m12l33 - Module 12 Yield Line Analysis for Slabs Version 2...

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Module 12 Yield Line Analysis for Slabs Version 2 CE IIT, Kharagpur
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Lesson 33 Numerical Examples Version 2 CE IIT, Kharagpur
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Instructional Objectives: At the end of this lesson, the student should be able to: apply the theory of segmental equilibrium for the solution of different types of slab problems, apply the theory of virtual work for the solution of different types of slab problems. 12.33.1 Introduction The theoretical formulations of the yield line analysis of slabs are explained in earlier Lessons 30 to 32, considering (i) the method of segmental equilibrium and (ii) the method of virtual work. Illustrative examples are solved in Lesson 30. This lesson includes different types of illustrative examples of slabs with different combinations of support conditions. Two types of loadings, viz., (i) point loads and (ii) uniformly distributed loads, are considered. In some cases, two or three possible yield patterns are examined to select the correct one and the corresponding loads are determined. The problems include different types of rectangular or square two-way slabs, triangular, quadrantal and circular slabs. Circular slabs also include one practical example where it is supported by a central column in addition to clamped support along the periphery. Some practice problems and test problem are also included. Understanding the numerical problems and solving practice and test problems will give a better understanding of the theories of yield line analysis and their applications. 12.33.2 Illustrative Examples Version 2 CE IIT, Kharagpur
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Version 2 CE IIT, Kharagpur
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Problem 1. Determine the uniformly distributed collapse load w kN/m 2 of a square slab ( L m × L m ), simply supported at three edges and the other edge is free. Assume M y = 2M x = 2M. Solution 1. The slab is shown in Fig.12.33.1a. Given data are: L x = L y = L and M y = 2M x = 2M. Step 1. To examine the possibility of yield patterns 1, 2 or both Equation12.45 of Lesson 32 stipulates that (M y /M x ) 4 (L y /L x ) 2 for yield pattern 1 and Eq.12.54 of Lesson 32 gives the condition that (M y /M x ) (4/3) (L y /L x ) 2 for yield pattern 2. Here, L y /L x = 1 and M y /M x = 2. So, ( M y /M x ) 4 ( L y /L x ) 2 and ( M y /M x ) (4/3) ( L y /L x ) 2 . Thus, both the yield patterns are to be examined. Step 2. Value of y for yield pattern 1 The value of y is obtained from Eq.12.43 of Lesson 32, which is 4 M x L y y 2 + 2 M y L x 2 y – 3 M y L x 2 L y = 0 (12.43) or 2 y 2 +2 L y – 3 L 2 = 0 (1) which gives y = 0.823 L (2) Step 3. Determination of collapse load w kN/m 2 Equations 12.41 and 12.42 of the method of segmental equilibrium of Lesson 32 and Eq.12.48 of the method of virtual work of Lesson 32 shall be used to determine w . The results are given below. (i)
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m12l33 - Module 12 Yield Line Analysis for Slabs Version 2...

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