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University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Extremely common structural element l In buildings majority of loads are vertical.

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Presentation on theme: "University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Extremely common structural element l In buildings majority of loads are vertical."— Presentation transcript:

1 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Extremely common structural element l In buildings majority of loads are vertical and majority of useable surfaces are horizontal 1/39

2 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman devices for transferring vertical loads horizontally action of beams involves combination of bending and shear 2/39

3 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Be strong enough for the loads l Not deflect too much l Suit the building for size, material, finish, fixing etc 3/39

4 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Checking a Beam l what we are trying to check (test) l adequate strength - will not break l stability - will not fall over l adequate functionality - will not deflect too much l what do we need to know l loads on the beam l span - how supported l material, shape & dimensions of beam l allowable strength & allowable deflection 4/39

5 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Designing a Beam l what we are trying to do l determine shape & dimensions l what do we need to know l loads on the beam l span - how supported l material l allowable strength & allowable deflection ? ? 5/39

6 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l A beam picks up the load halfway to its neighbours l Each member also carries its own weight spacing span this beam supports the load that comes from this area 6/39

7 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l A column generally picks up load from halfway to its neighbours l It also carries the load that comes from the floors above 7/39

8 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Code values per cubic metre or square metre l Multiply by the volume or area supported Length Height Thickness Load = Surface area x Wt per sq m, or volume x wt per cu m 8/39

9 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Code values per square metre l Multiply by the area supported Total Load = area x (Live load + Dead load) per sq m + self weight Area carried by one beam 9/39

10 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Point loads, from concentrated loads or other beams Distributed loads, from anything continuous Distributed Load 10/39 Point Load Reactions

11 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l The loads (& reactions) bend the beam, and try to shear through it BendingShear 11/39

12 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman 12/39 e Bending e e e C T Shear

13 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman in architectural structures, bending moment more important ●importance increases as span increases short span structures with heavy loads, shear dominant ●e.g. pin connecting engine parts beams in building designed for bending checked for shear 13/39

14 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l First, find ALL the forces (loads and reactions) l Make the beam into a freebody (cut it out and artificially support it) l Find the reactions, using the conditions of equilibrium 14/39

15 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Consider cantilever beam with point load on end W L R = W M R = -WL vertical reaction, R = W and moment reaction M R = - WL l Use the freebody idea to isolate part of the beam l Add in forces required for equilibrium 15/39

16 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Take section anywhere at distance, x from end Shear V = W constant along length (X = 0 -> L) Add in forces, V = W and moment M = - Wx V = W Shear Force Diagram Bending Moment BM = W.x when x = L BM = WL when x = 0 BM = 0 Bending Moment Diagram BM = WL BM = Wx x W V = W M = -Wx 16/39

17 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman For maximum shear V and bending moment BM L w /unit length vertical reaction, R = W = wL and moment reaction M R = - WL/2 = - wL 2 /2 Total Load W = w.L L/2 R = W = wL M R = -WL/2 = -wL 2 /2 17/39

18 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Take section anywhere at distance, x from end Shear V = wx when x = L V = W = wL when x = 0 V = 0 Add in forces, V = w.x and moment M = - wx.x/2 Bending Moment BM = w.x 2 /2 when x = L BM = wL 2 /2 = WL/2 when x = 0 BM = 0 (parabolic) V = wL = W Shear Force Diagram X/2 wx X/2 For distributed V and BM V = wx M = -wx 2 /2 Bending Moment Diagram BM = wL 2 /2 = WL/2 BM = wx /2 2 18/39

19 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l To plot a diagram, we need a sign convention “Positive” shear “Negative” shear L.H up R.H downL.H down R.H up l The opposite convention is equally valid, but this one is common l There is no difference in effect between positive and negative shear forces 19/39

20 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Shear Force Diagram l Starting at the left hand end, imitate each force you meet (up or down) Diagram of loading R1 R2 W1W2 W3 R1 R2 W1 W3 W2 2039

21 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Point loads produce a block diagram Shear force diagrams Diagrams of loading l Uniformly distributed loads produce triangular diagrams 21/39

22 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Although the shear forces are vertical, shear stresses are both horizontal and vertical Split in timber beamReo in concrete beam l Shear is seldom critical for steel l Concrete needs special shear reinforcement (45 o or stirrups) l Timber may split horizontally along the grain 22/39

23 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l To plot a diagram, we need a sign convention l This convention is almost universally agreed - Hogging NEGATIVE Sagging POSITIVE + 23/39

24 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Sagging bending moment is POSITIVE (happy) + Hogging bending moment is NEGATIVE (sad) - 24/39

25 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Simple beam + l Simple beams produce positive moments l Built-in & continuous beams have both, with negative over the supports Built-in beam -- + Cantilevers - - l Cantilevers produce negative moments 25/39

26 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Positive moments are drawn downwards (textbooks are divided about this) This way is normal coordinate geometry + This way mimics the beam’s deflection + 26/39

27 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Point loads produce triangular diagrams Diagrams of loading Bending moment diagrams 27/39

28 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Distributed loads produce parabolic diagrams Bending moment diagrams UDL Diagrams of loading 28/39

29 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Maximum value l We are mainly concerned with the maximum values 29/39

30 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Draw the Deflected Shape (exaggerate) l Use the Deflected shape as a guide to where the sagging (+) and hogging (-) moments are /39

31 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Cantilevered ends reduce the positive bending moment l Built-in and continuous beams also have lower maximum BMs and less deflection Simply supported Continuous 31/39

32 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Use the standard formulas where you can L Central point load Max bending moment = WL/4 Uniformly distributed load Max bending moment = WL/8 or w L 2 /8 where W = w L L W Total load = W ( w per metre length) 32/39

33 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman End point load Max bending moment = -WL Uniformly Distributed Load Max bending moment = -WL/2 or - w L 2 /2 where W = w L W L ( w per metre length) Total load = W L 33/39

34 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman W W Beam Cable BMD WW WW W WW W W W W /m 34/39

35 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman V = +W M max = -WL L WW L V = +W/2 V = -W/2 M max = WL/4 L W = w L V = +W/2 V = -W/2 M max = WL/8 = w L 2 /8 L W = w L V = +W M max = -WL/2 = - w L 2 /2 35/39

36 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l Causes compression on one face and tension on the other l Causes the beam to deflect How much deflection? How much compressive stress? How much tensile stress? 34/37

37 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman l It depends on the beam cross-section how big & what shape ? is the section we are using as a beam l We need some particular properties of the section 37/39

38 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman Codes give maximum allowable stresses l Timber, depending on grade, can take 5 to 20 MPa l Steel can take around 165 MPa l Use of Codes comes later in the course 38/39

39 University of Sydney – Structures BEAMS Peter Smith & Mike Rosenman we need to find the Section Properties next lecture 39/39


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