"CESE 656 – Advanced Mechanicsfor Structural EngineersCivil Structural EngineeringUniversity of Alabama at BirminghamCurved Beams Curved Beams with Thin Flanges2 Curved Beams CESE 656 Curved Beams with Thin Flanges? The radial forces developed in a curved beam cause thinflanges to deflect radially? This deflection causes the circumferential stressdistribution to be non-uniform across the flange? The circumferential stress predicted by the curved beamformula under-predicts the maximum stress3 Curved Beams CESE 656 Designing Curved Beam with Thin Flanges? Prevent radial distortion of the flanges by using stiffeners? Use correction factors to modify the cross-section into an “equivalent”section for use with the curved beam formula4 Curved Beams CESE 656 Bleich’s Correction Factors? Reduced flange width is' ' ? b = flange projectionp ? ? = 2 ? + ?? ? ? b' = reduced flange projection' p ? ? = ? ?? ? ? t = web width2 w ? ? ? ? = ?? t = flange thickness? ¯ ? f ? ? a = correction factor? ? ¯ = radius of curvature tocenter of flange5 Curved Beams CESE 656 Bending Stress in the Flange? The state of stress in a curved beam is not uniaxial? There exists a stress in x-direction as well? Bleich estimates this stress in the inner flange as? ? = - ? ? ??? ? ? ? ? ? = circumferential stress at mid-thickness of inner flange? ? x6 Curved Beams CESE 656 Example – I-beam with Correction? A curved I-shaped bracket is subjected to a moment of 85 kN ·m as shown. Use theBleich correction factors to modify the cross-sectiona. Determine the maximum circumferential stress in the beamb. Determine the radial stress in the web at the web-flange junction? Section Modifications? Bottom flange? ? ¯ = 410 ??2 2 ? 4 2. 5? ? = = 0.220? ¯ ? 4 1 0( 2 0 ) ? ? ? = 0.972 (Interpolation)' ? ? = 2 0.972 · 42.5 ?? + 15 ?? = 97.6 ??85 kN ·m? Top Flange? ? ¯ = 580 ??2 2 ? 4 2. 5? ? = = 0.156? ¯ ? 5 8 0( 2 0 ) ? ? ? = 0.989 (Extrapolation)' ? ? = 2 0.989 · 42.5 ?? + 15 ?? = 99.1 ??7 Curved Beams CESE 656?? ?? ?? ?? ?? ??Section Properties? Section Properties2 ? ? = 6,184 ? ?? ? = 495.41 ??420?? 5 70 59 0? ? = 97.6 ?? ln + 15 ?? ln + 99.1 ?? ln? 400?? 4 20 57099.1 mm? ? = 12.760 ??? 2 6, 184? ? ? ? = = 484.63 ??? 12. 760? ? - ? = 495.41 ?? - 484.63 ??? ? ? - ? = 10.78 ??? 97.6 mm85 kN ·m8 Curved Beams CESE 656?? ?? ?? ?? ??a) Circumferential Stress? Circumferential Stress at Point A? ? = 590??? ? ? - ? 8 5, 0 0 0, 0 0 0 ? · ( 4 8 4. 6 3 - 5 90 ) ? ? ? = =? 2 ? ? ? - ? 6, 1 8 4 ? ? ( 5 90 )( 1 0. 7 8 ) ? ? ? = 228? ( ? )? ? Circumferential Stress at Point B? ? = 400??? ? ? - ? 8 5, 0 0 0, 0 0 0 ? · ( 4 8 4. 6 3 - 4 0 0 ) ? ? ? = =? 2 ? ? ? - ? 6, 1 8 4 ? ? ( 4 0 0 )( 1 0. 7 8 ) ? ? ? = 270 ? ( ? )? 85 kN ·m? Stress DistributionA-228 MPaBB 270 MPa9 Curved Beams CESE 656?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ??b) Radial Stress at Web-Flange Junction? At the web-flange junction? ? = 420 ??99.1 mm? ? = 15 ??' 2 ? ? = 97.6 ?? 20 ?? = 1,952? ?4 2 0 ?? ' ? ? = 97.6 ?? ln = 4.762 ??? 4 0 0 ?? ' ' ? ? - ? ? ? 97.6 mm? ? = ?? ? - ? ? 2 4 8 4. 6 3?? ( 4. 7 6 2 ??) -( 1,952 ? ? ) ? ? = 85,000,000? · ??2 1 5 ?? 4 2 0 ?? ( 6, 1 8 4 ? ? ) 1 0. 7 8 ?? ? ? = 72.0? ??85 kN ·mAs = 77.1 MPa atrr-max r = 457.0 mmB10 Curved Beams CESE 656?? ?? ?? ??"