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Chapter 10 Stength Theories

Mechanics of Materials. Chapter 10 Stength Theories. CHAPTER THEORIES OF FAILURE. 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods

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Chapter 10 Stength Theories

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  1. Mechanics of Materials Chapter 10 Stength Theories LIU Jiemin @SJZU 2008

  2. CHAPTER THEORIES OF FAILURE 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  3. 第十章 压杆稳定 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  4. M P 9.1 Introduction 1.Complex of material failure Categories of failure Fracture（Brittle failure） Yield（Ductile failure） Factors of failure Material Stress state Basic assumption • Unit factor (stress, strain, strain energy). • Ultimate value of the factor is independent of stress state. Failure theories can be set up using results of simple test Any strength theories must pass through check of practice

  5. 9.1 Introduction Four strength theories 1、最大拉应力理论（Maximum tensile stress theory） （伽利略播下） 2、最大拉应变理论（Maximum tensile strain theory） （马里奥特关于变形过大引起破坏的论述是该理论萌芽) 3、最大切应力理论（Maximum shearing stress theory） （杜奎特：C.Duguet提出) 4、最大畸变能理论（Maximum distortion energy theory） （von Mises提出）

  6. 第十章 压杆稳定 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  7. 9.2 Failure theories of fracture 1. Maximum tensile stress theory（The first-F.T. ）： Suppose that fracture of materials is due to maximum tensile stress. As long as maximum tensile stress reaches to ultimate strength in the case of axial tension, failure takes place. 1、Failure criterion： 2、Failure condition： 3、Limit of application：members lost effectiveness in form of fracture.

  8. 9.2 Failure theories of fracture 2. Maximum tensile strain theory (Te second F.T.) Suppose that fracture of materials is due to maximum tensile strain. As long as maximum tensile strain reaches to ultimate tensile strain in the case of axial tension, failure takes place. 1、Failure criterion： 2、Strength condition： The second equivalent stress 3、Limit of application：members lost effectiveness in form of fracture.

  9. 第十章 压杆稳定 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  10. 9.3 Failure theories of yield 1.Maximum shear stress theory (The third F.T.) Suppose that fracture of materials is due to maximum shear stress. As long as maximum shear stress reaches to ultimate shear stress in the case of axial tension, failure takes place. 1、Failure criterion： 2、Strength condition： The third equivalent stress 3、Limit of application：members lost effectiveness in form of yield.

  11. 9.3 Failure theories of yield 4.Distorsion energy theory (The fourth F.T.) Suppose that fracture of materials is due to maximum distorsional strain energy density. As long as M.D.S.E.D. reaches to ultimate value of D.S.E.D.in the case of axial tension, failure takes place. 1、Failure criterion： Axial tension: 2、Strength condition： 3、Limit of application：members lost effectiveness in form of yield.

  12. s t s A t 9.3 Failure theories of yield 3. Strength condition under combined state of axial and shear stresses Principal stresses： Equivalent stresses fro the third and fourth strength theories：

  13. t t 9.3 Failure theories of yield Ex.9.2 Determine the allowable shear stress under pure shear according to the third and fourth strength theories. Solution： For pure shear stress state Thus, In general, [σ] is easily determined than [τ], thus, [τ] is evaluated usually by [σ] .

  14. 第十章 压杆稳定 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  15. Fl a M s s T M M ㈠ Me M ㈩ b F a tT tT 9.4 Combined bending and torsion 1. Strength calculation of combined bending and torsion A

  16. s s a tT tT 9.4 Combined bending and torsion 2. Combined bending-torsion and axial load A P F M Critical section: Fixed end Critical point: Upper edge

  17. T A F F T A 9.4 Combined bending and torsion Ex.9.1The cast iron circular rod of d=0.1m is subjected to T=7kNm, F=50kN. []=40MPa, Check the strength. Solution: (2) Principal stresses Safety

  18. s A 300 F C B 200 t 9.4 Combined bending and torsion Ex.9.2The circular rod of d=70mm is subjected to F =10kN. [σ] = 160MPa. Check the strength by the third strength theory. （σr3= 107 MPa） Solution： Combined bending and torsion。 Critical point: The upper edge of section A 300 200 Safety

  19. (a) Mz 300 200 My F1 F2 100 2.64 150 250 80° 7.04 F2x T 120 y Mx F1 Mx z x C F2y 40 71.3 9.4 Combined bending and torsion Ex. 9.8 The controlling rod of hollow circular section with d = 24 mm，D = 30 mm。Q235 steel，[σ] = 100 MPa。F1 = 600 N。Check the strength by the third strength theory. Solution: (1) Analysis of external forces ∑Mx=0: 0.3F2sin80 °- 0.2F1 = 0 F2 =(0.2F1) /(0.3sin80°) = 406N Mz F2y =F2sin80° = 40N Unit:Nm F2z =F2cos80° = 70.5N My (2) Internal diagrams (3) Critical sectionsC+ T T =120N.m,

  20. s s t t 9.4 Combined bending and torsion (4) Strength calculation The strength of the rod is OK.

  21. 第十章 压杆稳定 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  22. 9.5 Application of strength theory 一、强度计算的步骤： 1、外力分析：确定所需的外力值。 2、内力分析：画内力图，确定可能的危险面。 3、应力分析：画危面应力分布图，确定危险点并画出单元体， 求主应力。 4、强度分析：选择适当的强度理论，计算相当应力，然后进行 强度计算。

  23. 9.5 Application of strength theory 二、强度理论的选用原则：依破坏形式而定。 1、脆性材料：当最小主应力大于等于零时，使用第一理论； 当最小主应力小于零而最大主应力大于零时，使用莫尔理论。 当最大主应力小于等于零时，使用第三或第四理论。 2、塑性材料：当最小主应力大于等于零时，使用第一理论； 其它应力状态时，使用第三或第四理论。 3、简单变形时：一律用与其对应的强度准则。如扭转，都用： 4、破坏形式还与温度、变形速度等有关！

  24. 第十章 压杆稳定 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  25. 9.6 Thin-walled circular cylinder Pressure circular cylinder: p, average diameter D,Thicknessδ,length l。 1. longitudinal stress 2.hoop stress LIU Jiemin @SJZU 2008

  26. 1 2 外表面 9.6 Thin-walled circular cylinder 3.radial stress σr ≈ 0 4. Stress state at out-suefacr σ3 = 0 LIU Jiemin @SJZU 2008

  27. y A x s y s x A 9.6 Thin-walled circular cylinder Ex.9.3The thin-walled circular cylinder.x=1.8810-4, y=7.3710-4 under maximum pressure. Knowing: E=210GPa，[]=170MPa,=0.3. Check the strength by the third strength theory. Solution：From Hooke’s Law: No-safety! LIU Jiemin @SJZU 2008

  28. y A x sｔ s x A 9.6 Thin-walled circular cylinder EX.9.4The thin-walled circular cylinder is subjected to p and M. Knowing: D, δ, l. E，[],，M=πD3p/4. （1）Set up the strength condition by the third strength theory （2）Axial deformation； （3）Variation of in-diameter. Ｐ Ｍ Solution: (1) S.C. of the cylinder （2） Axial deformation LIU Jiemin @SJZU 2008

  29. p O 9.6 Thin-walled circular cylinder （3）Variation of in-diameter。 Suppose the Variation of in-diameter: Hoop strain： From Hooe’s law： LIU Jiemin @SJZU 2008

  30. 9.6 Thin-walled circular cylinder Ex.9.3The thin-walled circular cylinder.x=1.8810-4, y=7.3710-4 under maximum pressure. Knowing: E=210GPa，[]=170MPa,=0.3. Check the strength by the fourth strength theory. Solution: From the above example, 不安全！ LIU Jiemin @SJZU 2008

  31. 第九章 破坏理论 9.1 Introduction 9.2 Strength theories about fracture 9.3 Strength theories about yielding 9.4 Combined bending and torsion and axial load 9.5 General cases of combined loading for rods with rectangular cross-section 9.6 Thin-walled pressure circular cylinder 9.7 Experimental verification of strength theories 9.8 Mohr theory of failure LIU Jiemin @SJZU 2008

  32. 9.8 Mohr strength theory Mohr considers that Main factor causing failure is maximum shear stress, however, the normal stress on the section on which the maximum shear stress is has imortant effect on failure.

  33. 极限应力圆 极限应力圆的包络线 近似包络线 9.8 Mohr strength theory 1. Two concepts (1)Extreme stress circle：corresponding to failure stress circle 2、Extreme curves：envelope of extreme stress circle. t s O

  34. 3 t a M K L 莫尔理论危险条件的推导 P s N o a [L] [ y] O3 O2 O1 1 9.8 Mohr strength theory 2. Mohr strength theory：As long as stress circle at a point contacts with the extreme curve, yield or shear failure takes place. 1、Failure criteren： 2、Strength condition：

  35. 9.8 Mohr strength theory 3. Limit of appilication：yield failure and fracture of brittle materials whose tension and compression strength are not the same.(Rocks and concrets). 4. Equivalent stresses, Identical form of strength criteren

  36. 3 t a M K a L 莫尔理论危险图 P s N o a [L] [ y] O3 O2 O1 1 9.8 Mohr strength theory Ex.9.5Cast iron member, t= 400MPa，c= 1200MPa，Maximum shear stress at yielding according to Mohr teory is 450MPa. Determine the principal stress at this point. Solution：Plot Mohr circle 破坏判据：

  37. 9.8 Mohr strength theory Solve the above simultaniesely, gets:

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