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Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are

Beams. Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are horizontal. 1/39. devices for transferring vertical loads horizontally. Beams. action of beams involves combination of bending and shear. 2/39.

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Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are

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  1. Beams • Extremely common structural element • In buildings majority of loads are vertical and majority of useable surfaces are horizontal 1/39

  2. devices for transferring vertical loads horizontally Beams action of beams involves combination of bending and shear 2/39

  3. What Beams have to Do • Be strong enough for the loads • 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) • stability - will not fall over • adequatestrength - will not break • adequatefunctionality - will not deflect too much • what do we need to know • span - how supported • loads on the beam • 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 • span - how supported • loads on the beam • material • allowable strength & allowable deflection 5/39

  6. Tributary Areas • A beam picks up the load halfway to its neighbours • Each member also carries its own weight this beam supports the load that comes from this area span spacing 6/39

  7. Tributary Areas (Cont. 1) • A column generally picks up load from halfway to its neighbours • It also carries the load that comes from the floors above 7/39

  8. Length Thickness Load = Surface area x Wt per sq m, or volume x wt per cu m Height Dead Loads on Elements • Code values per cubic metre or square metre • Multiply by the volume or area supported 8/39

  9. Live Loads on Elements • Code values per square metre • Multiply by the area supported Area carried by one beam Total Load =area x (Live load + Dead load) per sq m +self weight 9/39

  10. Distributed Load Point Load Reactions Loads on Beams • Point loads, from concentrated loads or other beams • Distributed loads, from anything continuous 10/39

  11. Bending Shear What the Loads Do • The loads (& reactions) bend the beam, and try to shearthrough it 11/39

  12. e e e e C T Bending Shear What the Loads Do (cont.) 12/39

  13. beams in building designed for bending checked for shear Designing Beams • 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 13/39

  14. How we Quantify the Effects • 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 14/39

  15. W vertical reaction, R = W and moment reaction MR = - WL MR= -WL L R = W Example 1 - Cantilever Beam Point Load at End • Consider cantilever beam with point load on end • Use the freebody idea to isolate part of the beam • Add in forces required for equilibrium 15/39

  16. W M = -Wx x V = W V = W Shear Force Diagram BM = Wx BM = WL Bending Moment Diagram Example 1 - Cantilever Beam Point Load at End (cont1.) Take section anywhere at distance, x from end Add in forces, V = W and moment M = - Wx Shear V = W constant along length (X = 0 -> L) Bending Moment BM = W.x when x = L BM = WL when x = 0 BM = 0 16/39

  17. w /unit length L Total Load W = w.L MR = -WL/2 = -wL2/2 L/2 L/2 R = W = wL Example 2 - Cantilever Beam Uniformly Distributed Load Formaximum shear V and bending moment BM vertical reaction,R = W = wL and moment reactionMR = - WL/2 = - wL2/2 17/39

  18. wx M = -wx2/2 X/2 X/2 V = wx V = wL = W Shear Force Diagram BM = wx /2 2 BM = wL2/2 = WL/2 Bending Moment Diagram Example 2 - Cantilever Beam Uniformly Distributed Load (cont.) For distributed V and BM Take section anywhere at distance, x from end Add in forces,V = w.xand momentM = - wx.x/2 ShearV = wx when x = LV = W = wL when x = 0V = 0 Bending MomentBM = w.x2/2 when x = LBM = wL2/2 = WL/2 when x = 0BM = 0 (parabolic) 18/39

  19. “Positive” shear “Negative” shear R.H down L.H down L.H up R.H up Sign Conventions Shear Force Diagrams • 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 19/39

  20. W3 W1 W2 R1 Diagram of loading R2 W1 R1 W2 R2 W3 Shear Force Diagram Plotting the Shear Force Diagram • Starting at the left hand end, imitate each force you meet (up or down) 2039

  21. Shape of the Shear Force Diagram • Point loads produce a block diagram • Uniformly distributed loads produce triangular diagrams Diagrams of loading Shear force diagrams 21/39

  22. Split in timber beam Reo in concrete beam What Shear Force does to the Beam • Although the shear forces are vertical, shear stresses are both horizontal and vertical • Timber may split horizontally along the grain • Shear is seldom critical for steel • Concrete needs • special shear reinforcement • (45o or stirrups) 22/39

  23. Hogging NEGATIVE Sagging POSITIVE - + Sign Conventions Bending Moment Diagrams • To plot a diagram, we need a sign convention • This convention is almost universally agreed 23/39

  24. Saggingbending momentisPOSITIVE (happy) + Hoggingbending momentisNEGATIVE (sad) - Sign Conventions Bending Moment Diagrams (cont.) 24/39

  25. Cantilevers produce negative moments - - Cantilevers • Simple beams produce positive moments - - + + Simple beam Built-in beam • Built-in & continuous beams have both, with negative over the supports Positive and Negative Moments 25/39

  26. + + This way mimics the beam’s deflection This way is normal coordinate geometry Where to Draw the Bending Moment Diagram • Positive moments are drawn downwards (textbooks are divided about this) 26/39

  27. Diagrams of loading Bending moment diagrams Shape of the Bending Moment Diagram • Point loads produce triangular diagrams 27/39

  28. UDL UDL Diagrams of loading Bending moment diagrams Shape of the Bending Moment Diagram (cont1.) • Distributed loads produceparabolic diagrams 28/39

  29. Maximum value Shape of the Bending Moment Diagram (cont.2) • We are mainly concerned with the maximum values 29/39

  30. - - - - + + + Shape of the Bending Moment Diagram (cont.3) • Draw theDeflected Shape(exaggerate) • Use the Deflected shape as a guide to where the sagging (+) and hogging (-) moments are 30/39

  31. Simply supported Continuous Can we Reduce the Maximum BM Values? • Cantilevered ends reduce the positive bending moment • Built-in and continuous beams also have lower maximum BMs and less deflection 31/39

  32. Total load = W W (w per metre length) L L Central point load Max bending moment = WL/4 Uniformly distributed load Max bending moment = WL/8 or wL2/8 where W = wL Standard BM Coefficients Simply Supported Beams • Use the standard formulas where you can 32/39

  33. Total load = W W (w per metre length) L L End point load Max bending moment = -WL Uniformly Distributed Load Max bending moment = -WL/2 or -wL2/2 where W = wL Standard BM Coefficients Cantilevers 33/39

  34. W W W /m W W W W W W W W W /m W W Standard BM Coefficients Simple Beams Beam Cable BMD 34/39

  35. W W W = wL W = wL L L L L V = +W/2 V = +W/2 V = +W V = +W V = -W/2 V = -W/2 Mmax = WL/8 = wL2/8 Mmax = -WL Mmax = -WL/2 = -wL2/2 Mmax = WL/4 SFD & BMD Simply Supported Beams 35/39

  36. How much compressive stress? How much tensile stress? How much deflection? What the Bending Moment does to the Beam • Causes compression on one face and tension on the other • Causes the beam to deflect 34/37

  37. how big & what shape? How to Calculate the Bending Stress • It depends on the beam cross-section • We need some particular properties of the section is the section we are using as a beam 37/39

  38. What to do with the Bending Stress • 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 38/39

  39. we need to find the Section Properties Finding Section Properties next lecture 39/39

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