Slide1: 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
Slide5: 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
Slide6: 6 Lateral Stability of Beams
Slide7: 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
Slide8: 8 Limit states for LR beams Limit state of flexure Limit state of shear Limit state of bearing Limit state of serviceability
Slide9: 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
Slide10: 10 1 2 3 4 Plastic Hinge Simply supported beam and its deflection at various stages W
Slide11: 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
Slide12: 12 2.0 1.7 1.27 1.14 1.5 Some typical shape factor
Slide13: 13 EQUATIONS FOR SHEAR CAPACITY
Slide14: 14 Shear yielding near support Web buckling Web crippling
Slide15: 15 45 0 d / 2 d / 2 b 1 n 1 Effective width for web buckling
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Slide17: 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
Slide19: 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
Slide20: 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
Slide21: 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
Slide22: 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
Slide24: 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
Slide28: 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
Slide29: 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
Slide30: 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
Slide32: 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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