1 / 94

Chapter 5 –Part Ⅱ : S tresses in Beams 梁的应力

Chapter 5 –Part Ⅱ : S tresses in Beams 梁的应力. §1 Normal stresses in beams 梁的正应力 §2 Moment of Inertia 惯性矩 §3 Shear stresses in beams 梁的切应力 §4 Strength analysis of beams 梁的强度条件 §5 Rational design of beams 梁的合理设计 §6 Combined bending and axial load 弯拉 ( 压 ) 组合.

nelson
Download Presentation

Chapter 5 –Part Ⅱ : S tresses in Beams 梁的应力

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 5 –Part Ⅱ : Stresses in Beams梁的应力 §1Normal stresses in beams梁的正应力§2Moment of Inertia惯性矩§3 Shear stresses in beams梁的切应力§4 Strength analysis of beams梁的强度条件§5 Rational design of beams梁的合理设计§6Combined bending and axial load 弯拉(压)组合

  2. §1Normal stresses in beams梁的正应力 Introduction  Experiment & Assumptions  Normal stresses in beams

  3.  Introduction 历史回顾 伽利略像

  4. A C B P 伽利略指出: 如果杆件断裂,断口将发 生在B部位.原因:固接的边缘充当施力杠杆BC的支点,而杆的厚度BA则是杠杆的另一臂,沿BA作用有抗力。此抗力阻止墙内部分与墙外部分分离

  5. 马略特的研究(1680): 马略特作了伽利略所作的实验 假定断裂时梁的悬臂段绕B旋转,并得出纵向纤维所 受的拉力与其到B的距离成正比的结论。 发现有的纤维拉伸,有的纤维压缩 A C B P A C B P

  6. M FS dA dA 弯曲时横截面上的应力 弯曲正应力 弯曲切应力 Stresses in beams  Normal stresses -s  Shear stresses -t

  7. Symmetric bending 对称弯曲 能实现对称弯曲:矩形、圆形、工字型、T字型。 • 梁至少有一个纵向对称面(即横截面至少有一根对称轴),而外力就作用在此对称面内 ;此时,梁的变形必然对称于纵向对称面,变形后的梁轴线为一条平面曲线。 The applied loads act in the plane of symmetry and the bending of the beam occurs in that plane Pure bending 纯弯曲:横截面上只有弯矩 The condition of constant bending moment (no shear force)

  8. 对称纯弯曲的弯曲正应力分析 横截面上的内力与应力的关系: 弯曲应力问题是一个静不定问题 Experiment (Pure and symmetric bending)  Transverse line横线:仍为直线,仍与纵线正交; Longitudinal line纵线:变为曲线,上缩短,下伸长; Thickness上宽度变宽, 下宽度变窄。

  9.  Experiment & Assumptions Assumptions Plane Assumption(平面假设): The entire transverse section of the beam , originally plane , remains plane and normal to the longitudinal fibers of the beam after bending横截面变形后仍保持平面,仍与纵线正交 • Assumption on the fibers under uni-axial tension and compression各纵向”纤维” ,处于单向受力状态: • no interaction between longitudinal fibers

  10. 一侧伸长,一侧缩短 存在既不伸长,也不缩短的面  The top surface is in compression and the bottom in tension. Obviously, there must exist a surface parallel to the upper and lower faces of the member where no elongation and no stress exist. This surface is called the neutral surface (中性层).  Neutral axis(中性轴)– the intersection of neutral surface with any cross section, ⊥symmetric axis 有关中性轴位置的历史讨论:  在伽利略梁应力分析模型中不存在中性轴;  马略特的梁应力分析模型:由于计算错误,中性轴位于 截面的下边缘或位于截面中间得到了相同的结果;

  11. 1700年左右雅各布.伯努利认为自己首先发现梁弯曲时一 边受拉、另一边受压,但无法确定中性轴的位置。最后 提出“中性轴位置无关紧要”的结论。  1713年法国学者帕伦假定中性轴不通过截面型心,横截面 上拉力和压力呈不同的三角形分布。但他认识到了截面上 的内力必须与载荷平衡。  1819年,纳维提出可以由横截面上的拉力对中性轴的力 矩等于压力对该轴的力矩的条件来确定中性轴的位置。  1826年,纳维应用静力学三个平衡方程,得出了正确的 结论。

  12.  Normal stresses in symmetric bending beams Compatibility equation Constitutive equation ρ=? Where is the neutral axis?

  13. (d) Statics equation  (a)(b) The neutral axis passes through the centroid of cross section中性轴通过截面形心 parallel force system in space (a)(c)  Moment of inertia of cross section with respect to z EIz: flexural rigidity  (d)(a)

  14. section modulus in bending 结论  中性轴位置: 中性轴过截面形心  中性层曲率:  正应力公式:  应用条件: , 纯弯与非纯弯 ,对称弯曲

  15. 一些易混淆的概念 对称弯曲symmetric bending与纯弯曲pure bending 对称弯曲-对称截面梁,在纵向对称面承受横 向外力时 的受力与变形形式 纯 弯 曲-杆段各截面的弯矩为常数、剪力为 零的内力状态 中性轴neutral axis与形心轴Centroidal axis 中性轴-横截面受拉与受压区的分界线 形心轴-通过横截面形心的坐标轴 截面弯曲刚度flexural rigidity与抗弯截面系数section modulus 弯曲刚度EI-代表梁截面抵抗弯曲变形的能力 抗弯截面系数Wz-代表梁截面几何性质对弯曲 强度的影响

  16. §2 Moment of Inertia惯性矩 Moment of first order and moment of inertia Moment of inertia of simple section Parallel axis theorem Examples

  17. First moment of area (静矩,一次矩) Moment of first order and moment of inertia静矩与惯性矩 -the first moment of the area of cross section with respect to z(截面对z轴的静矩) Second moment of area (moment of inertia惯性矩) -the moment of inertia of cross section with respect to z(截面对 z 轴的惯性矩)

  18. Rectangular section  Moment of inertia of simple section(简单截面惯性矩) Circular section Polar second moment of area Moment of inertia

  19. 简单截面对形心轴的惯性矩——直接计算或查表  Parallel axis theorem  简单截面对任意轴的惯性矩——平行移轴定理 Parallel axis theorem (平行轴定理) C-centroid Similarly:

  20. 50 z 50 A1 60 A2 10 y 例:求下图所示截面的形心位置

  21. 100 A2 z 20 10 20 100 z0 A1 A3 A4 10 y 例: 求下图所示截面对z方向形心轴的惯性矩 1、求形心轴位置 2、求对形心轴的惯性矩

  22.  Examples A beam section with the dimensions shown is subjected to a bending moment of 1800 N-m about its centroidal axis, as shown. Determine the normal stress due to bending. T beam The centroid is determined using the table below The moment of inertia is determined from

  23. The stress at A is tensile. Point A is 63 - 24.75 = 38.25 mm (0.0382 m) from the neutral bending axis The stress at B is compressive. Point B is 24.15- 18= 6.75 mm (0.00675 m) from the neutral bending axis

  24.  Examples Beam AB is subjected to the two 6 kN forces shown. The cross section of the beam has the dimensions shown(channel beam). Determine the maximum tensile and compressive stresses in the beam. The centroid is determined using the table below. Note that section 2 is subtracted from section

  25. The moment of inertia is determined from

  26. The normal stress due to bending on the bottom (0.1155 m from the neutral bending axis) is compressive and is defined by The normal stress due to bending on the top of the beam (0.0645 m from the neutral bending axis) is tensile and is defined by

  27. 已知:梁用№18 工字钢( I beam)制成, Me=20 kN•m, E=200 GPa。计算:最大弯曲正应力smax, 梁轴曲率半径 r  Examples 解:1. 工字钢 一种规范化、系列化的工字形截面的标准钢材 (GB 706-88) №18 工字钢:

  28. Me=20 kN•m,E=200 Gpa,求 smax 与 r 2. 应力计算 3. 变形计算

  29. 已知:钢带厚d=2mm, 宽b=6mm, D=1400mm, E=200GPa。计算:带内的 smax 与 M 解:1.问题分析 已知钢带变形,求钢带应力与内力  应力~变形关系:  内力~变形关系:

  30. 带厚d=2 mm, 宽b= 6mm, D=1400mm, E=200GPa,求 smax 与 M 2. 应力计算 3.弯矩计算

  31. §3Shear stresses in beams弯曲切应力  Shear stresses in rectangular beam 矩形截面梁弯曲切应力 Shear stresses in a thin-walled beam 薄壁梁弯曲切应力 Comparison of normal and shear stresses in a beam 弯曲正应力与切应力比较 Examples

  32. 梁在非纯弯段,横截面上一般同时存在剪力和弯矩,梁在非纯弯段,横截面上一般同时存在剪力和弯矩, 此时,横截面上同时存在弯曲正应力和弯曲切应力。 历史 枕木开裂 D.J.Jourawski(俄国铁路工程师1821-1891): 材力分析法 梁横截面窄而高;

  33.  Shear stresses in rectangular beam Problem slender rectangular beam(h>b) Assumptions ① The shear stresses are parallel to the shear force Fs and the vertical sides of the cross section t (y)// 截面侧边 (Shear stress in pair ) ② The distribution of the shear stresses is uniform across the width of the beam t (y)沿截面宽度均匀分布

  34.   Shear stresses in a beam of narrow rectangular cross-section矩形截面梁弯曲切应力 Sz(w)-the first area moment with respect to the neutral axis z of that part of the cross section w located under the line m-n. 面积w对中性轴z 的静矩

  35.  h/b越大,解越精确。(h/b>2时,足够精确)

  36. (  When FS=Const,ab=a’b’。若相邻横截面的剪力相同,则翘曲变形也相同 Warping of cross sections of a beam due to shear 非纯弯梁截面翘曲 Non-uniform shear stressNon-uniform shear strain Warping of cross sections横截面翘曲 The deformation of longitudinal fibers is unaffected by Fs纤维的纵向正应变不受剪力的影响  When FS≠Const, 当l » h时,纯弯正应力公式仍然相当精确。 If l >5h,the formula of s derived for pure bending has enough precision in the case of non-uniform bending

  37. Thin-walled I-beam 工字形薄壁梁  Shear stresses in a thin-walled beam Assumptions ://腹板(翼缘)侧边  // sides of the web(or flange); and厚度方向均匀分布uniformly distributed across the thickness。 • - y 下侧部分截面对中性轴 z 的静矩(first area moment with respect to the neutral axis z of that part of the cross section located under y. ) Variation of shear stresses in an I-beam

  38. 腹板: 翼缘: 翼缘与腹板的交接处: 应力分布较复杂,有应力集中现象

  39. C z y 根据对称性,A、B端的剪应力为0 A box beam A B 腹板: 盖板与腹板的交接处: 应力分布较复杂,有应力集中现象

  40. Comparison of normal and shear stresses in a beam弯曲正应力与切应力比较 When l >> h ,smax >> tmax

  41.  Examples Cantilevered beam BC has a 14kN end load applied as shown. At section D, determine the shear stress at point A (42 mm below the top of the beam), knowing For this cantilevered beam the reactions at the fixed end are FB = 14kN , M = 16.8 kN-m

  42. Since the shear force is constant, the Fs and M diagrams are as shown To compute the shear stress at A, we will determine Sz for the flange and then the web. Sz,flange = (200)(28)(61) = 341,600 mm3 Sz,web = (100)(14)(40) = 56,000 mm3

  43. At A the first area moment is Sz,flange +Sz,web τA = 1.128 MPa

  44. 例 FS = 15 kN,Iz = 8.8410-6 m4, b = 120 mm,d = 20 mm, yC= 45 mm,求tmax、腹板与翼缘交接处切应力ta  Examples 解:

  45. 例 已知梁段剪力FS,试分析铆钉之受力 解:

  46. -上翼板对中性轴 z 的静矩

  47. Two beam sections are fastened by 16mm-diameter bolts spaced 25mm apart as shown. A 3.2kN shear force is applied in the direction indicated. Determine the shear stress in each bolt. Using the beam sections shown, we must determine the area moment of inertia for the composite beam and q at the interface of the two beams, where the bolts are located.

  48. The moment of inertia with respect to the neutral axis of the entire beam is The first area moment of inertia at the bolts is The shear flow is defined by The shear flow is related to the bolt spacing and force carried by each bolt through q=F/s

  49. Therefore, the force along the top row of bolts is F = qs = (6.165 kN/m)(0.025 m) = 0.154 kN The area of each bolt over which the force is distributed is The shear stress in each bolt is

  50. 思考题 B为材料常数。此关系式对于拉伸和压缩均适用。试推导在对称弯曲时的曲率公式和正应力公式(已知平面假设成立)。 仿照推导线弹性材料梁正应力公式的步骤进行。 线弹性材料梁几何方面的关系式e=y/r能否用于此处? 图示矩形截面纯弯梁,其应力-应变为非线性关系,用 表示。 物理方面的关系式如何?正应力是否仍沿梁高度线性分布? 静力学方面的两个关系式如何?中性轴是否仍通过截面形心?

More Related