2. THREE HINGED CIRCULAR ARCH
Theory of Structures 1
Submitted by: FAIZAN ALI
06-CE-10

3. CONTENTS HISTORY. ARCH. USE. TECHNICAL ASPECTS. HINGE INTRODUCTION.
CLASSIFICATION.
INTRODUCTION OF THREE HINGED CIRCULAR ARCH.
DERIVATIONS AND EXAMPLES.
CURRENT USE.

4. 1. HISTORY (contd.)
Arches were used by the Persian, Harappan, Egyptian, Babylonian, Greek and Assyrian civilizations for underground structures such as drains and vaults, but the ancient Romans were the first to use them widely above ground although it is thought that Romans learned it from the Etruscans. The arch has been used in some bridges in China since the Sui dynasty and in tombs since the Han Dynasty.

5. 1. HISTORY (contd.)
ARCH
A structure, usually curved, that when subjected to vertical loads causes its two end supports to develop reactions with inwardly directed horizontal components.
Parts:
The designations of the various parts of an arch are given in the illustration.

6. 1. HISTORY (contd.)
Use:
The commonest uses for an arch are as a bridge, supporting a roadway, railroad track, or footpath, and as part of a building, where it provides a large open space unobstructed by columns. Arches are usually built of steel, reinforced concrete, or timber.
Technical Aspects:
The arch is significant because, in theory at least, it provides a structure which eliminates tensile stresses in spanning an open space. All the forces are resolved into compressive stresses.
This is useful because several of the available building materials such as stone, cast iron and concrete can strongly resist compression but are very weak when tension, shear or torsional stress is applied to them. By using the arch configuration, significant spans can be achieved.

7. 1. HISTORY (contd.) Hinge Introduction:
Two unknowns. The reaction are two components of Force, or the magnitude and direction Ф of the resultant force .
At the hinge joint moment is zero means it can’t resist the bending moment produce by external force.

8. Classification a. Hinge less Arch:
The hinge-less arch uses no hinges and allows no rotation at the foundations.
b. Two Hinged Arch:
The two hinged arch uses hinged bearings which allow rotation. The only forces generated at the bearings are horizontal and vertical forces.

9. Classification (contd.)
c. Tree Hinged Arch:
The three-hinged arch adds an additional hinge at the top or crown of the arch. The three-hinged arch suffers very little if there is movement in either foundation (due to earthquakes, sinking, etc.)
d. Tied Arch:
The tied arch is a variation on the arch which allows construction even if the ground is not solid enough to deal with the horizontal forces.

10. 2. THREE HINGED CIRCULAR ARCH
A three hinged system consist of two plates, connected together by means of a hinge with two hinged supports A and B resting on the ground.
When the plates 1 and 2 consist of curved bars, the system is called a three- hinged arch.
The distance l between the centers of the hinges at the support is called the span of arch.
While distance f from the center of the crown hinge to the straight line passing through the former two is called its rise.

11. 2. THREE HINGED CIRCULAR ARCH (contd.)
A three hinged system may or may not have a vertical axis of symmetry. In the first case the central hinge c will lie on the axis of symmetry and the hinges at the support A and B are at same level.
Non symmetrical systems may have their supports at different level.

12. 2. THREE HINGED CIRCULAR ARCH (contd.)
Determinacy:
The reactions of a three hinged arch will be fully determined by four parameters, for instance, the amounts of reaction Ha, Hb, Va, and Vb.
Thus a three hinged system is always statically determinate

13. 2. THREE HINGED CIRCULAR ARCH (contd.)
Advantages:
One of the advantages of three hinged arch is that the bending moments and shears acting over cross sections of three hinged arches are considerably smaller than the corresponding stresses in a simple beam covering the same span and carrying the same load. Therefore three hinged arches are more economical than ordinary beams, particularly for large span structures.
Calculations are easier than other type of arches.
No bending moments are caused at the abutments and the crown because hinges cannot resist moments.

14. 2. THREE HINGED CIRCULAR ARCH (contd.)
Differential settlements of the supports do not appreciably affect stresses, since the pines or hinges enable the arch to take up the slightly different shape consequent upon settlement.
The pin joints enable the arch to adjust itself to expansions and contractions due to changes in temperature. 
A disadvantage is that bending moments away from the pins are larger than in the 2-pin and completely rigid arches.

15. 3. DERIVATIONS AND EXAMPLES.
Supports Reactions of a three-hinged Arch:
Analytical Method.
Graphical Method.

16. 3. DERIVATIONS AND EXAMPLES. (contd.)
Analytical Method:
When a system of vertical load is applied to a three hinged, a vertical and horizontal reactions will arise at each of two supports making four reactions to be determined at all. In addition of three equations of equilibrium, a fourth equation can be used in the case of a three hinged arch; this equation demonstrates that the bending moment at the hinge C equals zero both right or to the left of this hinge.
Left ξ Mc = 0 or right ξ Mc = 0.

17. Analytical Method. General Steps:
In the case of an ordinary arch as shown above, we may first write the equilibrium equation for the moments of all forces about hinge B which will contain only one vertical reaction Va.
Then we will use the equation left ξ Mc = 0. this equation may contain the reactions Va which has been just determined and the unknown reaction Ha.
Then we may proceed with the solution of an equation demonstrating that the moment of all the external forces about hinge A is zero which will give us the value of reaction Vb.
Then we obtain the magnitude of Hb by equating to zero the projection of all the external forces on the horizontal.
The computation just described may be checked using the equations
ξ Y = 0 and ξ Mc = 0.

18. Analytical Method. (contd.)
DERIVTION of REACTIONS.
The last formula shows that the; Thrust arising at both supports of three hinged symmetrical arches subjected to vertical loads are equal in magnitude and opposite in direction.

19. Analytical Method (contd.)
Calculation of Reactions when a UDL is applied.
Va = Vb = ½ q*l
Ha = Hb =q*l/4

20. Graphical Method Steps:
The graphical determination of the reactions requires that the resultants R1 and R2 of all the forces applied to the left and to the right of the central hinge should be found in the first place. The reactions induced by each of these resultants R1 and R2 will then are determined, their summation giving the final value of the reactions required.
Consider the following arch:

21. Graphical Method (contd.)
We may start with determining the reactions at the support caused by the application of the force R1. In this case the reaction at the right hand support B1 must pass through the hinge at this support and the hinge C at the crown as otherwise the right hand portion if the arch which is subjected solely to the reaction at B1 and the interaction of hinge C could not remain in equilibrium.
with reaction A1 arising at the left hand support, the arch as a whole will be in equilibrium under the action of three forces A1, B1, R1.

22. Graphical Method (contd.)
Theoretical Mechanics states that three coplanar forces acting on a body in equilibrium must concur at one and the same point.
The use of this theorem enables us to find the direction of reaction A1 after which the force polygon will give us the magnitude of both the support reactions A1 and B1.
The support Reactions A2 and B2 due to the application of right hand resultant R2 will be found in exactly the same way.

23. Graphical Method (contd.)
Some Examples:

24. 3. DERIVATIONS AND EXAMPLES
Determination of stresses in Three Hinged Circular Arches:
Analytical method is used here to calculate the internal forces (i.e; Axial force, shear force and bending moment) at a cross- section of a arch.

25. Analytical method Case 1:
When a semicircular three hinged arch is subjected to two loads on one side of the crown hinge.
First of all calculate the reactions by the method described above.
NORMAL FORCE: the normal force at x is obtained by resolving the forces to one side of the x in a direction tangential to arch at x .
Nx = -Va cosβ – Ha sinβ + P1 cosβ

26. Analytical method (contd.)
SHEAR FORCE: The shear at x can be found by resolving the forces to one side of the x in a direction perpendicular to the tangent at x. We shall take the positive shear force as acting radially inwards when it is to the left of the section.
Sx = - Va sinβ + Ha cosβ + P1 sinβ
BENDING MOMENT: Now taking moments about x for forces to the left of x and regarding a positive moment as causing tension on the outer side of the arch, we have;
Mx = Va( r – rcosβ ) – Ha*r*sinβ – P1( rcosą – rcosβ)

27. Analytical method (contd.)
Case 2:
To calculate the internal forces when the portion of a three hinged circular arch is subjected to a load.
First of all calculate all the reactions by the method described above.
Since the structure is symmetrical, we need to only analyze one half.
Consider the section of AC shown in the figure. Note that due to circular form, it is easiest to work in terms of the angle ą measured from the center, rather the distance along the arch.

28. Analytical method (contd.)
Resolve vertically:
S siną + T cosą + Va = 0
Resolve horizontally:
T siną + S cosą + Ha = 0
Solving simultaneously:
S = - Va siną + Ha cosą,
T = - Va cosą + Ha siną
Now the moment at the section x is given by:
M + Ha(r siną – r/2) – Va(l – rcosą ) = 0
This solution is only valid for the whole of AC.

29. Real Life Examples
To calculate the reactions of a three Hinged Circular Arch.
1. The Ayub arch, Pakistan.
Built in: 1962
Location: Sukkur-Rohri, Pakistan.
Type: Two hinged Circular arch

30. Example 1 Calculation of reactions: Take moment about A = 0:
ξ Ma = - 60*20 -40*70 + Vb*100 = 0
Vb*100 =
Vb = 4000/100 = 40N
Now take moment about B = 0:
ξ Mb = -Va* * *30 = 0
Va*100 =
Va = 6000/100 = 60N
Now take moment about C to its left equal to zero
ξ LM c = 60*30 -60*50 + Ha*30 = 0
Ha*30 = 3000 – 1800
Ha = 1200/30 = 40N
Now take the moment about C to its right equals to zero:
ξ RM c = -40*20 +40*50 –Hb*30 = 0
Hb*30 = 1200
Hb = 40 N

31. Example 2 The Iron Bridge, England: Carries: pedestrian traffic
Crosses: River Severn
Locale: Iron bridge Gorge near Coalbrookdale
Design: cast iron arch bridge
Longest span: 30.5 meters (100 ft)
Total length: 60 meters (200 ft)
Clearance below: 60 feet (18 m)
Beginning date of construction: 1775
Completion date: 1779
Opening date:

32. Example 2 (contd.) calculation of reactions: ξ MB = 0
- Ay* *100 = 0
Ay = /200 = 3000 N
ξ Fy = 0
Ay +By = 6000
By = 3000 N
Now in this case the reaction: Ax = Bx = H
So by taking sum of all the moments acting to the left of hinge C equal to zero, we get:
ξ L Mc = 0
Ay* l/2 – H*60 – 30*100*50 = 0
H*60 = 3000*100 –
H*60 = –
H = 2500 N

33. Example 2 (contd.) Calculation of Internal Forces at a section: where:
P1 = 6000 N, l = 100 ft , r = 120ft , Ф = 118 °
Ha = Hb = H = 2500 N
Va = Vb = 3000N
1. Normal force at x: Nx:
We will resolve the forces to the right of X in a direction tangential to the arch at x.
Considering tensile forces to be positive.
Nx = cos sin cos 60 =
= N

34. Example 2 (contd.) 2. Shear force at x: Sx:
The shear at x can be found by resolving the forces to one side of the x in a direction perpendicular to the tangent at x. We shall take the positive shear force as acting radially inwards when it is to the left of the section.
Sx = sin cos sin60 =
= N
3. Bending moment at x: Mx:
Now taking moments about x for forces to the right of x and regarding a positive moment as causing tension on the outer side of the arch, we have;
Mx = 3000(100 – 120 cos 60) – 2500 * 120* sin 60 – (120 cos 39 – 120 cos 60)
= – –
= N
= KN.ft = kN.m

35. Example 3 3. Salginatobel Bridge.
Salginatobel Bridge is a reinforced concrete arch bridge designed by renowned Swiss civil engineer Robert Maillart.
It was constructed across an alpine valley in Schiers, Switzerland between 1929 and 1930.
Technical Profile
Construction type: Three-hinged arch, developed as hollow-box girder
Construction material: Reinforced concrete
Total length: m
Width of roadway: 3.50 m
Slope of roadway: 3% or 3.97 m
Span of arch: m
Arch rise: m
Load bearing capacity: 8 t or 350 kg/m2
Height above water: over 90 m
Design Engineer: Robert Maillart, Geneva
Construction time: 1929/30

36. Example 3 (contd.) Calculation of reactions: ξ MB = 0
Va*90 – 308.7*45 = 0
Va = /90 = kN
ξ MA = 0
-Vb* *45 = 0
Va = kN
Now:
Ha = Hb = H
ξ L Mc = 0
-Va* 45 + H*13 = 0
H = /13 = kN

37. Example 3 (contd.) Calculation of internal forces at section x:
Resolve vertically all the forces:
S sin60 + T cos60 =
Resolve horizontally all the forces acting at section x:
T sin60 – S cos60 =
Now solving simultaneously, we get:
S = *sin *cos60
S = = kN
T = *cos60 – *sin60
T = – = kN
Now Bending moment at section x
M (52 sin60 – 39) – (45 – 52 cos 60) = 0
M = = kN.m
(this solution is valid for when 49° < Ф < 90°)

38. Example 4 4. The Transverse Arch: Location: Mahdia Mosque. Iraq.
Built in: 8th Century.

39. Example 4 (contd.) Calculation of Reactions: ξ MA = 0
- 10(5 – 5 cos45) – 10(5 + 5 cos 45) + Vb*10 = 0
Vb = ( )/10
Vb = 99.99/10 = 9.9 or 10N
ξ Fy = 0
Va – 10 – = 0
Va = 10N.
ξ L Mc = 0
-10*5 + Ha*5 + 10*3.535 = 0
Ha = /5 = 3N
And Hb = 3N

40. Example 4 (contd.) Calculation of internal forces: 1. At section x1:
Normal force = Nx1 = - 10 sin cos15 =
= -10.4N (comp)
Shear force = Sx1 = -10 sin cos 15 =
= 0.309N
Moment = Mx1 = 10(5-5 cos15) – 3*5*sin15 =
= N.m
2. At section x2:
Nx2 = - 10 cos30 – 3 sin30
= – 1.5
= N (comp)
Sx2 = -10 sin cos30
= = -2.4N
Mx2 = 10(5 -5 cos30) – 3*5*sin30
= – 7.5 = N.m

41. Example 4 (contd.) 2.At section x3:
Nx3 = -10 cos45 – 3 sin cos45
= N
Sx3 = - 10 sin cos sin45
= 2.12N
Mx3 = 10(5 – 5 cos45) – 3*5*sin45
– 10(5 cos45 – 5 cos45)
= – = 4N.m.
4. At section x4:
Nx4 = -10 cos60 – 3 sin cos60
= or -2.6N
Sx4 = - 10 sin cos sin60
= 1.5N
Mx4 = 10(5 - 5 cos60) – 3*5*sin60
– 10( 5 cos45 – 5 cos60)
= 25 – –
= 1.704N.m

42. Example 4 (contd.) 5. At section x5:
Nx5 = - 10 cos75 – 3 sin cos75
= N
Sx5 = - 10 sin cos sin75
= 0.776N
Mx5 = 10(5 - 5 cos75) – 3*5*sin75
– 10( 5 cos45 – 5 cos75)
= – – 22.41
= 0.16N.m
6. At section x6:
Nx6 = -10 cos90 – 3 sin cos90
= - 3N
Sx6 = - 10 sin cos sin90
= 0N
Mx6 = 10(5 - 5 cos90) – 3*5*sin90
– 10( 5 cos45 – 5 cos90)
= 50 – 15 – 10(3.5)
= 0N.m

43. Example 4 (contd.)
Note: Since the arch is symmetrical so we will analyse only one half of the arch.
Section
Ф , °
X ,m
Y ,m
N(NF), N
S(SF) ,N
M ,N.m X1 15
4.82
1.3
-10.4
0.31
-2.18 X2 30
4.33
2.5
-10.16
-2.4
-0.8 X3 45
3.54
-2.12
2.12
4.03 X4 60
-2.6
1.5
1.704 X5 75
-2.89
0.776
0.16 X6 90 5 -3

44. Example 4 (contd.)

45. Current Use
Now the three hinged arch bridges are rarely used. In the past we find very few examples of this type of bridges, of which I found out the following:
1. The Iron Bridge, England:
Design: cast iron arch bridge
Completion date: 1779
2. Salginatobel Bridge.
Construction type: Three-hinged arch, developed as hollow-box girder
Construction time: 1929/30
Reasons:
Following are the reasons due to which this type of bridges is not used nowadays:
1. The three-hinged arch experiences much more deflection and the hinges are complex and can be difficult to fabricate. The three-hinged arch is rarely used anymore.
2. Bending moments away from the pins are larger than in the 2-pin and completely rigid arches.

46. REFERENCES: Book: STRUCTURAL MECHANICS
Edited by: PROFESSOR A. DARKOV D. SC. (ENG.)
Book: THE ELEMENTS OF STRUCTURES
By W. MORGANS
Book: STRUCTURES (Theory and Analysis)
By: M S Williams and T J Todd.
Book: STRUCTURES AND STRESS ANALYSIS
By: T H G Megson.
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