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CE24 2018 DRCS Assignment With Solutions Week 03b

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  CE24 : Design of RC Structures Assignment-3 July-October, 2018 : Week-3 Q1:  The rectangular beam of width, 300 mm is having effective depth of 365 mm. Theconcrete grade is M20 and the grade of reinforcing steel is Fe415. As per limit statemethod, the area of steel in a balanced section is :(a)  1046 mm 2 (b)  1151 mm 2 (c)  1256 mm 2 (d)  1360 mm 2 Solution: (i) Concrete Grade= M  20 (ii) Steel Grade= Fe 415 (iii) Width,  b = 300  mm (iv) Effective depth,  d = 365  mm (v) Area of tensile reinforcement, balanced section,  A stb = k 1  f  ck  ( xubd  )  b d 0 . 87  f  y = 0 . 36 × 20 × 0 . 48 × 300 × 3650 . 87 × 415 = 1046 . 3  mm 2 Answer :  1. (a)  1046 mm 2  CE24 : Design of RC Structures Assignment-3 July-October, 2018 : Week-3 Q2:  The rectangular beam of width, 300 mm is having effective depth of 365 mm. Theconcrete grade is M20 and the grade of reinforcing steel is Fe415. As per limit statemethod, the lever arm in a balanced section is equal to :(a)  175 mm (b)  205 mm (c)  234 mm (d)  292 mm Solution: (i) Concrete Grade= M  20 (ii) Steel Grade= Fe 415 (iii) Width,  b = 300  mm (iv) Effective depth,  d = 365  mm (v) Lever arm,  Z  ub = d (1 − k 2 x ub d  ) = 365 . 0  ×  (1 − 0 . 42  ×  0 . 48) = 292 . 2 Answer :  2. (d)  292 mm  CE24 : Design of RC Structures Assignment-3 July-October, 2018 : Week-3 Q3:  The rectangular beam of width, 300 mm is having effective depth of 365 mm. Theconcrete grade is M20 and the grade of reinforcing steel is Fe415. The tensilereinforcement is provided by 3-20 mm dia bars. As per limit state method, themoment of resistance due to steel is equal to :(a) 51.010 kNm(b) 76.515 kNm(c) 102.020 kNm(d) 127.526 kNm Solution: (i) Concrete Grade= M  20 (ii) Steel Grade= Fe 415 (iii) Width,  b = 300  mm (iv) Effective depth,  d = 365  mm (v) Area of tensile reinforcement,  A st = n t  π φ 2 t / 4 = 3 × 3 . 14159 × 20 2 / 4 = 942 . 5  mm 2 (vi) Moment of resistance due to steel,  M  us = 0 . 87  fy A st  (1 − k 2 x u d  )  d = 0 . 87 × 415 × 942 . 5 × (1 − 0 . 42 × 0 . 43) × 365 . 0 = 102 . 020  kNm Answer :  3. (c) 102.020 kNm  CE24 : Design of RC Structures Assignment-3 July-October, 2018 : Week-3 Q4:  The width of a rectangular beam is 300 mm and the effective depth is 365 mm. Theconcrete grade is M20 and the grade of reinforcing steel is Fe415. The position of neutral axis from compression side is(a)  306 mm (b)  262 mm (c)  219 mm (d)  175 mm Solution: (i) Characteristics strength of concrete,  f  ck = 20  N/mm 2 (ii) Characteristics strength of reinforcing steel,  f  y = 415  N/mm 2 (iii) Width,  b = 300  mm (iv) Effective depth,  d = 365  mm (v) Neutral axis factor,  x ub d =   c  c +  s =  0 . 00350 . 0035+0 . 0038 = 0 . 48 (vi) Position of neutral axis from the compression side,  x ub = x ub d  d = 0 . 48  ×  365 . 0 = 174 . 9 Answer :  4. (d)  175 mm
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