Wk1 2 Beams

[object Object],[object Object],Beams 1/39

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

[object Object],What Beams have to Do ,[object Object],[object Object],[object Object],3/39

Checking a Beam ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],4/39

Designing a Beam ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],5/39 ? ?

[object Object],[object Object],spacing span this beam supports the load that comes from this area Tributary Areas 6/39

[object Object],[object Object],Tributary Areas (Cont. 1) 7/39

[object Object],[object Object],Dead Loads on Elements 8/39 Length Height Thickness Load = Surface area x Wt per sq m, or volume x wt per cu m

[object Object],[object Object],Total Load = area x (Live load + Dead load) per sq m + self weight Live Loads on Elements Area carried by one beam 9/39

[object Object],[object Object],Loads on Beams 10/39 Distributed Load Point Load Reactions

[object Object],What the Loads Do 11/39 Bending Shear

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

[object Object],[object Object],[object Object],[object Object],Designing Beams 13/39 beams in building designed for bending checked for shear

[object Object],[object Object],[object Object],How we Quantify the Effects 14/39

[object Object],[object Object],[object Object],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

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

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

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

[object Object],[object Object],[object Object],[object Object],[object Object],Sign Conventions Shear Force Diagrams 19/39 “ Positive” shear “ Negative” shear L.H up R.H down L.H down R.H up

[object Object],Plotting the Shear Force Diagram 2039 Shear Force Diagram Diagram of loading R1 R2 W1 W2 W3 R1 R2 W1 W3 W2

[object Object],[object Object],Shear force diagrams Diagrams of loading Shape of the Shear Force Diagram ,[object Object],[object Object],21/39

[object Object],[object Object],What Shear Force does to the Beam ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],22/39 Split in timber beam Reo in concrete beam

[object Object],[object Object],Sign Conventions Bending Moment Diagrams 23/39 - Hogging NEGATIVE Sagging POSITIVE +

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

Positive and Negative Moments 25/39 Simple beam + ,[object Object],[object Object],Built-in beam - - + Cantilevers - - ,[object Object]

[object Object],[object Object],Where to Draw the Bending Moment Diagram 26/39 This way is normal coordinate geometry + This way mimics the beam’s deflection +

[object Object],Shape of the Bending Moment Diagram 27/39 Diagrams of loading Bending moment diagrams

[object Object],Shape of the Bending Moment Diagram (cont1.) 28/39 Bending moment diagrams UDL UDL Diagrams of loading

[object Object],[object Object],Shape of the Bending Moment Diagram (cont.2) 29/39 Maximum value

[object Object],[object Object],Shape of the Bending Moment Diagram (cont.3) 30/39 - + - - - + +

[object Object],[object Object],Can we Reduce the Maximum BM Values? 31/39 Simply supported Continuous

[object Object],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)

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

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

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

[object Object],[object Object],What the Bending Moment does to the Beam 34/37 How much deflection ? How much compressive stress ? How much tensile stress ?

[object Object],is the section we are using as a beam How to Calculate the Bending Stress ,[object Object],[object Object],37/39 how big & what shape ?

[object Object],[object Object],[object Object],[object Object],What to do with the Bending Stress 38/39

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

Wk1 2 Beams

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