Wk1 2 Beams
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Wk1 2 Beams

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    Wk1 2 Beams Wk1 2 Beams Presentation Transcript

      • Extremely common structural element
      • In buildings majority of loads are vertical and majority of useable surfaces are horizontal
      Beams 1/39
    • action of beams involves combination of bending and shear Beams 2/39 devices for transferring vertical loads horizontally
      • Be strong enough for the loads
      What Beams have to Do
      • Not deflect too much
      • Suit the building for size, material, finish,
      • fixing etc
      3/39
    • Checking a Beam
      • what we are trying to check (test)
      • adequate strength - will not break
      • stability - will not fall over
      • adequate functionality - will not deflect too much
      • what do we need to know
      • loads on the beam
      • span - how supported
      • material, shape & dimensions of beam
      • allowable strength & allowable deflection
      4/39
    • Designing a Beam
      • what we are trying to do
      • determine shape & dimensions
      • what do we need to know
      • loads on the beam
      • span - how supported
      • material
      • allowable strength & allowable deflection
      5/39 ? ?
      • A beam picks up the load halfway to its neighbours
      • Each member also carries its own weight
      spacing span this beam supports the load that comes from this area Tributary Areas 6/39
      • A column generally picks up load from halfway to its neighbours
      • It also carries the load that comes from the floors above
      Tributary Areas (Cont. 1) 7/39
      • Code values per cubic metre or square metre
      • Multiply by the volume or area supported
      Dead Loads on Elements 8/39 Length Height Thickness Load = Surface area x Wt per sq m, or volume x wt per cu m
      • Code values per square metre
      • Multiply by the area supported
      Total Load = area x (Live load + Dead load) per sq m + self weight Live Loads on Elements Area carried by one beam 9/39
      • Point loads, from concentrated loads or other beams
      • Distributed loads, from anything continuous
      Loads on Beams 10/39 Distributed Load Point Load Reactions
      • The loads (& reactions) bend the beam, and try to shear through it
      What the Loads Do 11/39 Bending Shear
    • What the Loads Do (cont.) 12/39 e Bending e e e C T Shear
      • 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
      Designing Beams 13/39 beams in building designed for bending checked for shear
      • First, find ALL the forces (loads and reactions)
      • Make the beam into a freebody (cut it out and artificially support it)
      • Find the reactions , using the conditions of equilibrium
      How we Quantify the Effects 14/39
      • Consider cantilever beam with point load on end
      • Use the freebody idea to isolate part of the beam
      • Add in forces required for equilibrium
      Example 1 - Cantilever Beam Point Load at End 15/39 W L R = W M R = -WL vertical reaction, R = W and moment reaction M R = - WL
    • 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 Bending Moment BM = W.x when x = L BM = WL when x = 0 BM = 0 Example 1 - Cantilever Beam Point Load at End (cont1.) 16/39 V = W Shear Force Diagram Bending Moment Diagram BM = WL BM = Wx x W V = W M = -Wx
    • For maximum shear V and bending moment BM vertical reaction, R = W = wL and moment reaction M R = - WL/2 = - wL 2 /2 Example 2 - Cantilever Beam Uniformly Distributed Load 17/39 L w /unit length Total Load W = w.L L/2 L/2 R = W = wL M R = - WL/2 = - wL 2 /2
    • 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) For distributed V and BM Example 2 - Cantilever Beam Uniformly Distributed Load (cont.) 18/39 V = wL = W Shear Force Diagram X/2 wx X/2 V = wx M = -wx 2 /2 Bending Moment Diagram BM = wL 2 /2 = WL/2 BM = wx /2 2
      • To plot a diagram, we need a sign convention
      • The opposite convention is equally valid,
      • but this one is common
      • There is no difference in effect between
      • positive and negative shear forces
      Sign Conventions Shear Force Diagrams 19/39 “ Positive” shear “ Negative” shear L.H up R.H down L.H down R.H up
      • Starting at the left hand end, imitate each force you meet ( up or down )
      Plotting the Shear Force Diagram 2039 Shear Force Diagram Diagram of loading R1 R2 W1 W2 W3 R1 R2 W1 W3 W2
      • Point loads produce
      • a block diagram
      Shear force diagrams Diagrams of loading Shape of the Shear Force Diagram
      • Uniformly distributed loads
      • produce triangular diagrams
      21/39
      • Although the shear forces are vertical, shear stresses are both horizontal and vertical
      • Shear is seldom critical for steel
      What Shear Force does to the Beam
      • Concrete needs
      • special shear reinforcement
      • (45 o or stirrups)
      • Timber may split
      • horizontally along
      • the grain
      22/39 Split in timber beam Reo in concrete beam
      • To plot a diagram, we need a sign convention
      • This convention is almost universally agreed
      Sign Conventions Bending Moment Diagrams 23/39 - Hogging NEGATIVE Sagging POSITIVE +
    • Sign Conventions Bending Moment Diagrams (cont.) 24/39 Sagging bending moment is POSITIVE (happy) + Hogging bending moment is NEGATIVE (sad) -
    • Positive and Negative Moments 25/39 Simple beam +
      • Simple beams produce positive moments
      • Built-in & continuous beams have both, with negative over the supports
      Built-in beam - - + Cantilevers - -
      • Cantilevers produce negative moments
      • Positive moments are drawn downwards
      • (textbooks are divided about this)
      Where to Draw the Bending Moment Diagram 26/39 This way is normal coordinate geometry + This way mimics the beam’s deflection +
      • Point loads produce triangular diagrams
      Shape of the Bending Moment Diagram 27/39 Diagrams of loading Bending moment diagrams
      • Distributed loads produce parabolic diagrams
      Shape of the Bending Moment Diagram (cont1.) 28/39 Bending moment diagrams UDL UDL Diagrams of loading
      • We are mainly concerned
      • with the maximum values
      Shape of the Bending Moment Diagram (cont.2) 29/39 Maximum value
      • Draw the Deflected Shape (exaggerate)
      • Use the Deflected shape as a guide to where the sagging ( + ) and hogging ( - ) moments are
      Shape of the Bending Moment Diagram (cont.3) 30/39 - + - - - + +
      • Cantilevered ends reduce the positive bending moment
      • Built-in and continuous beams also have lower maximum BMs and less deflection
      Can we Reduce the Maximum BM Values? 31/39 Simply supported Continuous
      • Use the standard formulas where you can
      Standard BM Coefficients Simply Supported Beams 32/39 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)
    • Standard BM Coefficients Cantilevers 33/39 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
    • Standard BM Coefficients Simple Beams Beam Cable BMD 34/39 W W W W W W W W W W W W W /m W /m
    • SFD & BMD Simply Supported Beams 35/39 V = +W M max = -WL L W W 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
      • Causes compression on one face and tension on the other
      • Causes the beam to deflect
      What the Bending Moment does to the Beam 34/37 How much deflection ? How much compressive stress ? How much tensile stress ?
      • It depends on the beam cross-section
      is the section we are using as a beam How to Calculate the Bending Stress
      • We need some particular properties of the
      • section
      37/39 how big & what shape ?
      • Codes give maximum allowable stresses
      • Timber, depending on grade, can take 5 to 20 MPa
      • Steel can take around 165 MPa
      • Use of Codes comes later in the course
      What to do with the Bending Stress 38/39
    • next lecture Finding Section Properties 39/39 we need to find the Section Properties