# University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The size and shape of the cross- section of the piece of material used l For timber,

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University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The size and shape of the cross- section of the piece of material used l For timber, usually a rectangle l For steel, various formed sections are more efficient l For concrete, either rectangular, or often a Tee A timber and plywood I-beam 1/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l What shapes are possible in the material? l What shapes are efficient for the purpose? l Obviously, bigger is stronger, but less economical Some hot-rolled steel sections 2/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Beams are oriented one way l Depth around the X-axis is the strong way l Some lateral stiffness is also needed l Columns need to be stiff both ways (X and Y) 3/28 Timber post Hot-rolled steel Steel tube Y Y Cold-formed steel Timber beam XX

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l ‘ Stress is proportional to strain’ l Parts further from the centre strain more l The outer layers receive greatest stress Most shortened Most lengthened Unchanged length 4/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The stresses developed resist bending l Equilibrium happens when the resistance equals the applied bending moment C T All the tensile stresses add up to form a tensile force T All the compressive stresses add up to form a compressive force C a M R = Ca = Ta Internal Moment of Resistance 5/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Simple solutions for rectangular sections For a rectangular section d b l Doing the maths (in the Notes) gives the Moment of Inertia I = bd 3 12 mm 4 6/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The bigger the Moment of Inertia, the stiffer the section l It is also called Second Moment of Area l Contains d 3, so depth is important l The bigger the Modulus of Elasticity of the material, the stiffer the section l A stiffer section develops its Moment of Resistance with less curvature 7/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Simple solutions for rectangular sections b d l Doing the maths (in the Notes) gives the Section Modulus For a rectangular section Z  bd 2 6 mm 3 8/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The bigger the Section Modulus, the stronger the section l Contains d 2, so depth is important 9/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Strength --> Failure of Element l Stiffness --> Amount of Deflection depth is important 10/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The area tells how much stuff there is ●used for columns and ties ●directly affects weight and cost r x = d/√12 r y = b/√12 A = bd The radius of gyration is a derivative of I ● used in slenderness ratio 11/28 Y b d XX Y

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Can be calculated, with a little extra work l Manufacturers publish tables of properties 12/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman 12/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Checking Beams l Designing Beams ●given the beam section ●check that the stresses & deflection are within the allowable limits ●find the Bending Moment and Shear Force ●select a suitable section 13/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Go back to the bending moment diagrams l Maximum stress occurs where bending moment is a maximum f = M Z M is maximum here 14/28 Bending Moment Section Modulus Stress =

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Given the beam size and material Z = bd 2 / 6 M = max BM Actual Stress = M / Z Allowable Stress (from Code) b d l Find the maximum Bending Moment l Use Stress = Moment/Section Modulus l Compare this stress to the Code allowable stress Actual Allowable? < 15/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Given a softwood timber beam 250 x 50mm Section Modulus Z = bd 2 / 6 Actual Stress f = M / Z 50 250 l Given maximum Bending Moment = 4kNm l Given Code allowable stress = 8MPa 4 kNm = 4 x 10 3 x 10 3 / 0.52 x 10 6 = 50 x 250 2 / 6 = 0.52 x 10 6 mm 3 = 7.69 MPa < 8MPa Actual Stress < Allowable Stress 16/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Given the maximum Bending Moment l Given the Code allowable stress for the material l Use Section Modulus = Moment / Stress l Look up a table to find a suitable section b? d? M = max BM Allowable Stress (from Code) required Z = M / Allowable Stress a) choose b and d to give Z >= than required Z or b) look up Tables of Properties 17/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Given the maximum Bending Moment = 4 kNm l Given the Code allowable stress for structural steel = 165 MPa b? d? required Z = 4 x 10 6 / 165 = 24 x 10 3 mm 3 looking up a catalogue of steel purlins we find C15020 - C-section 150 deep, 2.0mm thickness has a Z = 27.89 x 10 3 mm 3 (steel handbooks give Z values in 10 3 mm 3 ) (smallest section Z >= reqd Z) 18/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Both E and I come into the deflection formula (Material and Section properties) Depth, d Span, L W 19/28 l The load, W, and span, L 3 l Note that I has a d 3 factor l Span-to-depth ratios (L/d) are often used as a guide

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman 20/28  WL 3 48EI 8d Central point load W L  5WL 3 384EI 5d Uniformly Distributed Load where W is the TOTAL load (w per metre length) Total load = W L

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman W L Central point load  WL 3 8EI 48d  WL 3 3EI 128d 21/28 where W is the TOTAL load Uniformly Distributed Load (w per metre length) L Total load = W

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l The deflection is only one-fifth of a simply supported beam l Continuous beams are generally stiffer than simply supported beam where W is the TOTAL load  WL 3 384EI d (w per metre length) L Total load = W Uniformly Distributed Load 22/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman Given load, W, and span, L Given Modulus of Elasticity, E, and Moment of Inertia, I Use deflection formula to find deflection Be careful with units (work in N and mm) Compare to Code limit (usually given as L/500, L/250 etc) 23/28 l Given the beam size and material l Given the loading conditions l Use formula for maximum deflection l Compare this deflection to the Code allowable deflection

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Check the deflection of the steel channel previously designed for strength l The maximum deflection <= L / 500 W = 8kN L = 4m Loading Diagram Section = C15020E = 200 000 MPa I = 2.119 x 10 6 mm 4  = (5/384) x 8000 x 4000 3 / (200000 x 2.119 x 10 6 )  = (5/384) x WL 3 /EI mm ( Let us work in N and mm ) Maximum allowable deflection = 4000 / 500 = 16 mm = 8 mm deflects too much - need to chose stiffer section 24/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman Need twice as much I design for strength check for deflection 65 150 75 200 l Could use same section back to back 100% more material l A channel C20020 (200 deep 2mm thick) has twice the I but only 27% more material strategy for heavily loaded beams 25/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Given the loading conditions l Given the Code allowable deflection Use deflection formula to find I l Look up a table to find a suitable section Given load, W, span, L, and Modulus of Elasticity, E Use the Code limit — e.g., turn L/500 into millimetres Use deflection formula to find minimum value of I Look up tables or use I = bd 3 /12 and choose b and d 26/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman Beams need large I and Z in direction of bending Need stiffness in other direction to resist lateral buckling Some sections useful for both Columns usually need large value of r in both directions = better sections for beams 27/28

University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Deep beams are economical but subject to lateral buckling 28/28

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