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1 DESIGN OF FLEXURAL MEMBER AND BENDING WITH HIGH SHEAR Prepared by Er.C.MAKENDRAN M.E.,(Ph.D)

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Flexural members Laterally supported beam:

Flexural members Laterally supported beam Elastic Analysis Plastic Analysis When factored design shear ≤ 0.6V d and 4

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5 Conditions to Qualify as a Laterally Restrained Beam It should not laterally buckle None of its element should buckle until a desired limit state is achieved Limit state of serviceability must be satisfied Member should behave in accordance with the expected performance of the system

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6 Lateral Stability of Beams

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7 Local Buckling In IS:800 (1984) the local buckling is avoided by specifying b/t limits. Hence we don’t consider local buckling explicitly However in IS:800(2007) limit state design, the local buckling would be the first aspect as far as the beam design is concerned How do we consider? By using section classification

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8 Limit states for LR beams Limit state of flexure Limit state of shear Limit state of bearing Limit state of serviceability

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9 Stress 1 strain 2 3 4 f y Plastic range Elastic range Idealized elasto- plastic stress stain curve for the purpose of design Idealised stress strain curve f

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10 1 2 3 4 Plastic Hinge Simply supported beam and its deflection at various stages W

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11 Moment ‘M’ Curvature M Y Moment curvature characteristics of the simply supported beam Yield moment M P Plastic moment Effect of strain hardening may occur after large rotation

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12 2.0 1.7 1.27 1.14 1.5 Some typical shape factor

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13 EQUATIONS FOR SHEAR CAPACITY

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14 Shear yielding near support Web buckling Web crippling

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15 45 0 d / 2 d / 2 b 1 n 1 Effective width for web buckling

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17 b 1 n 2 1:2.5 slope Root radius Effective width of web bearing Web Crippling in beams

Design of Laterally Supported Beam:

Design of Laterally Supported Beam Limit State Method – As per IS: 800 - 2007. Example No : 1 Design a suitable I beam for a simply supported span of 5 m. and carrying a dead load of 20 kN/m and imposed load of 40 kN/m. Take fy = 250 MPa Design load calculations : Factored load = γ LD x 20 + γ LL x 40 Using partial safety factors for D.L γ LD = 1.50 and for L.L γ LL = 1.5 ( Cl. 5.3.3 Table 4, Page 29) 18

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Total factored load = 1.50 x 20 + 1.5 x 40 = 90 kN/m Factored Bending Moment M = 90 x 5 x 5 / 8 = 281.25 kN.m Zp required for value of fy = 250 MPa and γ mo = 1.10 ( Table 5, Page 30 ) Zp = (281.25 x 1000 x 1000 x 1.1) / 250 = 1237500 mm 3 = 1237.50cm3 Using shape factor = 1.14, Ze = 1237.50/1.14 =1085.52 cm3 Options ISWB 400 @ 66.7 kg/m or ISLB 450 @ 65.3 kg/m Try ISLB 450 Ze = 1223.8 cm 3  1085.52 19

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20 Geometrical Properties : ISLB 450 D = 450 mm , B = 170 mm , tf = 13.4 mm , tw = 8.6 mm , h1 = 384 mm , h2 = 33 mm Ixx = 27536.1 cm4 As fy = 250 MPa , Section Classification : B/2tf = 85 / 13.4 = 6.34  9.4ε h1 / tw = 384/8.6 = 44.65 < 83.9 ε Section is Classified as Plastic Zp = 1.14 x 1223.8 = 1395.132 cm3

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21 Design Bending Strength: Md > 281.25 kN.m β b = 1.0 for plastic section (Cl. 8.2.1.2, Page 53) Check for Serviceability – Deflection Load factor = γ LD and γ LL = 1.00 both , (Cl. 5.6.1, Page 31) Design load = 20 + 40 = 60 kN/m

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22 Limiting deflection = Span/360 (Table. 5.3, Page 52) = 5000/360 = 13.889 mm…. OK Hence Use ISLB 450

Working Stress Method IS : 800 - 1984:

Working Stress Method IS : 800 - 1984 23 Max Bending Moment = 60 x 5 x 5/8 = 187.5 kN.m Max Shear Force = 60 x 5/2 = 150 kN Select ISLB 450 Zxx = 1223.8 Moment Capacity = 201.927 kN.m Check for Shear < 100 MPa

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24 Check for Deflection Limiting deflection = Span/325 = 5000/325 = 15.38 mm… OK

Comparison of ISLB 450 Section:

Comparison of ISLB 450 Section Working Stress Method Limit State Method Moment Capacity 201.927 kN.m > 187.5 KNm 317.075 KNm > 281.25 KNm Shear Capacity 387 KN > 150 KN 507.497KN > 225 KN Section Designed ISLB 450@ 65.3 Kg/m ISLB 450 @ 65.3 kg/m The Section designed as per LSM is having more reserve capacity for both BM and SF as compared to WSM 25

Design of Beam with High Shear LSM:

Design of Beam with High Shear LSM Example No. 2 Factored Load 100 KN/m A B C ________ 5m_______________ 5m_________ 26

Plastic Analysis:

Plastic Analysis Degree of Redundancy = r = 1 No. of plastic hinges required to transform structure into mechanism = r + 1 = 2 Failure of any span is failure of continuous beam. Failure mechanism of AB & BC is identical due to symmetry & this is similar to failure mechanism of propped cantilever beam with udl. wp = 11.656 Mp / l 2  Mp = wp.l 2 / 11.656 = 100 x 25 / 11.656 = 214.48 KNm. 27

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As both spans fail simultaneously actual no of plastic hings are three – two hinges each at 0.414 l from A & C & third at B.  as n = 3  2 required Collapse is over complete Zp = 214.48 x 10 6 x 1.10 / 250 mm 3 = 943.72 cm 3 Ze = 943.72 / 1.14 = 827. 82 cm 3 Select ISLB 400 Zxx = 965.3 cm 3 Md = 1.0 x 1.14 x 965.3 x 250 / 1.10 = 250.1 KNm  214.48 28

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Reaction at A Considering free body of AB Mp = 214.48 KNm Mp + RA x 5 = 100 x 5 x 5/2  RA = 207.1 KN RB1 = 500 – 207.1 = 292.9 KN Due to symmetry in loading Maximum shear is at B = 292.9 KN= V 29

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Vd = 0.577 x 400 x 8 x 250 / 1.1 = 419.636 KN Where 400 x 8 = D.tw of ISLB 400 As V/Vd = 292.9 / 419.636 = 0.697  0.6 As per C1.9.2.2 Page No. 70 Effect of shear is to be considered for reduction in moment capacity Mdv = Md – β(Md – Mfd) β= (2V/Vd – 1)2 = 0.156 Mfd = Plastic moment capacity of flanges only = 165 x 12.5 (400 – 12.5) x 250 / 1.1 = 181.64 KNm  Mdv = 250.1 – 0.156 (250.1 – 181.64) = 239.42 KNm As Mdv = 239.42  Mp = 214.48 ------- Ok Select ISLB 400 @ 56.9 kg / m 30

Laterally supported beam :

Laterally supported beam Design of Beams with High Shear by WSM Factored load in LSM is 100 KN/m  Working load in WSM = 100 / 1.5 = 66.67 KN/m 66.67 KN/m A 5m B 5m C 31

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Reactions - RB = 5/8 x 66.67 x 10 = 416.66 kN , RA = RC = 125.0 kN Maximum Bending Moment At continuous support = 125.0 x 5 – 66.67 x 5 x 5/2 = -208.33 kN.m Design Shear = 208.33 kN Design Moment = 208.33 kN.m As per IS:800 – 1984, 6bc = 0.66fy = 0.66 x 250 = 165 MPa Z required = (208.33 x 106) / 165 = 1262.62 cm3 Try ISMB 450 @ 72.4 kg/m. Zxx = 1350 cm2  1262.62 Cheak for shear tw = 9.4 mm qav = (208.33 x 1000) / (450 x 9.4) = 49.25 N/mm2  0.4fy i.e. 100 N/mm2 32

Comparison of WSM vs LSM :

Comparison of WSM vs LSM Working Stress Method Limit State Method Moment Capacity 222.75 KNm  208.33 KNm 239.42 KNm  214.48 Shear Capacity 423 KN  208.33 KN 419.636 KN  292.90 KN Section Designed ISMB 450 @ 72.4 kg/m ISLB 400 @ 56.9 kg/m Design of beam by LSM is more economical 33

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THE END 34

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