Bending stress课件.ppt

2、正应力和剪应力强度条件： ?带翼缘的薄壁截面，最大正应力与最大剪应力的情况与上述相同；还有一个可能危险的点，在Fs和M均很大的截面的腹、翼相交处。（以后讲） 3、强度条件应用：依此强度准则可进行三种强度计算： s M Fs t t s 6.7 Non-symmetric bending Bending centor of open-thin-walled sections 4.Special cases necessary for shear stress chesk ? Non-standard combined section members, ratio of the web to flange is smaller. ? Span of a beam is small，M smaller，and Fs larger。 ?For materials （wood）whose capability of resisting shear is bad. Solution：?Plot M diagram Ex.1 Shown is a Rectangular section beam (b?h=0.12m?0.18m). [?]=7MPa，[?]=0. 9 MPa，Determine ratio of σmax to τmax, and check strength of the beam. q=3.6kN/m A B L=3m Fs – + x x + ?Find maximum stresses ?Ratio of stresses Fs – + x x + Strength is OK. q=3.6kN/m A y1 y2 C A1 A2 A3 A4 Solution：? Plot M-diagram Ex.6.2 Cast iron beam of T-section. [?t]=30MPa，[?c]=60 MPa，C is centriod，y1=52mm， y2=88mm，Iz=763cm4 ，Check the strength of the beam. How does the beam lie rationally？ ? Critical points P1=9kN 1m 1m 1m P2=4kN A B C D M x 2.5kNm 4kNm M FAy FBy ?Check the strength y1 y2 C y1 y2 C A3 A4 2.5kNm 4kNm M A1 A2 ?”T” is rational, “ ”is not rational Ex.6.3 The cantilever beam of rectangular section is subjected to Fy and Fz . Knowing Fy = Fz = F = 1.0 kN, h = 80 mm，b = 40 mm ，a = 800 mm, [σ]=160 MPa. Check the strength of the beam。 Solution: ①Internal forces ②Stress and Strength MyA MzA 危险点：d d ③Discussion k z y Two-dimensional bending, neutral axis passes through the centriod of the section Chapter 6 Bending stress 6.1 Introduction 6.2 Normal stresses in beams 6.3 Shear stresses in beams 6.4 Strength criteria of beams 6.5 Optimum design of beams for stength 6.6 Combined bending and axial load; core of sections 6.5 Optimum design of the beams for strength Two ways of enhancing strenth of beams wZ Mmax 6.5.1 Optimum sections of the beam 1. Rectangular section R b h Proof：Make maximum stress to get extreme value。 From theory of M.M： Optimum strength Optimum stiffness 6.5 Optimum design of th