chapter4 pure bending

Department of Chemical Engineering Strength of Materials for Chemical Engineers (0935381 ) Chapter 4 Pure Bending 1) Center of gravity and neutral axis. 2) Moment of inertia. 3) Theory of simple bending. 4) Bending of composite or flitched beam. 5) Combined bending and direct stress. 1) Center of Gravity Centroid of an area is the point about which the area c ould be balanced if it was supported from that point. a) Simple shapes b) Area has an axis of symmetry c) Two axis of symmetry do not occur Tiiii T AyAyyA yA ∑∑ == Where AT = total area o f the composite shape y= distance to the centroid of the co mposite shape measured form some reference axis OX. Ai = area of one component part the shape. yi = distance the centroid at the co mponent part the reference axis. Or Tiiii T AxAxxA xA ∑∑ == Where x= distance to the centroid of the co mposite shape measured form some reference axis OY. xi = distance the centroid at the co mponent part the reference axis. 2) Moment of Inertia Moment of inertia is a measure of the resistan ce of the section to a pplied moment or load that tends to bend it. Moment of inertia depends on the shape and not material. It is a derived property. ∫∑= = dAy yda IXX . .2 2is called the moment of inertia or the second moment of area about the axis XX. ∫∑= = dAx xda IYY . .2 2is called the moment of inertia or the second moment of area about the axis YY. Parallel axis Theorem This theorem is mainly used to transfer the moment of inertia of a body from its individual axis to another reference line. 22 .. xA I IyA I I YY OYXX OX +=+= 3) Assumption for the Simple Bending Theory 1. The beam is initially straight and unstressed. 2. The material is homogeneous, same density and elastic properties. 3. The elastic limit is nowhere exceeded. 4. Young’s modulus for the materials is the same in tension and compression. 5. Plane cross–section remains pl ane before and after bending. 6. No resulted force perpendicu lar to any cross – section. Bending theory • If we apply a constant B.M. • If will bend to radius R. • The top fibers of the beam will be subjected to tension. • The bottom fibers of the beam will be subjected to compression. • Somewhere between the two there are points at which the stress is zero. Locus of all points is te rmed the neutral axis. The radius of curvature R is then measured to this axis. The neutral axis will always pass th rough the centre of area of centroid. • The maximum tension and compression will be at the outer surface or the farthest distance from the neutral axis. Consider fiber AB distance y from the N. A. when the beam is bent this will stretch to A ′B′. ABAB BA Length OriginalExtensionAB−′′= = in Strain DC CDCD AB ′′== DCDC-BA ′′′′′′=ε Strain ()θθ yR BAR DC +=′′=′′ () Ry RR yR=−+= ∴θθθε Strain Within the elastic region: Eσ= =Modulus s Young'StressStrain yRERE yRy E ===∴ σσσ If the strip of area δA AyREA FyRE δδσσ .. .=== This force has a moment about the N.A of AyREyF δ.. .2= The total moment for the whole cross section is: ∑∑∑ == AyREMAyREyF δδ ... . 22 ∑ Ayδ.2 is the second moment of area of the cross-section. y RE IMIREM σ===∴ If the beam is of uniform section, the material of the beam is homogeneous and the applied moment is constant, I, E, M remain constant and hence the radius of curvature of the beam will also be constant. IMER= EI is known as the flexural rigidity. • For larger value of R smaller deflection greater the rigidity. max max Stress Maximum yIM=σ max max maxmaxσσ yI yIM == maxyIis termed the section modulus Z. 22 11maxmax and sections cal unsymmetrifor beam l symmetricafor yIZyIZZ MyIZ = === σ 4) Bending of Composite or Flitched Beams Composite beam is one which is constructed from a combination of material (flitched beam). ()()equivalent steel tdy tdy ′′= = .. .. Forcesteel σ σ Since the moment at any section must be the same in the equivalent section as in the original, so that the force at given dy in the equivalent beam must be equal to that at the strip it replaces. tE tEt t ′′′=′′= .. ... . εεσσ Strain must be equal tEtE′′=′= . .εε tEEtEE tt ′=′′=′ Thus to replace the steel strip by an equiva lent wooden strip the thickness must be multiplied by the modular ratioEE ′. σσσσ ′′=′=′=′ EEtt EE Where: σ= the stress in the steel section. σ′= the stress in the equivalent section. t= the thickness of the steel section. t′= the thickness of the equivalent 5) Combined Bending and Direct Stress – Eccentric Loading a) Eccentric loading on one axis yIeP APeP MyIM AP xxxx .. mm === σσ The equation of the N.A can be obtained by setting σ equal to zero. yIeP APyIeP AP xxxx ..0 ±==m eAIyxx N.m= The larger the eccentricity of the load the cl oser the N.A. will be to the axis of symmetry. y N is the distance between the N.A. and the axis of symmetry. b) Eccentric loading on two axis If the applied load will not be applied on eith er of the axis of symmetry, there will be a direct stress effect plus si multaneous bending about both axis. The total stress at any point ( x, y): yIkPxIhP AP xx yy. .mm=σ Equation of the N.A. is obtained by equating to zero: yIkAxIhAyIkPxIhP AP xx yyxx yy . .1. .0 ±±== mm This is linear equation in x and y. The line may or may not cut the section. (Middle–Quarter) and (Middle–Third) Rules For rectangular and circular cross sections, provided that the load is applied within certain defined areas, no tension will be produced whatever th e magnitude of the applied compressive load.