Assignment-5.CIVL232.Spring_2006.solution

Assignment-5.CIVL232.Spring_2006.solution - CIVL 232 Design...

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CIVL 232 Design of Structural Concrete Spring 2006 Solutions of Assignment 5 Question 5.1 (a) The bond coefficient for bars in tension (Type 2 deformed bars) β = 0.5 The ultimate anchorage bond stress 2 0.5 30 2.74 N/mm bu cu ff β == = Hence the ultimate anchorage length 0.95 0.95 460 20 797 mm 4 4 2.74 y bu f l f φ × × = × [ Alternative method: From Table 3.3.-2, K A = 40. By using Eq. (3.26), the design anchorage length l = K A φ = 40 φ = 40 × 20 = 800 mm] (b) For high yield bars, r 3 φ if φ 25 mm. Take r = 3 φ = 3 × 20 = 60 mm. The effective anchorage length for a standard 180° hook l e = 8r = 8 × 60 = 480 mm, but no greater than 24 φ = 24 × 20 = 480 mm. OK The total anchorage length required is ( ) () ( ) 2 ( / 2) / 2 4 800 480 2 60 20 / 2 / 2 4 20 620 mm e ll r πφ π −+ + + =−+ + + × = [ Alternative : From Table 3.3-2 (p.132), the design anchorage length l = 40 φ and, from Table 3.3-3 (p.135), the effective anchorage length for the standard hooks l e = 24 φ . Thus the total anchorage length required is 800 Design anchorage length
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() ( ) 2( / 2 ) / 2 4 40 24 2 / 2 / 2 4 16 / 2 4 16 20 60 20/ 2 4 20 620 mm] e ll r rr πφ φ φφ π −+ + + =−+ + + = + ++ =×+ + +×= (c) The design anchorage length (see Table 3.3-2, p.132)
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.

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Assignment-5.CIVL232.Spring_2006.solution - CIVL 232 Design...

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