(4)Checktheboundaryconditions:Principle boundary conditions: , = O, t,, = 0Secondary boundaryconditions:h[3,(α,)x=0,dy = 0Satisfyx direction: 6dy·dy= 02m2M[3,0.dy=03,(t,)x=0, dy= 0y direction:(或dh22hdMoment of force:[(ox)x=0,y.dy = MM2. 6dy2dy = M222MM12M(5)Stress components:9o191hI(h3 /12)11hConclusion: It can be seen thatthis result is the same as that in2M=-3dhQMthemechanicsofmaterialsindicating that the stress result ofxthe pure bending beam in themechanicsofmaterialsiscorrect.h=3dhOmraxy112
11 （4）Check the boundary conditions： = 0, = 0 y xy Principle boundary conditions： Satisfy Secondary boundary conditions： − = = 2 2 0, ( ) 0 h h x x l x direction： dy 6 0 2 2 − h h dy dy Moment of force: y dy M h h x x l = − = 2 2 0, ( ) − = 2 2 2 6 h h dy dy M h M d = 3 2 ) 2 ( 3 h M 或d = y I M y x = h M x 3 12 = y h M x ( /12) 3 = y direction： − = = 2 2 0, ( ) 0 h h xy x l dy 0 0 2 2 − h h dy （5）Stress components: x 1 y 2 h 2 h l l 3dh min = − 3dh max = M M Conclusion: It can be seen that this result is the same as that in the mechanics of materials, indicating that the stress result of the pure bending beam in the mechanics of materials is correct
Explanation:11(1)The surface force of the forcehcouple of beam end must bedistributed linearly and the2M=-3dhMcenter must be zero. Only ifthe result is accurate.x(2)If the surface forceh=3dhdistributes in other forms.Omaythe results will be2inaccuracy.HoweverTx, = 0accordingtothe Saint-,=6dy ,=0Venant's principle, the errorMis only large at both endsa.yand small at thefar end.x1(3) When I is much more larger than h, the error is small.On the contrary, the error is large.12
12 = 0 xy dy x = 6 = 0 y x 1 y 2 h 2 h l l M M y I M x = 3dh min = − 3dh max = Explanation： (1) The surface force of the force couple of beam end must be distributed linearly and the center must be zero. Only if the result is accurate. (2) If the surface force distributes in other forms, the results will be inaccuracy. However, according to the SaintVenant’s principle, the error is only large at both ends and small at the far end. (3) When l is much more larger than h, the error is small. On the contrary, the error is large
PlaneProblems in Rectangular CoordinatesS3-3Determination ofDisplacement ComponentsTake the pure bending of rectangular beam for example toexplain how to determine the displacement components by thestress components1.In the State of Two-dimensional StressMPut the stress component=0,t,=Tx=0toy,o.1into the physical equationHE2(1 + μ)2之xyE13
13 §3-3 Determination of Displacement Components Take the pure bending of rectangular beam for example to explain how to determine the displacement components by the stress components. 1.In the State of Two-dimensional Stress Put the stress component into the physical equation x = y, y = 0, xy = yx = 0 I M + = = − = − xy xy y y x x x y E E E 2(1 ) ( ) 1 ( ) 1
PlaneProblems in RectangularCoordinatesThenthedeformationcomponentsare:MuM=0(a)8.y,8Y,YEIEIau6xOxThen put (a) into the geometricOvequation:YayOvOuYayaxauMOvuMOvou(b)0SO:VaxaxEIayEIayIntegrate the twoMuMformer equationsxy+f(y),v7+ f2(x)u=(c)EI2EIabove:where f and f, is the arbitrary function. Put (c) into the third equation of (b)14
14 Then the deformation components are： x = , y = − y, xy = 0 EI M y EI M （a） Then put (a) into the geometric equation： + = = = y u x v y v x u xy y x so： , , = 0 + = − = y u x v y EI M y v y EI M x u （b） Integrate the two former equations above: ( ) 2 ( ), 2 2 1 y f x EI M x y f y v EI M u = + = − + （c） where and is the arbitrary function. f 1 f 2 Put（c）into the third equation of（b）
PlaneProblemsin RectangularCoordinatesMdf(y)df2(x)SO:xdxEIdyOn the left side of the equation is a function of y, on the right side of theequation is a function of x. So both sides should equal to a constant w, and:Mdf.(y)df.(x)x+00dxEIdyMAfter the integral:fi(y) =-のy + uo, f,(x+ox+v2EIMu=+uXPut into the formula (c), thentheEI(d)displacementcomponentsare:MuMV+ox+v2EI2EIThe arbitrary constants ,uo, Voabove must be obtained through the constraintconditions.15
15 so： x EI M dx df x dy df y − = + ( ) ( ) 1 2 = − = − x + EI M dx df x dy df y ( ) , ( ) 1 2 After the integral： 0 2 1 0 2 2 ( ) , ( ) x x v EI M f y = −y + u f x = − + + The arbitrary constants , , above must be obtained through the constraint conditions. u0 0 v Put into the formula (c), then the displacement components are： = − − + + = − + 0 2 2 0 2 2 x x v EI M y EI M v x y y u EI M u （d） On the left side of the equation is a function of , on the right side of the equation is a function of . So both sides should equal to a constant , and: y x w