Pure Bending. A beam satisfying above given requirements are shown below: Why this surface is called neutral will be explained later in the lecture.

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2 In this section we will derive a formula to analyze a the deformation and stress distribution of a beam under flexural action. Theformulatobederivedinthis section will be used for straight beams with sections having at least one-axis of symmetry and moments acting perpendicular to this symetry axis.

3 A beam satisfying above given requirements are shown below: (*) Why this surface is called neutral will be explained later in the lecture.

4 Straight Members Let s consider the straight prismatic member made of highly deformable material given below. This member is under the action of bending moments acting at its ends.

5 BENDING DEFORMATION OF A STRAIGHT MEMBER Assumptions: 1. Plane section remains plane 2. Length of longitudinal axis remains unchanged 3. Plane section remains perpendicular to the longitudinal axis 4. In-plane distortion of section is negligible Copyright 2014 Pearson Education, All Rights Reserved

6 Straight Members Notice that based on the sign of the moment, zones on the beam which are subjected to tension and compression switch. Also note that where the neutral axis appear in both cases. It is called the neutral axis since no deformation under flexural action takes place. +M +M -M -M

7 Straight Members Let s analyze this deformation. To do that we need toanalyzeasegmentofthebeamfromanarbitrary distance x from the support. The undeformed thicknessof thissegment isδx. Belowwe cansee the undeformed and deformed views of the segment. Notice that on the neutral surface there is no deformation x

8 Straight Members Under this moment action we can say that above the neutral axis there is shortening and belowthereiselongation.atadistanceyabove the neutral axis, normal strain on a material fiber can be written as follows: ε = lim s s s s 0 (1) Let s find a relationship between this normal strain and radius of curvature (ρ) and the distance y. After Deformation

9 Straight Members Before the deformation Δs = Δx. After the deformationδxwillhaveacenterato and radius of curvature ρ. Δθ is the angle between the two cross-sections. Therefore x= s= ρ θ Similarly, at distance y above the neutral axis, the shortened length of the segment canbefoundasfollows: ( ρ y) s = θ (2) (3) Boyuna eksen

10 Straight Members If we substitute expressions (2) and (3) into (1), and simplify, we can get the following ( ρ ) y θ ρ θ y ε = lim = s 0 ρ θ ρ Thisisaveryimportantresult:Thisgivesthenormalstrainatpointon thebeamduetobending.thenormalstrainisafunctionofradiusof curvature and that point on the beam and distance y which is the distance on the cross-section from the neutral axis.

11 Inotherwords,thenormalstrainislinearlychangingwiththedistancey from the neutral axis. Notice that positive distance +y is associate with shortening (negative sign), negative distance -y is associated with elongation(positive sign). Straight Members ε ε max = y/ ρ c/ ρ y ε = ε c max Normal strain distribution on the cross-section

12 Straight Members Now, let s derive a relationship between the normal stresses and the internal moment along the beam. For this, we will assume that the material is linear elastic, therefore the Hooke s law is applicable. If this the case, normal stresses on the section will be linearly related with normal stresses: y ε = ε c max y σ = σ c max Similar triangles appear

13 Failure Mode under Bending

14 Bending Formula Here, the positive sign convention is important: positive moment(along +z direction) would produce negative stresses(compression) along +y direction, and along y directions positive stresses (tension) would occur. Let s consider the beam under positive bending below: + For example, on the crosssection at a point which is at distance y from the neutral axis, compression stresses would occur.

15 So, where does the neutral axis occur? To find that we need to consider the forces acting on the cross-section. Due to equilibrium the resultant force due to normal stresses must be zero(notice that only moment exists). By reference to the below figure: Bending Formula 0 y = df = A σda= A σ A maxda c σmax = yda c A Inorderthisexpressiontobezero: ya= A yda=0 Thisshowsthatthefirstareamomentmustbezero.Thiscanonlybesatisfiedif the neutral axis passes through the centroid of the cross-section. In other words, if thecentroidofthesectionisknown,thelocationoftheneutralaxisisalsoknown.

16 Bending Formula Stresses on the section can be found considering moment equilibrium on the section: internal moment = moment due to stress distribution along the section(only moment exists). ( M ) R = y( σda) = M ; M = dm = ydf z z A A ( ) ; 0 R y A y = y A σ c σmax M= c A M M z σ da = y = A max Notice that since the section is symettic wrt the y-axis, the following condition is automatically satisfied: 2 y da

17 Bending Formula σ M= c max y 2 da A Intheaboveequation,theintegrandisequaltothemomentofinertia of the cross-section with respect to the z-axis passing through the centroid.ifwerepresentthiswiththeletteri,σ max canbewrittenas: σ max M = c I By using the expression σ max /c = - σ/y, we can obtain the bending formulathatcanbeusedtocalculatestressatanypointonthecrosssection. This is called the bending formula which is very important. M σ=- y I (-)sign is important, since positive moment based on the right-hand-rule would generate compressive stresses above the neutral axis, and tensile stresses below this axis.

18 Bending Formula σ max M = c I Notethatthebendingformulacanbe used only if (i) the section is symmetric wrt an axis perpendicular to the neutral axis, and (ii) the bending moment is acting along this axis.

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23 In the exam, a 2D drawing would be enough!

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26 The location of the centroid of the cross-section, note that this is also the location of the neutral axis: Neutral axis z

27 Moment of inertia wrt to the neutral axis: