§6-3求梁的挠度与转角的共轭梁法 方法的用途:求梁上指定点的挠度与转角。 二、方法的理论基础:相似比拟 梁的挠曲线微分方配厂"(x)=-M(x) 梁的外载与内力的为"(x)=q(x) 上二式形式相同,用类比法,将微分方程从形式上转化为 外载与内力的关系方程。从而把求挠度与转角的问题转化为求 弯矩与剪力的问题
§6-3 求梁的挠度与转角的共轭梁法 梁的挠曲线微分方程: EIf (x) = −M (x) 一、方法的用途：求梁上指定点的挠度与转角。 二、方法的理论基础：相似比拟。 梁的外载与内力的关为:系M(x) = q(x) 上二式形式相同，用类比法，将微分方程从形式上转化为 外载与内力的关系方程。从而把求挠度与转角的问题转化为求 弯矩与剪力的问题
DEFORMATIONOF BEAMS DUE TO BENDING 3 Conjugate beam (relations between the real beam and the imaginary beam) (The direction of the axis x and the origin of coordinates 2Having same geometric shapes Corresponding equation of the real beam: Elf (x)=M(x) Corresponding equation of the imaginary M"(x)=g(x) Alet: q(x)=+M(), auording to this equation establish distributed loads on the imaginary beam E!"(x)=M"(x) SIntegral of the"force differential M"(x)=g(x) equation of the imaginary beam O(x)=M(x)=S9(dx+Oo M(x)= g(xddx dx+Oox+n 0
3、Conjugate beam（relations between the real beam and the imaginary beam）： ①The direction of the axis x and the origin of coordinates ②Having same geometric shapes. ③Corresponding equation of the real beam： EIf (x) = −M (x) M (x) = q(x) EIf (x) = M (x) ⑤Integral of the “force” differential equation of the imaginary beam. 0 0 Q(x) M (x) q(x)dx Q x = = + 0 0 0 0 M(x) ( q(x)dx)dx Q x M x x = + + ( ) ( ), q x M x auording to this equation establish distributed loads on the imaginary beam let： = − 。 ④ M (x) = q(x) Corresponding equation of the imaginary beam：
、共轭梁(实梁与虚梁的关系) ①x轴指向及坐标原点完全相同。 ②几何形状完全相同 ③实梁对应方程:E"(x)=-M(x) 虚梁对应方程: M"(x)=q(x) ④令:q(x)=-M(x)依此建立虚梁上的分布载荷 E"(x)=M"(x) ⑤虚梁“力”微分方程的积分Mf(x)=q(x) O(x)=M'(x)=q(x)dx+O M(x)= g(xddx dx+Oox+n
三、共轭梁（实梁与虚梁的关系）： ①x轴指向及坐标原点完全相同。 ②几何形状完全相同。 ③实梁对应方程： EIf (x) = −M (x) M (x) = q(x) EIf (x) = M (x) ⑤虚梁“力”微分方程的积分 0 0 Q(x) M (x) q(x)dx Q x = = + 0 0 0 0 M(x) ( q(x)dx)dx Q x M x x = + + ④ 令： q(x) = −M (x)依此建立虚梁上的分布载荷。 M (x) = q(x) 虚梁对应方程：
DEFORMATIONOF BEAMS DUE TO BENDING Integral of the displacement differential equation of the real beam Ef(x)==M(x) Eb=E∥(x)=(-M(x)b+EO EI(x)=G(M(x)dx)dx+E100x+Elo The quantities with subscripts 0'dentte they are in the coordinate origin coordinate origin ∵.EJ(x)=M(x) E1O()=Q(x) 6Set up the "force"" boundary conditions of the imaginary beam according to the displacement boundary conditions of the real beam EL=M ElBA=Oa
EIf (x) = M (x) The quantities with subscripts “0”dentte they are in the coordinate origin coordinate origin. Integral of the“displacement” differential equation of the real beam EIf (x) = −M (x) 0 0 EI EIf (x) ( M(x))dx EI x = = − + 0 0 0 0 EIf (x) ( ( M(x))dx)dx EI x EIf x x = − + + EI (x) = Q(x) ⑥Set up the “force” boundary conditions of the imaginary beam according to the “displacement” boundary conditions of the real beam. EIf A = M A EI A = QA ;
实梁“位移”微分方程的积分 EIf()==M(x) EIG=EIf(x)=S(M()dx+ Eleo Elf(x) ∫ GM(D)dxdx+ E10ox+EIfo 下脚标带“0°的量均为坐标原点的量。 ∵.EJ(x)=M(x) E16(x)=o(x) ⑥依实梁的“位移”边界条件建立虚梁的“力”边界条件。 EI=M:E1O=o
EIf (x) = M (x) 下脚标带“0”的量均为坐标原点的量。 实梁“位移”微分方程的积分 EIf (x) = −M (x) 0 0 EI EIf (x) ( M(x))dx EI x = = − + 0 0 0 0 EIf (x) ( ( M(x))dx)dx EI x EIf x x = − + + EI (x) = Q(x) ⑥依实梁的“位移”边界条件建立虚梁的“力”边界条件。 EIf A = M A EI A = QA ;