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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 and majority of useable surfaces are horizontal 1/39

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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

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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

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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

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

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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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

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