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Wk1 2 Beams

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Wk1 2 Beams

  1. 1. <ul><li>Extremely common structural element </li></ul><ul><li>In buildings majority of loads are vertical and majority of useable surfaces are horizontal </li></ul>Beams 1/39
  2. 2. action of beams involves combination of bending and shear Beams 2/39 devices for transferring vertical loads horizontally
  3. 3. <ul><li>Be strong enough for the loads </li></ul>What Beams have to Do <ul><li>Not deflect too much </li></ul><ul><li>Suit the building for size, material, finish, </li></ul><ul><li>fixing etc </li></ul>3/39
  4. 4. Checking a Beam <ul><li>what we are trying to check (test) </li></ul><ul><li>adequate strength - will not break </li></ul><ul><li>stability - will not fall over </li></ul><ul><li>adequate functionality - will not deflect too much </li></ul><ul><li>what do we need to know </li></ul><ul><li>loads on the beam </li></ul><ul><li>span - how supported </li></ul><ul><li>material, shape & dimensions of beam </li></ul><ul><li>allowable strength & allowable deflection </li></ul>4/39
  5. 5. Designing a Beam <ul><li>what we are trying to do </li></ul><ul><li>determine shape & dimensions </li></ul><ul><li>what do we need to know </li></ul><ul><li>loads on the beam </li></ul><ul><li>span - how supported </li></ul><ul><li>material </li></ul><ul><li>allowable strength & allowable deflection </li></ul>5/39 ? ?
  6. 6. <ul><li>A beam picks up the load halfway to its neighbours </li></ul><ul><li>Each member also carries its own weight </li></ul>spacing span this beam supports the load that comes from this area Tributary Areas 6/39
  7. 7. <ul><li>A column generally picks up load from halfway to its neighbours </li></ul><ul><li>It also carries the load that comes from the floors above </li></ul>Tributary Areas (Cont. 1) 7/39
  8. 8. <ul><li>Code values per cubic metre or square metre </li></ul><ul><li>Multiply by the volume or area supported </li></ul>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. 9. <ul><li>Code values per square metre </li></ul><ul><li>Multiply by the area supported </li></ul>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. 10. <ul><li>Point loads, from concentrated loads or other beams </li></ul><ul><li>Distributed loads, from anything continuous </li></ul>Loads on Beams 10/39 Distributed Load Point Load Reactions
  11. 11. <ul><li>The loads (& reactions) bend the beam, and try to shear through it </li></ul>What the Loads Do 11/39 Bending Shear
  12. 12. What the Loads Do (cont.) 12/39 e Bending e e e C T Shear
  13. 13. <ul><li>in architectural structures, bending moment more important </li></ul><ul><ul><li>importance increases as span increases </li></ul></ul><ul><li>short span structures with heavy loads, shear dominant </li></ul><ul><ul><li>e.g. pin connecting engine parts </li></ul></ul>Designing Beams 13/39 beams in building designed for bending checked for shear
  14. 14. <ul><li>First, find ALL the forces (loads and reactions) </li></ul><ul><li>Make the beam into a freebody (cut it out and artificially support it) </li></ul><ul><li>Find the reactions , using the conditions of equilibrium </li></ul>How we Quantify the Effects 14/39
  15. 15. <ul><li>Consider cantilever beam with point load on end </li></ul><ul><li>Use the freebody idea to isolate part of the beam </li></ul><ul><li>Add in forces required for equilibrium </li></ul>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. 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. 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. 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. 19. <ul><li>To plot a diagram, we need a sign convention </li></ul><ul><li>The opposite convention is equally valid, </li></ul><ul><li>but this one is common </li></ul><ul><li>There is no difference in effect between </li></ul><ul><li>positive and negative shear forces </li></ul>Sign Conventions Shear Force Diagrams 19/39 “ Positive” shear “ Negative” shear L.H up R.H down L.H down R.H up
  20. 20. <ul><li>Starting at the left hand end, imitate each force you meet ( up or down ) </li></ul>Plotting the Shear Force Diagram 2039 Shear Force Diagram Diagram of loading R1 R2 W1 W2 W3 R1 R2 W1 W3 W2
  21. 21. <ul><li>Point loads produce </li></ul><ul><li>a block diagram </li></ul>Shear force diagrams Diagrams of loading Shape of the Shear Force Diagram <ul><li>Uniformly distributed loads </li></ul><ul><li>produce triangular diagrams </li></ul>21/39
  22. 22. <ul><li>Although the shear forces are vertical, shear stresses are both horizontal and vertical </li></ul><ul><li>Shear is seldom critical for steel </li></ul>What Shear Force does to the Beam <ul><li>Concrete needs </li></ul><ul><li>special shear reinforcement </li></ul><ul><li>(45 o or stirrups) </li></ul><ul><li>Timber may split </li></ul><ul><li>horizontally along </li></ul><ul><li>the grain </li></ul>22/39 Split in timber beam Reo in concrete beam
  23. 23. <ul><li>To plot a diagram, we need a sign convention </li></ul><ul><li>This convention is almost universally agreed </li></ul>Sign Conventions Bending Moment Diagrams 23/39 - Hogging NEGATIVE Sagging POSITIVE +
  24. 24. Sign Conventions Bending Moment Diagrams (cont.) 24/39 Sagging bending moment is POSITIVE (happy) + Hogging bending moment is NEGATIVE (sad) -
  25. 25. Positive and Negative Moments 25/39 Simple beam + <ul><li>Simple beams produce positive moments </li></ul><ul><li>Built-in & continuous beams have both, with negative over the supports </li></ul>Built-in beam - - + Cantilevers - - <ul><li>Cantilevers produce negative moments </li></ul>
  26. 26. <ul><li>Positive moments are drawn downwards </li></ul><ul><li>(textbooks are divided about this) </li></ul>Where to Draw the Bending Moment Diagram 26/39 This way is normal coordinate geometry + This way mimics the beam’s deflection +
  27. 27. <ul><li>Point loads produce triangular diagrams </li></ul>Shape of the Bending Moment Diagram 27/39 Diagrams of loading Bending moment diagrams
  28. 28. <ul><li>Distributed loads produce parabolic diagrams </li></ul>Shape of the Bending Moment Diagram (cont1.) 28/39 Bending moment diagrams UDL UDL Diagrams of loading
  29. 29. <ul><li>We are mainly concerned </li></ul><ul><li>with the maximum values </li></ul>Shape of the Bending Moment Diagram (cont.2) 29/39 Maximum value
  30. 30. <ul><li>Draw the Deflected Shape (exaggerate) </li></ul><ul><li>Use the Deflected shape as a guide to where the sagging ( + ) and hogging ( - ) moments are </li></ul>Shape of the Bending Moment Diagram (cont.3) 30/39 - + - - - + +
  31. 31. <ul><li>Cantilevered ends reduce the positive bending moment </li></ul><ul><li>Built-in and continuous beams also have lower maximum BMs and less deflection </li></ul>Can we Reduce the Maximum BM Values? 31/39 Simply supported Continuous
  32. 32. <ul><li>Use the standard formulas where you can </li></ul>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. 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. 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. 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. 36. <ul><li>Causes compression on one face and tension on the other </li></ul><ul><li>Causes the beam to deflect </li></ul>What the Bending Moment does to the Beam 34/37 How much deflection ? How much compressive stress ? How much tensile stress ?
  37. 37. <ul><li>It depends on the beam cross-section </li></ul>is the section we are using as a beam How to Calculate the Bending Stress <ul><li>We need some particular properties of the </li></ul><ul><li>section </li></ul>37/39 how big & what shape ?
  38. 38. <ul><li>Codes give maximum allowable stresses </li></ul><ul><li>Timber, depending on grade, can take 5 to 20 MPa </li></ul><ul><li>Steel can take around 165 MPa </li></ul><ul><li>Use of Codes comes later in the course </li></ul>What to do with the Bending Stress 38/39
  39. 39. next lecture Finding Section Properties 39/39 we need to find the Section Properties