Lecture Strength of Materials I: Chapter 7 - PhD. Tran Minh Tu

Lecture Strength of Materials I - Chapter 7: Bending. The following will be discussed in this chapter: Introduction, bending stress, shearing stress in bending, strength condition, sample problems, deflections of beam, statically indeterminate beams.
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Lecture Strength of Materials I: Chapter 7 - PhD. Tran Minh Tu STRENGTH OF MATERIALS1/10/2013 TRAN MINH TU - University of Civil Engineering, 1 Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, VietnamCHAPTER 7 BENDING 1/10/2013 Contents 7.1. Introduction 7.2. Bending stress 7.3. Shearing stress in bending 7.4. Strength condition 7.5. Sample Problems 7.6. Deflections of beam 7.7. Statically indeterminate beams1/10/2013 37.1. Introduction In previous charters, we considered the stresses in the bars caused by axial loading and torsion. Here we introduce the third fundamental loading: bending. When deriving the relationship between the bending moment and the stresses causes, we find it again necessary to make certain simplifying assumptions. We use the same steps in the analysis of bending that we used for torsion in chapter 6.1/10/2013 47.1. IntroductionClassification of Beam Supports1/10/2013 57.1. Introduction  Limitation1/10/2013 67.1. Introduction  Segment BC: Mx≠0, Qy=0 => Pure Bending  Segments AB,CD: Mx≠0, Qy≠0 => Nonuniform Bending1/10/2013 77.1. IntroductionPure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane1/10/2013 87.2. Bending stress  Simplifying assumptions1/10/2013 97.2. Bending stress The positive bending moment causes thematerial within the bottom portion of the beamto stretch and the material within the top portionto compress. Consequently, between these tworegions there must be a surface, called theneutral surface, in which longitudinal fibers of Neutral axisthe material will not undergo a change inlength.1/10/2013 107.2. Bending stress Compatibility 1 2 Consider a segment of the beam a bbounded by two cross-sections that c y dare separated by the infinitesimaldistance dz. dz 2 1 Due to bending moment Mx caused d Neutral fiberby the applied loading, the cross-  1 2section rotate relatively to each other a bby the amount of d. y c d The Normal strain of the longitudinal 1 2fiber cd that lies distance y below theneutral surface. dz c d  cd    y  d   d y y z     z  dz cd  d  1/10/2013  – radius of curvature of the neutral fiber. 117.2. Bending stress y 1 Following Hooke’s law, we have. z  E  ????   Equilibrium Mx x Because of the loads applied in the plane yOz, thus: Nz=My=0 and Mx≠0. x K z z y E dA N z    z dA   yd A  0 A  A y  yd A  S A x 0 x – neutral axis (the neutral axis passes through the centroid C of the E  M y  x z dA  A  A xyd A  0 cross-section).  xyd A  I A ...