Chapt-5-Stress-due-to-bending：5章受力弯曲教学课件

Chapt 5.Stress due to bending 1 5.1 introductionStress distribution&the resultantwhat about bending?2 Geometry of deform.strainstressRadius of curvature-(1)3 5.2 Geometry of deformationConsidering a straight beam with the sym.cross-section of homogeneous material under pure bending 4 Sym.of ABED&BCFE plane surfaces long.axis remain plane&plane AD,BE,CF have a common point of intersection OAssume no-deform.in y-dir.AD=A1D1,BE=B1D15 For linear elements on the plane of sym./long.axisshorten ”Neutral axis.”()neutral surface or planeelongate6-(2)(1)7 Assume(2)holds for every line element/long.axis(2)independent of materialfor transverse dirs.8 Ex 5.1)A steel bar of rectangular section under a constant moment.If ,min.permissible&angle change?9 5.3 stress distribution&equil.requirementsFor a line element/the long.axis,using Hookes law like a tensile bar comp.for y tension for y requirements are10 where(2)(2)11(2)-(3)12 Ex 5.2)4 point bending of pure alumina beam.If the fracture stress=240MPa,P at fracture?13 BMD(3)14 Ex 5.3)(a)(b)Location of centroid&Iz?15(a)(Total Area)c=Ac+2Ac must holdwdt16“parallel axis theorem”centroid17 Iz=(Iz)+(Iz)=(Ic)I+AI(c-c)2+(Ic)+A(c-c)2(b)similarly,c&Iz program“MOMENT OF INERTIA”18 Fig 5.14 see Appendix C,D,E for I19 5.4 stress in symmetric elastic beams with variable bending momentEven for non-pure bending,(1)(3)are still good approx.for slender&sym.beams20 Ex 5.4)21)(3)22 Ex 5.5)P=7kN,L=6mMax.tensile stress?23)if neglecting weight24 due to weight25 Ex 5.6)Max tensile&compressive stress?26)SFDFor c&IAc=AIcI+AIIcIIAIcI+AIIcIIAI+AIIOr Program“moment of inertia”MBD27 I=II+III=ICI+(c-cI)2AI+ICII+(c-cII)2AIIat x=a+b check ()28 Ex 5.7)29)neglecting the weightuse BENDING OF BEAMS to get SFD&BMDFig.5.20(a)&(b)30 For C&IC center31 Ex 5.8)32)t=50mm using MOMENT OF INERTIA33 Using BENDING OF BEAMS(see Fig.5.23&24)M=-173.1KNm at x=10m&M=97.38KNm at x=3.75m34 If t=pressivemax.tensile35 Ex 5.9)a=b=36)FBDR37 section modulusFor given load,we can select the cross section on the basis of S(3)dependent on geometry only38 Ex 5.10)39)Not good!OK40 5.6 Shear stress distribution in symmetric beams with variable bending momentvariable M shear force shear stress on the section cf)normal stress given by(3)41 Moment&force equil.Fig 5.3242 Fig 5.33 43 For A1For total A44-(4)In a case of a rectangular beam(4)3/2At the neutral surface45 Ex 5.12)P=1500lb B shear stress?(neglect the weight)46)“V ”47 Ex 5.13)oak48)1549 650 Ex 5.14)51)Slenderness 1/1052 circular cross-sectionAssuming uniform distribution through bA But you must consider the free surface-effect!53&for a thin hollow shaft54 Ex 5.15)For the same material length,weight55 for beams of arbitrary cross-sectionEx)I-beam 56 5.6 Built-up beamsEx)for shear&slide Q,.57 Ex 5.16)h58)Fig 1.18 59 Fig 1.1860 Ex 5.17)Ex 5.16)61 Ex 5.18)htb62)shear flowBolt 63 谢谢你的阅读v知识就是财富v丰富你的人生谢谢