英汉双语材料力学9完整版

1、 2 91 SUMMARY 92 SKEW BENDING 93 COMBINATION OF BENDING AND TORSION 9-4 9-4 BENDING AND TENSION OR COMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION 3 91 概述概述 92 斜弯曲斜弯曲 93 弯曲与扭转的组合弯曲与扭转的组合 9-4 9-4 拉拉( (压压) )弯组合弯组合 偏心拉（压）偏心拉（压） 截面核心截面核心 4 91 SUMMARY M P R z x y P P 1 1、Composite defo

2、rmation：Structural members will produce several types of simple deformations when subjected to complex external loads .The stress corresponding to each simple deformation can not be neglected when the magnitude of each stress has the same order .This kind of deformation is called the composite defor

3、mation . 5 一、组合变形一、组合变形 ：在复杂外载作用下，构件的变形会包含几种简 单变形，当几种变形所对应的应力属同一量级时，不能忽略 之，这类构件的变形称为组合变形。 91 概概 述述 M P R z x y P P 6 P hg g 7 P hg g 8 Dam q P hg g 9 水坝水坝 q P hg g 10 2 2、Methods to study composite deformation sPrinciple of superposition Analysis of external forces：External forces are reduced along

4、the centroid of section and resolved along principal axes of inertia. Analysis of internal forces：Determine the internal-force equation and its diagram corresponding to each external force component and the critical section. Analysis of stresses ，do the superposition of the stresses and establish th

5、e strength condition of the critical point. 11 二、组合变形的研究方法二、组合变形的研究方法 叠加原理叠加原理 外力分析：外力向形心(或弯心)简化并沿形心主惯性轴分解 内力分析：求每个外力分量对应的内力方程和内力图，确 定危险面。 画危险面应力分布图，叠加，建立危险点的强 度条件。 12 92 SKEW BENDING 1 1、Skew bending：After bending deformation , the deflection curve and the external forces（transversal forces） of the

6、 rod are not in the same plane 2 2、Methods to study the skew bending： 1).Resolve：Resolve the external load along two centroid principal axes of inertia of the cross section and get two perpendicular planar bending . Pz Py y z P j j x y z P Py Pz 13 92 斜弯曲斜弯曲 一、斜弯曲一、斜弯曲：杆件产生弯曲变形，但弯曲后，挠曲线与外力（横 向力）不共面。

7、 二、斜弯曲的研究方法二、斜弯曲的研究方法 ： 1.分解：将外载沿横截面的两个形心主轴分解，于是得到两个正交 的平面弯曲。 Py Pz Pz Py y z P j j 14 x y z P Py Pz 13 2).Sum：Analyze bending in two perpendicular planes and sum the results of the calculation. x y z Py Pz P Pz Py y z P j j 15 2.叠加：对两个平面弯曲进行研究；然后将计算结果叠加起来。 x y z Py Pz P Pz Py y z P j j jsinPPyjcosP

8、P z Solution：1.Resolve the external force along the centroid principal axis of inertia of the cross section 2.Study the bending in two planes : jjsinsin)()(MxLPxLPM yz jcosMM y x y z Py Pz P Pz Py y z P j j L m m x 17 jsinPPy jcosPP z 解：1.将外载沿横截面的形心主轴分解 2.研究两个平面弯曲 j j sin sin)( )( M xLP xLPM yz jcos

9、MM y 18 Pz Py y z P j j jcos yy y I M I zM z jsin zz z I M I yMy )sincos(jj zy I y I z M Stress due to My : Stress due to M z ： Resultant stress： L Pz Py y z P j jx y z Py Pz P L m m x 19 jcos yy y I M I zM z jsin zz z I M I yMy )sincos(jj zy I y I z M My引起的应力： M z引起的应力： 合应力： 20 L Pz Py y z P j jx y

10、 z Py Pz P L m m x 0)sincos( 00 jj zy I y I z M jctgtg 0 0 y z I I z y It is obvious that only as Iy = Iz the neutral axis is perpendicular to the external force. 2maxDL 1maxDy 22 zy fff z y f f tg As As j j = ，it is the planar bending. it is the planar bending. Pz Py y z P j j D1 D2 Neutral axis f

11、fz fy The maximum normal stress of tension or compression occurs in the points that lie in two sides of the neutral axis and have the farthest distance to the neutral axis. 21 0)sincos( 00 jj zy I y I z M jctgtg 0 0 y z I I z y 可见：只有当Iy = Iz时，中性轴与外力才垂直。 在中性轴两侧，距中性轴最远的点为拉压最大正应力点。 22 zy fff z y f f tg

12、 当当j j = 时，即为平面弯曲。时，即为平面弯曲。 D1 D2 中性轴中性轴 22 2maxDL 1maxDy Pz Py y z P j j D1 D2 f fz fy Example 1 1 Force P is through the center of section and makes an angle the j with axis z in the beam as shown in the figure. Determine the maximum stress and deflection of the beam. 2 max max 1maxD y y z z DL W

13、M W M 2 3 2 3 22 ) 3 () 3 ( y z z y zy EI LP EI LP fff jtgtg z y z y I I f f As Iy = Iz the beam produce planar bending. Solution：Analysis of the critical point is shown in the figure f fz fy y z L x Py Pz P h b Pz Py y z P j j D2 D1 Neutral axis 23 例例11结构如图，P过形心且与z轴成j角，求此梁的最大应力与挠 度。 2 3 2 3 22 ) 3

14、() 3 ( y z z y zy EI LP EI LP fff jtgtg z y z y I I f f 当Iy = Iz时，即发生平面弯曲。 解：危险点分析如图 24 2 max max 1maxD y y z z DL W M W M 中性轴中性轴 f fz fy y z L x Py Pz P h b Pz Py y z P j j D2 D1 Example 2 A wood purline is shown in the figure. Its span is L=3m and a uniformly distributed load q=800N/m is acting on

15、 it. The permissible stress and deflection of it is respectively =12MPa and L/200 ，E=9GPa，Try to determine the dimension of the section and check the rigidity of the beam. N/m358447. 0800sinqqy Solution：Analysis of external force resolve q y y z z W M W M max N/m715894. 0800cosqqz Nm403 8 3358 8 2 2

16、 max Lq M y z Nm804 8 3715 8 22 max Lq M z y 2634 h b y zq q L A B 25 例例2 矩形截面木檩条如图，跨长L=3m,受集度为q=800N/m的 均布力作用， =12MPa，许可挠度为：L/200 ，E=9GPa，试 选择截面尺寸并校核刚度。 N/m358447. 0800sinqqy 解：外力分析分解q y y z z W M W M max N/m715894. 0800cosqqz Nm403 8 3358 8 2 2 max Lq M y z Nm804 8 3715 8 22 max Lq M z y 2634 h b

17、y zq q L A B 26 93 COMBINATION OF COMBINATION OF BENDING AND TORSION 80 P2 z y x P1 150200100 A BC D 27 93 弯曲与扭转的组合弯曲与扭转的组合 80 P2 z y x P1 150200100 A BC D 28 Solution：Reduce the external force to the centroid of the section and resolve it Establish the strength condition of the rod shown in the fig

18、ure. Bending and torsion 80 P2 z y x P1 150200100 A BC D 150200100 A B CD P1 Mx z x y P2y P2z Mx 29 解：外力向形心 简化并分解 建立图示杆件的强度条件 弯扭组合变形 80 P2 z y x P1 150200100 A BC D 30 150200100 A B CD P1 Mx z x y P2y P2z Mx Sum the bending moments and plot the diagram of the resultant bending moment. )()()( 22 xMxM

19、xM zy )( ; )( ; )(xMxMxM nzy M Z (N m) X (Nm)Mz x My (N m) X My (Nm) x （Nm） x Mn MnMn (Nm) x M (N m) X Mmax M (Nm) Mmax x Internal equations and diagrams corresponding to each external force component 31 每个外力分量对应 的内力方程和内力图 叠加弯矩，并画图 )()()( 22 xMxMxM zy 确定危险面 )( ; )( ; )(xMxMxM nzy M Z (N m) X (Nm)Mz

20、x My (N m) X My (Nm) x （Nm） x Mn MnMn (Nm) x M (N m) X Mmax M (Nm) Mmax x 32 W M xB max 1 P n B W M 1 22 3 1 ) 2 ( 2 22 313 4 r 2 2 2 2 max 4 P n W M W M 1 xB 1 B 1 xB 1 B 33 xB1 B2My Mz Mn M x 1 xB 2 xB M 1 B 画危险面应力分布图，找危险点 W M xB max 1 P n B W M 1 22 3 1 ) 2 ( 2 建立强度条件 2 2 2 2 max 4 P n W M W M 1

21、xB 1 B 1 xB 1 B 34 xB1 B2My Mz Mn M x 1 xB 2 xB M 1 B 22 313 4 r 2 13 2 32 2 214 2 1 r 1 xB 1 B 22 3 W MM n 22 75. 0 W MMM nzy 222 75. 0 W MMM nzy r 222 4 75. 0 W MMM nzy r 222 3 22 313 4 r2 2 2 2 max 4 P n W M W M W MMM nzy 222 35 36 22 3 W MM n 22 75. 0 W MMM nzy 222 75. 0 W MMM nzy r 222 4 75. 0

22、W MMM nzy r 222 3 22 313 4 r2 2 2 2 max 4 P n W M W M W MMM nzy 222 35 1 xB 1 B 2 13 2 32 2 214 2 1 r Analysis of external forces：Reduce the external forces to the centroid of section and resolve them. Analysis of stresses Establish strength conditions. Steps of solving the problem of composite defo

23、rmation of bending and torsion: Analysis of internal forces ：Determine the internal equation and its diagram corresponding toeach external force component and critical section. 37 W MMM nzy r 222 3 W MMM nzy r 222 4 75. 0 外力分析：外力向形心简化并分解。 内力分析：每个外力分量对应的内力方程和内力图，确定危 险面。 建立强度条件。 W MMM nzy r 222 3 W MM

24、M nzy r 222 4 75. 0 弯扭组合问题的求解步骤：弯扭组合问题的求解步骤： 38 Example 3 A hollow circular shaft is shown in the figure.Its inside diameter is d=24mm and its outside diameter is The diameters of pulley B and D are respectively D=30mm D1 =400mm and D2 =600mm， P1=600N，=100MPa. Try to check the strength of the shaft

25、with the third strength . Analysis of external forces： Bending and torsion 80 P2 z y x P1 150200100 A BC D 150200100 A B CD P1 Mx z x y P2y P2z Mx Solution: 39 例例3 图示空心圆轴， 内径d=24mm，外 径D=30mm，B 轮 直径D 1 400mm， D轮直径 D 2 600mm，P1=600N， =100MPa，试用 第三强度理论校核 此轴的强度。 外力分析： 弯扭组合变形 80 P2 z y x P1 150200100 A B

26、C D 150200100 A B CD P1 Mx z x y P2y P2z Mx 解： 40 Analysis of internal forces：Internal forces in the critical section are： W MM n r 22 max 3 Nm3 .71 max M Nm120 n M )8 . 01 (03. 014. 3 1203 .7132 43 22 MPa5 .97 It is safe M Z (N m) X (Nm)Mz x My (N m) X My (Nm) x （Nm） x Mn MnMn (Nm) x M (N m) X Mmax

27、 M (Nm) 71.3 x 71.25 40 7.05 120 5.5 40.6 41 内力分析：危 险面内力为： Nm3 .71 max M Nm120 n M )8 . 01 (03. 014. 3 1203 .7132 43 22 MPa5 .97 安全 M Z (N m) X (Nm)Mz x My (N m) X My (Nm) x （Nm） x Mn MnMn (Nm) x M (N m) X Mmax M (Nm) 71.3 x 71.25 40 7.05 120 5.5 40.6 42 W MM n r 22 max 3 94 BENDING AND TENSION OR C

28、OMPRESSION ECCENTRIC TENSION OR COMPRESSION KERNEL OF THE SECTION P R P x y z P My x y z P My Mz 1 1、Composite deformation of Composite deformation of bending and tension or compression： Deformation of the rod due to simultaneous action of transversal and axial forces. 43 94 拉拉( (压压) )弯组合弯组合 偏心拉（压）偏

29、心拉（压） 截面核心截面核心 一、拉一、拉( (压压) )弯组合变形：弯组合变形：杆件同时受横向力和轴向力的作用而产 生的变形。 P R 44 P x y z P My x y z P My Mz A P xP z z xM I yM z y y xM I zM y y y z z x I zM I yM A P 2、Analysis of stress： 45 P My Mz P MZMy x y z z y A P xP z z xM I yM z y y xM I zM y y y z z x I zM I yM A P 二、应力分析二、应力分析： 46 P My Mz P MZMy x

30、 y z z y 0 0 0 y y z z x I zM I yM A P 4、Critical point （Farthest point from the neutral axis） 3、Equation of the neutral axis For the problem of eccentric tension or compression 0)1 ( 2 0 2 0 2 0 2 0 y P z P y P z P i zz i yy A P Ai zPz Ai yPy A P y y z z L W M W M A P max y y z z y W M W M A P max

31、01 2 0 2 0 y P z P i zz i yy 47 P（z P， ， yP） y z y z ),( PP yzP Neutral axis 0 0 0 y y z z x I zM I yM A P 四、危险点四、危险点 （距中性轴最远的点） 三、中性轴方程三、中性轴方程 对于偏心拉压问题 0)1 ( 2 0 2 0 2 0 2 0 y P z P y P z P i zz i yy A P Ai zPz Ai yPy A P 01 2 0 2 0 y P z P i zz i yy 中性轴中性轴 48 y y z z L W M W M A P max y y z z y W

32、M W M A P max P（z P， ， yP） y z y z ),( PP yzP y z 5、Kernel of section in the problem of the eccentric tension、 compression： ay az 01 2 z yP i ay 01 2 y zP i az After knowing ay and az , The action range of the compressive force. As the compressive force is acted in this range there are no tensile st

33、resses in the section. May determine an action point of the force P. ),( PP yz 01 2 0 2 0 y P z P i zz i yy Neutral axis ),( PP yzP Kernel of section 49 y z 五、（偏心拉、压问题的）截面核心：五、（偏心拉、压问题的）截面核心： ay az 01 2 z yP i ay 01 2 y zP i az 已知 ay， az 后 ， 压力作用区域。 当压力作用在此区域内时，横截面上无拉应力。 可求P力的一个作用点),( PP yz 01 2 0 2

34、 0 y P z P i zz i yy 中性轴中性轴 ),( PP yzP 截面核心 50 MPa75. 8 2 . 02 . 0 350000 max2 A P 11 max1 z W M A P MPa7 .11 3 .02 .0 650350 3 .02 .0 350000 2 Solution：The stresses in the cross sections of the two poles are both compressive ones. Example 4 4 Two poles subjected to the force P=350kN are shown in th

35、e figure. The section of one pole is unequal and the section of the other pole is equal .Try to determine the normal stress with maximum absolute value in the poles. M P P d 51 Fig .Fig . P 300 200 200 P 200 200 MPa75. 8 2 . 02 . 0 350000 max2 A P 11 max1 z W M A P MPa7 .11 3 .02 .0 650350 3 .02 .0

36、350000 2 解：两柱横截面上的最大正 应力均为压应力 例例4 4 图示不等截面与等截面柱，受力P=350kN，试分别求出两 柱内的绝对值最大正应力。 图（1）图（2） M P P d 52 . P 300 200 200 P 200 200 mm5 102010100 201020 C z 2 3 510010 12 10010 C y I 45 2 3 mm1027. 7 252010 12 2010 Solution：Analysis of the internal force is shown in the figure. Centroid of the slot in the c

37、oordinates shown in the figure Nm500105 3 PM P P Slot Example 5 A steel plate shown in the figure is subjected to forces P=100kN. Try to determine the maximum normal stress；If the slot is moved to the middle of the plate and the maximum normal stress is kept constant , how much should the width of t

38、he slot be ? 53 P P M N 20 100 20 y z yC 10 mm5 102010100 201020 C z 2 3 510010 12 10010 C y I 45 2 3 mm1027. 7 252010 12 2010 例例5 图示钢板受力P=100kN，试求最大正应力；若将缺口移至板 宽的中央，且使最大正应力保持不变，则挖空宽度为多少？ 解：内力分析如图 坐标如图，挖孔处的形心 Nm500105 3 PM P P 54 P P M N 20 100 20 y z yC 10 P P M N yc I zM A N max max MPa8 .1628 .37

39、125 Analysis of stress is shown in the figure. 7 3 6 3 1027.7 1055500 10800 10100 As the hole is moved to the middle of the plate )100(10mm9 .631 108 .162 10100 2 6 3 max x N A mm8 .36 sox 20 100 20 y z yC 55 P P M N MPa8 .1628 .37125 应力分析如图 7 3 6 3 1027.7 1055500 10800 10100 孔移至板中间时 )100(10mm9 .631

40、 108 .162 10100 2 6 3 max x N A mm8 .36 x 20 100 20 y z yC 56 yc I zM A N max max MPa7 .35 1 . 0 700016 3 n W T MPa37. 610 1 . 0 504 3 2 A P Solution：For the composite deformation of tension and torsion ,the stressed state at the critical point is shown in the figure . Example 6 A circular rod which

41、 the diameter d =0.1m is subjected to forces T=7kNm and P=50kN as shown in the figure, =100MPa. Try to check the strength of the rod with the third strength theory. Therefore, the rod is safe. 22 3 4 r MPa7 .71 7 .35437. 6 22 A A PP T T 57 MPa7 .35 1 . 0 700016 3 n W T MPa37. 610 1 . 0 504 3 2 A P 解：拉扭组合,危险点应力状态如图 例例6 直径为d=0.1m的圆杆受力如图,T=7kNm,P=50kN, =100MPa,试按第三强度理论校核此杆的强度。 故，安全。 MPa7 .71 7 .35437. 6 22 A A PP T T 58 22 3 4 r 59 Chapter 9 Exercises 1. A circular shaft of steel is deformed under tensi