"CESE 656 – Advanced Mechanicsfor Structural EngineersCivil Structural EngineeringUniversity of Alabama at BirminghamBending Deflections for Non-symmetric Bending2 Bending CESE 656 Deflections for Non-symmetric Beams? Consider the intersection of y-z plane with a beam asshown, and assume deflections are small? The geometry of bending gives2 1 ? ? ? ? ? ? - = ?2 ? ? ? ? ? ? The bending formula with x=0 and Hooke’s Law give? ? + ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? = - ? + ??? ?? 2 2 ? ? - ? ? ? - ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? =2 ? ? ( ? ? - ? ) ? ? ? ? 2 ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? - = -2 2 ? ? ? ? ? - ? ? ( ? - ? ) ? ? ? ? ? t an ? ? ? 3 Bending CESE 656 Similarity to Symmetric Bending? For symmetric bending 2 ? ? ? ? =2 ? ? ? ? ? For non-symmetric bending2 ? ? ? ? ? = -2 ? ? ? ( ? - ? ) ? ? ? t an ? ? Any solution for symmetrical bending (beam deflectiontables) can be used by replacing I with {I – I tan( a)} andx xy M with M = M sin( f)x 4 Bending CESE 656 Direction of Displacement? Solution to differential equation gives verticaldisplacement, v? Total displacement ? neutral axis so,? ? = -? tan ?? 2 2 ? ? = ? + ? =co s ? 5 Bending CESE 656 Example – Purling on Sloping Roof? An I-beam is used as a simply-supportedpurlin on a sloping roof as shown.? The beam has an allowable normalstress of s = 160 MPa and w actsallow through the centroid of the beam? The beam has a maximum allowabletotal deflection of L/180? E = 200 GPaa. Determine the largest w that can beapplied without overstressing thebeamb. Determine the largest w that can beapplied without exceed thedeflection limit of the beam6 Bending CESE 656 Section Properties and NA Location3 3 100200 901 706 4 ? ? = - = 29.8 10 ? ?? 12 12 3 3 15100?? 170106 4 ? ? = 2 + = 2.51 10 ? ?? 12 12 ? ? = 0 (symmetric)?? ? - ? co t ? - 29. 8 co t( 75°) ? ? ? ? tan ? = =? - ? co t ? 2. 51 ? ? ? ? ? = -72.5°yNAx7 Bending CESE 656?? ?? ?? ?? ?? ?? ??a) Maximum w based on stress? Solution based on bending formula for arbitrary axes2 2 ? ? ? 6, 0006 2 ? ? = = = 4.5 10 ? ? · ? , axis? plane of loads, compr?8 8 6 2 ? ? = -? cos 15° = ? -4.347 10 mm? ? ? - ? ta n ? ? ? ? =? - ? ta n ? ? ? ? 6 - 4. 347 10 ? [ - 100 -( - 50 ta n - 72. 5°)] y? 160 ? ?? =NA6 4 29. 8 10 ? ? x? ? ? = 4.24M?? ? ? ? ? = 4.24? 8 Bending CESE 656?? ?? ?? ??a) Maximum w based on stress? Solution based on bending formula for Principal axes2 2 ? ? ? 6, 0006 2 ? ? = = = 4.5 10 ? ? · ? , axis? plane of loads, compr?8 8 6 2 ? ? = -? cos 15° = ? -4.347 10 mm? 6 2 ? ? = ? sin 15° = ? 1.165 10 mm? ?? ?? ? ? ? ? = -? ? y? ? NA6 6 - 4. 347 10 ? ( - 10 0 ) 1. 165 10 ? ( - 50 ) ? 160 ? ?? = -6 4 6 4 29. 8 10 ? ? 2. 51 10 ? ? xM? ? ? = 4.23?? ? ? ? ? = 4.23? 9 Bending CESE 656?? ?? ?? ??b) Maximum w based on deflection4 5? ? ? Deflection of simple span with uniform load is ? = , so here384 ? ? 4 4 2 5? ? si n ? 5? 6, 0 00 sin 75° ? ? ? ? = = = 2,735 · ?6 4 384? ( ? - ? ta n ? ) 384 20 0? ( 29. 8 10 ? ? ) ? ? ? ? ? 2 ? ? ? 2, 735 ? 6, 0 00?? ? ? = = = = 33.3 ??co s ? co s( - 72. 5°) 180 y? ? NA- 3 ? ? = 3.67 10?? x? ? M? ? = 3.67? ? With w = 3.67 kN/m? ? = 10.0??? ? = 31.8 ??10 Bending CESE 656?? ?? ??"