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

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

Checking a Beam what we are trying to check (test) stability - will not fall over adequate strength - will not break adequate functionality - 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

Designing a Beam what we are trying to do what do we need to know ? determine shape & dimensions what do we need to know span - how supported loads on the beam material allowable strength & allowable deflection 5/39

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

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

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

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

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

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

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

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 beams in building designed for bending checked for shear 13/39

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

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

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

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

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

Sign Conventions Shear Force Diagrams To plot a diagram, we need a sign convention “Positive” shear “Negative” shear L.H up R.H down L.H down R.H up 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

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

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

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 Split in timber beam Shear is seldom critical for steel Reo in concrete beam Concrete needs special shear reinforcement (45o or stirrups) 22/39

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

Bending Moment Diagrams (cont.) Sign Conventions Bending Moment Diagrams (cont.) Sagging bending moment is POSITIVE (happy) + Hogging bending moment is NEGATIVE (sad) - 24/39

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

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

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

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

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

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

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 Simply supported Continuous 31/39

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

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

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

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

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

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

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

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