东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A18 Combined Loading

Combined Loadingmi@seer.cn

Combined Loading mi@seu.edu.cn

Contents·Unsymmetric Bending（不对称弯曲）·Tension&Bending（拉弯组合）·EccentricCompression（偏心压缩）·Core of Cross-sections（截面核心区域）·CoreofRectangularCross-sections（矩形截面核心区域）·CoreofCircularCross-sections（圆形截面核心区域）·Tension&Torsion（拉扭组合）·Bending＆Torsion（弯扭组合）·Tension,Bending&Torsion（拉弯扭组合）2

• Unsymmetric Bending（不对称弯曲） • Tension & Bending（拉弯组合） • Eccentric Compression（偏心压缩） • Core of Cross-sections（截面核心区域） • Core of Rectangular Cross-sections（矩形截面核心区域） • Core of Circular Cross-sections（圆形截面核心区域） • Tension & Torsion（拉扭组合） • Bending & Torsion（弯扭组合） • Tension, Bending & Torsion（拉弯扭组合） Contents 2

Introduction: A circular bar subjected to a single type of loadStressStresses Produced by Each Load IndividuallyStressesDistributionsBATorsionalTorsional shearLoadstressX(Torque )Txo = Tp/I,CTAxialB白ATensile averageFLoadAnormal stressOavg?(Force F)Cang=F/ADDBOMBending normalAPN.A--ACBstressBending门Loado, = My/lN.A(Transverse1ForceP)Transverseshear stressNATx = FsS*/1.b3

• A circular bar subjected to a single type of load 3 Introduction  x z z  M y I * xy z z F S I b S   x p T I    

Introduction: Prismatic bars are frequently subjected to several loadssimultaneously: The principle of superposition is used to determine theresultant stress & strain? Conditions for the principle of superposition- Linear elasticity & small deformation- No interaction between variously loads=100mm=100m=50mmRT-Pa1L4

• The principle of superposition is used to determine the resultant stress & strain • Prismatic bars are frequently subjected to several loads simultaneously • Conditions for the principle of superposition - Linear elasticity & small deformation 4 Introduction - No interaction between variously loads

Unsymmetric Bending:Analysis of pure bending has beenlimitedtomembers subjected to bending momentsVacting in a plane of symmetry.: Members remain symmetric and bend inthe plane of symmetryV.: The neutral axis of the cross sectioncoincides with the axis of the couple: Will now consider situations in which thebending couples do not act in a plane ofsymmetry..Cannot assumethat the memberwill bendin the plane of the couples.. In general, the neutral axis of the section willnot coincide with the axis of the couple5

Unsymmetric Bending • Analysis of pure bending has been limited to members subjected to bending moments acting in a plane of symmetry. • Will now consider situations in which the bending couples do not act in a plane of symmetry. • In general, the neutral axis of the section will not coincide with the axis of the couple. • Cannot assume that the member will bend in the plane of the couples. • The neutral axis of the cross section coincides with the axis of the couple • Members remain symmetric and bend in the plane of symmetry. 5

Unsymmetric Bending. If neutral axis passes through centroidrM.ydA0=F=|LgdA=/or 0=|LydA. Stress distributionWishtodeterminethe conditionsunderM=Mwhich the neutral axis of a cross sectionof arbitrary shape coincides with the· Moment vector must be directedaxisofthemomentasshownalong a principal centroidal axisM.Y(z)dA0=M.: The resultant force and momentfrom the distribution ofor O =[yz dA= I = product of inertiaelementary forces in the sectionmust satisfy: Superposition is applied to determinestresses in the most general case ofF, =0= M, M, = M= applied coupleunsymmetric bending.6

Wish to determine the conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the moment as shown. • Moment vector must be directed along a principal centroidal axis 0   or 0 product of inertia z y z yz M y M z dA I yz dA I        • The resultant force and moment from the distribution of elementary forces in the section must satisfy 0 applied couple F M M M x y z     • If neutral axis passes through centroid 0 or 0 z x x z M y F dA dA I y dA         • Stress distribution   z z z M y M M y dA I    • Superposition is applied to determine stresses in the most general case of unsymmetric bending. Unsymmetric Bending 6

Unsymmetric BendingF·SignConventionax? Construct a coordinate system.: In x-y plane: positive M, results in compression for y< 0· Bending stress and deflection: o,= M,y/I,, w"=-M,/El,. The same sign conventions can be used for bending in x-z plane? Positive M, results in compression for z < 0.· Bending stress and deflection: O, = M,z/I,, w" =-M,/EIy7

, .  x z z y z z    M y I w M EI  • Construct a coordinate system. • In x-y plane: positive Mz results in compression for y < 0. • Bending stress and deflection: • The same sign conventions can be used for bending in x-z plane. • Positive My results in compression for z < 0. • Bending stress and deflection: , .  x y y z y y    M z I w M EI  • Sign Convention 7 z x yb h x F z  y z,  Fy a Unsymmetric Bending

Unsymmetric BendingFaNxM1a≤x<LM, = F,(x-CMOM1M.=Fx0<x<≤L8

 , , , , 0 z z y x z y z x x x y z y y z x y M y M F x a a x L I M y M z M z I I M F x x L I                              8 z x yb h x F z  y z,  Fy a Unsymmetric Bending

Unsymmetric Bending? Equation for Neutral axis:MM.yoZ0=0,1FMM20FZ= tanの:BFIM.I.xyoML11xtan p= tanα(y0,-0)1.1x0±αB,± 1.a±0α±VThe neutral axispasses throughthecentroid ofthecross-section.Themaximum stresses occur at the two farthest points from theneutral axis.. With the exception ofI,= I., i.e. for circular/square cross-section, the bendingstressanddeflectioncannotbecalculatedfromtheresultantmoment Bending in a single plane occurs if and only if the orientation of neural axis staysthe same for every cross-section, i.e. M./M,=constant for a prismatic beam: In general, superposition should be resorted to determine both the bending stress9and the deflection

• Equation for Neutral axis:     0 0 0 0 0 tan tan ta 0 n z y x z y y y z z zy y z y z y y z z M y M z I I z I I x a y I x I I I x a I x M M I F F I a I                           • The neutral axis passes through the centroid of the cross-section. • The maximum stresses occur at the two farthest points from the neutral axis. • With the exception of Iy = Iz , i.e. for circular/square cross-section, the bending stress and deflection cannot be calculated from the resultant moment. • Bending in a single plane occurs if and only if the orientation of neural axis stays the same for every cross-section, i.e. Mz /My=constant for a prismatic beam. • In general, superposition should be resorted to determine both the bending stress and the deflection. 9  y z 0 0 ,    M y M z Fy F z  z y Unsymmetric Bending

Sample ProblemSOLUTION:1600 lb -in.30°. Resolve the couple vector intocomponents along the principlecentroidal axes and calculate the3.5 in.correspondingmaximumstresses.Combinethe stressesfrom thecomponent stressdistributions1.5in. Determine the angle of the neutral axisA 1600 lb-in couple is applied to arectangular wooden beam in a planeforming an angle of 30 deg. with thevertical.Determine (a)themaximumstressinthebeam,(b)theanglethattheneutral axis forms with the horizontalplane.10

A 1600 lb-in couple is applied to a rectangular wooden beam in a plane forming an angle of 30 deg. with the vertical. Determine (a) the maximum stress in the beam, (b) the angle that the neutral axis forms with the horizontal plane. SOLUTION: • Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses. • Combine the stresses from the component stress distributions. • Determine the angle of the neutral axis. Sample Problem 10