Bending Beam With Double Reinforcement

Faculty of Civil Engineering and PlanningCivil Engineering Department Petra Christian University BEND-DoubleReinf 1 WHEN IS COMPRESSION REINFORCEMENT NEEDED ? DESIGN OF DOUBLE REINFORCEMENT TABLES FLOW CHART OF DOUBLE REINFORCEMENT DESIGN USING TABLE FOR RECTANGULAR SECTION PROBLEMS EXERCISES BEND-DoubleReinf 2 . . (SKSNI T15-1991-03 sec.WHEN IS COMPRESSION REINFORCEMENT NEEDED ? 1. 3. more than its maximum value. for a certain bending loading. Mu. If a beam section with single reinforcement has a reinforcement ratio. BEND-DoubleReinf 3 . To accomplish the seismic design provision.2(1)).3. 3. 2.14. Practically reasons. maks. fs1 = As max .½a) = As max . fy (d .85 fc’ ’cu = 0. (d .DOUBLE REINFORCEMENT If single reinforcement has  > maks 0.003 d’ c As ’ a d Mu d-½a As b CONDITION 1 : T1 s fs = fy Cs Cc 1 + Mn1 d-d’ T2 Mn2 2 T1 = As1 .½a) BEND-DoubleReinf Mu >  Mn1 4 . fy Mn1 = T1 . 003 c s’ c . is : T2 s’ d’ s 5 .d’ = 0.. Check.Mn1 = Cs (d .003 c s’ = …….d’) … (2)  Check whether compression steel has yielded ? BEND-DoubleReinf s’  y ? Mn2 2 ’cu = 0.CONDITION 2 : Cs For compatibility condition : d-d’ Mu  = (Mn1 + Mn2) … (1) Mn2 = Mu . fy Cs As’ = fy Compression steel is yield (s’ >y) Cs = A s ’ f s ’ IF “NO” : Compression steel is not yield (s’ y) fs’ = s’ .IF “YES” : Cs = As’ . Es Cs = As’ . fs’ Cs As’ = fs’ d-d’ T2 = As2 fy Mn2 2 The value of As needed for the section is : Where : As = As1 + As2 As1 = As maks As2 = T2 : fy BEND-DoubleReinf 6 . fy As2 BEND-DoubleReinf then : fs = As’ . fy As’ . fy = As2 . fy As2 = fy then : As2 = As’ * If compression steel is not yield : As’ . fs = As2 .See Condition 2 Cs = As’ fs’ d-d’ Mn2 2 T2 = As2 fy Cs = T2 * If compression steel is yield : As’ . f y 7 . 003 As ’ d’ c Mu As Cs Cc a d d-½a T1 s b 0.  See Table 2.j (CUR IV) for designing the double reinforcement in rectangular section with ’ = 0.85 fc’ ’s + Mn1 d-d’ T2 Mn2 fs = fy 1 2  See Table 5.a .5.DESIGN OF DOUBLE REINFORCEMENT ’cu = 0.5  .4 for double reinforcement with several ratio of ’/ from Copy Sheet BEND-DoubleReinf 8 .3.3. 15 0.0012 0.0214 0.0090 0.0064 2200 0.0034 0.0011 0.0046 0.0128 0.0014 0.0027 0.3.0114 0.0106 0.0021 800 0.0102 3000 0.0117 3400 0.0057 1800 0.0047 0.0035 0.CUR IV CONCRETE STRENGTH f’c 15 with compression reinforcement  = 0.10 0.0007 0.0050 1600 0.0103 0.0071 0.0220 0.0091 0.0132 0.0182 0.0172 0.0208 0.0048 0.0124 0.0104 0.0152 0.0084 0.5  Mu/bd2 fy = 240 MPa with d’/d = fy = 400 MPa with d’/d = 0.0132 0.0056 0.0078 0.0221 0.0116 0.0075 0.0129 0.0136 0.TABLE 5.0023 0.20 0.0136 0.0041 0.0042 1400 0.0047 0.0011 0.0082 0.0195 0.0118 0.0079 2400 0.0035 1200 0. ’= ratio of compression reinf.0119 0.0247 0.0144 0.0140 0.a .0144 0.0070 0.0067 0.0109 3200 0.0062 0.0160 0.0007 400 0.0057 0.0170 0.0023 0.0083 0.0007 0. = 0.0033 0.0165 0.0208 0.0092 0.0055 0.0125 0.0110 0.0061 0.0121 0.0133 3800 0.0020 0.0033 0.0202 0.20 200 0.0099 0.0227 0.0234 0.0054 0.0094 2800 0.0089 0.0040 0.0069 0.0044 0.0014 600 0.0177 0.0013 0.0068 0.8  = Ratio of tension reinf .0148 0.0140 0.0020 0.0087 2600 0.0034 0.0027 0.0113 0.0022 0.0157 0.0059 0.0233 0.0095 0.0125 3600 0.0096 0.0189 0.10 0.15 0.0101 0.0072 2000 0.0240 0.0080 0.0184 0.0028 1000 0.0077 0.0196 0.0141 9 BEND-DoubleReinf 4000 0.0117 0.0148 . 21 1.17 1..37 1.09 1.32 1.43 1.38 1.06 2.19 0.3 0.93 2.69 0.68 1.5 0.89 1.8 0.22 0.72 1.84 1.44 1.98 1.36 3.12 1.62 1.09 2.1 0.48 0.99 2.54 1.74 0.24 1.24 2.68 0.66 1.79 1.2 0.16 2.55 1.85 1.85 BEND-DoubleReinf 0.COPY SHEET RECTANGULAR BEAM WITH DOUBLE REINFORCEMENT RATIO OF REINFORCEMENT :  (%) Rn = Mn / (b.89 2.86 1.21 1.TABLE 2.76 1.82 2.39 1.62 1.03 2.61 0.32 1.79 0.70 1.72 1. fy Concrete strength.9 0.48 1..60 1.4 0.93 1.29 1.13 2.02 2.64 1.62 0.33 1.80 1.13 1.46 1.51 1.74 1.80 0.82 0.10 10 .15 0.74 1.14 1.48 1. b and d (m) Steel yield stress.51 1.96 2.02 1.65 0.57 2.77 0. 0.95 2.42 0.40 0.89 1.00 .91 1.96 1.09 1.16 1.05 1.36 1.0 0.38 2.95 1.94 2.49 1.69 1.29 1.82 1.43 1.d2) Units : Mn (kNm).58 1.21 2.56 1.21 1.92 2.52 0.26 0.44 0.13 1.40 1.29 0.58% ’/  Rn 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800 5000 5200 .66 1.7 0.15 1.40 1.90 2.91 1.57 0.25 1.89 3.59 1.89 1.35 1.99 2.87 3.17 1.00 1.28 1.64 1.54 1.79 0.19 2.6 0.11 2.83 1.74 0.04 2.47 1.08 1.25 1.22 0.27 1. fc’ d’/d min = = = = 240 MPa 20 MPa 0.85 1.04 1.4 .84 1.09 2.79 1.00 2.04 1.86 1. d’ (trial & errors) Calculate loads and Mu As =  . b .FLOW CHART OF DOUBLE REINFORCEMENT DESIGN USING TABLE FOR RECTANGULAR SECTION START 1 Define : b. As Finish Calculate : Mn = Mu /  Mn Rn = bd2 Select appropriate ’/ See Table to find  BEND-DoubleReinf 1 Not found found 11 . d. d As’ = (’/) . b. is it in “UNDER-REINFORCED” condition ? BEND-DoubleReinf 12 .SOLVE THESE PROBLEMS Time frame : 15 Minutes The solutions are on the last pages of this presentation 1. Check the section. cu’ = 0.003 d’ As’ As b c h d s ’ c = 120 mm d = 720 mm d’ = 80 mm fc’ = 25 MPa fy = 400 MPa b = 400 mm h = 800 mm a. Check the compression steel whether it is yield or not. As’ h Mu As b BEND-DoubleReinf fc’ d’/d h As As’ = 20 MPa = 0. As. needed for this rectangular beam with single reinforcement as drawn so that the depth of beam.2 = 500 mm = 4 D22 = 2 D22 fy = 400 MPa b = 200 mm 13 . and Mu are known. fy. h.2. If b. calculate the reinforcement area. is minimum. 3. h Mu As b Use design table to count the ultimate moment that can be carried by this section. fc’. M u 300 150 150 200 2. fy = 400 MPa Compression steel = 2 D16 Tension steel = 4 D16 How much Mu that could be carried by this section ? fc’ = 20 MPa . Mu = 125 kNm 500 fc’ = 20 MPa .EXERCISES 1. fy = 240 MPa Tension steel = 4  16 How much Mu that could be carried by this triangular section ? fc’ = 25 MPa .5 How much are As and As’ needed ? 300 BEND-DoubleReinf End of presentation 14 . fy = 400 MPa ’/ = 0. Mu 400 400 3. 002 y = y = Es 200.003 d’ As ’ s’ 40 h d 120 600 As s b s ’ 0.003 s ’ 40 = 120 = 0.000 BEND-DoubleReinf s’ < y Compression steel is not yield 15 . a. Is compression steel yield yet ? cu’ = 0.SOLUTION 1.001 f 400 = 0. As max = max .015 > y ( = 0..75 bal = ……. See CUR 1 p. Is section in UNDER-REINFORCED Condition ? s 0... h = minimum. b . The section is in UNDER-REINFORCED Condition.8 . 2. 54 16 .003 = 600 120 s = 0. 0.588  ) y bd2 fc ’ BEND-DoubleReinf .b.002) Tension steel is yield.. max = 0. d Mu fy =  . f (1 0. if As = maximum. Mu =  Mn = ……. ’/ fc’ fy d’/d As = = = = = 0.. kNm BEND-DoubleReinf 17 .4 (Copy Sheet) or Table 5.5 20 MPa 400 MPa 0.20 …… mm2  = As = ….3b (CUR 4) is found : Mn bd2 = ……. Mn = …….3. bd From Table 2.  #%  &# #%$ &$% ##%&#$%  $%#% 0130
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