ssppaatteerrSTRUCTURAL MECHANICS Lecture 6 : Simple Bending Theory and Beam Deflection When a beam is subjected to a bending moment M : M M Bending theory relates the moment to the beam’s curvature, deflection and stress. Assumptions: 1) 2) 3) 4) Small deflections Linear-elastic behaviour Plane sections remain plane - line ab straight before and after bending Pure bending - no shear or axial forces - in practice stress and deflections due to shear and axial forces can be calculated separately and added on using superposition. Consider a section of a beam subjected to pure bending: C ′D ′ = ( R - y) AB = CD = A ′B′ = δx 14/09/2006 1 δx R SM lecture 6 which is zero when there is pure bending.y ) δx R .e. i. the neutral axis position has to be calculated by considering the total axial force acting across the section. For beams that are not symmetric. the neutral axis is at mid height of the section. 14/09/2006 2 SM lecture 6 .C ′D ′ . rectangular and circular cross-section beams and I beams with flanges of equal size : For all beams of these types.CD = Strain in C ′D ′ = CD ( R .δx δx y = R y ∴ ε = R ssppaatteerr Hence strain distribution : +ve y gives -ve ε -ve y gives +ve ε (tension) (compression) If linear-elastic : σ = Eε (Stress ∝ Strain) ∴ σ = - Ey R Hence stress distribution: Neutral axis position We will only consider beams that have cross-sections that are symmetric about both the horizontal and vertical axes. = yσδA ∴ total moment M = - ∫A yσδA (∫ A indicates integration over the whole area) (-ve due to sign convention.ssppaatteerr Moment equilibrium y -σ Area δA Moment M x (N. I I is a property of a cross-section that indicates how effective it will be in resisting bending. +ve moment when there is a stress decrease in y direction) Now σ = - ∴ M = ∫A y 2 Ey R E R ∫Ay2dA dA is called the SECOND MOMENT OF AREA about N.) +σ Cross-section of beam Magnitude of stress σ down section Axial force on area δA = σδA ∴ moment of force on area δA about N.A.A. ∴ M E = I R Since σ = - Ey R Then σ = - My I or overall bending equation : 14/09/2006 M E σ = = I R y 3 SM lecture 6 . (x-axis).A. π r4 4 Derivation of second moment of area equations Rectangular cross-section : y b/2 b/2 d I = ∫ y 2 dA = ∫ 2d by 2 dy δA d/2 − A ∫ x A since dA = bdy 2 indicates integration over the whole area d ⎡ y3 ⎤ 2 bd 3 ∴ I = b⎢ ⎥ = ⎣ 3 ⎦ − d 12 d/2 2 Circular cross-section : y rcosθ r θ δA θ δy = rcosθδθ rsinθ δθ x r θ I = ∫ y 2 dA where y = rsinθ and dA = 2rcosθdy = 2r2cos2θdθ A π π ⎛ sin 2θ ⎞ ∴ I = ∫ π 2 r sin θ cos θ dθ = 2r ∫ π ⎜ ⎟ dθ − − ⎝ 2 ⎠ 2 2 4 2 = r4 4 r4 = 4 2 2 π ∫ π (1 − cos 4θ )dθ 2 − 4 2 since sin 2 2θ = sin 2θ 2 (1 − cos 4θ ) 2 π sin 4θ ⎤ 2 ⎡ θ − ⎢⎣ 4 ⎥⎦ − π since sin θ cos θ = 2 ∴ I= 2 π r4 4 2 14/09/2006 4 SM lecture 6 .A. I= 3 bd 12 I= N.ssppaatteerr Second Moment of Area of Common Shapes Rectangular cross-section Circular cross-section b r d N.A. I = 1 − 1 12 12 3 This gives same result as above. To use this method both centroids must be in the same position. 14/09/2006 5 SM lecture 6 .ssppaatteerr Parallel Axis Theorem I x′ = ∫ ( y + a) 2 dA = ∫ y 2 dA + 2a ∫ ydA + a 2 ∫ dA A But A A A ∫ ydA = 0 if G is centroid A ∴ I x′ = ∫y 2 dA + a 2 A A i.e. I x ′ = I x + Aa 2 Application to an I section : For web about x-axis: Iw = bd 3 12 For top flange about x-axis using parallel axis theorem: bd 3 ⎛d d ⎞ I f = 1 1 + b1d1 × ⎜ + 1 ⎟ ⎝2 2⎠ 12 2 Bottom flange similar to top flange ∴ for whole section : I = Iw + 2 I f = bd 3 b1d1 3 b1d1 + + (d + d1 ) 2 12 6 2 Note : I could also be calculated from: b ( d + 2d1 ) (b − b) d 3 i.e. ssppaatteerr Deflections of Beams Basic bending equation : ∴ M E σ = =− I R y 1 M = R EI It can be shown that 1 = R (d 2 v dx 2 ) 3 2⎤2 ⎡ ⎛ dv ⎞ ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ dx ⎠ ⎥⎦ where : Sign convention : + ve v upwards [some books use opposite] 2 Small deflections assumed ∴ ∴ dv ⎛ dv ⎞ small and ⎜ ⎟ → 0 ⎝ dx ⎠ dx 1 d 2v = R dx 2 Hence 14/09/2006 beam curvature d 2v M = dx 2 EI beam slope dv M =∫ dx dx EI beam deflection ⎛ M ⎞ v = ∫ ⎜∫ dx ⎟ dx ⎝ EI ⎠ 6 SM lecture 6 . ssppaatteerr Beam deflection example ω R= ωl l 2 R= ωl 2 Determine the maximum deflection and slope of the beam shown. 14/09/2006 7 SM lecture 6 . For the beam EI is constant along its length.