Muller-Breslau’s Principle, Two hinged and Fixed Arches and Stiffening Girders

Introduction- Muller Breslau principle , Arches i.e. Two hinged, fixed hinged arches and Stiffening Girder.

Explanation of procedure for all methods.

Illustrate some examples to understand the concept better

1. Noida Institute of Engineering and Technology, Greater Noida Muller-Breslau’s Principle, Two hinged and Fixed Arches , and Stiffening Girders Aayushi Assistant Professor Civil Engg . Department 5/18/2020 1 Unit: 2 Aayushi RCE-502, DOS 1 Unit 2 Subject Name : Design of Structure I Course Details : B Tech 5 th Sem

2. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 2 Content Course Objective Course O utcome CO-PO & PSO Mapping Prerequisite & Recap Muller Breslau Principle Procedure for Muller Breslau principle Numerical & Deflected shapes Arches Types of arches Two hinged arches Numerical on two hinged arches Syllabus of unit 2 Topic outcome and mapping with PO

3. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 3 Content You tube Video Links Daily Quiz Weekly Assignment MCQs Old Question Papers Expected Questions in University Examination Summary References Stiffening girder Fixed Arches & numerical Introduction &Numerical

4.   Objective 1 To impart the principles of elastic structural analysis and behavior of indeterminate structures. 2 To impart knowledge about various methods involved in the analysis of indeterminate structures. .. 3 To apply these methods for analyzing the indeterminate structures to evaluate the response of structures . 4 To enable the student get a feeling of how real-life structures behave 5 To make the student familiar with latest computational techniques and software used for structural analysis. . 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 4 Course Objective

5.   Students will be able CO1 To Identify and analyze the moment distribution in beams and frames by Slope Deflection Method, Moment Distribution Method and Strain Energy Method. CO2 To provide adequate learning of indeterminate structures with Muller’s Principle; Apply & Analyze the concept of influence lines for deciding the critical forces and sections while designing.. CO3 To learn about suspension bridge, two and three hinged stiffening girders and their influence line diagram external loading and analyze the same. CO4 To Identify and analyze forces and displacement matrix for various structural. CO5 To understand the collapse load in the building and plastic moment formed. CO6 Apply the concepts of forces and displacements to solve indeterminate structure. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 5 Course Outcome

6. CO PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 RCE502.1 3 3 1 2 1 - - - 2 - - 3 RCE502.2 3 2 2 2 3 - - 1 2 - 2 2 RCE502.3 3 - 2 2 1 - - - 2 - 2 2 RCE502.4 3 1 2 2 1 - - - 2 - 2 2 RCE502.5 3 2 2 2 1 - - - 2 - 2 2 RCE502.6 3 2 3 2 1 - - - 2 - 2 3 AVG. 3 2 2 2 1 - - - 2 - 2 2 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 6 CO-PO and PSO Mapping COs PSO1 PSO2 PSO3 PSO4 RCE502.1 2 1 2 3 RCE502.2 2 1 2 2 RCE502.3 2 2 2 2 RCE502.4 2 1 2 3 RCE502.5 2 1 2 3 RCE502.6 2 1 2 3 Avg. 2 1 2 3

7. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 7 Prerequisite and Recap Basics of strength of material Basics of engineering mechanics

8. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 8 Syllabus of Unit 2 Muller-Breslau’s Principle and its applications for drawing influence lines for indeterminate beams, Analysis of two hinged and fixed arches, Influence line diagrams for maximum bending moment, Shear force and thrust in two hinge arches. Analysis of two and three hinged stiffening girders.

9. 5/18/2020 9 Objective of Unit 2 The Muller-Breslau principle for influence lines for BM and Shear force in beams and truss.Derivation of the principle for different types of internal forces . Aayushi RCE-502, DOS 1 Unit 2

10. The Müller-Breslau principle for influence lines . Derivation of the principle for different types of internal forces . Example of application of this principle . 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 10 Topic Objective

11. To Identify the moment distribution in beams and Draw ILD for different sections. 5/18/2020 11 Topic Outcomes Once the student has successfully completed this unit, he/she will be able to: To analyze different beams and frames. Apply the concepts of forces and displacements to solve indeterminate structure . Aayushi RCE-502, DOS 1 Unit 2

12. 5/18/2020 12 Objective of Topic Topic-1 Name The Muller-Breslau principle Objective of Topic-1: To Construct influence line diagram (ILD) for BM and SF in beams and truss. Aayushi RCE-502, DOS 1 Unit 2

13. 5/18/2020 13 Topic Outcome and mapping with PO Programme Outcomes (POs) PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 TO1 1 1 1 1 1 1 1 1 1 Outcome of Topic-1: After the successfully competition of this topic students will be able t o Construct influence line diagram (ILD) for BM and SF in beams and truss. Aayushi RCE-502, DOS 1 Unit 2

14. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 14 Prerequisite and Recap Basic concept of Influence line diagram Influence line Diagram. Understanding of bending moment and shear force . Understanding of forces.

15. Heinrich Franz Bernhard Müller was born in Wroclaw (Breslau) on 13 May1851 . In 1875 he opened a civil engineer‘s office in Berlin. Around this time he decided to add the name of his hometown to his surname, becoming known as Muller-Breslau . In 1883 Muller-Breslau became a lecturer and in 1885 a professor in civil engineering at the Technische Hochschule in Hanover . In 1886, Heinrich Müller-Breslau develop a method for rapidly constructing the shape of an influence line . 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 15 T opic : History

16. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 16 Topic: Muller-Breslau principle The influence line for a function (reaction, shear, moment ) is to the same scale as the deflected shape of the beam when the beam is acted on by the function. To draw the deflected shape properly, the ability of the beam to resist the applied function must be removed .

17. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 17 Topic: Procedure for Muller-Breslau’s Step-1 : T o draw ILD for any support remove that support. Step-2 : Apply unit load at that support. Step-3 : Draw bending moment diagram for that support. Step-4 : Construct conjugate beam. Step-5 : Find deflection at some specified intervals ( for a conjugate beam, deflection at any point = BM at that point ). Step-6 : Divide each deflection by deflection corresponding to the point of application of unit load. Step-7 : We obtain the ordinates for the influence for that particular support.

18. Step-8 : For other support repeat the same procedure. Step-9 : For ILD of bending moment, construct a static equations from the beam and substitute values at different interval and you will get ordinates of ILD for BMD. Step-10 : F or shear force diagram construct static equations and solve. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 18 Continuous…

19. Consider the following simply supported beam. Let’s try to find the shape of the influence line for the vertical reaction at A . Remove the ability to resist movement in the vertical direction at A by using a guided roller 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 19 Topic: Numerical

20. Remove the ability to resist movement in the vertical direction at A by using a guided roller 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 20 Continuous…

21. Consider the following simply supported beam . Let’s try to find the shape of the influence line for the shear at the mid-point (point C ). Remove the ability to resist shear at point C 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 21 Continuous…

22. Consider the following simply supported beam . Let’s try to find the shape of the influence line for the moment at the mid-point (point C ). Remove the ability to resist moment at C by using a hinge 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 22 Continuous…

23. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 23 Topic: Explanation The Muller-Breslau principle uses Betti's law of virtual work to construct influence lines. To illustrate the method let us consider a structure AB (Figure a ). Let us apply a unit downward force at a distance x from A , at point C . Let us assume that it creates the vertical reactions R A and R B at supports A and B , respectively (Figure b). Let us call this condition “System 1 .” In “System 2” (figure c), we have the same structure with a unit deflection applied in the direction of R A . Here Δ is the deflection at point C .

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25. According to Betti's law, the virtual work done by the forces in System 1 going through the Corresponding displacements in System 2 should be equal to the virtual work done by the forces in System 2 going through the corresponding displacements in System 1. For these two systems, we can write: The right side of this equation is zero, because in System 2 forces can exist only at the supports, corresponding to which the displacements in System 1 (at supports A and B ) are zero. The negative sign before Δ accounts for the fact that it acts against the unit load in System 1. Solving this equation we get: R A = Δ. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 25 Continuous… (R A )(1) + (1)(- Δ) =0

26. In other words, the reaction at support A due to a unit load at point C is equal to the displacement at point C when the structure is subjected to a unit displacement corresponding to the positive direction of support reaction at A . Similarly, we can place the unit load at any other point and obtain the support reaction due to that from System 2. Thus the deflection pattern in System 2 represents the influence line for R A . 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 26 Continuous…

27. ILD FOR SIMPLY SUPPORTED BEAM: 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 27 Topic: Deflected Shape

28. ILD FOR TWO SPAN CONTINOUS BEAM: 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 28 Continuous…

29. ILD FOR THREE SPAN CONTINOUS BEAM : 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 29 Continuous…

30. In this section, we looked at main applications for influence lines. The first is the use of an influence line to determine the influence of a single point load . The second is the use of an influence line to determine the effect of a distributed load or patterned distributed load . The last is the use of an influence line to determine the effect of a moving pattern of loads. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 30 Summary of Muller- Breslau

31. Compute horizontal reaction in two-hinged arch by the method of least work Write strain energy stored in two-hinged arch during deformation. Analyse two-hinged arch for external loading. Compute reactions developed in two hinged arch due to temperature loading. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 31 Topic Objective

32. To Identify the horizontal thrust and forces in the arches. 5/18/2020 32 Topic Outcomes Once the student has successfully completed this unit, he/she will be able to: To analyze different Arches . Apply the concepts of forces and displacements to solve indeterminate structure . Aayushi RCE-502, DOS 1 Unit 2

33. 5/18/2020 33 Objective of Topic Topic-2 Name Arches Objective of Topic-2: To analysis the strain energy stored in the two hinged arch during deformation. Aayushi RCE-502, DOS 1 Unit 2

34. 5/18/2020 34 Topic Outcome and mapping with PO Programme Outcomes (POs) PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 TO1 1 1 1 1 1 1 1 1 Outcome of Topic-1: After the successfully competition of this topic students will be able to analysis the strain energy stored in the two hinged arch during deformation. Aayushi RCE-502, DOS 1 Unit 2

35. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 35 Prerequisite and Recap Briefing about Arches Basic of force system Basic of bending moment and shear force

36. Arches and arched structures have a wide range of uses in bridges, arched dams and in industrial, commercial, and recreational buildings . They represent the primary structural components of important and expensive structures, many of which are unique . Current trends in architecture heavily rely on arched building components due to their strengths and architectural appeal. Complex structural analysis of arches is related to the analysis of the arches strength , stability, and vibration. This type of multidimensional analysis aims at ensuring the proper functionality of an arch as one of the fundamental structural elements . 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 36 Topic: Arches

37. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 37 Topic: Types of Arches Hinge less arch two-hinged arch one-hinged arch three-hinged arch

38. 1. Material of the arch obeys Hooke’s law (physically linear statement) 2. Deflections of the arches are small compared with the span of the arch ( geometrically linear statement). The cases of nonlinear statement are specifically mentioned . 3. All constraints, which are introduced into the arched structure are two-sided, i.e ., each constraint prevents displacements in two directions. The case of one-sided constraints is specifically mentioned. 4. In the case of elastic supports the relationship between deflection of constraint and corresponding reaction is linear. 5. The load is applied in the longitudinal plane of symmetry of the arch. The case of out-of-plane loading is specifically mentioned. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 38 Topic: Assumptions

39. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 39 Topic: Shape of the Arches Circular arch : Ordinate y of any point of the central line of the circular arch is calculated by the formula Where:- x is the abscissa of the same point of the central line of the arch; R is the radius of curvature of the arch; f and l are the rise and span of the arch .

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41. Analysis of two-hinged arch 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 41 Topic: Two hinged Arch

42. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 42 Continuous… The strain energy due to bending The total strain energy of the arch is given by, according to the principle of least work ,where H is chosen as the redundant reaction

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45. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 45 Continuous… Temperature effect

46. A semicircular two hinged arch of constant cross section is subjected to a concentrated load as shown. Calculate reactions of the arch and draw bending moment diagram. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 46 Topic: Numerical

47. Taking moment of all forces about hinge B leads to , 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 47 Continuous…

48. Now, the horizontal reaction H may be calculated by the following expression 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 48 Continuous…

49. Now the bending moment at any cross section of the arch when one of the hinges is replaced by a roller support is given by , 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 49 Continuous… Integrating the numerator in equation

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51. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 51 Continuous… The value of denominator in equation, after integration is, Hence, the horizontal thrust at the support is, Bending moment diagram Bending moment M at any cross section of the arch is given by,

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53. The deflection and the moment at the center of the hinge less arch are somewhat smaller than that of the two-hinged arch. However, the hinge less arch has to be designed for support moment . A hinge less arch (fixed–fixed arch) is a statically redundant structure having three redundant reactions. In the case of fixed–fixed arch there are six reaction components; three at each fixed end. Apart from three equilibrium equations three more equations are required to calculate bending moment, shear force and horizontal thrust at any cross section of the arch. These three extra equations may be set up from the geometry deformation of the arch. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 53 Topic : Fixed Arch

54. Analysis of Symmetrical Hinge less Arch 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 54 Continuous… T he strain energy due to axial compression and bending

55. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 55 Continuous… C onsider bending moment and the axial force at the crown as the redundant.

56. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 56 Continuous… Temperature stresses T he moment at any cross-section of the arch

57. S train energy stored in the arch 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 57 Continuous… Solving equations , M t and H t and may be calculated

58. A semicircular fixed-fixed arch of constant cross section is subjected to symmetrical concentrated load as shown .Determine the reactions of the arch. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 58 Continuous…

59. Since, the arch is symmetrical and the loading is also symmetrical , Now the strain energy of the arch is given by, 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 59 Continuous… The bending moment at any cross section is given by

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61. Since the arch is symmetrical, integration need to be carried out between limits 0 to π/2 and the result is multiplied by two. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 61 Continuous…

62. Solving equations 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 62 Continuous…

63. Two-hinged arch is the statically indeterminate structure to degree one. Usually , the horizontal reaction is treated as the redundant and is evaluated by the Kharagpur method of least work. Towards this end, the strain energy stored in the two hinged arch during deformation is given. The reactions developed due to thermal loadings are discussed. Finally , a few numerical examples are solved to illustrate the procedure. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 63 Summary of Arches

64. Differentiate between rigid and deformable structures. Define funicular structure. State the type stress in a cable. Analyse cables subjected to uniformly distributed load. Analyse cables subjected to concentrated loads. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 64 Topic Objective

65. To Identify Forces and BM in girders 5/18/2020 65 Topic Outcomes Once the student has successfully completed this unit, he/she will be able to: To analyze different Girder Apply the concepts of forces and displacements to solve indeterminate structure . Analyze moments to joint rotations and support settlements. Aayushi RCE-502, DOS 1 Unit 2

66. 5/18/2020 66 Objective of Topic Topic-3 Name The Stiffening Girder Objective of Topic-3: To Differentiate between rigid and deformable structures. Analysis of cables subjected to different loads. Aayushi RCE-502, DOS 1 Unit 1

67. 5/18/2020 67 Topic Outcome and mapping with PO Programme Outcomes (POs) PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 TO1 1 1 1 1 Outcome of Topic-3: After the successfully competition of this topic students will be able to d differentiate between rigid and deformable structures. Analysis of cables subjected to different loads. Aayushi RCE-502, DOS 1 Unit 2

68. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 68 Prerequisite and Recap Basic of forces in cables Diagram of bending moment and shear stress

69. The  stiffening girder  transmits the dead weight of the roadway and the live traffic loads acting on the roadway in the transverse direction of the bridge to the suspension points of the cables where these loads are removed by the cables. As a result, horizontal compressive forces are present in the  stiffening girder 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 69 Topic : Stiffening Girder

70. The 3-hinged stiffening girder of a suspension bridge of span 120m is subjected to two point loads of 240KN and 300KN at distance 25m and 80m from the left end . Find the SF and BM for the girder at a distance of 40m from the left end. The supporting cable has a central dip of 12m. Find , also the maximum tension in the cable and draw the BMD for the girder . 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 70 Topic : Stiffening Girder

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72. 1.To find support reaction: Considering the stiffening girder as a suspension beam supporting the given external load system, ΣMa =0 240 ∗25+300∗80−V B ∗120=0 V B =250KN ΣF Y =0 V A −240− 300+250=0 V A =290KN 2.To find H: Beam moment at C= BMc =290∗60−240∗ 35 BMc =9000KNm Beam moment under the 240KN load=(290∗25)= 7250KNm 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 72 Continuous…

73. Beam moment under the 300KN load= (250∗40)= 1000KNm Horizontal reaction at each end of the cable,H =Mc/h H=9000/12=750KN 3.Find equivalent UDL: 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 73 Continuous…

74. Let   we/unit= UDL transferred to the cable. H=We∗ l 2 /8 ∗h=We∗ 120 2 /8 ∗ 12=750 We=5KN/m Each vertical reaction for the cable= V=We∗l 2 =5∗ 120 2 =300KN Max tension in the cable= Tmax =(V 2 +H 2 ) 1/2 (300 2 +750 2 ) 1/2 Tmax =807.8KN 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 74 Continuous…

75. 4. S.F calculation : For the girder, S.F at any section =  SFx =( Beamshear−Htan θ ) For the cable at any point,tan θ=4 h/l 2 ∗(l− 2x) At 40m from the left end,tan θ=4∗ 12 / 120 2 ∗(120−2∗40)= 21 / 5=0.1333 Beam shear at 40m from left end =(290−240)= 50KN Actual SF at 40m from the left end =(50−750∗ 2/15 )=− 50KN 5.BM calculation: For the girder, BM at any section, M=( Beammoment−Hmoment )= Beammoment – Hy Beam moment at 40m from the left end =(290∗40−240∗15)=8000KNm 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 75 Continuous…

76. At 40m from the left end, for the cable ,   y=4h/l2 ∗x(l−x)=4∗ 12/1202 ∗40∗ 80=32/3m Actual BM at 40m from left end  =( Beammoment − Hmoment ) =( 8000−750∗323)= 6.Actual BMD for the girder: Dip of the cable 25m from left end  =4h/l2∗x(l−x)= frac4∗121202∗25∗95=7.92m Actual BM at 25m from left end  =( Beammoment−Hmoment )=(7250−750∗7.92) =1310KNm 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 76 Continuous…

77. Dip of the cable 80m from the left end =4∗ 12/1202 ∗80∗ 40=32/3m Actual BM at 80m from left end =(1000−750∗ 32/3 )=2000KNm 7.Actual BMD for the stiffening girder 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 77 Continuous…

78. The girder is defined as the structure in pure tension having the funicular shape of the load. The procedures to analyse cables carrying concentrated load and uniformly distributed loads are developed. A few numerical examples are solved to show the application of these methods to actual problems. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 78 Summary of Stiffening Girder

79. Youtube /other Video Links 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 79 Youtube Video Links Topic Links Process Muller Breslau Principle https://www.youtube.com/watch?v=jAuT2qaIszw&t=732s Click on the link Arches https://www.youtube.com/watch?v=jAuT2qaIszw&t=732s Click on the link Stiffening Girder https://www.youtube.com/watch?v=cPgKkWK7q28 Click on the link

80. Top most part of an arch is called ________ a) Sofit b) Crown c) Center d) Abutment Which of the following is true in case of stone brick? a) They are weak in compression and tension b) They are good in compression and tension c) They are weak in compression and good in tension d) They are good in compression but weak in tension 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 80 Daily Quiz

81. Explain Müller Breslau principle. Draw the influence line diagram for reaction R, for the beam shown in the Fig. Compute the ordinate at 1 m interval. The flexural rigidity is constant throughout. Derive the influence line diagram for reactions and bending moment at any section of a simply supported beam. Using the ILD, determine the support reactions and find bending moment at 2 m, 4 m and 6 m for a simply supported beam of span 8 m subjected to three point loads of 10 kN, 15 kN and 5 kN placed at 1 m, 4.5 m and 6.5 m respectively. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 81 Daily Quiz

82. Prove that horizontal thrust developed due to a point load W acting at crown in a two hinged semicircular arch of radius R' is independent of its radius. Consider EI as constant. A two hinged parabolic arch of span 30 m and rise 6 m carries two point loads, each 60 kN, acting at 22.5 m and 15 m from the right end respectively. Determine the horizontal thrust and maximum positive and negative moment in the arch. A parabolic two hinged arch has a span of 32 meters and a rise of 8 m. A uniformly distributed load of 1 kN/m covers 8.0 m horizontal length of the left side of the arch. If I = I, sec 0, where 0 is the inclination of the arch of the section to horizontal, and, I, is the moment of the inertia of the section at the crown, find out the horizontal thrust at hinges and bending moment at 8.0 m from the left hinge. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 82 Daily Quiz

83. What is Indeterminate structure? What is a Procedure for ILD? What is difference between Two hinge and Fixed arches ? Define stiffening girder. List out the assumptions made for Arches.  Define Fixed arches and find out its degree of freedom. Define types of arches. Mention the formula for circular Arch. Define Muller- Breslau Principle. What is an influence line diagram? Explain its importance in structural analysis.   5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 83 Weekly Assignment

84. Draw the schematic diagrams for horizontal thrust, bending moment at any section, radial shear and normal thrust at any given section for a typical two-hinged symmetrical parabolic arch. A two hinged parabolic arch has a span of 30 m and a central rise of 5.0 m . Calculate the maximum positive and negative bending moment at a section distant 10 m from the left support , due to a single point load of 10 kN rolling from left to right. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 84 Weekly Assignment

85. Shape of three hinged arch is always :- a) Hyperbolic b) Circular c) Parabolic d) Can be any arbitrary curve Internal bending moment generated in a three hinged arch is always:- a) 0 b) Infinite c) Varies d) Non zero value but remains constant 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 85 MCQ s

86. What is the degree of indeterminacy of a fixed arch? a) 1 b) 2 c) 3 d) 4 What is the degree of indeterminacy of a two hinged arch ? a) 1 b) 2 c) 3 d) 4 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 86 MCQ s

87. Which of the following is true in case of stone brick ? a) They are weak in compression and tension b) They are good in compression and tension c) They are weak in compression and good in tension d) They are good in compression but weak in tension 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 87 MCQ s

88. Identify the FALSE statement from the following, pertaining to the methods of structural analysis. Influence lines for stress resultants in beams can be drawn using Muller Breslau's Principle. The Moment Distribution Method is a force method of analysis, not a displacement method. The Principle of Virtual Displacements can be used to establish a condition of equilibrium. The Substitute Frame Method is not applicable to frames subjects to significant side sway. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 88 MCQ s

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100. The 3-hinged stiffening girder of a suspension bridge of span 120m is subjected to two point loads of 240KN and 300KN at distance 25m and 80m from the left end. Find the SF and BM for the girder at a distance of 40m from the left end. The supporting cable has a central dip of 12m. Find, also the maximum tension in the cable and draw the BMD for the girder . A semicircular two hinged arch of constant cross section is subjected to a concentrated load as shown. Calculate reactions of the arch and draw bending moment diagram. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 100 Expected Questions for University Exam

101. Using Muller Breslau Principle, compute the influence line ordinates at 2 m intervals for moment at mid span of BC of the continuous beam ABC shown in Fig. A two hinged semicircular arch of radius R' carried l a load "W' at a section the radius vector corresponding to which makes an angle 'a' with the horizontal. Find the horizontal thrust at each support. Assume uniform flexural rigidity. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 101 Expected Questions for University Exam

102. Introduction- Muller Breslau principle , Arches i.e. Two hinged, fixed hinged arches and Stiffening Girder. Explanation of procedure for all methods. Illustrate some examples to understand concept better. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 102 Summary

103. Jain, A. K., “Advanced Structural Analysis “, Nem Chand & Bros., Roorkee . Hibbeler , R.C., “Structural Analysis”, Pearson Prentice Hall, Sector - 62, Noida-201309 C . S. Reddy “Structural Analysis”, Tata Mc Graw Hill Publishing Company Limited,New Delhi. Timoshenko , S. P. and D. Young, “ Theory of Structures” , Tata Mc- Graw Hill BookPublishing Company Ltd., New Delhi. Dayaratnam , P. “ Analysis of Statically Indeterminate Structures”, Affiliated East- WestPress . Wang, C. K. “ Intermediate Structural Analysis”, Mc Graw -Hill Book PublishingCompany Ltd. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 103 References

104. 5/18/2020 Aayushi RCE-502, DOS 1 Unit 2 104 References Thank You