1. CED, UET, Taxila 1 Arches Theory of Structure - I

2. CED, UET, Taxila 2 Contents Introduction to Three Hinged Arches General Proof Parabolic Arch Circular Arch Practice Problems

3. CED, UET, Taxila 3 Introduction Arch is a curved structure or humped beam, primarily bears the applied loads by compression. The hinge introduced anywhere in the arch makes the structure determinate as Both supports are assumed to be hinged Hinge introduced in the arch provides a further equation to analyze the arch i.e., moment of all forces about hinge is equal to zero.

4. CED, UET, Taxila 4 Introduction Normally the third hinge is introduced at the top most point on the arch curve known as crown. y c = Rise Hinge at Crown l = Span

5. CED, UET, Taxila 5 Introduction One can solve the arch as beam if we know the horizontal reactions at various supports. With the help of third hinge, we can easily determine the horizontal reactions and hence the arch can be analyzed.

6. CED, UET, Taxila 6 Arch Formula’s Consider an arch ACB, hinged at A, B and C. l is the horizontal span and y c is central rise. y c = Rise Hinge at Crown l = Span AB C

7. CED, UET, Taxila 7 Arch Formula’s Obtain the vertical reactions V A and V B at the ends as usual. To find the horizontal thrust, MC the moment at central hinge must be zero.

8. CED, UET, Taxila 8 General Derivation To find H the horizontal thrust, M c the moment at central hinge C must be known. We can find moment at any cross-section X of the arch whose coordinates are (x, y).

9. CED, UET, Taxila 9 General Derivation The vertical and horizontal actions on the section, considering the portion AX arc, Vertical reaction can be find out by vertical shear force at the section as for a straight horizontal beam. Horizontal thrust at both ends is same.

10. CED, UET, Taxila 10 General Derivation ycyc X y W1 W2  A C H V P H  H cos  H sin  F  Vcos  V sin  V

11. CED, UET, Taxila 11 General Derivation Draw the tangent at X to the centre-line of the arch and let its inclination to the horizontal must be . Resolving V and H normally to the section and tangentially, ( i.e., along the tangent at X) If the resultant T is required, use eq.5

12. CED, UET, Taxila 12 Parabolic Arch If the three hinged arch is parabolic in shape and if it carries a uniformly distributed load over the entire span, every action of the arch will be purely in compression, resisting only a normal thrust; there will be no shear force nor B. M. at the section.

13. CED, UET, Taxila 13 Parabolic Arch The linear arch for a given load system on an arch represents the  -diagram. With a uniformly distributed load over the entire span, the  -diagram is a parabola. The linear arch which is parabolic will, then, have three points (at the hinges) in common with the centre-line of the actual arch, which is also parabolic.

14. CED, UET, Taxila 14 Parabolic Arch The linear arch will therefore be identical with the actual arch. For any other loading on a parabolic arch, there will be three straining actions, P, F and M at any section. To obtain the bending moment, it will be necessary to calculate the rise at any section of the arch.

15. CED, UET, Taxila 15 Parabolic Arch It may also be noted that at quarter points, where x=l/4, the rise = ¾ y c

16. CED, UET, Taxila 16 Circular Arch A B C X y ycyc   D Y O x

17. CED, UET, Taxila 17 Circular Arch If the centre-line of the arch is a segment of a circle of radius R, it is more convenient to have the origin at D, the middle of the span. Let (x, y) be the coordinates of a section.

18. CED, UET, Taxila 18 Circular Arch

19. CED, UET, Taxila 19 Example 01 A three hinged Parabolic arch, at the crown and springings has a horizontal span of 48ft. It carries UDL of.75 Ton/ft run over the left hand half of the span. Rise = 10’. Calculate the Reactions Normal Thrust Shear force and BMat 6, 12, 30ft from left hinge.

20. CED, UET, Taxila 20 y c = 10ft l = 48ft 0.75 ton/ft A C B

21. CED, UET, Taxila 21 Solution V A = 13.5 T; V B = 4.5 T; H = 10.8 T y = 5/288*x(48-x),  = 5/288*(48-2x) xy  cos  sin  MVPF 64.37532 o 0.8480.529-20.25913.931.91 127.522 o 37’0.9230.348-274.511.70 309.6611 o 46’0.9790.20320.254.511.492.2

22. CED, UET, Taxila 22 Example 02 Circular arch of span 80 ft with central rise 16 ft is hinged at crown and supports. Carries a point load of 10 tons 20ft from left support. Reactions Normal Thrust Maximum and Minimum BM

23. CED, UET, Taxila 23 Solution V A = 7.5 T; V B = 2.5 T; H = 6.25 T y c (2R-y c )=l 2 /4, R 2 =x 2 +{y+(R-y c )} 2

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33. CED, UET, Taxila 33 Practice Problems Mechanics of Structures By S. B. Junnarkar Chapter 5, Examples 1 to 4.

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