Download presentation

Presentation is loading. Please wait.

Published byKristopher Strickland Modified over 4 years ago

1
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

2
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

3
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

4
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

5
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

6
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

7
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

8
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

9
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

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

11
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

12
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

13
University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman 12/28

14
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

15
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 =

16
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

17
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

18
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

19
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

20
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

21
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

22
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

23
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

24
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

25
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

26
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

27
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

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

29
University of Sydney – Structures SECTIONS Peter Smith & Mike Rosenman l Deep beams are economical but subject to lateral buckling 28/28

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google