DEFORMATIONOF BEAMS DUE TO BENDING For the straight beam with the same shape and equal section area, the approximate differential equation of the deflection curve may be written as the following form Elf (x)=-M() 2- Determine the equation of the deflection curve(elastic curve 1). Integration of the differential equation EJ"(x)=-M(x) EIf(x)=GM(x)dx+C E/(x)=∫(∫(-Mx)dx)dx+Cix+C2 2) Boundary conditions of the displacement P P C B D )●
EIf (x) = −M (x) For the straight beam with the same shape and equal section area, the approximate differential equation of the deflection curve may be written as the following form： 2、Determine the equation of the deflection curve (elastic curve) EIf (x) = −M (x) d 1 EIf (x) = (−M (x)) x +C d 1 2 EIf (x) = ( (−M (x))dx) x +C x +C 1).Integration of the differential equation 2).Boundary conditions of the displacement P A C B P D
对于等截面直梁,挠曲线近似微分方程可写成如下形式: E"(x)=-M(x) 二、求挠曲线方程(弹性曲线) 1.微分方程的积分 E/(x)=-M(x)E∥/(x)=f M(xdx+C E∥(x)=J小(M(x)dkx+Cx+C2 2.位移边界条件 P P C B D )●
EIf (x) = −M (x) 对于等截面直梁，挠曲线近似微分方程可写成如下形式： 二、求挠曲线方程（弹性曲线） EIf (x) = −M (x) d 1 EIf (x) = (−M (x)) x +C d 1 2 EIf (x) = ( (−M (x))dx) x +C x +C 1.微分方程的积分 2.位移边界条件 P A C B P D
DEFORMATIONOF BEAMS DUE TO BENDING O Displacement conditions at the supports f4=0fB=0 fD=0=0 @ Continuity conditions: f-=fa+or Cleftcright 3 Smooth conditions O O O Clef三 Cright Discussion: Applying to the planar bending of slender beams with elastic materials and small deformations 2 May be applied to determine the displacements of beams with equal or variable sections under all types of loads SIntegral constants may be determined by the geometric compatible conditions (boundary conditions continuity conditions) 4Advantages: Wide applications and accurate solutions by the direct-solving method Disadvantages Complicated calculation
Discussion： ①Applying to the planar bending of slender beams with elastic materials and small deformations . ②May be applied to determine the displacements of beams with equal or variable sections under all types of loads . ③Integral constants may be determined by the geometric compatible conditions（boundary conditions、continuity conditions). ④Advantages：Wide applications and accurate solutions by the direct-solving method Disadvantages Complicated calculation. Displacement conditions at the supports： Continuity conditions： Smooth conditions： f A = 0 f B = 0 f D = 0 D = 0 − = + C C f f − = + C C or Cleft = Cright Cleft Cright or f = f
o支点位移条件 f=0f=0D=0D=0 e连续条件:-=或写成=厂C右 0光滑条件:日C-=日 或写成O左=O,右 讨论: ①适用于小变形情况下、线弹性材料、细长构件的平面弯曲。 ②可应用于求解承受各种载荷的等截面或变截面梁的位移。 ③积分常数由挠曲线变形的几何相容条件(边界条件、连续条 件)确定。 ④优点:使用范围广,直接求岀较精确;缺点:计算较繁
讨论： ①适用于小变形情况下、线弹性材料、细长构件的平面弯曲。 ②可应用于求解承受各种载荷的等截面或变截面梁的位移。 ③积分常数由挠曲线变形的几何相容条件（边界条件、连续条 件）确定。 ④优点：使用范围广，直接求出较精确； 缺点：计算较繁。 支点位移条件： 连续条件： 光滑条件： f A = 0 f B = 0 f D = 0 D = 0 − = + C C f f − = + C C 或写成 左 右 C C = 或写成 左 右 C C f = f
DEFORMATIONOF BEAMS DUE TO BENDING Example 1 Determine the elastic curves maximum deflections and maximum angles ofrotation of the following equal-section straight beams Solution O Set up the coordinates and write out the bending-moment equation f M(x)=P(x-L 3 Determinate the integral constants by 2 Write out the differential the boundary conditions equation and integrate it Elf=M(x=P(L-x EJ(0)=:PD+C2=0 =-P(L-x)2+C1 E 1EB(O)=E(0)=-P+C1=0 2 2 Elf-P(L-x)+Cx+C,.C=PL; C2=-PLS
Example 1 Determine the elastic curves 、maximum deflections and maximum angles of rotation of the following equal-section straight beams. Set up the coordinates and write out the bending-moment equation M (x) = P(x − L) Write out the differential equation and integrate it Determinate the integral constants by the boundary conditions EIf = −M (x) = P(L − x) 1 2 ( ) 2 1 EIf = − P L − x +C 1 2 3 ( ) 6 1 EIf = P L − x +C x +C 0 6 1 (0) 2 3 EIf = PL +C = 0 2 1 (0) (0) 1 2 EI = EIf = − PL +C = 3 2 2 1 6 1 ; 2 1 C = PL C = − PL Solution： P L x f x