Arch Structures

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Definition of a Three-Hinged Arch:

An arch is a structure that spans a certain distance and supports the structure and weight above it.[1] They are classified as three-hinged, two-hinged or hingeless. As shown in Fiqure 1.1, if the arch is three-hinged, two hinges are at the supports and the third hinge is anywhere within arch. This type of arch is commonly used in steel structures. Compared to fixed-end connections, three hinged arches are a better choice. This type of arch is also a statically derminate stucture, therefore it can be easily solved by using the laws of equilibrium and statics.[2]
Figure 1.1: Three hinged arch [3]

Application of Three-Hinged Arches in Construction:

Compared to other arch structures, the three-hinged arch has two hinges at the base and an extra connection at the mid-span. This allows the arch to move in two opposite directions and compensate for any expansion and contraction. Therefore, "thermal expansion and contraction of the arch will cause vertical movements at the peak pin joint but will have no appreciable effect on the bases".[4] The three-hinged arch is therefore said to be statically determinate.It is most often used for medium-span structures, such as bridges and building's roofs.[5]
Three-Hinged Arches--Bridges
Figure 1.2: Three hinged arched bridge‎ [6]
As an example, the railway support structure, Royal Gorge is a simple 3-hinged structure consisting of two straight struts, pinned to each side of the gorge. The center pin supports the side of a girder bridge for the railway line (the other side being supported directly on the cliff face). This is a good example for illustrating boundary forces in a 3-hinged arch. (Colorado)[7]
Figure 1.3: Three hinged arched bridge‎ [8]
In a reinforced concrete highway bridge, as seen in Fig. 1.3, where the arch rib itself is three-hinged and the spandrel walls are integral with both the arch and the continuous road level girder, the behaviour of the bridge highly redundant. The mid-span hinge in the arch can be seen together with the split spandrel wall at mid-span. Note that both the arch rib and the spandrel columns are continuous over the width of the bridge. [7]

Three-Hinged Arches--Roof Structures
Figure 1.4: Three hinged arched roof structures‎ [9]

In the railway station roof structure in Fig. 1.4, we can see the building is constructed with two intersecting laminated 3-hinged arches. The building is square in plain view with the arches in the middle.

Figure 1.5: Three hinged roof structures

In Fig. 1.5, there is a small glue-laminated structure over a swimming-pool. It is hexagonal in plain view and the structure consists of six radial arch ribs forming three intersecting 3-hinged arches.[7]

Types of Three Hinged Arches

The Parabolic Arch
The parabolic arch is a curve which has an eccentricity that is equal to one and is obtained by slicing a cone, with a plane parallel to one side. The point on the axis of symmetry that intersects the parabola is called the vertex and is the point where the curvature is greatest.  In the parabolic arch, only normal forces and bending moments occurs, but no shear forces. That means when the arch is subjected to a uniform distributed load and both ends are mounted in fixed bearings, loads acting upon the inner portion of the arch are primarily compressive forces acting in the direction of the normal force at every point of the arch. [10]
Figure 1.6: The Parabolic Arch [11]
The Catenary Arch
The curved arch described by a uniform chain hanging from two supports in a uniform gravitational field is called a Catenary Arch. If the sag is small so that the weight is approximately uniformly distributed, the curve is close to a parabola (a quadratic curve). However, the Catenary is a hyperbolic cosine curve, $ y = a cosh(x/a) $, where x is measured from the lowest point. The tension in the chain increases from the lowest point of the arch up to the arch supports. Because all of the load is vertical, the horizontal component of the tension is constant. A uniform cable also hangs in a Catenary; the essential thing is that there is negligible transverse stiffness.[12]. The Catenary Arch is the optimum for constant dead load of the arch. It is used in Architecture and Engineering, for the design of bridges and arches so that forces do not result in bending moments.[13]
Figure 1.7: The Catenary Arch [14]
The Circular Arch
Arches with a circular form were commonly employed by the builders of ancient, heavy masonry arches. Ancient Roman builders relied heavily on the rounded arch to span large, open areas. Several rounded arches placed in-line (end-to-end) form an arcade, such as the Roman aqueduct.[15]
Figure 1.8: The Circular Arch‎ [16]

Basic Concepts and Structural Analysis:

Concepts and Equations to Know Before Analysis:

Before starting you should be familiar with the following topics:
Also, you should be familiar with the following equations:
Name Equation Variables
Equilibrium on the x-axis $ \ \Sigma F_x = 0 $ $ \ F_x = $ Forces on the Horizontal Direction
Equilibrium on the y-axis $ \ \Sigma F_y = 0 $ $ \ F_y = $ Forces on the Vertical Direction
Moment Equilibrium $ \ \Sigma M_i = 0 $ $ \ M_i = $ Moment at a Point "i" due to Forces on the System

Analysis of Static Point Loads on a Three-Hinged Arch:

Three-hinged arch structures are statically determinate. There is one hinge at each of the two supports and the third hinge is located in the center of the arch.[17] This position is also called the crown.[17] To solve arch systems, you can use the equations of equilibrium.
In this system, there are four reaction components. They are all located at the supports. Due to the shape of an arch structure, horizontal reaction forces do occur, unlike a beam. [17] There are no moments at the supports due to the fact that the arch structure is supported by pin connections (resultant of hinges at the supports).

General Steps

NOTE: For all steps of analysis, take into account the point load/uniformly distributed load (UDL) acting on the structure.
To solve the system:
  • First take the moment about one support and solve for the reaction component at the other support, acting in the y-direction.
  • Next, apply the equilibrium equation for the sum of forces in the y-direction to solve for the reaction component in the y-direction at the other support.
  • Now, to solve for the x-direction reaction components, split the arch structure at the center hinge (crown). Looking at just one side of the split (left OR right, take the moment about the crown hinge to solve for the x-direction component at the one support.
  • To solve for the last reaction component, "re-assemble" the arch structure and apply the equation of equilibrium for the sum of forces in the x-direction.

Vertical Distance to Point Load

Knowing the vertical distance to a point load helps when solving for the shear and moment of a three hinged arch system.
There are three different types of arch shapes: Parabola, Circular, and Catenary. For practical applications, we will be looking at just the Parabolic and Circular arch.
Equation of a Parabola: $ y = c x [L - x] $ ~ Eq.1 [18]
(See Figure 2.1)
Figure 2.1: Typical parabola three hinged arch with point load
- c is a constant
- x is a varying distant that starts at one support (A)
  • indicates horizontal location of the point load
- L is the horizontal span (length) of the arch
  • When x is at the middle of the arch (0.5L), y is equal to the horizontal distance from Datum (zero-line) to the crown hinge of the arch, h
- Substitute these variable definitions into Eq. 1 to solve for c
- Resulting Equation: $ y = (4 h x / L^2)*(L x - x^2) $ ~ Eq.2 [18]
When you substitute in the corresponding values for your system and solve for y, you will get the vertical distance to the location of the point load.
Radius of Curvature: $ R = (L^2 / 8 h ) + (h / 2) $ ~ Eq.3 [18]
(See Figure 2.2)
Figure 2.2: Typical circular three hinged arch with point load
To find the vertical distance to a point load on a circular arch:
  • First, substitute the corresponding variables into Eq.3 to solve for R .
  • Next, you need to solve for the angle between R and the tangent to the arch, at the location of the point load. This angle is called phi.
Equation: $ sin (\Phi) = L-2x / 2R $ [18]
  • Now, to find the vertical distance, use the equation below with the corresponding variables to solve for y.
Equation: $ cos (\Phi) = (R - h + y) / R $ [18]

Shear and Bending Moment Diagrams for Point/Continuous Loads

Note: This entire section,"Shear and Bending Moment Diagrams for Point/Continuous Load", is adapted from [19], but has been re-analysed independently and explained differently.
In order to simplify the calculations for the Shear and Bending Moment diagram for a three-hinged arch, the next steps must be followed:
  • Step 1: Construct a simple supported beam, also called "reference beam", with the same loading and loading locations of the three-hinged arch in the x and y direction. Ignore the middle hinge when doing so.
  • Step 2: Draw a free body diagram of the reference beam and find the shear and bending moment. As shown below (see example), the bending moment for the three-hinged arch does not reach a zero value at the location of the mid-hinge. This is because both, shear and moment diagrams, are for a simple supported beam and not for a three-hinged arch just yet.
  • Step 3: In order to obtain the real values of the shear and bending moment for the three-hinged arch, we need to take in consideration the thrust component or horizontal force component at any given point. To do so, we need to subtract the thrust component from the shear and bending moment value of the reference beam at any specific section in order to obtain the actual shear and bending moment value for a three-hinges arch. We also need to know that the thrust in an arch is inversely proportional to the rise of the arch. This calculation can be obtain by:
Thrust for shear values: $ H = M_c/h $. Where, $ H = $ thrust component for shear, $ M_c = $ moment at location of the middle hinge in the reference beam, and $ h = $ height of the arch.
Thrust for bending moment values: $ H_y = H*y $. Where, $ H_y = $ thrust component for bending moment, $ H = $ thrust component for shear, and $ y = $ y-component of any given point.
Once we have obtain the thrust values, we can calculate the actual shear and bending moment values for the three-hinged arch by following the next equations:
$ M_x = M_(x,0)-H_y $
where: $ M_x = $ is the actual bending moment value of the three-hinged arch., $ M_(x,0)= $ is the moment value at any given point in the reference beam.
$ V_x = V_(x,0)-(H*Sin \varphi) $
where: $ V_x = $ is the actual shear value of the three-hinged arch., $ V_(x,0)= $ is the shear value at any given point in the reference beam.
For a more detailed and comprehensive explanation of the application of the afore mentioned formulas, please refer to the "Example" section.

Analysis of a Moving Point Load on a Circular Three-Hinged Arch:

For a full explanation on Influence Lines and its applications, please refer to the Beam and Frame Influence Lines and Muller Breslau Principle sections of this website.
Note: This entire section,"Analysis of a Moving Point Load on a Circular Three-Hinged Arch", is adapted from [20], but has been re-analysed independently and explained differently.

Influence Lines for Support Reactions:

Figure 2.3: Influence Diagrams for the Reaction Supports of a Three-hinged Arch
Consider the three-hinged arch in Fig. 2.3. The arch is subjected to a 1 KN unit load in the y-direction that moves from point A to point B, through C, in the x-direction. From the figure, we want to draw the influence lines for the vertical and horizontal reactions of the system.
  • The first step is to determine the value of $ A_y $ as function of "x" (changing distance that the 1 KN unit load travels from point A to B) by taking the moment at point B.
For $ 0 \le x \le L $:
$ \Sigma M_B = 0 $
$ -A_y(L)+1.0(L-x) = 0 $
$ A_y = \frac{(L-x)}{L} $
  • Similarly, we can find the expressions for $ B_y $, $ A_x $, and $ B_x $ by taking the moments at point A, at point C with respect to AC, and at point C with respect to BC, respectively.

For $ 0 \le x \le L $:
$ \Sigma M_A = 0 $ $ \Sigma M_C^{AC} = 0 $ $ \Sigma M_C^{BC} = 0 $
$ B_y(L)-1.0(x) = 0 $ $ -A_y(a) + A_x(h) = 0 $ $ B_y(b)-B_x(h) = 0 $
$ B_y = \frac{x}{L} $ $ A_x = \frac{A_y*a}{h} $ $ B_x = \frac{B_y*b}{h} $
  • Once we have the expressions for all the supports reactions, we need to plug in values into the equations to obtain the influence diagram for the support. The values we are using are "0" and "L" since they are the limits of the equations; and point "a" since it is the location of the middle hinge. Next, you can see the calculation for the support reaction $ A_y $.
For $ x=0 $ : $ A_y = \frac{(L-x)}{L} = \frac{(L-0)}{L} = \frac{L}{L} = 1 $
For $ x=L $ : $ A_y = \frac{(L-x)}{L} = \frac{(L-L)}{L} = \frac{0}{L} = 0 $
For $ x=a $ : $ A_y = \frac{(L-a)}{L} = \frac{(L-(L-b))}{L} = \frac{b}{L} $
  • After we obtain the three points for the support $ A_y $, we can connect them using a straight line, as seen in Figure 2.3. This line will lead us to create the Influence diagram for the reaction $ A_y $. Similarly, we can find the values for the other supports reactions, as shown below:
$ B_y $ $ A_x $ $ B_x $
$ x = 0 $ : $ 0 $ $ \frac{(A_y*a)}{h} $ $ 0 $
$ x = L $ : $ 1 $ $ 0 $ $ \frac{(B_y*b)}{h} $
$ x = a $ : $ \frac{a}{L} $ $ \frac{(a*b)}{(L*h)} $ $ \frac{(a*b)}{(L*h)} $
  • Finally, we can see the Influence Diagrams for all the reaction supports for a three-hinged arch. See Fig. 2.3.

Influence Lines for Internal Forces:

Let's say we want to find the Influence Line equations for the shear and moment at a point "m" with distance $ X_m $ from the origin "O". The point "m" is also located at a height of "z" value, as see on the Figure below.
Figure 2.4: Point "m" at the Circular Three-hinged Arch
  • The first step is to make a cut on the "m" point, as seen on the Figure below, and identify the shear and moment at each end.
Figure 2.5: Point "m" at the Circular Three-hinged Arch
  • The next step is to analyse two possible scenarios, one with the unit point load at the left of "m" and another one at the right of "m". This is because the shear and moment behave differently with respect to the position of the unit point load.
For the first case, we take the moment around point "m" to find $ M_m $, and the sum of vertical forces in the x'-y' plane to find $ V_m $:
Figure 2.6: Point "m" at the Circular Three-hinged Arch
Case 1: For $ X < X_m $:
$ \Sigma M_m = 0 $ $ \Sigma F_y = 0 $
$ B_y(L - X_m)-B_x(z)- M_m = 0 $ $ V_m + B_xSin(\alpha) + B_yCos(\alpha) = 0 $
$ M_m = B_y*(L-X_m)-B_x(z) $ $ V_m = -B_yCos(\alpha)-B_xSin(\alpha) $
Similarly, for the second case, we take the moment around point "m" to find $ M_m $, and the sum of vertical forces in the x'-y' plane to find $ V_m $:
Figure 2.7: Point "m" at the Circular Three-hinged Arch
Case 2: For $ X \ge X_m $:
$ \Sigma M_m = 0 $ $ \Sigma F_y = 0 $
$ B_y(L - X_m)-B_x(z)- 1.0(X-X_m)-M_m = 0 $ $ V_m + B_xSin(\alpha) + B_yCos(\alpha)-1.0Cos(\alpha) = 0 $
$ M_m = B_y*(L-X_m)-B_x(z)-X+X_m $ $ V_m = -B_yCos(\alpha)-B_xSin(\alpha)+Cos(\alpha) $
  • Finally, after obtaining the shear and moment equations for both cases, it is possible to determine the Influence Diagram for both. Unfortunately, since we are dealing only with variables without magnitude, it is not possible to graph such diagrams unless we have actual values for the variables.

Sample Problem:

Example of Point Load and Uniform Load on a Circular Three-Hinged Arch:

Figure 3.1: Point Load and Uniform Load on Three-Hinged Arch Example

  • Step 1: The reactions on the reference beam shown are determined using equilibrium equations:
The moment at each point is calculated by multiplying each force by the perpendicular distance from that point to the line of action of the force.
  • Step 2:
Solve for the vertical reactions by calculating the moment at hinges B and A to get Ay and By:
Calculate the Moment about hinge B:
$ \Sigma M_B=0 $
$ -A_y(32)+P_1(24)+q(8)(12)+P_2(4)=0 $
$ A_y=\frac{(16)(24)+(3)(8)(12)+(12)(4)}{32}=22.5 KN $
Moment about hinge A:
$ \Sigma M_A=0 $
$ B_y(32)-P_1(8)-q(8)(20)-P_2(28)=0 $
$ B_y=\frac{(-16)(8)-(3)(8)(20)-(12)(28)}{32}=29.5 KN $
  • Step 3:
Moment about hinge C:
We calculate the moment of hinge C from the right side of point C to get the positive moment value.
$ \Sigma M _C=0; -q(8)(4)-P_2(12)+B_y(16)=0 $
$ M_C=(-3)(8)(4)-(12)(12)+(29.5)(16)=232 KN.m $
The shear and moment diagrams of the calculated reactions are shown in the figure, at point C ($ x = 16m $), the bending moment is $ M_C=232 KN.m $
From the diagram shown, the shear diagram started with a value of 22.5 as we have a vertical force acting on hinge A, $ A_y = 22.5 kN $, and then went down to 6.5 since we have an opposite point load force $ P_1=16 KN $, the diagram then goes down to a negative value of 17.5 caused by the uniform load $ q=3 KN/m $ and so on.
The moment diagram was drawn by calculating the moment for each load using the area of the shear diagram under each load (depending on the shape of the shear diagram between each load).
The moment at (P1 = length * width) since we have a rectangle shape in the shear diagram

*$ M_P_1=22.5 KN*8 m=180 KN.m $
*$ M_C=180+(6.5 KN*8 m)=232 KN.m $
  • Step 4: Adjust the Shear and Moment Diagrams from the reference beam to the three-hinged arch.
In order to simplify this example, the arch has been divided into 4m sections in the x-direction along with its respective y-direction value.
The following table has been calculated using the formulas described in the "Analysis of a point load on a Three-hinged arch" section. Please refer to it for more details.
Figure 3.2: Table showing the actual values for the Shear and Bending Moment diagram for a Three-Hinged Arch subjected to two point loads and an uniform load
Sample Calculations for Section 3:
$ H = M_c/h $ $ R=(h/2)+(L^2/(8*h)) $ $ Sin \varphi = L-2x/2R $
$ H = 232/8 $ $ R=(8/2)+(32^2/(8*8)) $ $ Sin \varphi = 32-2(12)/2(20) $
$ H = 29 KN $ $ R= 20m $ $ Sin \varphi = 0.2 $

$ Cos \varphi = (y+R-h)/R $ $ H_y = H*y $ $ M_x = M_(x,0)-H_y $
$ Cos \varphi = (7.596+20-8)/20 $ $ H_y = 29*7.596 $ $ M_x = 206-220.284 $
$ Cos \varphi = 0.9798 $ $ H_y = 220.284 KNm $ $ M_x = -14.284 KNm $
Where $ M_(x,0) $ is the moment in the Ref. Beam at that section.

$ V_x = V_(x,0)-(H*Sin \varphi) $
$ V_x = 6.5 -(29*0.2) $
$ V_x = 0.5687 KN $
Where $ V_(x,0) $ is the shear in the Ref. Beam at that section.

  • Step 5: Once we have obtained the actual values for the shear and bending moment diagram we can re-draw them, as seen in Fig. 3.3.
Figure 3.3: Actual Shear and Bending Moment for a Three-Hinged Arch


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