Bending II: Equilibrium of Beams Equilibrium of Beams and Shear

Bending Moment Along a Beam• In this lecture, we will learn how to find the bending moment M alongthe beam.• Mi is not necessarily constant; t sometimes Mi is a function of x.• We will also solve for the shear force V(x), which will be used inChapter 6.• As before, we need a new FBD every time the loading changes.1 2 3 12Slide 2

Review of Beam Supports• 3 equilibrium equations: Σ F Y = 0, Σ F V = 0, Σ M = 0• Ignore the horizontal (x-direction) components, because these areaxial loading.R 1 R 3 R 1 R 3 R 1 M 1R 2 R 2R 2RR 1M R 11 R R RR 3 R 1 MR 3 1 3 M 242R22 R 4Slide 3

Sign Convention• Recall the applied loading results in both a bending moment M and ashear force V.Positive shear and bending momentSlide 4

Example Problem 1SOLUTION:• Treating the entire beam as a rigidbody, determine the reaction forcesFor the timber beam and loadingshown, draw the shear and bendmomentdiagrams and determine themaximum normal stress due tobending.• Section the beam at points nearsupports and load application points.Apply equilibrium analyses onresulting free-bodies to determineinternal shear forces and bendingcouples• Identify the maximum shear andbending-moment from plots of theirdistributions.• Apply the elastic flexure formulas todetermine the correspondingmaximum normal stress.

SolutionSOLUTION:• Treating the entire beam as a rigid body, determinethe reaction forcesfrom F = 0 = M : R = 46kN R 14kN∑ ∑ =yB• Section the beam and apply equilibrium analyseson resulting free-bodies∑ = 0∑ M = 01− 20 kN −V1= 0( 20kN)( 0m) + M = 0 M = 0B1V1D= −20kN1F y∑ =0F y( )( ) m− 20 kN −V2=0V2= −20kN∑ M =2020kN2.5m+ M2=0M2= −50kN⋅VV34VV56= + 26kN= + 26kN= −14kNMMM= −14kNM3456= −50kN⋅ m= + 28kN ⋅ m= + 28kN⋅ m= 0

Solution• Identify the maximum shear and bendingmomentfrom plots of their distributions.Vm= 26 kNM=M=50kN ⋅m= m B• Apply the elastic flexure formulas todetermine the correspondingmaximum normal stress.S( 0.080m)( 0.250 ) 21 2b =16 6m−63= h= 833.33×10mσ mM3B50×10N ⋅m=833.3310−6 m3×= Sσ m=60.0×106Pa

Example Problem 2Draw the shear and bending moment diagrams for the beam andloading shown, and determine the maximum absolute value (a) of theshear and (b) of the bending moment.Slide 8

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Relations Between F, V, and M• For beams with more complicated loading, it is helpful to develop arelationship between load, shear and bending moment.Following our signconvention, W ispositive downwards!• Sum forces in the vertical direction.∑ F = − ( + Δ ) − Δ xyV V V w = 0dVdx−w⇒ΔV= −wΔx= or V −V= − wdxDCxD∫xCSlide 10

Relations Between F, V, and M cont’d• Sum forces in the vertical direction.∑Δx2 =( M + ΔM) − M −VΔx+ wΔx0M C=1ΔM= VΔx−2( Δx) 2• Neglect (Δx) 2 term since it is much smaller thanΔx term.wdM =dxVorMD−MC=xxD∫CVdxSlide 11

Example Problem 3Determine (a) the equations of the shear and bending moment curvesfor the beam and loading shown, and (b) the maximum absolute valueof the bending moment in the beam.Slide 12

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• Sample problem 5.3 inBeer’s book (v 5)Example 4Slide 14

Shear - Review• So far only dealt with normal stresses caused by bending moments.• This chapter deals with shear stress caused by shear forces.Line of failureSlide 15

Shear Stress in BeamsSlide 16• Consider the effects of shear force (V).• Already know how to find resulting axial force andmoment due to stress σ x from Chapter 5.• We have two more equations for shear stress:– Total shear force in the y-direction:– Total shear force in the z-direction: ∫ τ xz dA = 0∫τ xydA∫ 0= V

Shear Stress in Beams cont’d• Consider a cantilever beam composed of separate planks clampedat one end:Pure bendingShear force• Shear force causes tendency to “slide.”• Stresses are equal in horizontal andvertical directions.Slide 17

Shear Stress: Horizontal• Let us consider the horizontal component (τ xz = τ zx ).• Cut a section with cross-sectionalarea a at a distance y 1 above thecentroid.FBD →Slide 18

Shear Stress: Horizontal cont’d• ΔH is the horizontal shearing force.• Element width is Δx.• Sum forces in x-direction:∑ Fx= ΔH+ ∫ σC−• Recall:∫ ( σ) Dda = 0aσ =MyI• Solve for ΔH and use equation for σ:ΔH=⎛M−MI⎛MD CD C∫ ( σD−σC) da =∫ ⎜ ⎟yda= ⎜∫aa⎝⎞⎠⎝−MI⎞⎟⎠aydaSlide 19

Shear Stress: Horizontal cont’d• Recall first moment, Q, is defined as:Q = ∫ yda• The term M D -M C can be rewritten as:aMD− MC= ΔM=dMdxΔx= VΔx• Applying this to our equation for ΔH:ΔH=VQIΔx• We can rearrange this to define horizontal shear per unit length, q,called shear flow.ΔHVQq = =ΔxISlide 20

Side Note on Q• Q is the definition of the centroid for the area above y 1 ,Q= ∫ayda=a ywhere y bar is the distance between the centroid of the shadedsection and the centroid of beam cross-section.Slide 21

Shear Stress: Vertical• Now, let us consider the vertical component (τ xy = τ yx ).• We can calculate the average vertical shear stress on the cross-section.ΔH⎛VQ⎞⎛1 ⎞ VQτAVE= = ⎜ Δx⎟⎜⎟ = =τΔA⎝ I ⎠⎝tΔx⎠ Itτ AVE=VQItAVESlide 22

Shear Stress: Vertical cont’d• So, where is τ AVE maximum and minimum?– Use Q to find out.– Q = 0 at top and bottom surfaces– Q = maximum somewhere in betweenmax normal stressshear stress = 0max shear stressnormal stress = 0max normal stressshear stress = 0Slide 23

Shearing Stress in Common Shapes• Rectangular cross-section ( ) ( ) (2 2Q = Ay = b c − y c + y = b c − y )1212τxy=VQIb=2 2( c − y ) 3V2 2= ( c − y )V b2 2b 4bc3( b( 2c)/12) 43τ xy3V⎛ y⎜1−2A⎝ c2=2⎞⎟ ⎠τ max=3VV2ASlide 24

Shearing Stress in Common Shapes cont’d• Beams with flanges– Vertical shear stresses are larger in the web than in the flange.– Usually only calculate the values in the web.– Ignore the effects of the small fillets at the corners.– Flanges have large horizontal shear stresses, which we will learn how tocalculate later on.FlangeWebSlide 25

Example Problem 5A beam having the cross section shown is subjected to a vertical shearV. Determine the horizontal line along which the shearing stress ismaximum.Slide 26

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Example Problem 6For the beam and loading shown, consider section n-n and determinethe shearing stress at (a) point a, (b) point b.Slide 28

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