5 Stress and strain analysis

1. 5 Stress and strain analysis • structure—components—inner force—stress—strength; • Stresses, distribution of inner force, are used to determine the dangerous location in the dangerous cross section; • Stress distribution, mechanical properties, and strength theories are introduced in this chapter. Kylinsoft

2. 5 Stress and strain analysis 5.1 Normal stress and shear stress Divide body as part A and B with plane C Decompose stress alone nominal and tangent direction Normal stress Shear stress n is the normal Kylinsoft

3. If normal of cross section C is the direction of coordinate y Subscripts of normal stress： • direction of the stress component Subscripts of shear stress： • first indicating the direction of the normal to the plane • the second indicating the direction of the component of the stress Kylinsoft

4. y t t x z Notification for stresses • Normal stress: same as axial load • Shear stress: it is positive if its moment to every point of element is clockwise. Kylinsoft

5. t t dy t dx Conclusions for shear stress • Shear stresses on opposite face of an element are equal in magnitude and opposite in direction. • Shear stresses on perpendicular faces of an element are equal in magnitude and have directions such that both stresses point toward, or both point away from, the line of intersection of the faces. S mz = 0t(dy t )dx = t’(dx t )dy t = t’ Kylinsoft

6. 应力分析 Stress state in a point • There are infinite planes passing through a point • Stresses in different plane passing through a point are called stress state of a point Stress components Kylinsoft

7. n n P n P n Pin connection n P/2 V n P P P/2 Shear Stresses Kylinsoft

8. Discussion about strains • Normal Strain • Shear Strain Kylinsoft

9. Discussion about strains ex= ey= gxy= ex= ey= gxy= ex= ey= gxy= ex= ey= gxy= Kylinsoft

10. Plan assumption There is no any wrap in cross section of loaded members considered in this course. Cross sections remain plane and normal to longitudinal axis. Kylinsoft

11. Tension and Compression P P L d P P L+d Prismatic bar loaded by axial force at the centroid of ends (Axially loaded members). Axial force produce a uniform stretching of the bar. Prismatic bar in tension If the forces are reversed in direction, the bar is in compression Kylinsoft

12. Stress σ: the intensity of internal force Assuming: that axial force acts through the centroid of the cross sectional area, cross section will keep flat during tension, stress has a uniform distribution. resultant of stress N For stress acts in a direction perpendicular to the cuted surface, it is referred to as anormal stress Sign convention for inner force and normal stress： tensile as positive, and compressive as negative Kylinsoft

13. Saint-Venant’s Principle Kylinsoft

14. x A(x) x N(x) For case of that the bar has a straight axis and varied cross section area How to get N(x)? Kylinsoft

15. Example The diameter of bar AB is d=20mm, the load Q=15kN. Determine the normal stress in the bar AB. P A  B Solution: 1. Internal force: P=Q/sin  sin =0.8/(0.82+1.02)1/2=0.388 N=P=15/0.388=38.7 kN 2. Stress Kylinsoft

16. P P Dl L P P L+Dl L1 Normal Strain Kylinsoft

17. Normal Strain normal strainεis a dimensionless quantity. Tensile strain (positive) bar is in tension, representing an elongation or stretching of material Compressive strain (negative) bar is in compression, representing a shorten of material Kylinsoft

18. k P P a k k pa P Pa k sa P pa ta k Stresses On Inclined Sections Normal stresses acting on cross section Area of the inclined section k-k Distributed stresses of the inclined section k-k Kylinsoft

19. sa k sa P pa ta k ta Stresses On Inclined Sections Due to Normal stress on inclined section Shear stress on inclined section Kylinsoft

20. l P P b1 b Dl l+Dl Deflections of axially loaded members Strain Stress Hooke’s law When s<sp s = Ee Elongation Kylinsoft

21. A C D B P2 P1 5kN N 50 50 50 5kN Solution: N1=-5kN; N2=-5kN; N3=5kN example Bar with changes in cross section, A1=2cm2, A2=4cm2, P1=5kN, P2=10kN. E=120×103MPa. Determine its elongation. shortened Kylinsoft

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23. l P P b1 b l+Dl Dl Lateral strain and Poisson’s ratio 横向应变Lateral Strain m：泊松比Poisson’s Ratio If s<sp m, E are material properties E is about 200 GPa for steel m is about 0.25 to 0.35 for many metal Kylinsoft

24. y aε cμε bμε o x σ a c z b Volume change Initial volume Final volume Unit volume change Kylinsoft

25. materials Structural steel Low alloy steel(16Mn)cast ironaluminum alloy(LY12)rubberwoodconcreterock 2.092.00.6-1.620.710.0050.09-0.120.152-0.360.46-0.58 0.280.25-0.300.23-0.270.330.450.09-0.120.16-0.180.21-0.26 Eand μ of common materials μ E/105(MPa) Kylinsoft

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27. Calculation of truss displacement Kylinsoft

28. Mechanical Properties of Material The mechanical properties of materials are the behaviors of loaded materials in strength and deformation. It can be showed with- plot. Kylinsoft

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30. P d l P Material testing system Tensile testing specimen Tensile test of low carbon steel Gage length l Standard specimen(GB) ：l=5d；l=10d Static test: A cylindrical test specimen is placed in the middle of the load assembly. As the specimen is pulled very slowly, the load and the elongation over the gage length are measured and recorded. Kylinsoft

31. Stress  equals P/A After performing a tension test and determine the stress and strain at various magnitudes of load, we can plot a diagram of stress versus strain, that is Stress-Strain Diagram Kylinsoft

32. D E B C A Stress-strain diagram of low carbon steel Nominal Stress P s True Stress Dl e Kylinsoft

33. F P(s) ultimate stress sb yield stressss B E A C proportional limit sP Dl (e) elastic Stress-strain diagram for soft steel D linear elastic fracture elastic limit hardening necking yielding Kylinsoft

34. Stress-strain diagram for soft steel 弹性 elasticity 线性弹性 linear elastic 比例极限 sP proportional limit 强度极限 sb ultimate stress 弹性极限 seelastic limit 屈服 yielding 屈服应力 ss yield stress 强化 hardening 颈缩 necking 断裂 fracture Kylinsoft

35. P d l P Procedural of tensile test of low carbon steel elastic hardening necking fracture Kylinsoft

36. 延伸率 Percent elongation d l: original gage length l1: distance between the gage marks at fracture 截面收缩率 Percent reduction in area  A0: original cross-sectional area Af: final area at the fracture section Kylinsoft

37. s sb sp e O • Ductility of a material in tension can be characterized by its percent elongation and by the reduction in area at the cross section where fracture occurs. • Brittle Materials fail in tension at relatively low values of strain after the proportional limit is exceeded. • Steel may have an elongation in the range of 10% to 40%. For structural steel, values of 25% or 30% are common, and its reduction in area is about 50%. • Ductile materials ：d>5% such as: carbon steel，copper，aluminum alloy • Brittle materials ：d<5% such as: cast iron，glass，stone, concrete, ceramic materials Kylinsoft

38. 无屈服阶段 20Cr s 无屈服 无颈缩阶段 16Mn T10A 四个阶段 B C A3 四个阶段 H62 A 无屈服阶段 e mechanical properties of other ductile materials in tension Kylinsoft

39. s s0.2 A 0.002 offset e O 0.2% offset yield stress s0.2 名义屈服极限 When a material does not have an obvious yield point and undergoes large strain after the proportional limit is exceed, a line is drawn on the stress-strain diagram parallel to initial linear part of the curve but is offset by some standard amount of strain,such as 0.2%. offset yield stress s0.2 The intersection of the offset line and the stress-strain curve defines the yield stress Kylinsoft

40. Reloading: material has a higher yield stress D s G B C A loading unloading e H O Residual strain Unloading Elasticity: It returns to its original dimension during unloading. Plasticity:For a higher loading level, unloading line is parallel to the initial portion of the loading curve. When the load has been entirely removed, a residual strain remains in the material. Kylinsoft

41. s D B C E A e Stress-strain Curve for Steel in Compression Nominal Stress True Stress Kylinsoft

42. s sb B sb e O Stress-Strain Curve for Cast Iron There are no yielding in Stress-Strain Curve for Cast Iron. Break face is orient at 45~55° with axis in compression and 90 ° in tension Brittle materials usually reach much higher ultimate stresses in compression than in tension. Kylinsoft

43. Mechanical properties of materials Kylinsoft

44. Which one has best ductility and which one breaks first among the three loaded prismatic bars with different materials.? Kylinsoft

45. Which one has highest ultimate limit and which one breaks first among the three loaded prismatic bars with different materials.? Kylinsoft

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47. s B C A sP e O eP Hooke’s Law In linear elastic stateOA (s<sP), stress is proportional to strain, E：Modulus of Elasticity (MPa; GPa) or Young’s Modulus One of the properties of material Kylinsoft

48. Hook’s Law in Shear The properties of a material in shear can be determined experimentally from direct-shear tests or torsion tests. Stress-strain diagrams in shear are similar in shape to that in tension. Kylinsoft

49. t g The initial part of the shear stress-strain diagram is a straight line also. Hook’s Law in Shear G: Shear Modulus of Elasticity, GPa Kylinsoft

50. D E B C A Stress-strain diagram of low carbon steel Nominal Stress P s True Stress Dl e Kylinsoft