Direct Integration from Distributed Loads: For a beam subjected to distributed loadsd'MdMdFs=q(x)= Fs (x),Axdx?dxdxB: Equation for beam displacement[yA=0][VB = 0]becomesd'wd'M[A = 0][Mβ = 0]EIg(x)dr4dx?? Integrating four times yieldsEI w(x)= - dx[ dxJ dxJ q(x)dxBA++C,x3 +C,x? +C,x+C4: Constants are determined from[yB= 0][YA=0]conditions on the shear forces and[MA=0][M= 0]bending moments as well as conditionson the slopes and deflections11
Direct Integration from Distributed Loads • Equation for beam displacement becomes 4 2 4 2 d w d M EI q x dx dx 3 4 2 2 2 3 1 6 1 1 C x C x C x C EI w x dx dx dx q x dx • Integrating four times yields • For a beam subjected to distributed loads • Constants are determined from conditions on the shear forces and bending moments as well as conditions on the slopes and deflections. 2 S S 2 , dM d M dF F x q x dx dx dx 11
Direct Integration from Transverse Loads For a beam subjected to transverse loads(without distributed loads)Equation for beam displacementbecomesd'wdM-Fs(xEIdr3dx Integrating three times yieldsEI w(x) = -[ dx[ dx[ Fs(x)dxB+C2x*+C,x+C4Constants are determined fromconditions on the bending moments as?well as conditions on the slopes anddeflections12
Direct Integration from Transverse Loads • Equation for beam displacement becomes F x dx dM dx d w EI S 3 3 3 4 2 2 2 1 S C x C x C EI w x dx dx F x dx • Integrating three times yields • For a beam subjected to transverse loads (without distributed loads) • Constants are determined from conditions on the bending moments as well as conditions on the slopes and deflections. 12
Deformations in a Transverse Cross Section: Deformation due to bending moment isquantified by the curvature of the neutral surface8 (y)_ 0(y)1NeutralFEypysurface: Although cross sectional planes remain planarwhen subjected to bending moments, in-planedeformations are nonzero,ε,()=-ve,(n)=-, 8 ()=-ve,()=-PO: For a rectangular cross-section, no change in theNeutral axis ofvertical dimension will be observed.transverse section.Horizontal expansion above the neutral surfaceand contraction below it cause an in-planecurvature16.(y)V-= anticlastic curvaturep'yp13
Deformations in a Transverse Cross Section 1 x x y y y Ey y x z x , y y y y y y 1 anticlastic curvature z y y 13 • Deformation due to bending moment is quantified by the curvature of the neutral surface • Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero, • For a rectangular cross-section, no change in the vertical dimension will be observed. • Horizontal expansion above the neutral surface and contraction below it cause an in-plane curvature
Curvature Shortening. When a beam is bent, the ends ofBAthe beam move closer together.: It is common practice to disregardthese longitudinal displacementsB"Bds - dx =dx22A= LAB - LAB'd40.2: For immovable supports, ahorizontal reaction will一develop at each end.=HL/EA H=^EA/L. This equation gives a close estimate of the tensile stress produced bythe immovable supports of a simple beam14
Curvature Shortening 2 2 2 0 1 1 1 2 1 2 L AB AB ds dx w dx w dx L L w dx • When a beam is bent, the ends of the beam move closer together. • It is common practice to disregard these longitudinal displacements. • For immovable supports, a horizontal reaction will develop at each end. HL EA H EA L • This equation gives a close estimate of the tensile stress produced by the immovable supports of a simple beam. 14
Sample Problem: Given: flexural rigidity (ED) of a simply supported beam under auniformly distributed load of density q: Find: eguations of deflections and slopes, and their maximum values0.max,Wmax)C15
• Given: flexural rigidity (EI) of a simply supported beam under a uniformly distributed load of density q • Find: equations of deflections and slopes, and their maximum values (θmax, wmax) q l Sample Problem 15