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# Wk1 2 Beams

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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
• 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
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