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 c<O ) or compressive force (if c>O)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, )