Assignment-3.unprintable_solution

Assignment-3.unprintable_solution - CIVL 232 Design of...

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1 CIVL 232 Design of Structural Concrete Spring 2006 Solutions of Assignment 3 Question 3.1 Let d = h – 50 = 550 – 50 = 500 mm. 6 22 330 10 0.176 0.156 250 490 30 cu M K bd f × == = > ×× Hence compression steel is required. With x = d / 2 , d' / d = 50/500 = 0.1 < 0.185; compression steel will therefore have yielded. Thus 2 2 2 () (0.176 0.156) 30 250 490 187 mm 0.95 ( ) 0.95 460 (490 50) cu s y KKfb d A fdd −× × × = × Provide 2T16 ( A’ s = 402 mm 2 ); 2 2 2 0.156 250 490 30 187 1879 mm 0.95 0.95 460 0.775 490 cu ss y Kbd f AA fz × =+ = + = × Provide 3T20+2T25 ( A s = 1923 mm 2 )
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2 Question 3.2 (a) It is assumed that both compression and tension steel will have yielded at failure; then f s = f sc = 0.95 f y Hence, 0.95 0.45 0.95 ys c u f Af b s f A =+ 0.95 ( ) 0.95 460 (2454 942) 163 mm 0.45 0.45 30 300 ys s cu fA A s fb ×× == = / 0.9 163/ 0.9 181 mm xs = . Since x / d = 181/500 = 0.362 < 0.615 and d’ / x = 50/181 = 0.276 < 0.37, both tension and compression steel will have yielded, as assumed. Taking moments about the centroid of A s , the ultimate moment of resistance of the cross-section is () 0.45 0.95 2 Rc u y s s M sd fA d d ⎛⎞ =− + ⎜⎟ ⎝⎠ ) 50 500 ( 942 460 95 . 0 ) 2 163 500 ( 163 300 30 45 . 0 × × × + × × × × = = 461.5 kNm (b) It is assumed that both compression and tension steel will have yielded at failure, f s = f sc = 0.95 f y Hence, 0.95 0.45 0.95 c u f b s f A 0.95 ( ) 0.95 460 (2454 1472) 106 mm 0.45 0.45 30 300 cu s = / 0.9 106 / 0.9 118 mm = Therefore, x / d = 118/500 = 0.236 < 0.5, tension steel will have yielded, but d’ / x = 50/118 = 0.424 > 0.37, compression steel will not have yielded at failure.
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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-3.unprintable_solution - CIVL 232 Design of...

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