{{{I}}_0}}}\left( {{\beta_{n\left( {n - 1} \right)}}{M_{n - 1}} + {\beta_{nn}}{M_n} + {\beta_{n\left( {n + 1} \right)}}{M_{n + 1}}} \right). $$

(5.2)

In the case of approximating the bending moment diagram by parabolas, the coefficients $\beta$ may be presented in terms of length and moment of inertia of each element (s, I) as well as length and moment of inertia in terms of some basic element (S ₀, I ₀)

$$ {\beta_{n\left( {n - 1} \right)}} = \frac{{{s_n} + 2{s_{n + 1}}}}{{2\left( {{s_n} + {s_{n + 1}}} \right)}}{\rho_n} - \frac{{s_{n + 1}^2}}{{2{s_n}\left( {{s_n} + {s_{n + 1}}} \right)}}{\rho_{n + 1}}, \hfill \\{\beta_{nn}} = \left( {2 + \frac{{{s_n}}}{{2{s_{n + 1}}}}} \right){\rho_n} + \left( {2 + \frac{{{s_{n + 1}}}}{{2{s_n}}}} \right){\rho_{n + 1}}, \hfill \\{\beta_{n\left( {n + 1} \right)}} = - \frac{{s_n^2}}{{2{s_{n + 1}}\left( {{s_n} + {s_{n + 1}}} \right)}}{\rho_n} + \frac{{{s_{n + 1}} + 2{s_n}}}{{2\left( {{s_n} + {s_{n + 1}}} \right)}}{\rho_{n + 1}},\quad {\rho_n} = \frac{{{s_n}{I_0}}}{{{I_n}{S_0}}}. \hfill \\ $$

(5.3)

The vector of elastic loads is expressed by a matrix of elastic loads ${B_W}$ and the vector of moments at the nodal points

$$ \overrightarrow W = \frac{{{S_0}}}{{6{{E}}{{{I}}_0}}}{B_W}\overrightarrow M. $$

(5.4)

Vector of bending moments of any broken rod during buckling is presented as

$$ \overrightarrow M = {\left\lfloor {\begin{array}{*{20}{c}} {{M_1}} & {{M_2}} & \ldots & {{M_n}} \\ \end{array} } \right\rfloor^{\text{T}}}. $$

The matrix of elastic loads reduces to a Jakobi-like tri-diagonal matrix

$$ {B_W} = \frac{{{S_0}}}{{6{\text{EI}}}}\left[ {\begin{array}{*{20}{c}} {{\beta_{11}}} & {{\beta_{12}}} & 0 & 0 & \ldots & \ldots & 0 \\ {{\beta_{21}}} & {{\beta_{22}}} & {{\beta_{23}}} & 0 & \ldots & \ldots & 0 \\ 0 & {{\beta_{32}}} & {{\beta_{33}}} & {{\beta_{34}}} & \ldots & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & {{\beta_{n\left( {n - 1} \right)}}} & {{\beta_{nn}}} \\ \end{array} } \right]. $$

(5.5)

2.1.1 Special Case

Assume that the rod is divided into elements of same length $({s_n} = {s_{n + 1}} = {S_0} = {\hbox{const)}}$. In this case, expressions for parameters $\beta$, according to formulas (5.3) may be simplified

$$ \begin{gathered} {\beta_{n\left( {n - 1} \right)}} = \frac{3}{4}{\rho_n} - \frac{1}{4}{\rho_{n + 1}}; \hfill \\ {\beta_{nn}} = \frac{5}{2}\left( {{\rho_n} + {\rho_{n + 1}}} \right); \hfill \\ {\beta_{n\left( {n + 1} \right)}} = - \frac{1}{4}{\rho_n} + \frac{3}{4}{\rho_{n + 1}}. \hfill \\ \end{gathered} $$

(5.6)

Let us make an additional assumption ${I_n} = {I_{n + 1}} = {I_0}$. In this case ${\rho_n} = 1$ and for parameters $\beta$ we get

$$ \begin{gathered} {\beta_{21}} = {\beta_{32}} = \ldots = 0.5, \hfill \\ {\beta_{11}} = \frac{5}{2}\left( {{\rho_1} + {\rho_2}} \right) = 5;\quad {\beta_{22}} = \frac{5}{2}\left( {{\rho_2} + {\rho_3}} \right) = 5, \ldots \hfill \\ {\beta_{12}} = {\beta_{23}} = \ldots = 0.5, \hfill \\ \end{gathered} $$

(5.7)

Matrix of elastic loads becomes

$$ {B_W} = \frac{{{S_0}}}{{6{\text{E}}{{\text{I}}_0}}}\left[ {\begin{array}{*{20}{c}} 5 & {0.5} & 0 & 0 & \ldots & \ldots & 0 \\ {0.5} & 5 & {0.5} & 0 & \ldots & \ldots & 0 \\ 0 & {0.5} & 5 & {0.5} & \ldots & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & {0.5} & 5 \\ \end{array} } \right]. $$

(5.8)

If curvilinear bending moment diagram M _P within portion s _n and s _n+1 are replaced by straight elements, then

$$ {\beta_{n\left( {n - 1} \right)}} = {\rho_n};\quad \quad {\beta_{nn}} = 2\left( {{\rho_n} + {\rho_{n + 1}}} \right);\quad \quad {\beta_{n\left( {n + 1} \right)}} = {\rho_{n + 1}}. $$

(5.9)

In a special case, if the length of all the elements is ${s_n} = {s_{n + 1}} = {S_0} = {\hbox{const}}$ and ${I_n} = {I_{n + 1}} = I$ then we get following matrix of elastic loads:

$$ {B_W} = \frac{{{S_0}}}{{6{\text{EI}}}}\left[ {\begin{array}{*{20}{c}} 4 & 1 & 0 & 0 & \ldots & \ldots & 0 \\ 1 & 4 & 1 & 0 & \ldots & \ldots & 0 \\ 0 & 1 & 4 & 1 & \ldots & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & 1 & 4 \\ \end{array} } \right]. $$

(5.9a)

The matrix ${B_W}$ is a square matrix of (n − 1)th order, where n is the number of elements of the arch.

Matrices (5.8) and (5.9a) are as presented assuming that the moments at the ends points are equal to zero [Smi47], [Smi84].

2.2 Moment Influence Matrix

Let us consider two-hinged circular symmetrical arch of radius R and central angle $2\alpha$. The arch is loaded by uniform radial load q. In case of antisymmetric form of loss of stability the initial arch is replaced by its equivalent half-arch with rolled support at the axis of symmetry (Fig. 5.1) [Kar10]. Moment influence matrix L is constructed for a fictitious (conjugate) structure. For design diagram in Fig. 5.1 fictitious structure coincides with real structure. Divide the axis of the half-arch into n equals segments and number of all nodal points from 0 to n. The central angle for each segment is $\beta$, therefore the length of each portion is ${S_0} = R\beta$.

Fig. 5.1

Equivalent half-arch of entire symmetric two-hinged circular arch, elastic loads at joints 1, …, k and the angles notation

Each column of the moment influence matrix L contains the moments at the nodal points 1, 2, … caused by unit load which act at these arch points.

Let a unit radial force $P = 1$ be applied at arbitrary nodal point k; this point is defined by the angle $k\beta$ (Fig. 5.1). The vertical reaction ${R_C}$ and radial reaction ${R_A}$ are

$$ {R_A} = \frac{{\sin \,k\beta }}{{\sin \,\alpha }},\quad {R_C} = \frac{{\sin \left( {\alpha - k\beta } \right)}}{{\sin \,\alpha }}. $$

Bending moments at points i, k, j are

$$ \begin{gathered} {m_{ik}} = R\frac{{\sin \,i\beta \,\sin \left( {\alpha - k\beta } \right)}}{{\sin \alpha }}\left( {i < k} \right) \hfill \\ {m_{kk}} = R\frac{{\sin \,k\beta \,\sin \left( {\alpha - k\beta } \right)}}{{\sin \,\alpha }} \hfill \\ {m_{jk}} = R\frac{{\sin \,k\beta \,\sin \left( {\alpha - j\beta } \right)}}{{\sin \,\alpha }}\left( {i > k} \right). \hfill \\ \end{gathered} $$

(5.10)

If half-arch is divided into two equal parts, then k = 1 and matrix ${L_{1m}}$ according to the second formula (5.10) becomes

$$ {L_{1m}} = R\frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }}\left[ 1 \right]. $$

(5.10a)

If half-arch is divided into three equal parts, then for matrix ${L_{2m}}$ we get

$$ {L_{2m}} = \left[ {\begin{array}{*{20}{c}} {{l_{11}}} \hfill & {{l_{12}}} \hfill \\ {{l_{21}}} \hfill & {{l_{22}}} \hfill \\ \end{array} } \right] = R\frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }}\left[ {\begin{array}{*{20}{c}} {2\cos \,\beta } \hfill & 1 \hfill \\ 1 \hfill & {2\cos \,\beta } \hfill \\ \end{array} } \right]. $$

(5.10b)

If half-arch is divided into four equal parts, then for matrix ${L_{3m}}$ we get

$$ {L_{3m}} = R\frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }}\left[ {\begin{array}{*{20}{c}} {4{{\cos }^2}\,\beta - 1} \hfill & {2\cos \,\beta } \hfill & 1 \hfill \\ {2\cos \,\beta } \hfill & {4{{\cos }^2}\,\beta } \hfill & {2\cos \,\beta } \hfill \\ 1 \hfill & {2\cos \,\beta } \hfill & {4{{\cos }^2}\,\beta - 1} \hfill \\ \end{array} } \right]. $$

(5.10c)

The matrix L _m is symmetric and square order (n − 1).

2.3 Stability Equation in Matrix Form

Bending moment at any section is $M = qR\upsilon$. Vector of elastic loads is

$$ \overrightarrow \upsilon = {L_m}\overrightarrow W = qR{L_m}{B_W}\overrightarrow \upsilon. $$

Thus, the stability equation becomes

$$ \det \left( {C - \lambda E} \right) = 0, $$

where E is identity matrix, ${L_m}$ is moment influence matrix,

$$ C = {L_m}{B_W},\quad \lambda = 1/(qR). $$

2.3.1 Matrix Procedures

In the case of a two-hinged circular arch subjected to uniform radial load q the following procedure may be applied [Smi47], [Smi84]:

• 1. Divide the arch into n equal curvilinear segments. Larger values of n ensure greater numerical accuracy.

• 2. Construct the moment influence matrix ${L_m}$ and matrix ${B_W}$ of elastic loads [Smi84].

• 3. Compute the matrix product $C = {L_m}{B_W}$ and forms the stability equation $\det \left( {C - \lambda E} \right) = 0$, where $\lambda = 1/(qR)$ is eigenvalue of the stability problem.

• 4. Find the greatest eigenvalue $\lambda$ and calculate the smallest critical load ${q_{\min }} = 1/(\lambda R)$.

Application of this procedure for stability analysis of symmetrical arches of two different shapes is discussed in the following sections.

3 Two-Hinged Symmetrical Arches

Two types of two-hinged arches are considered. They are circular and parabolic symmetrical arches. Circular uniform and nonuniform arches are loaded by uniform radial load, while the parabolic arch is loaded by uniform vertical load and simple force at crown. Assume the antisymmetrical form of the loss of stability occurs. In this case, equivalent half-arch has a rolled support on the axis of symmetry and half-arch itself is statically determinate structure.

3.1 Circular Uniform Arch

This arch is loaded by uniform radial load. A half-arch is divided into two equal portions with one nodal point (1). In this case, the moment influence matrix according to (5.10a) has a single entry [Smi84]

$$ {L_{1m}} = R\frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }}\left[ 1 \right]. $$

Elastic load is applied at point 1 and have radial direction. Matrix of elastic load according to (5.8) is

$$ {B_W} = \frac{{5{S_0}}}{{6{{EI}}}}\left[ 1 \right]. $$

The matrix product

$$ C = {L_m}{B_W} = R\frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }} \frac{{5{S_0}}}{{6{{EI}}}}\left[ 1 \right]. $$

Stability equation is $\det \left( {C - \lambda E} \right) = 0$.

We get the following expression for the required eigenvalue

$$ \lambda = \frac{{5{S_0}R}}{{6{{EI}}}} \frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }}. $$

Since ${S_0} = R\beta$, then expression for $\lambda$ may be rewritten as

$$ \lambda = \frac{{5{R^2}\beta }}{{6{{EI}}}} \frac{{{{\sin }^2}\,\beta }}{{\sin \,\alpha }}. $$

The critical load becomes

$$ {q_{\rm{cr}}} = \frac{1}{{\lambda R}} = \frac{{12}}{{5\beta \,\tan \,\beta }}\frac{{EI}}{{{R^3}}}. $$

If $\alpha = \pi /2,\quad \beta = \pi /4$, then

$$ {q_{\rm{cr}}} = 3.055\frac{{EI}}{{{R^3}}}. $$

Exact result according to Levy’s formula is ${q_{\rm{cr}}} = 3.0({{EI}}/{R^3})$ (see Sect. 4.2.1). Relative error is 1.83%.

If half-arch is divided into three equal parts the critical load becomes [Smi47]

$$ {q_{\rm{cr}}} = \frac{{12\left( {2\,\cos \,\beta - 1} \right)}}{{11\beta \,\sin \,\beta }}\frac{{EI}}{{{R^3}}}. $$

If $\alpha = \pi /2,\quad \beta = \pi /6$, then ${q_{\rm{cr}}} = 3.0504({{EI}}/{R^3})$. Thus, approximating of the uniform circular arch by only two segments leads to very good results.

3.2 Circular Nonuniform Arch

Design diagram of symmetrical two-hinged circular arch ACB is shown in Fig. 5.2a. The radius of the arch is R and central angle 2α = 120º. The relative moments of inertia for special portions are underlined, i.e., ${I_{C - 1}} = 1.0{I_0},\ {I_{1 - 2}} = 0.8{I_0},<$> <$> {I_{2 - 3}} = 0.6{I_0}, \ {I_{3 - A}} = 0.5{I_0}$. The arch is subjected to uniform radial load q; this load is not shown. In case of antisymmetrical form of buckling the equivalent design diagram of the half-arch is shown in Fig. 5.2b [Smi47], [Smi84].

Fig. 5.2

Design diagram of symmetrical two-hinged circular arch and its equivalent design diagram for antisymmetrical buckling

The half-arch is divided into four equal portions with central angle $\beta = \alpha /4 = \pi /12$ for each portion. For given presentation of the arch we get

$$ \eqalign{ \hskip 5pc\sin \,\beta = 0.2588,\quad \cos \beta = 0.9659, \hfill \\ {\sin^2}\,\beta = 0.0670,\quad {\cos^2}\,\beta = 0.9330,\quad {\sin^2}\,\beta /\sin \,\alpha = 0.07737. \hfill \\}<!endgathered> $$

Matrix of moments according to (5.10c) becomes

$$ {L_m} = 0.07737R\left[ {\begin{array}{*{20}{c}} {2.732} & {1.9318} & {1.0} \\ {1.9318} & {3.732} & {1.9318} \\ {1.0} & {1.9318} & {2.732} \\ \end{array} } \right] = R\left[ {\begin{array}{*{20}{c}} {0.2114} & {0.1495} & {0.07737} \\ {0.1495} & {0.2887} & {0.1495} \\ {0.07737} & {0.1495} & {0.2114} \\ \end{array} } \right]. $$

Since the arch is divided into equal segments, one must use (5.6) for construction of the elastic loads matrix B _W . The entries ${b_{kk}}$ of matrix B _W are as follows:

$$ \eqalign{ {b_{11}} = \frac{5}{2}\left( {{\rho_1} + {\rho_2}} \right) = \frac{5}{2}\left( {1 \times \frac{1}{1} + \frac{1}{{0.8}}} \right) = 5.625; \hfill \\{b_{22}} = \frac{5}{2}\left( {1 \times \frac{1}{{0.8}} + \frac{1}{{0.6}}} \right) = 7.2917;\quad {b_{33}} = \frac{5}{2}\left( {1 \times \frac{1}{{0.6}} + \frac{1}{{0.5}}} \right) = 9.1667. \hfill \\}<!endgathered> $$

For entries ${b_{k\left( {k + 1} \right)}}$ of matrix B _W we get:

$$ \eqalign{ {b_{12}} = - \frac{1}{4}{\rho_1} + \frac{3}{4}{\rho_2} = - \frac{1}{4} \times \frac{1}{1} + \frac{3}{4} \times \frac{1}{{0.8}} = 0.6875; \hfill \\{b_{13}} = 0;\quad {b_{23}} = - \frac{1}{4} \times \frac{1}{{0.8}} + \frac{3}{4} \times \frac{1}{{0.6}} = 0.9375. \hfill \\}<!endgathered> $$

The entries ${b_{k\left( {k - 1} \right)}}$ of matrix B _W are

$$ \eqalign{ {b_{21}} = \frac{3}{4}{\rho_2} - \frac{1}{4}{\rho_3} = \frac{3}{4} \times \frac{1}{{0.8}} - \frac{1}{4} \times \frac{1}{{0.6}} = 0.5208; \hfill \\{b_{31}} = 0;\quad {b_{32}} = \frac{3}{4} \times \frac{1}{{0.6}} - \frac{1}{4} \times \frac{1}{{0.5}} = 0.7500. \hfill \\}<!endgathered> $$

Elastic loads matrix B _W becomes

$$ {B_W} = \frac{{{S_0}}}{{6{\text{E}}{{\text{I}}_0}}}\left[ {\begin{array}{*{20}{c}} {5.6250} & {0.6875} & {0.0} \\ {0.5208} & {7.2917} & {0.9375} \\ {0.0} & {0.7500} & {9.1667} \\ \end{array} } \right]. $$

For matrix C we get

$$ C = {L_m}{B_W} = \frac{{R{S_0}}}{{6{\text{E}}{{\text{I}}_0}}}\left[ {\begin{array}{*{20}{c}} {1.2669} & {1.2935} & {0.8494} \\ {0.9913} & {2.3200} & {1.6418} \\ {0.5130} & {1.3018} & {2.0780} \\ \end{array} } \right]. $$

Maximum eigenvalue of the stability equation

$$ \det \left( {{L_m}{B_W} - \lambda E} \right) = 0 $$

is $\lambda = 4.2241(R{S_0}/6{{E}}{{{I}}_0})$. Corresponding eigenvector is $\left[ {\begin{array}{*{20}{c}} { - 0.4627} & { - 0.7046} & { - 0.5380} \\ \end{array} } \right]$. The form of the loss of stability is shown in Fig. 5.2b by a dotted line.

Since $\lambda = (1/qR) = 4.2244(R{S_0}/6{{E}}{{{I}}_0})$ and ${S_0} = \beta R = (\pi R/12)$ then for radial critical load we get ${q_{\rm{cr}}} = 5.4251{{EI}}/{R^3}.$

3.3 Parabolic Uniform Arch

This section is devoted to stability analysis of two-hinged parabolic arch subjected to two types of loading: vertical uniformly distributed load along the span of the arch and concentrated force applied at the crown of the arch. In both cases, arches are symmetric. In case of antisymmetric form of the loss of stability the initial two-hinged arch is replaced by its equivalent half-arch. This scheme contains rolled support at the axis of symmetry and half-arch itself presents a statically determinate structure. Therefore, computation of elastic loads is easily evaluated using only the equilibrium conditions.

3.3.1 Uniformly Distributed Load

The arch of span l and rise f is loaded by a uniform vertical load distributed within the entire span (Fig. 5.3a). The equivalent half-arch, assuming the antisymmetric form of the loss of stability, is shown in Fig. 5.3b. The half-arch is a statically determinate structure.

Fig. 5.3

Design diagram of symmetrical two-hinged parabolic arch and equivalent half-arch for antisymmetrical form of the loss of stability and its equivalent design diagram for antisymmetrical buckling

The axis of the arch is divided into equal curvilinear portions and the nodal points are denoted as 1, 2, …

The nodal point i has coordinates ${x_i} = {\xi_i}l$ and ${y_i} = {\eta_i}l$.

3.3.1.1 Geometrical Relationships

Equation of the axial line of the arch is $y = 4fx\left( {l - x} \right)(1/{l^2})$.

The general expressions for slope and slope at the support A are

$$ \tan \,\varphi = 4\frac{f}{{{l^2}}}\left( {l - 2x} \right),\quad \tan \,{\varphi_0} = 4m,\quad m = \frac{f}{l}. $$

Length of half-arch is determined as

$$ S = \frac{l}{4}\left[ {\sec \,{\varphi_0} + \frac{1}{{4m}}\ln \left( {4m + \sec \,{\varphi_0}} \right)} \right]. $$

(5.11)

The length ${S_k}$ of axis of the arch from origin (point A) to arbitrary point $\mathcal{K}$ with coordinates ${x_k} = {\xi_k}l$, ${y_k} = {\eta_k}l$ is

$$ {S_k} = S - \frac{l}{{16m}}\left( {\frac{{\tan \,{\varphi_k}}}{{\cos \,{\varphi_k}}} + \ln \frac{{1 + \sin \,{\varphi_k}}}{{\cos \,{\varphi_k}}}} \right). $$

(5.12)

Stability equation is $\det \left( {C - \lambda E} \right) = 0$, where $C = {L_m}{B_W}G$. The eigenvalue and thrust are related by the formula $\lambda = 1/H.$

For constructing the influence matrix L _m it is necessary to apply the unit force at point k and to determine the bending moment at the neighboring points (Fig. 5.3b); expressions for the entries of this matrix can be found in [Smi84]. The matrix B _W is constructed in the same manner as for circular arch. The diagonal matrix G contains the entries $\sec \,{\varphi_k}$.

Let us show this procedure and evaluate numerical results if the length of half-arch is divided only into two equal portions; the nodal point 1 in Fig. 5.3 is not shown. Assume $m = f/l = 0.5$.

3.3.1.2 Geometrical Parameters

Slope at the support A is $\tan \,{\varphi_0} = 4m = 2.0$ and $\sec \,{\varphi_0} = 2.2361$.

The length of half-arch with adopted parameter m is

$$ S = \frac{l}{4}\left[ {2.2361 + \frac{1}{{4 \cdot 0.5}}\ln \left( {4 \cdot 0.5 + 2.2361} \right)} \right] = 0.740l. $$

(5.12a)

The length of each curvilinear segment is ${S_0} = S/2 = 0.370l$ .

The length ${S_0}$ of the segment A − 1 and slope at point 1 are related by

$$ 0.370l = \frac{l}{{16m}}\left( {\frac{{\tan \,{\varphi_1}}}{{\cos \,{\varphi_1}}} + \ln \frac{{1 + \sin \,{\varphi_1}}}{{\cos \,{\varphi_1}}}} \right). $$

(5.12b)

Solution of this equation determines the slope at the point 1

$$ {\varphi_1} = 50.7^\circ, \quad \tan \,{\varphi_1} = 1.2217,\quad \sin \,{\varphi_1} = 0.7738,\quad \cos \,{\varphi_1} = 0.6334. $$

Coordinates of joint 1 Dimensionless coordinates of point 1 are ${\xi_1} = {x_1}/l,\quad {\eta_1} = {y_1}/l$. For this point 1, we have $\tan \,{\varphi_1} = 4m\left( {1 - 2{\xi_1}} \right)$ or $1.2217 = 4 \cdot 0.5\left( {1 - 2{\xi_1}} \right)$. Solution of this equation is ${\xi_1} = 0.1945$, so ${x_1} = 0.1945l$. For ordinate of point 1 we get ${y_1} = 0.3133l$.

Stability equation Since the arch is divided into two portions, then each matrix from $C = {L_m}{B_W}G$ will be a scalar.

Entry of matrix ${L_m}$ is

$$ \eqalign{ \left[ {{L_m}} \right] = l\left( {{\xi_1}\,\cos \,{\varphi_1} + {\eta_1}\,\sin \,{\varphi_1}} \right) \times \left( {1 - 2{\xi_1}} \right) \hfill \\ = l\left( {0.1945 \times 0.6334 + 0.3133 \times 0.7738} \right) \times \left( {1 - 2 \times 0.1945} \right) = 0.2234l\left[ 1 \right]. \hfill \\}<!endgathered> $$

Entries of matrices ${B_W}$ and G are

$$ \left[ {{B_W}} \right] = \frac{{5{S_0}}}{{6{{EI}}}}\left[ 1 \right] = \frac{{5 \cdot 0.370l}}{{6{{EI}}}}\left[ 1 \right] = \frac{{0.3083l}}{{EI}}\left[ 1 \right], $$

$$ \left[ G \right] = \sec \,{\varphi_1}\left[ 1 \right] = 1.5788\left[ 1 \right]. $$

For matrix C we get $C = {L_m}{B_W}G = 0.2234l \times (0.3083l/{{EI}}) \times 1.5788\left[ 1 \right] = <$> <$>(0.1087{l^2}/{{EI)}}.$

Stability equation $\det \left( {C - \lambda E} \right) = 0$ leads to the following eigenvalue $\lambda = 0.1087{l^2}/{{EI}}.$

Since $\lambda = 1/H$, while thrust $H = q{l^2}/(8f)$, then for critical load we get ${q_{\rm{cr}}} = 36.8{{EI}}/{l^3}$.

For different parameters m the critical load may be presented as q _G = K(EI)/l ³. Parameter K is presented in Table 5.1; coefficients K obtained by Dinnik [Din46] are shown in parentheses. This table also contains the abscissa ${x_1}$ for point 1 and length ${S_0}$.

Table 5.1 Parameter K for critical load [Smi84], [Din46]

3.3.1.3 Analysis of Results

With the half-arch divided only into two segments, Smirnov’s method leads to acceptable results. Even for arches with parameters (m = 0.8–1.0) the relative error is not more than 6%. If the axis of the arch is divided into arbitrary number of the segments, the analytical expressions for entries of C can be found in [Smi84].

3.3.2 Concentrated Load

The arch AB of span l and rise f is loaded by the single force P at a crown C. The form of the loss of stability is shown in Fig. 5.4 by dotted line.

Fig. 5.4

Design diagram of symmetrical two-hinged parabolic arch subjected to load P and form of the loss of stability and its equivalent design diagram for antisymmetrical buckling

If half-arch is divided into two segments, then Smirnov’s method leads to the following critical load

$$ {P_{\rm{cr}}} = \frac{{{K_{\rm{P}}}{{EI}}}}{{{l^2}}}. $$

Parameter ${{ K}_{\rm{P}}}$ in terms of $m = f/l$ is presented in Table 5.2.

Table 5.2 Parameter ${\mathcal{K}_{\text{P}}}$ for critical load [Smi84]

4 Hingeless Symmetrical Arches

In case of symmetrical hingeless arch we can replace it by equivalent half-arch. However, in this case, in contrast to two-hinged arch, the half-arch now presents a redundant structure. This fact adds additional features in Smirnov’s procedure: the matrix C in the stability equation may be constructed after solution of corresponding canonical equation [Smi47].

4.1 Duality of Bending Moment Diagram and Influence Line

For stability analysis of hingeless symmetrical arch we use the following theorem:

Bending moment diagram in the real structure caused by unit load (force or couple) which acts at point k coincides with influence line of corresponding fictitious load factor at the same point k for fictitious (conjugate) structure [Smi47].

This correspondence for some structures is shown in Table 5.3

Table 5.3 Correspondence of bending moment diagram for real structure and influence line for fictitious structure

We explain the concept of “corresponding load.” In the fictitious structure, the bending moment in the point k due to the fictitious load is a linear displacement at the same point of the real structure; this displacement and unit force P in the real structure correspond to each other, so P = 1 corresponds to M _fict. Similarly, M = 1 corresponds to fictitious shear force Q _fict.

This theorem allows us to determine the displacements of the real structure by applying elastic loads to the bending moment diagram and treating them as the influence line [Smi47], [Kle80]. This theorem will also be used for stability analysis of the hingeless arch.

Let us demonstrate the application of this theorem for computation of displacement at the free end of the cantilever uniform beam of the span l (Fig. 5.5a).

Fig. 5.5

(a) Design diagram of the beam; (b) bending moment diagram of the real beam; (c) fictitious beam; (d) unit state and corresponding bending moment diagram; (e) influence line for bending moment at the clamped support of the fictitious beam

Subdivide the beam into two equal parts (0–1 and 1–2). The specified points are labeled as 0, 1, and 2. The bending moment diagram for actual beam is shown in Fig. 5.5b. Fictitious beam and elastic loads W ₀ and W ₁ are shown in Fig. 5.5c.

For calculation of W ₀ we need to know bending moments at three consecutive points; dotted line shows additional portion of the beam with end points −1 and 0; the length and stiffness of this portion are λ ₀ and EI ₀ = ∞. The elastic loads are

$$ {W_0} = \frac{{{\lambda_0}}}{{6{{E}}{{{I}}_0}}}\left( {{M_{ - 1}} + 2{M_0}} \right) + \frac{{{\lambda_1}}}{{6{{E}}{{{I}}_1}}}\left( {2{M_0} + {M_1}} \right)= \frac{{{\lambda_0}}}{\infty }\left( {{M_{ - 1}} + 2{M_0}} \right) + \frac{l}{{12{{EI}}}}\left( {2Pl + \frac{{Pl}}{2}} \right) = \frac{{5P{l^2}}}{{24{{EI}}}}, $$

$$ {W_1} \hskip -3pt= \frac{{{\lambda_1}}}{{6{{E}}{{{I}}_1}}}\left( {{M_0} + 2{M_1}} \right) + \frac{{{\lambda_2}}}{{6{{E}}{{{I}}_2}}}\left( {2{M_1} + {M_2}} \right) = \frac{l}{{12{{EI}}}}\left( {Pl + 2\frac{{Pl}}{2}} \right) + \frac{l}{{12{{EI}}}}\left( {2\frac{{Pl}}{2} + 0} \right) = \frac{{P{l^2}}}{{4{{EI}}}}. $$

Now these elastic loads should be applied to fictitious beam. Since the bending moment diagram is traced on the tensile fibers and ordinates of M diagram are located above the axis, then the elastic loads should be directed upward. Unit state which corresponds to the required displacement and corresponding bending moment diagram is shown in Fig. 5.5d. Influence line for bending moment at the clamped support of the fictitious beam is shown in Fig. 5.5e. To determine vertical displacement at point 2 of the real structure we need to load the influence line $M_2^{\rm{fict}}$ by the elastic loads

$$ {y_2} = {W_0} \times 1 \times l + {W_1} \times 0.5l = \frac{{5P{l^2}}}{{24{{EI}}}} \times l + \frac{{P{l^2}}}{{4{{EI}}}} \times \frac{l}{2} = \frac{{P{l^3}}}{{3{{EI}}}}. $$

4.2 Parabolic Uniform Arch

Design diagram of hingeless parabolic uniform arch is shown in Fig. 5.6a. Let us show the application of the theorem considered in the previous section for the stability analysis by the Smirnov’s method. If the loss of stability occurs according to the antisymmetrical form, then the equivalent half-arch is shown in Fig. 5.6b.

Fig. 5.6

Design diagram of symmetrical hingeless parabolic arch and equivalent half-arch for antisymmetrical buckling and its equivalent design diagram for antisymmetrical buckling

Stability equation, as before, is $\det \left( {C - \lambda E} \right) = 0,$ where E is identity matrix, eigenvalue $\lambda = 1/H$, and thrust $H = q{l^2}/(8f).$ However, since the half-arch is a redundant structure, then matrix C should be determined by a different method [Smi84]. The structure in Fig. 5.6b has one redundant constraint. Canonical equation of the Force method and primary unknown are

$$ {\delta_{11}}{X_1} + {\Delta_{1{\rm{P}}}} = 0 \to {X_1} = - \frac{{{\Delta_{1{\rm{P}}}}}}{{{\delta_{11}}}}. $$

Let the primary unknown ${X_1}$ be the moment at the fixed support. The axis of the arch is divided into n equal curvilinear parts. The nodal points are denoted by 1, 2, …, n−1, the coordinates of which need to be calculated. After that we show the bending moment diagram for unit state (Fig. 5.7a). The matrix moments in the unit state and its transposed matrix are

Fig. 5.7

Primary system, bending moment diagram due X ₁ = 1 and influence line R _A

$$ {\overline M_1} = \left[ {\begin{array}{*{20}{c}} {{M_{1A}}} \\ {{M_{11}}} \\ \vdots \\ {{M_{1\left( {n - 1} \right)}}} \\ \end{array} } \right],{ }\overline M_1^{\text{T}} = \left[ {\begin{array}{*{20}{c}} {{M_{1A}}} & {{M_{11}}} & \cdots & {{M_{1\left( {n - 1} \right)}}} \\ \end{array} } \right], $$

where first subscript 1 represents the primary unknown X ₁, while the second subscript represents the index of the nodal point.

It can be shown [Smi84] that the unit displacement is

$$ {\delta_{11}} = \frac{1}{{128{m^2}{{EI}}}}\left( {{{\sec }^3}\,{\varphi_0} - 2\overline S } \right), $$

where

$$ m = \frac{f}{l};\quad {\varphi_0} = \arctan \,4m,\ \ { }\overline S = \frac{S}{l} = \frac{1}{4}\left[ {\sec \,{\varphi_0} + \frac{1}{{4m}}\ln \left( {4m + \sec \,{\varphi_0}} \right)} \right]. $$

Free term ${\Delta_{1{\rm{P}}}}$ of canonical equation presents an angular displacement at support A. This displacement may be presented in terms of shear at point A of a fictitious structure. For given supports of the primary system (Fig. 5.7a) the fictitious structure and real half-arch coincides. Fictitious shear at support A equals to the reaction at A. Influence line of this reaction due to load P = 1, which is directed normally to the axis of the arch, coincides with bending moment diagram ${\overline M_1}$ (see Table 5.3).

The final results are as follows [Smi84]. Matrix C of stability equation is determined by formula

$$ C = {L_m}{B_W}\left( {E - \frac{1}{{{\delta_{11}}}}{{\overline M }_1}\overline M_1^{\rm{T}}B_W^*} \right)G. $$

(5.13)

Moment influence matrix ${L_m}$ will be the same as the one for a two-hinged parabolic arch.

The matrices of elastic load for primary system and given structure are

$$ B_W^* = \frac{{{S_0}}}{{6{\text{EI}}}}\left[ {\begin{array}{*{20}{c}} 2 & 1 & 0 & . & {} & {} & {} \\ 1 & 4 & 1 & . & {} & {} & {} \\ . & . & . & . & . & . & . \\ {} & {} & {} & . & 1 & 4 & 1 \\ {} & {} & {} & . & 0 & 1 & 4 \\ \end{array} } \right],\quad {B_W} = \frac{{5{S_0}}}{{6{\text{EI}}}}\left[ {\begin{array}{*{20}{c}} {0.5} & {0.1} & 0 & . & {} & {} & {} \\ {0.1} & {1.0} & {0.1} & . & {} & {} & {} \\ . & . & . & . & . & . & . \\ {} & {} & {} & . & {0.1} & {1.0} & {0.1} \\ {} & {} & {} & . & 0 & {0.1} & {1.0} \\ \end{array} } \right] $$

(5.14)

At point A only one portion (A − 1) exist. Therefore, the first entry of the matrices ${B_W}$ and $B_W^*$ are twice as small as the rest of the diagonal entries. Last diagonal entry of these matrices remains unchanged since at the last point C the bending moment is zero. Diagonal matrix

$$ G = {\text{diag}}\left[ {\begin{array}{*{20}{c}} {\sec \,{\varphi_0}} & {\sec \,{\varphi_1}} & \ldots & {\sec \,{\varphi_{\left( {n - 1} \right)}}} \\ \end{array} } \right]. $$

Detailed stability analysis of uniform parabolic arch with ratio $f/l = 1/6$ is presented in [Smi84]. If half-arch is divided into four equal parts then the critical load becomes

$$ {q_{\rm{cr}}} = \left( {89.5{{EI}}} \right)/{l^3}. $$

The critical uniformly distributed load may be determined by the formula (4.29) ${q_{\rm{cr}}} = { K}({{EI}}/{l^3})$. Parameter ${ K}$ for different types of parabolic uniform arches in terms of f/l is presented in Table 4.6 [Din46], [Mor73].

5 Arch with Complex Tie

A tie of the arch may be represented as the single member at the level of supports (or the elevated tie), as well as a complex tie. One example of such complex tie is shown in Fig. 5.8a. If a load is applied to the tie, then internal forces in the hangers create the effect of elastic foundation for the arch. If load acts on the arch itself, then with the aid of hangers the load is transferred onto the tie.

Fig. 5.8

(a) Design diagram of two-hinged arch with complex tie; (b) half-arch and geometrical notation

The stability problem for arch with complex tie may be effectively solved by Smirnov’s matrix method as shown below.

Design diagram of the uniform symmetric parabolic arch of span l and rise f is shown in Fig. 5.8. The complex tie includes a tie at the level of supports and vertical hangers. Connections of the hangers with arch itself are realized by means of simple hinges, while the connections with horizontal part of the complex tie are realized by means of multiple hinges. The tie of the arch is subjected to uniformly distributed load (this load is not shown). Distances d between all hangers are equal. The span of the arch is divided into 2n equal portions, so the length s _k of each curvilinear portion of the arch are different; in our case $n = 4$.

Assume that parameter $f/l = 0.4$; coordinates x and y for each joint (points k = 0, 1, 2, 3, 4), the trigonometric functions for these joints, and the length of curvilinear portions of the arch are shown in Table 5.4.

Table 5.4 Geometrical parameters of the parabolic arch [Smi84]

As before, the stability equation is $\det \left( {C - \lambda E} \right) = 0$. However, matrix C is calculated by taking into account the features of design diagram.

Numerical procedure is based on the following additional structural analysis [Smi84]:

• 1. Calculation of the vertical displacements of joints of a tie at the moment loss of stability occurs and corresponding change of internal force in each hanger.

• 2. Calculation of additional forces which prevent the loss of stability and corresponding additional bending moments in the arch.

This analysis can be presented in matrix form [Smi84]. For antisymmetrical loss of stability the first stage of numerical procedure for given arched structure includes the construction of the following matrices:

5.1 Initial Matrices

(The total number of panels for half-arch is n = 4)

• 1. Matrix of elastic loads according to (5.9) is

  $$ {B_{WV}} = \frac{l}{{6{\text{EI}}}}\left[ {\begin{array}{*{20}{c}} {2\left( {{\rho_1} + {\rho_2}} \right)} & {{\rho_2}} & 0 \\ {{\rho_2}} & {2\left( {{\rho_2} + {\rho_3}} \right)} & {{\rho_3}} \\ 0 & {{\rho_3}} & {2\left( {{\rho_3} + {\rho_4}} \right)} \\ \end{array} } \right] = \frac{l}{{6{\text{EI}}}}\left\{ {{b_{ik}}} \right\}, $$

  where ${\rho_k} = {s_k}I/{I_k}l$ $\left( {k = 1, \ldots, 4} \right)$, so the relative flexibility ${\rho_k}$ of a member k of an arch equals to its relative length, i.e., ${\rho_1} = {s_1}/l = 0.21506,\quad {\rho_2} = 0.17678, \ldots$. For entry ${b_{11}}$ of matrix ${B_{WV}}$ we get

  $$ {b_{11}} = 2\left( {0.21506 + 0.17678} \right) = 0.78368. $$

• 2. For primary system the matrix (3 by 3) of elastic loads

  $$ B_W^0 = \frac{l}{{6{{EI}}}}\left\{ {b_{ik}^0} \right\} $$

  is determined according to (5.3). For entry $b_{11}^0$ of matrix $B_W^0$ we get

  $$ b_{11}^0 = \left( {2 + \frac{{{s_1}}}{{2{s_2}}}} \right){\rho_1} + \left( {2 + \frac{{{s_2}}}{{2{s_1}}}} \right){\rho_2} = \left( {2 + \frac{{0.21506}}{{2 \cdot 0.17678}}} \right)0.21506 + \left( {2 + \frac{{0.17678}}{{2 \cdot 0.21506}}} \right)0.17678 = 0.98715. $$

• 3. The moment influence matrix ${L_m}$ is determined by radial forces

  $$ {L_m} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} {3{r_1}} & {3{r_2} - 4{c_{12}}} & {3{r_3} - 4{c_{13}}} \\ {2{r_1}} & {2{r_2}} & {2{r_3} - 4{c_{23}}} \\ {{r_1}} & {{r_2}} & {{r_3}} \\ \end{array} } \right] = \frac{1}{4}\left\{ {{l_{ik}}} \right\}, $$

  where segments, according to Fig. 5.8b, are

  $$ \begin{array}{*{20}{c}} {{r_1} = \overline d \,\cos \,{\varphi_1} + {\eta_1}\,\sin \,{\varphi_1},}{{c_{12}} = \overline d \,\cos \,{\varphi_2} + \left( {{\eta_2} - {\eta_1}} \right)\sin \,{\varphi_2},} \\ {{r_2} = 2\overline d \,\cos \,{\varphi_2} + {\eta_2}\,\sin \,{\varphi_2},}{{c_{13}} = 2\overline d \,\cos \,{\varphi_3} + \left( {{\eta_3} - {\eta_1}} \right)\sin \,{\varphi_3},} \\ {{r_3} = 3\overline d \,\cos \,{\varphi_3} + {\eta_3}\,\sin \,{\varphi_3},}{{c_{23}} = \overline d \,\cos \,{\varphi_3} + \left( {{\eta_3} - {\eta_2}} \right)\sin \,{\varphi_3}.} \\ \end{array} $$

  For entry ${l_{12}} = 3{r_2} - 4{c_{12}}$ of matrix ${L_m}$ we get

  $$ \eqalign{ {r_2} = 2\overline d \,\cos \,{\varphi_2} + {\eta_2}\,\sin \,{\varphi_2} = 2 \times 0.125l \times 0.78087 + 0.300l \times 0.62470\cr = 0.38263l, \hfill \\{c_{12}} = \overline d \,\cos \,{\varphi_2} + \left( {{\eta_2} - {\eta_1}} \right)\sin \,{\varphi_2} = 0.125l \times 0.78087 + \left( {0.30 - 0.175} \right)l\cr \quad \times 0.62470 = 0.17570l, \hfill \\{l_{12}} = 3{r_2} - 4{c_{12}} = 3 \times 0.38263l - 4 \times 0.17570l = 0.44509l. \hfill \\}<!endgathered> $$

  In our case we get [Smi84]

  $$ {L_m} = \frac{l}{4}\left[ {\begin{array}{*{20}{c}} {0.64338} & {0.44509} & {0.23675} \\ {0.42892} & {0.76526} & {0.39926} \\ {0.21446} & {0.38263} & {0.48745} \\ \end{array} } \right]. $$

• 4. The moment influence matrix ${L_{mV}}$ is determined by vertical forces for simply supported beam of length $l/2$.

  $$ {L_{mV}} = \frac{l}{{32}}\left[ {\begin{array}{*{20}{c}} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \\ \end{array} } \right]. $$

• 5. Diagonal matrices G and G _V are

  $$ G = \left[ {\begin{array}{*{20}{c}} {\sec \,{\varphi_1}} & 0 & 0 \\ 0 & {\sec \,{\varphi_2}} & 0 \\ 0 & 0 & {\sec \,{\varphi_3}} \\ \end{array} } \right],\quad {G_V} = \left[ {\begin{array}{*{20}{c}} {\cos \,{\varphi_1}} & 0 & 0 \\ 0 & {\cos \,{\varphi_2}} & 0 \\ 0 & 0 & {\cos \,{\varphi_3}} \\ \end{array} } \right], $$

  (5.15)

  where ${\varphi_k}$ is the angle between tangent at point k and horizontal line.

5.2 Matrix Procedures

• 1. Matrix C ₀ relates to the two-hinged arch without tie ${C_0} = {L_m}B_W^0G$.

• 2. Matrix C ₁ takes into account effect of tie ${C_1} = {L_m}{B_{WV}}{L_{mV}}{G_V}.$

• 3. Stability equation of the arched structure is $\det \left( {C - \lambda E} \right) = 0$. Stability matrix $C = {C_0} - {C_1} = ({l^2}/24{{EI)}}\left\{ {{c_{ik}}} \right\}$.

• 4. Maximum stability parameter ${\lambda_{\max }}$ should be calculated numerically. For given arch ${\lambda_{\max }} = 0.895{l^2}/24{{EI}}$. Corresponding thrust is ${H_{\min }} = 1/{\lambda_{\max }} = <$> <$>24{{EI}}/0.895{l^2}$ and smallest critical load is

  $$ {q_{\min }} = \frac{{8{H_{\min }}f}}{l} = 85.81\frac{{EI}}{{{l^3}}}. $$

According to the formula $C = {C_0} - {C_1}$, the critical load for arch with tie is more than the critical load for arch without tie. For arch with tie in Fig. 5.8 the critical load is almost twice as large as the critical load for the same arch without tie.

More stability problems of the arches (with elastic supports and with overarched structure, out-of-plane stability, etc.) are presented in [Smi47], [Mor61], [Smi84].

6 Displacement Method

If an arch has certain features (e.g., nonsymmetrical or skew arch) then easiest approach for stability analysis is based on substituting the arch by frame. The frame is constructed from chords of the arch. Stability analysis of this substitute structure may be performed by any classical method, in particular, by Displacement method in canonical [Kar10], [Uma72-73] or expanded form [Sni66]. The Displacement method is exact for stability analysis of frames. However, it leads to approximate results when analyzing arches because the initial design diagram of arch is replaced by its modified scheme. This method has also some disadvantages – the modified design diagram contains the inclined members and diagram itself has sidesway. Increasing the number of substitute straight members leads to an increase in the overall computational complexity.

6.1 General

The distributed load which acts on the arch may be replaced by the set of concentrated forces P. Position of the nodal points and their number depend on the position of the flexural rigidity changes of the arch and required computational numerical accuracy. Increasing the number of substitute straight members leads to a increase in the numerical accuracy. Thus, this approximation may be performed by several different ways.

Different variations of approximating the arch by straight members in the vicinity of the crown C are shown in Fig. 5.9.

Fig. 5.9

Approximation of initial design diagram of the arch by substitute frame

In all cases in Fig. 5.9 we obtained a framed structure and for its stability analysis we can apply the Displacement method. In case of a symmetrical structure we can replace the substitute frame by its equivalent half-frame.

Canonical equations of the Displacement method for structure with n unknowns Z _j (j = 1,2, …, n) are

$$ \eqalign{ {r_{11}}{Z_1} + {r_{12}}{Z_2} + \cdots + {r_{1n}}{Z_n} = 0 \hfill \\{r_{21}}{Z_1} + {r_{22}}{Z_2} + \cdots + {r_{2n}}{Z_n} = 0 \hfill \\\vdots \hfill \\{r_{n1}}{Z_1} + {r_{n2}}{Z_2} + \cdots + {r_{nn}}{Z_n} = 0. \hfill \\}<!endgathered> $$

(5.16)

6.1.1 Features of (5.16)

• 1. Since the forces P _i are applied only at the joints, then the canonical equations are homogeneous ones.

• 2. Bending moment diagrams, caused by unit displacements of introduced constrains, within the compressed members are curvilinear. Reactions of constraints depend on axial forces in the members of the frame, i.e., contain parameter $\upsilon$ of critical load. If a frame is subjected to different forces P _i , then critical parameters should be formulated for each compressed member $\upsilon_i^2 = {P_i}{l_i}^2/{\left( {{EI}} \right)_i}$ and after that all of these parameters should be expressed in terms of parameter $\upsilon$ for specified basic member. Thus, the unit reactions are functions of parameter, i.e., ${r_{ik}}\left( \upsilon \right)$.

The trivial solution $\left( {{Z_i} = 0} \right)$ of (5.16) corresponds to initial nondeformable design diagram. Nontrivial solution $\left( {{Z_i} \ne 0} \right)$ corresponds to the new form of equilibrium. This occurs if the determinant, which is consisting of coefficients of unknowns, equals zero, i.e.,

$$ \det \left[ {\begin{array}{*{20}{c}} {{r_{11}}\left( \upsilon \right)} & {{r_{12}}\left( \upsilon \right)} & \cdots & {{r_{1n}}\left( \upsilon \right)} \\ {{r_{21}}\left( \upsilon \right)} & {{r_{22}}\left( \upsilon \right)} & \cdots & {{r_{2n}}\left( \upsilon \right)} \\ \vdots & \vdots & \vdots & \vdots \\ {{r_{n1}}\left( \upsilon \right)} & {{r_{n2}}\left( \upsilon \right)} & \cdots & {{r_{nn}}\left( \upsilon \right)} \\ \end{array} } \right] = 0. $$

(5.17)

Condition (5.17) is called the stability equation of a structure in a form of the Displacement method. For practical engineering, it is necessary to calculate the smallest root of the above equation. This root defines the smallest parameter $\upsilon$ of critical force, or smallest critical force. Canonical equation of the Displacement method in the form (5.17) for stability analysis of the frames was developed by Leites (1937) [Uma72-73, Chapters 17.1–17.10].

It is obvious that condition (5.17) leads to a transcendental equation with respect to parameter $\upsilon$. The functions $\varphi \left( \upsilon \right)$ and $\eta \left( \upsilon \right)$ are tabulated (Table A.33). Since the determinant is very sensitive with respect to parameter $\upsilon$, it is recommended to solve (5.17) using a computer. The functions $\varphi \left( \upsilon \right)$ and $\eta \left( \upsilon \right)$ may be presented in approximate form [Bol64]; in this case (5.17) leads to an algebraic equation.

The only limitation for applying this method is that the flexural rigidity and axial compressed force should be constant within the each member.

Let us derive the stability equation and determine the critical load for frame shown in Fig. 5.10a. This frame has one unknown of the Displacement method. The primary unknown is the angle of rotation of rigid joint. Figure 5.10b shows the primary system, elastic curve, and bending moment diagram caused by unit rotation of introduced constrain. The bending moment diagram for compressed vertical member of the frame is curvilinear. The ordinate for this member is taken from Table A.32, row 1.

Fig. 5.10

(a) Design diagram; (b) primary system of the Displacement method and unit bending moment diagram

The bending moment diagram yields ${r_{11}} = 4{i_1}{\varphi_2}\left( {{\upsilon_1}} \right) + 4{i_2}$, where ${i_1} = {{E}}{{{I}}_1}/{l_1},<$> <$>{i_2} = {{E}}{{{I}}_2}/{l_2}$ and parameter of critical load ${\upsilon_1} = {l_1}\sqrt {P/{{E}}{{{I}}_1}}$. Note that subscript 2 at function ϕ is related to the clamped–clamped member subjected to angular displacement of one support (Table A.32), while the subscript 1 at the parameter $\upsilon$ is related to the compressed-bent member 1. Canonical equation of the Displacement method is ${r_{11}}\left( {{\upsilon_1}} \right)Z = 0.$

Nontrivial solution of this equation leads to equation of stability ${r_{11}} = 0$ or in expanded form

$$ {r_{11}} = \frac{{4{{E}}{{{I}}_1}}}{{{l_1}}}{\varphi_2}\left( {{\upsilon_1}} \right) + \frac{{4{{E}}{{{I}}_2}}}{{{l_2}}} = 0. $$

Note, that Displacement method allows us to take into account some changes in the design diagram; for example, if a fixed support A is replaces by a pinned support, then stability equation becomes

$$ {r_{11}} = 3{i_1}{\varphi_1}\left( {{\upsilon_1}} \right) + 4{i_2} = 0. $$

6.1.2 Special Cases

• 1. Assume that ${l_2} \to 0$. In this case the second term $(4{{E}}{{{I}}_2}/{l_2}) \to \infty$, rigid joint is transformed to clamped support and the initial frame is transformed into the vertical clamped–clamped column. Stability equation becomes ${\varphi_2}\left( {{\upsilon_1}} \right) = - \infty$. The smallest root of this equation is ${\upsilon_1} = 2\pi$ and critical force becomes

  $$ {P_{\rm{cr}}} = \frac{{\upsilon_1^2{{EI}}}}{{l_1^{12}}} = \frac{{4{\pi^2}{{EI}}}}{{l_1^2}} = \frac{{{\pi^2}{{EI}}}}{{{{\left( {0.5{l_1}} \right)}^2}}}, $$

  where $\mu = 0.5$ is effective-length factor for clamped–clamped column.

• 2. Assume ${{E}}{{{I}}_2} \to 0$. In this case, the rigid joint is transformed to hinge and the initial frame is transformed into the vertical clamped–pinned column. Stability equation becomes ${\varphi_2}\left( {{\upsilon_1}} \right) = 0$. The smallest root of this equation is ${\upsilon_1} = 4.488$ and for critical force we get

  $$ {P_{\rm{cr}}} = \frac{{\upsilon_1^2{{EI}}}}{{l_1^2}} = \frac{{{{4.488}^2}{{EI}}}}{{l_1^2}} = \frac{{{\pi^2}{{EI}}}}{{{{\left( {0.7{l_1}} \right)}^2}}},\quad \mu = 0.7. $$

• 3. If ${l_1} = {l_{2,\quad }}{{E}}{{{I}}_1} = {{E}}{{{I}}_2}$, then stability equation becomes ${\varphi_2}\left( {{\upsilon_1}} \right) + 1 = 0$.

  The smallest root of this equation is ${\upsilon_1} = 5.3269$ and critical load equals ${P_{\rm{cr}}} = (\upsilon_1^2{{EI}}/l_1^2) = (28.397{{EI}}/l_1^2).$

6.1.3 Modified Approach of the Displacement Method

In general case, the Displacement method requires introducing constraints, which prevent angular displacement of rigid joints and independent linear displacements of joints. However, in stability problems of a frame with sidesway, it is possible for some modification of the classical Displacement method. Using modified approach, we can introduce a new type of constraint, mainly the constraint, which prevents angular displacement, but simultaneously has a linear displacement (Table A.32, row 3). Modified approach of the Displacement method is presented in [Kar10].

6.2 Two-Hinged Arch

Let us demonstrate application of the Displacement method in canonical form for stability analysis of two-hinged uniform symmetric arch loaded by two forces P (Fig. 5.11a). Simplest version of the substituted frame is shown by solid lines; it is constructed in such way so that forces are applied at the rigid joints of the frame. The unavoidable disadvantage of this system is that the substituted frame is with sidesway.

Fig. 5.11

(a) Design diagram of symmetrical two-hinged uniform arch and its substituted frame; (b) primary system of Displacement method and group unknowns

To obtain the primary system of Displacement method we need to introduce two rigid joints 1 and 2 and support 3 into the design diagram; they are shown in bold in Fig. 5.11b. Constraints 1 and 2 prevent angular displacements of joint 1 and 2 and constraint 3 prevent linear displacement of the cross bars 1–2.

For stability analysis of this symmetrical frame it is very effective to adopt the group unknowns of the Displacement method as shown in Fig. 5.11b; Z ₁ represents the simultaneous angular displacement rotation of introduced constraints 1 and 2 in one direction, while Z ₂ represents the simultaneous angular displacement of same introduced constraints in opposite directions (Bresse’s method, 1854) [Bre54, 59]; Z ₃ represents a linear displacement of cross bars 1–2.

By using the group unknowns we can separate the full system of canonical equations into two separate independent subsystems. First subsystem allows us to determine the critical load for symmetrical form of loss of stability and second for antisymmetrical form of loss of stability.

Stability analysis consists of the following steps:

• 1. Determine some parameters of the frame

  $$ {l_{0 - 1}} = 13.4164m,\quad \tan \,\beta = 0.5,\quad \sin \,\beta = 0.44721,\quad \cos \,\beta = 0.89443. $$

• 2. For a given frame we find the thrust H by any analytical approach, for example, by the Force method or using tabulated data. Axial loads in members 0–1 and 1–2 in terms of H are ${N_{0 - 1}} = H/\cos \beta; \quad {N_{1 - 2}} = H.$

• 3. For each member of length ${l_i}$ of the frame, calculate the parameter ${\vartheta_i} = {l_i}\sqrt {{N_i}/{{EI}}}$ of critical force. For one specific member of the frame the parameter of a critical load is adopted as the base parameter and all remaining parameters are expressed in terms of the base parameter.

  Parameter of critical loads for members 0–1 and 1–2 are

  $$ {\vartheta_{0 - 1}} = {l_{0 - 1}}\sqrt {\frac{{{N_{0 - 1}}}}{{EI}}}, \quad {\vartheta_{1 - 2}} = {l_{1 - 2}}\sqrt {\frac{{{N_{1 - 2}}}}{{EI}}} = 8\sqrt {\frac{H}{{EI}}}. $$

  Assume the parameter ${\vartheta_{1 - 2}} = \vartheta$ is a base parameter. Then for ${\vartheta_{0 - 1}}$ we get

  $$ {\vartheta_{0 - 1}} = 13.4164\sqrt {\frac{H}{{0.89443{{EI}}}}} = 1.7733 \times 8\sqrt {\frac{H}{{EI}}} = 1.7733\vartheta. $$

  Figure 5.11b contains the length l, stiffness per unit length i axial forces N, and critical parameter ϑ for each member of the substituted frame.

• 4. Now we need to construct bending moment diagrams caused by unit primary unknowns of Displacement method and calculated unit reactions.

First unit state (Z ₁ = 1). Bending moment diagram caused by the unit rotation of introduced constraints Z ₁ = 1 (antisymmetrical unknown) is shown in Fig. 5.12. This diagram is antisymmetrical. Elastic curve is shown by dotted line.

Fig. 5.12

Bending moment diagram in the first unit state

$$ \eqalign{ M_{1 - 0}^{(1)} = 3{i_{0 - 1}}{\varphi_1}\left( {{\vartheta_{0 - 1}}} \right) \hfill \\M_{2 - 3}^{(1)} = M_{1 - 0}^{(1)} \hfill \\M_{1 - 2}^{(1)} = 4{i_{ 1 - 2}}{\varphi_2}\left( \vartheta \right) + 2{i_{ 1 - 2}}{\varphi_3}\left( \vartheta \right) \hfill \\M_{2 - 1}^{(1)} = M_{1 - 2}^{(1)} \hfill \\}<!endgathered> $$

Top subscript (1) at M denotes the first state. Subscript 1 at function ϕ ₁ relates to pinned–fixed beam in case of angular displacement of fixed support while subscripts 2 and 3 relates to fixed–fixed beam (member 1–2) in case of angular displacement of fixed support (Table A.32).

Unit reaction

$$ {r_{11}} = 2\left( {M_{1 - 0}^{(1)} + M_{1 - 2}^{(1)}} \right) = 2\left[ {3{i_{0 - 1}}{\varphi_1}\left( {{\vartheta_{0 - 1}}} \right) + 4{i_{ 1 - 2}}{\varphi_2}\left( \vartheta \right) + 2{i_{ 1 - 2}}{\varphi_3}\left( \vartheta \right)} \right]. $$

After substituting the corresponding quantities we get

$$ {r_{11}} = 2\left[ {0.2236{\varphi_1}\left( {1.7733\vartheta } \right) + 0.5{\varphi_2}\left( \vartheta \right) + 0.25{\varphi_3}\left( \vartheta \right)} \right]{{EI}}{.} $$

Secondary reactions will be calculated later.

Second unit state (Z ₂ = 1). Bending moment diagram caused by the unit rotation of introduced constraints Z ₂ = 1 (symmetrical unknown) is shown in Fig. 5.13. This diagram is symmetrical.

Fig. 5.13

Bending moment diagram in the second unit state

$$ \eqalign{ M_{1 - 0}^{(2)} = 3{i_{0 - 1}}{\varphi_1}\left( {{\vartheta_{0 - 1}}} \right) \hfill \\M_{2 - 3}^{(2)} = M_{1 - 0}^{(2)} \hfill \\M_{1 - 2}^{(2)} = 4{i_{ 1 - 2}}{\varphi_2}\left( \vartheta \right) - 2{i_{ 1 - 2}}{\varphi_3}\left( \vartheta \right) \hfill \\}<!endgathered> $$

Unit reaction

$$ {r_{22}} = 2\left( {M_{1 - 0}^{(2)} + M_{1 - 2}^{(2)}} \right) = 2\left[ {3{i_{0 - 1}}{\varphi_1}\left( {{\vartheta_{0 - 1}}} \right) + 4{i_{ 1 - 2}}{\varphi_2}\left( \vartheta \right) - 2{i_{ 1 - 2}}{\varphi_3}\left( \vartheta \right)} \right]. $$

After substituting the corresponding quantities we get

$$ {r_{22}} = 2\left[ {0.2236{\varphi_1}\left( {1.7733\vartheta } \right) + 0.5{\varphi_2}\left( \vartheta \right) - 0.25{\varphi_3}\left( \vartheta \right)} \right]{{EI}}{.} $$

Since ${\overline M_1}$ and ${\overline M_2}$ diagrams are antisymmetrical and symmetrical, respectively, then ${r_{12}} = {r_{21}} = 0$. This is an important result due to application of group unknowns. Other secondary reactions are discussed below.

Third unit state (Z ₃ = 1). Hinged scheme of the frame and its position caused by Z ₃ = 1 is shown in Fig. 5.14. The displacement of point 1 is directed perpendicularly to member 0–1 and its new position is denoted as 1′. Point 2 is moved perpendicularly to member 2–3 and its new position is denoted as 2′.

Fig. 5.14

Hinged scheme of the frame and displacement of the joints in the third unit state

$$ \eqalign{ {\Delta_{0 - 1}} = \csc \,\beta, \hfill \\{\Delta_{1 - 2}} = \cot \,\beta, \hfill \\{\Delta_{2 - 3}} = \csc \,\beta, \hfill \\{\Delta_{2 - 1}} = \cot \,\beta. \hfill \\}<!endgathered> $$

For member 1–2 relative vertical displacement of the ends is

$$ \Delta = {\Delta_{1 - 2}} + {\Delta_{2 - 1}} = 2\,\cot \,\beta. $$

Bending moment diagram in the third unit state is shown in Fig. 5.15.

Fig. 5.15

Bending moment diagram in the third unit state

$$ \eqalign{ M_{1 - 0}^{(3)} = 3\frac{{{i_{0 - 1}}}}{{{l_{0 - 1}}}}{\Delta_{0 - 1}}\times{\varphi_1}\left( {{\vartheta_{0 - 1}}} \right) \hfill \\M_{2 - 3}^{(3)} = M_{1 - 0}^{(3)} \hfill \\M_{1 - 2}^{(3)} = 6\frac{{{i_{ 1 - 2}}}}{{{l_{1 - 2}}}}\Delta \times{\varphi_4}\left( \vartheta \right) \hfill \\M_{2 - 1}^{(3)} = M_{1 - 2}^{(3)} \hfill \\}<!endgathered> $$

Design diagram for calculation of r _33. is shown in Fig. 5.16. The section passing through member 0–1 is infinitely close to joint 1. Bending moment M _1–0 is directed according to position of extended fibers, which are shown by dotted line; the top subscript “3” is omitted. This moment is equilibrated by shear force ${Q_{1 - 0}} = (3{{EI}}/l_{0 - 1}^3){\eta_1}\left( {{\vartheta_{0 - 1}}} \right){\Delta_{0 - 1}}$. Then this force is transferred on the part which is adjacent to joint 1. Also we need to show the axial force N _1–0.

Fig. 5.16

Calculation of r ₃₃

Similar procedure is applied to portions 1–2 and 2–3. Shear force for fixed–fixed member 1–2 subjected to relative vertical displacement Δ of the ends equals ${Q_{1 - 2}} = {Q_{2 - 1}} = ({{EI}}/l_{0 - 1}^3){\eta_2}\left( \vartheta \right)\Delta$. Equilibrium equations $\sum {X = 0}$ and $\sum {Y = 0}$ for joints 1 and 2 allow us to calculate the normal forces. Relationship ${N_{1 - 2}} = {N_{2 - 1}}$ can be used for verification of computed results. Moreover, ${N_{0 - 1}} = - {N_{2 - 3}}$. Therefore, equation $\sum {X = 0}$ for cross bar in whole leads to the following result

$$ {r_{33}} = 2Q_{1- 0}^{(3)}\,\sin \,\beta = 2 \times \frac{{3{{EI}}}}{{l_{0 - 1}^3}}{\Delta_{0 - 1}}{\eta_1}\left( {{\vartheta_{0 - 1}}} \right)\,\sin \,\beta = 2 \times \frac{{3{{EI}}}}{{l_{0 - 1}^3}}{\eta_1}\left( {{\vartheta_{0 - 1}}} \right). $$

Performing a similar procedure over the bending moment diagram ${\overline M_1}$ we get

$$ {r_{31}} = - 2\left[ {\frac{{3{{EI}}}}{{l_{0 - 1}^2\,\sin \,\beta }}{\varphi_1}\left( {{\vartheta_{0 - 1}}} \right) - \frac{{12{{EI}}}}{{l_{1 - 2}^2\,\tan \,\beta }}{\varphi_4}\left( \vartheta \right)} \right]. $$

This result may be obtained by considering equilibrium of joints 1 and 2 from diagram ${\overline M_3}$. In this case, moment is

$$ {r_{31}} = \underbrace { - M_{1 - 0}^{(3)} + M_{1 - 2}^{(3)}}_{{\rm{Joint}}\;{1}} + \underbrace {M_{2 - 1}^{(3)} - M_{2 - 3}^{(3)}}_{{\rm{Joint}}\;2}. $$

Substituting the numerical data, we get

$$ \eqalign{ {r_{11}} = 2\left[ {0.2236{\varphi_1}\left( {1.7733\vartheta } \right) + 0.5{\varphi_2}\left( \vartheta \right) + 0.25{\varphi_3}\left( \vartheta \right)} \right]{{EI}}, \hfill \\{r_{22}} = 2\left[ {0.2236{\varphi_1}\left( {1.7733\vartheta } \right) + 0.5{\varphi_2}\left( \vartheta \right) - 0.25{\varphi_3}\left( \vartheta \right)} \right]{{EI}}, \hfill \\{r_{33}} = 2 \times 0.001242{\eta_1}\left( {1.7733\vartheta } \right){{EI}}, \hfill \\{r_{13}} = {r_{31}} = - 2\left[ {0.037476{\varphi_1}\left( {1.7733\vartheta } \right) - 0.375{\varphi_4}\left( \vartheta \right)} \right]{{EI}}. \hfill \\}<!endgathered> $$

Since the bending moment diagrams ${\overline M_1}$ and ${\overline M_3}$ are antisymmetrical while ${\overline M_2}$ is symmetrical, then

$$ {r_{12}} = {r_{21}} = 0\quad {\hbox{and}}\quad {r_{32}} = {r_{23}} = 0. $$

Now we can form the stability equation, calculate the critical parameter, and determine the critical load.

In our case, the set of canonical equations of stability

$$ \eqalign{ {r_{11}}{Z_1} + {r_{12}}{Z_2} + {r_{13}}{Z_3} = 0 \hfill \\{r_{21}}{Z_1} + {r_{22}}{Z_2} + {r_{23}}{Z_3} = 0 \hfill \\{r_{31}}{Z_1} + {r_{32}}{Z_2} + {r_{33}}{Z_3} = 0. \hfill \\}<!endgathered> $$

are separated into two independent systems

$$ \begin{array}{*{20}{c}} {{r_{11}}{Z_1} + {r_{13}}{Z_3} = 0} \\ {{r_{31}}{Z_1} + {r_{33}}{Z_3} = 0} \\ \end{array} \quad {\text{and}}\quad {r_{22}}{Z_2} = 0. $$

Stability equations become

$$ \left| {\begin{array}{*{20}{c}} {{r_{11}}} & {{r_{13}}} \\ {{r_{31}}} & {{r_{33}}} \\ \end{array} } \right| = 0\quad {\text{and}}\quad {r_{22}} = 0. $$

The first subsystem describes the antisymmetrical loss of stability, while second equation describes symmetrical loss of stability. For solution of these equations we need to take into account expressions for functions ${\varphi_i},\eta$ according to Table A.33. For stability parameters we get

$$ {\vartheta_{\rm{ant}}} = 2.0344\quad {\hbox{and}}\quad {\vartheta_{\rm{sym}}} = 2.6007. $$

Critical thrust becomes

$$ {H_{\rm{asym}}} = {2.0344^2}\frac{{EI}}{{{8^2}}},{ }{H_{\rm{sym}}} = {2.6007^2}\frac{{EI}}{{{8^2}}}. $$

After that we can calculate the critical load P. Note, these results corresponds to a crude model approximating entire arch.

Numerical results for three-hinged and two-hinged arches subjected to uniformly distributed vertical load within the whole span are presented in [Sni66]. If entire arch is represented as five chords with equal horizontal projections then relative error of the Displacement method is no greater than 2%.

7 Comparison of the Smirnov’s and Displacement Methods

7.1 Advantages and Disadvantages of Smirnov’s Method

In Smirnov’ method the axis of the arch is replaced by a set of curvilinear segments which coincides with arch itself. Therefore, errors of the method arise from numerically inaccurate calculation of nodal point coordinates obtained from the solution of transcendental equation. It means that the results which are obtained from Smirnov’ method should be treated as ground truth.

According to Smirnov’s method, the arch should be presented as a set of curvilinear segments of equal length and constant stiffness within each segment. Meeting these requirements is not always possible in case of arches with variable stiffness (of course, with a reasonable number of the finite portions). The other disadvantage is related to the difficulties of analyzing nonsymmetric arches.

Any change in the design diagram of the arch (addition of the overarched structure, combined tie, elevated tie, fixed supports, etc.) leads to a procedure that is not easily generalizable. Even if the governing general stability equation $\det \left( {C - \lambda E} \right) = 0$ holds true, the matrix C depends on the features of the arched structure. These peculiarities limit the scope of the method.

We note that the span of a parabolic arch can be divided into a set of equal lengths portions. In this case, we can use the concept of the parabolic chain [Rab54b], [Rab58, Vol. II]. Its properties and important relationships are considered for free vibration analysis in Chap. 6.

7.2 Advantages and Disadvantages of the Classical Methods in Canonical Form

Stability problems of the arches may be solved using the classical Force and Displacement methods in canonical form. Approximate stability analysis by both of these methods requires the construction of a substitute frame, and thus these methods lead to the approximate results. A poor choice for the primary system of the Force method can lead to significant difficulties of computational nature [Smi47]. Displacement method does not have similar disadvantages, because the primary system is constructed according to strong rules. These rules allow considering nonsymmetrical arches, take into account the elastic supports as well as any type of loading. It is very convenient that the Displacement method in canonical form is easily generalizable. As in case of Smirnov’s method, computational procedure becomes more cumbersome with increasing the number of elements which approximate an arch.

Displacement method, generally, allows constructing the stability equation for the full spectra of critical forces. Given this, it is unnecessary to assume the initial form of loss of stability. The group unknowns allow simplifying numerical procedure and separate stability analysis for symmetrical and antisymmetrical forms of the loss of stability.