This is the first example with a hinge. We need to divide the beam into simple beams and solve them separately. For beam I we will have 1 boundary condition, for beam II - two conditions. The missing condition is the so-called "stitching condition" - the deflection at the joint of the beams must be the same and result from the equation of deflection of beam I and II.

\begin{aligned} &I\\ &\sum{M_C}=0\\ &-20\cdot 2+R_{AY}\cdot 4=0\\ &R_{AY}=10 kN\\ &\sum{Y}=0\\ &R_{CY}=10 kN\\ \\ &II\\ &\sum{M_D}\\ &-R_{EY}\cdot 3-R_{CY}=\cdot 3=0\\ &R_{EY}=-10 kN\\ &\sum{Y}=0\\ &R_{DY}=20 kN\\ \end{aligned}

Clebsch's method

\begin{aligned} &M_{g1}=R_{AY}x-20(x-2)\\ &EIw_1''=-M_{g1}=-10x+20(x-2)\\ &EIw_1'=-10\frac{x^2}{2}+20\frac{(x-2)^2}{2}+C_1 & (1)\\ &EIw_1=-10\frac{x^3}{6}+20\frac{(x-2)^3}{6}+C_1x+D_1 &(2)\\ \end{aligned}

Boundary condition

\begin{aligned} &w_1(0)=0\\ &D_1=0\\ \end{aligned}

Second beam

\begin{aligned} &M_{g2}=-R_{CY}x+R_{DY}(x-3)=-10x+20(x-3)\\ &EIw_2''=-M_{g2}=10x-20(x-3)\\ &EIw_2'=10\frac{x^2}{2}-20\frac{(x-3)^2}{2}+C_2 & (1)\\ &EIw_2=10\frac{x^3}{6}-20\frac{(x-3)^3}{6}+C_2x+D_2 & (2)\\ \end{aligned}

Boundary conditions

\begin{aligned} &w_2(3)=0\\ &w_2(6)=0\\ \end{aligned} \begin{aligned} 10\cdot \frac{3^3}{6}+C_2\cdot 3+D_2=0\\ 10\cdot \frac{6^3}{6}-20\cdot \frac{8^3}{6}+6C_2+D_2=0\\ \end{aligned}

Therefore:

\begin{aligned} &C_2=-75\\ &D_2=180\\ \end{aligned}

Stitching condition:

\begin {aligned} &w_1(4)=w_2(0)\\ &-10\frac{4^3}{6}+20\frac{2^3}{6}+4C_1+D_1=D_2 to C_1=65\\ \end{aligned}

Substituting into the appropriate equation for the point of interest on the beam, we obtain the desired displacements:

\begin{aligned} &w_1(2)=\frac{1}{EI}(-10\cdot \frac{2^3}{6}+65\cdot 2)=\frac{350}{3EI}\\ &w_2'(3)=\frac{1}{EI}(10\cdot \frac{3^2}{2}-75)=\frac{-30}{EI}\\ \end{aligned}

Deflections

Strength of materials - Deflections - Example 13

Example 13