m5l34

m5l34 - Module 5 Cables and Arches Version 2 CE IIT,...

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Module 5 Cables and Arches Version 2 CE IIT, Kharagpur
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Lesson 34 Symmetrical Hingeless Arch Version 2 CE IIT, Kharagpur
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Instructional Objectives: After reading this chapter the student will be able to 1. Analyse hingeless arch by the method of least work. 2. Analyse the fixed-fixed arch by the elastic-centre method. 3. Compute reactions and stresses in hingeless arch due to temperature change. 34.1 Introduction As stated in the previous lesson, two-hinged and three-hinged arches are commonly used in practice. The deflection and the moment at the center of the hingeless arch are somewhat smaller than that of the two-hinged arch. However, the hingeless arch has to be designed for support moment. A hingeless 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. 34.2 Analysis of Symmetrical Hingeless Arch Consider a symmetrical arch of span L and central rise of Let the loading on the arch is also symmetrical as shown in Fig 34.1. Consider reaction components c h Version 2 CE IIT, Kharagpur
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at the left support A i.e., bending moment , vertical reaction and horizontal thrust as redundants. a M ay R a H Considering only the strain energy due to axial compression and bending, the strain energy U of the arch may be written as + = s s EA ds N EI ds M U 0 2 0 2 2 2 (34.1) where M and are respectively the bending moment and axial force of the arch rib. Since the support N A is fixed, one could write following three equations at that point. 0 = a M U (34.2a) 0 = a H U (34.2b) 0 = ay R U (34.2c) Knowing dimensions of the arch and loading, using the above three equations, the unknown redundant reactions and may be evaluated. a a H M , ay R Since the arch and the loading are symmetrical, the shear force at the crown is zero. Hence, at the crown we have only two unknowns. Hence, if we take the internal forces at the crown as the redundant, the problem gets simplified. Version 2 CE IIT, Kharagpur
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Hence, consider bending moment and the axial force at the crown as the redundant. Since the arch and the loading is symmetrical, we can write from the principle of least work c M c N 0 = c M U (34.3a) 0 = c N U (34.3b) 0 0 0 = + = s c s c c ds M N EA N ds M M EI M M U (34.4a) 0 0 0 = + = s c s c c ds N N EA N ds N M EI M N U (34.4b) Where, is the length of centerline of the arch, s I is the moment of inertia of the cross section and A is the area of the cross section of the arch. Let and be the bending moment and the axial force at any cross section due to external loading. Now the bending moment and the axial force at any section is given by 0 M 0 N Version 2 CE IIT, Kharagpur
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0 M y N M M c c + + = (34.5a) 0 cos N N N c + = θ (34.5b) 1 = c M M ; y N M
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This note was uploaded on 06/30/2009 for the course CE 358 taught by Professor Trifunac during the Fall '07 term at USC.

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m5l34 - Module 5 Cables and Arches Version 2 CE IIT,...

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