# m12l32 - 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 32 Two-way Rectangular, Square, Triangular and Circular Slabs Version 2 CE IIT, Kharagpur
Instructional Objectives: At the end of this lesson, the student should be able to: analyse rectangular slabs simply supported at three edges and free at the other edge considering the two possible yield patterns, employing (i) the method of segmental equilibrium and (ii) the method of virtual work, analyse square slab with forking yield patterns when the corners are having inadequate reinforcement, predict yield lines of fan pattern for slabs in case this may be a possibility, analyse the fan pattern of yield lines to determine the collapse loads of triangular and circular slabs with different support conditions, analyse the fan pattern of yield lines to determine the collapse loads of circular slabs clamped along the circumference and having a column support at the centre. 12.32.1 Introduction Rectangular / square slabs may have different yield patterns depending on the support conditions and type of loads. Simply supported slabs at three edges and free at the other edge may have two types of yield patterns depending on the ratio of moment resisting capacities and the aspect ratio. This lesson first takes up such slabs to determine the condition for selecting a particular one out of the two possible yield patterns. The case of a square slab having forking yield pattern is explained when the corner reinforcement is inadequate. Several cases of triangular and circular slabs with or without a central column support are taken up to explain the yield lines of fan pattern. All the expressions are derived by employing the method of either segmental equilibrium or virtual work. In some cases both the methods are taken up to compare the values. 12.32.2 Rectangular Slabs Simply Supported at Three Edges and Free at the Other Edge Considering Yield Pattern 1 Version 2 CE IIT, Kharagpur

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Figure 12.31.4c of Lesson 31 shows the yield pattern 1 of such slabs involving one additional unknown y . So, there are two unknowns – y and w to be determined. The yield pattern divides the slab into three segments marked by 1, 2 and 3 (Fig.12.32.1a). The slab carrying uniformly distributed load of w kN/m 2 , undergoes deflection of Δ at point E. The free body diagrams of segments 1 and 2 are shown in Figs.12.32.1b and c, respectively. Due to symmetry, segments 1 and 3 are identical. Segment 1 is subdivided into two parts as 11 and 12. We employ the method of segmental equilibrium first. (A) Method of segmental equilibrium Here, the nodal forces are zero since the moment capacities of three intersecting yield lines are identical. The equilibrium equation of segment 1 is developed taking the moments of loads and moments of segment 1 about AB, Version 2 CE IIT, Kharagpur
and equating the same to zero. Thus, we get: ( L x /2) ( L y y ) w ( L x / 4) + ( L x /4) y w ( L x / 6) – M x L y = 0, or w = (24 M x L y ) / {3 Lx 2 ( L y y ) + L x 2 y } (12.41)

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## This note was uploaded on 03/17/2012 for the course CENG 3012 taught by Professor Prof.j.n.bandopadhyay during the Summer '01 term at Indian Institute of Technology, Kharagpur.

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m12l32 - Module 12 Yield Line Analysis for Slabs Version 2...

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