# m6l14 - Module 6 Shear Bond Anchorage Development Length...

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Module 6 Shear, Bond, Anchorage, Development Length and Torsion Version 2 CE IIT, Kharagpur

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Lesson 14 Limit State of Collapse in Shear – Numerical Problems Version 2 CE IIT, Kharagpur
Instructional Objectives: At the end of this lesson, the student should be able to: solve specific numerical problems of rectangular and T -beams for the complete design of shear reinforcement as per the stipulations of IS 456. 6.14.1 Introduction Lesson 13 explains the three failure modes due to shear force in beams and defines different shear stresses needed to design the beams for shear. The critical sections for shear and the minimum shear reinforcement to be provided in beams are mentioned as per IS 456. In this lesson, the design of shear reinforcement has been illustrated through several numerical problems including the curtailment of tension reinforcement in flexural members. 6.14.2 Numerical Problems Problem 1: Determine the shear reinforcement of the simply supported beam of effective span 8 m whose cross-section is shown in Fig. 6.14.1. Factored shear force is 250 kN. Use M 20 and Fe 415. Version 2 CE IIT, Kharagpur

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Solution 1: Here, A st = 2-25T + 2-20T gives the percentage of tensile reinforcement 1.43 (450) 250 (1609) 100 = = p From Table 6.1 of Lesson 13, τ c = 0.67 + 0.036 = 0.706 N/mm 2 (by linear interpolation). Employing Eq. 6.1 of Lesson 13, τ v = 2 3 N/mm 2.22 (450) 250 ) (10 250 = = d b V u and τ cmax = 2.8 N/mm 2 (from Table 6.2 of Lesson 13). Hence, τ c < τ v < τ cmax . So, shear reinforcement is needed for the shear force (Eq. 6.4 of Lesson 13). V us = V u τ c b d = 250 – 0.706 (250) (450) (10 -3 ) = 170.575 kN Providing 8 mm, 2 legged vertical stirrups, we have A sv = 2 (50) = 100 mm 2 Hence, spacing of the stirrups as obtained from Eq. 6.5 of Lesson 13: mm, 95.25 170575 (450) (100) (415) 0.87 0.87 = = = us sv y v V d A f s say 95 mm. For 10 mm, 2 legged vertical stirrups, ( A sv = 157 mm 2 ), spacing mm 149.54 170575 (450) (157) (415) 0.87 = = v s According to cl. 26.5.1.5 of IS 456, the maximum spacing of the stirrups = 0.75 d = 0.75 (450) = 337.5 mm = 300 mm (say). Minimum shear reinforcement (cl. 26.5.1.6 of IS 456) is obtained from (Eq.6.3 of sec. 6.13.7 of Lesson 13): Version 2 C E IIT, Kharagpur
) ( 0.87 0.4 ) ( y v sv f s b A From the above, A sv(minimum) = 2 mm 40.16 (415) 0.87 (145) (250) 0.4 87 . 0 4 . 0 = = y v f s b . So, we select 10 mm, 2 legged stirrups @ 145 mm c/c. Problem 2: Design the bending and shear reinforcement of the tapered cantilever beam of width b = 300 mm and as shown in Fig. 6.14.2 using M 20 and Fe 415 (i) without any curtailment of bending reinforcement and (ii) redesign the bending and shear reinforcement if some of the bars are curtailed at section 2-2 of the beam. Use SP-16 for the design of bending reinforcement. Solution 2: kNm 459.375 2 (3.5) (3.5) 75 1 - 1 section at = = u M 2 6 2 N/mm 6.125 ) 500 ( ) 500 ( 300 ) (10 459.375 = = d b M u Version 2 CE IIT, Kharagpur

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Solution 1: Here, A st = 2-25T + 2-20T gives the percentage of tensile reinforcement 1.43 (450) 250 (1609) 100 = = p From Table 6.1 of Lesson 13, τ c = 0.67 + 0.036 = 0.706 N/mm 2 (by linear interpolation). Employing Eq. 6.1 of Lesson 13, τ v = 2 3 N/mm 2.22 (450) 250 ) (10 250 = = d b V u and τ cmax = 2.8 N/mm 2 (from Table 6.2 of Lesson 13).
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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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m6l14 - Module 6 Shear Bond Anchorage Development Length...

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