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  • 1.
    • Extremely common structural element
    • In buildings majority of loads are vertical and majority of useable surfaces are horizontal
    Beams 1/39
  • 2. action of beams involves combination of bending and shear Beams 2/39 devices for transferring vertical loads horizontally
  • 3.
    • 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
  • 4. 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
  • 5. 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 ? ?
  • 6.
    • 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
  • 7.
    • 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
  • 8.
    • 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
  • 9.
    • 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
  • 10.
    • Point loads, from concentrated loads or other beams
    • Distributed loads, from anything continuous
    Loads on Beams 10/39 Distributed Load Point Load Reactions
  • 11.
    • The loads (& reactions) bend the beam, and try to shear through it
    What the Loads Do 11/39 Bending Shear
  • 12. What the Loads Do (cont.) 12/39 e Bending e e e C T Shear
  • 13.
    • 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
  • 14.
    • 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
  • 15.
    • 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
  • 16. 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
  • 17. 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
  • 18. 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
  • 19.
    • 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
  • 20.
    • 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
  • 21.
    • 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
  • 22.
    • 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
  • 23.
    • 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 +
  • 24. Sign Conventions Bending Moment Diagrams (cont.) 24/39 Sagging bending moment is POSITIVE (happy) + Hogging bending moment is NEGATIVE (sad) -
  • 25. 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
  • 26.
    • 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 +
  • 27.
    • Point loads produce triangular diagrams
    Shape of the Bending Moment Diagram 27/39 Diagrams of loading Bending moment diagrams
  • 28.
    • Distributed loads produce parabolic diagrams
    Shape of the Bending Moment Diagram (cont1.) 28/39 Bending moment diagrams UDL UDL Diagrams of loading
  • 29.
    • We are mainly concerned
    • with the maximum values
    Shape of the Bending Moment Diagram (cont.2) 29/39 Maximum value
  • 30.
    • 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 - + - - - + +
  • 31.
    • 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
  • 32.
    • 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)
  • 33. 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
  • 34. 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
  • 35. 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
  • 36.
    • 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 ?
  • 37.
    • 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 ?
  • 38.
    • 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
  • 39. next lecture Finding Section Properties 39/39 we need to find the Section Properties