《弹性力学》课程教学课件（英文讲稿）Chapter 3 Plane Problems in Rectangular Cordinates

ElasticityPlaneProblemsChapter3in Rectangular Coordinates

1 Elasticity

Plane Problems in Rectangular CoordinatesChapter3Plane Problem in Rectangular CoordinatesS 3-1 Inverse Solution Method and Semi-inverseMethod.Polynomial SolutionS 3-2Pure Bending ofRectangularBeamS 3-3 Determination ofDisplacement Components 3-4 Simply SupportedBeam Loadedby UniformLoad$ 3-5 Wedge Loaded by Gravity and HydraulicPressureExercises2

2 Chapter 3 Plane Problem in Rectangular Coordinates §3-1 Inverse Solution Method and Semi-inverse Method. Polynomial Solution §3-3 Determination of Displacement Components §3-4 Simply Supported Beam Loaded by Uniform Load §3-5 Wedge Loaded by Gravity and Hydraulic Pressure Exercises §3-2 Pure Bending of Rectangular Beam

Plane Problems in Rectangular CoordinatesS 3-1 Inverse Solution Method and Semi-inverse Method.Polynomials SolutionI .Inverse solution method and semi-inverse method1. Inverse solution method: the first step is to set up multiform stress functiongwhich satisfy the compatible equation (2),and get the stress component withthe formula (1), then investigate it according to the stress boundary conditionsOn the elastic body in every kind of shape, these stress componentscorresponding to what kind of surface force, from which we know that thestressfunctioncansolvewhatkindofproblemThe basic step of inverse solution method:MakingSubstitutionSubstitutionWhat kind ofObtainedtheObtained theSetting upsureproblem canbestresssurface forceStressFormula(1)solvedcomponents(Resultantforce)boundarycondition3

3 §3-1 Inverse Solution Method and Semi￾inverse Method. Polynomials Solution Ⅰ. Inverse solution method and semi-inverse method 1. Inverse solution method: the first step is to set up multiform stress function which satisfy the compatible equation (2), and get the stress component with the formula (1), then investigate it according to the stress boundary conditions. On the elastic body in every kind of shape, these stress components corresponding to what kind of surface force, from which we know that the stress function can solve what kind of problem.  The basic step of inverse solution method: Setting up  Obtained the stress components Obtained the surface force (Resultant force) What kind of problem can be solved Substitution Substitution Formula(1) Stress boundary condition Making sure

Plane Problems in Rectangular Coordinates2. Semi-inverse method: Aiming at the problem to solve, according to boundaryshape and force circumstance of the elastic body, supposing the partial and thewhole stress component as a certain form function, from which we conclude thestress function, then we investigate whether this stress function satisfy compatibleequation or not. And the original stress component for supposing and the reststress weight from this stress function are whether to satisfy stress boundarycondition and single worth condition of displacement or not. If both compatibleequation and all aspects of conditions can be satisfied; it is natural to get the rightanswer: if some aspect can not be satisfied, we should establish assumption onthe other hand and investigate it againThebasicstepsofsemi-inversemethod:FormulaDeducing theSatisfyingStressSetting up(1)Satisfying theGetting thestressright solutionpboundarytermYesVp=0BoundaryYesexpressiontermNoNo

4 2. Semi-inverse method: Aiming at the problem to solve, according to boundary shape and force circumstance of the elastic body, supposing the partial and the whole stress component as a certain form function, from which we conclude the stress function, then we investigate whether this stress function satisfy compatible equation or not. And the original stress component for supposing and the rest stress weight from this stress function are whether to satisfy stress boundary condition and single worth condition of displacement or not. If both compatible equation and all aspects of conditions can be satisfied; it is natural to get the right answer; if some aspect can not be satisfied, we should establish assumption on the other hand and investigate it again. The basic steps of semi-inverse method: Setting up Getting the right solution Deducing the stress expression Satisfying the boundary term Satisfying 0 4   = Yes Yes No No Formula (1) Stress Boundary term 

Plane Problems in Rectangular CoordinatesII.Polynomials solution1.TheStressFunctionintheformof a LinearPolynomial@ = a+bx+cyThe Stress Components:O,=0,0, =0,tx, =Tx=0The Stress Boundary Condition:X-Y=0Conclusion: （1) The linear stress function is correspondingto the state of nosurface force and no stress. (2) There'sno effect to the stress to add a linearfunctiontoanystressfunctionoftwo-dimensionalproblem2.The StressFunction in the form of a QuadraticPolynomialβ = ax? + bxy+cyCorrespondingto @ = ax2, the stress components :O,=0,0,=2a,Tx,= Tx=05

5 1.The Stress Function in the form of a Linear Polynomial  = a + bx + cy The Stress Components:  x = 0, y = 0, xy = yx = 0 The Stress Boundary Condition： X = Y = 0 Conclusion:（1）The linear stress function is corresponding to the state of no surface force and no stress.（2）There’s no effect to the stress to add a linear function to any stress function of two-dimensional problem. 2.The Stress Function in the form of a Quadratic Polynomial 2 2  = ax +bxy+ cy Corresponding to , the stress components : 2  = ax  x = 0, y = 2a, xy = yx = 0 Ⅱ. Polynomials solution

Plane Problems in Rectangular CoordinatesConclusion: (1). The stress function @ = ax? can be used to solve the problemof uniformly distributed tensile (if a >0) or compressive force (if a<0) ony-axis of rectangular plate. (Fig3-1a)2abxX2a2c2cbLyy(a)(c)(b)Fig.3-1(2). Corresponding to β = bxy, stress components :O, =0,0, =0,Tx,=Tx=-bConclusion: The stress function @ = bxycan used to solve the problemofuniformly distributed shearing stresses of rectangular plate. (Fig3-1b)6

6 (2). Corresponding to , stress components :  = bxy  x = 0, y = 0, xy = yx = −b Conclusion: The stress function can used to solve the problem of uniformly distributed shearing stresses of rectangular plate. (Fig3-1b)  = bxy Fig.3-1 x y o 2a 2a （a） x y o b b b b （b） x y o 2c 2c （c） Conclusion：(1). The stress function can be used to solve the problem of uniformly distributed tensile (if ) or compressive force (if ) on y-axis of rectangular plate. (Fig3-1a) a  0 a  0 2  = ax

Plane Problems in Rectangular Coordinates(3). The stress function β = cy can be used to solve the problemof uniformlydistributed tensile(if cO)on x-axis ofrectangular plate. (Fig3-lc)3.The Stress Functionin the form ofa Cubic Polynomialp =ay3TheCorrespondingStressComponents(a)Ox= 6ay,O, = 0, tx, = Tx = 0Conclusion: The stress function(a) can be used to solve the problem of purebending of rectangular beam. The rectangular beam is shown in Fig3-2.MMh2hxyh20xa1yFig.3-2

7 3.The Stress Function in the form of a Cubic Polynomial 3  = ay The Corresponding Stress Components Conclusion：The stress function (a) can be used to solve the problem of pure bending of rectangular beam. The rectangular beam is shown in Fig3-2.  x = 6ay, y = 0, xy = yx = 0 （a） − + M M h l 2 h 2 y h  x  x x y Fig.3-2 1 (3). The stress function can be used to solve the problem of uniformly distributed tensile (if ) or compressive force (if ) on x-axis of rectangular plate. (Fig3-1c) 2  = cy c  0 c  0

PlaneProblems in RectangularCoordinatesThe solutionisas follows:In fig 3-2, considering the unit width beam, named the moment of couple on per unitwidth M. The dimension of M is [force][length]/[length], the result is [force]On the left or right end, the two-dimensional force should be combined tocouples,asthemomentofdoubleforceisM,thisrequeststhat:hh6a[ ydy = 0,6a[h y2dy= M2CPut ,in equation(a) into the formula above, then:hho,d=0,,o,ydy= MThe first one always cansatisfy, and the second one requests that :2Mah312MPut into formula (a), then:.=0,...=T.=0ay,o.KX8

8 The solution is as follows: In fig 3-2, considering the unit width beam, named the moment of couple on per unit width . The dimension of is [force][length]/[length], M M the result is [force]. On the left or right end, the two-dimensional force should be combined to couples, as the moment of double force is , this requests that M ：   − − = = 2 2 2 2 2 6 0,6 h h h a ydy a h y dy M Put into formula (a), then： , 0, 0 12 3 x = y y = xy = yx = h M       − − = = 2 2 2 2 0, h h h  x dy h  x ydy M Put in equation (a) into the formula above  x ，then： The first one always cansatisfy, and the second one requests that ： 3 2 h M a =

Plane Problems in Rectangular CoordinatesBecause the inertia moment ofthe beam's cross section is I-Ih12so the formula above can be written into :M=y,0, =0, t,=Tx=0OThe result is same with the corresponding part in material mechanicsNote:The result above is useful to the beam which the length I is moregreatly than the depth h ; to the beam which the length I is equal tothe depth h , this result is useless9

9 Because the inertia moment of the beam’s cross section is ， so the formula above can be written into ： 12 1 3 h I  = x = y, y = 0, xy = yx = 0 I M     The result is same with the corresponding part in material mechanics. Note： The result above is useful to the beam which the length is more greatly than the depth ; to the beam which the length is equal to the depth , this result is useless. l h l h

Plane Problems in Rectangular CoordinatesS3-2Pure Bending of RectangularBeamNow we discuss the bending of the beam byopposite pairs of forces acting on the twohends of the beam, and the body force is not2considered. Our research object is the beamMMwith unit width. and the moment of theLxforce couple per unit width is M. Thedimension of the M is LMT2hWe can choose @ = dy3 as the stress12function to solve the above problem.(1) β = dy , where d is the undetermined coefficient.(2) Check that if β(x, y) satisfies the biharmonic function, so we haveatpatoatp0Vβ = 0 (The (x, y) can be usedax4as the stress function)(3)) The stress components can be calculated from (2-26): (f=f, = 0)a~02ap=00S:06dyaOtOy?ax?xyOxoy10

10 x 1 y 2 h 2 h l o M M Now we discuss the bending of the beam by opposite pairs of forces acting on the two ends of the beam, and the body force is not considered. Our research object is the beam with unit width, and the moment of the force couple per unit width is M. The dimension of the M is LMT-2 . We can choose as the stress function to solve the above problem. 3  = dy (1) , where d is the undetermined coefficient. (2) Check that if satisfies the biharmonic function, so we have 0, 0, 0 2 2 4 4 4 4 4 =    =   =   x y x y    0 4   = (The can be used as the stress function ) (fx = f (3) The stress components can be calculated from (2-26): y = 0) 0 2 =    = − x y xy  dy  y x 6 2 2 =   =   0 2 2 =   = x y   §3-2 Pure Bending of Rectangular Beam 3  = dy  ( x y, )  ( x y, )