Before starting this example, please read a short Theoretical Introduction

We calculate the moment functions for the beam in the initial state of loading and separately for the unit load.

Deflection at point B

State with external load - state P

\begin{aligned} &\sum M_C=0\\ &V_A\cdot 7+20-5\cdot 4\cdot 2=0\\ &V_A=\frac{20}{7}\\ \\ &\sum F_y=0\\ &V_C+V_A-20=0\\ &V_C=\frac{120}{7}\\ \end{aligned}

Moment functions:

\begin{aligned} &x_1\in (0; 3)\\ &Mp(x_1)=V_A\cdot x_1=\frac{20}{7}\cdot x_1\\ \\ &x_2 \in (0; 4)\\ &Mp(x_2)=V_C\cdot x_2-\frac{1}{2}\cdot 5x^2=\frac{120}{7}\cdot x_2-2.5x_2^2\\ \end{aligned}

Unit load state

\begin{aligned} &\sum M_C=0\\ &V_A\cdot 7-1\cdot 4=0\\ &V_A=\frac{4}{7}\\ \\ &\sum F_y=0\\ &V_C=\frac{3}{7}\\ \end{aligned}

Moment functions:

\begin{aligned} &x_1\in (0; 3)\\ &M(x_1)=V_A\cdot x_1=\frac{4}{7}\cdot x_1\\ \\ &x_2\in (0; 4)\\ &M(x_2)=V_C\cdot x_2=\frac{3}{7}\cdot x_2\\ \\ &EIw_B=\int_{x_{1p}}^{x_{1k}}(Mp(x_1)\cdot M(x_1)) dx_1+\int_{x_{2p}}^{x_{2k}}(Mp(x_2)\cdot M(x_2)) dx_2\\ \\ &EIw_B=\int_{0}^{3}\frac{20}{7}x\cdot \frac{4}{7}x dx+\int_{0}^{4}(\frac{120}{7}x-2.5x^2)\cdot \frac{3}{7}x dx=\\ &\frac{80}{49}\cdot \frac{1}{3}\cdot 3^3+\int_{0}^{4}(\frac{360}{49}x^2-\frac{15}{14}x^3) dx=\\ &\frac{720}{49}+\frac{360}{49}\cdot \frac{1}{3}\cdot 4^3-\frac{15}{14}\cdot \frac{1}{4}\cdot 4^4=\frac{720}{7}\\ &w_B=\frac{720}{7EI}\\ \end{aligned}

Rotation angle at point B

Unit load state

\begin{aligned} &\sum M_C=0\\ &V_A\cdot 7+1=0\\ &V_A=-\frac{1}{7}\\ \\ &\sum F_y=0\\ &V_C=\frac{1}{7}\\ \end{aligned}

Moment functions:

\begin{aligned} &x_1\in (0; 3)\\ &M(x_1)=V_A\cdot x_1=-\frac{1}{7}\cdot x_1\\ \\ &x_2\in (0; 4)\\ &M(x_2)=V_C\cdot x_2=\frac{1}{7}\cdot x_2\\ \\ &EI\varphi_B=\int_{x_{1p}}^{x_{1k}}(Mp(x_1)\cdot M(x_1)) dx_1+\int_{x_{2p}}^{x_{2k}}(Mp(x_2)\cdot M(x_2)) dx_2\\ \\ &EI\varphi_B=\int_{0}^{3}(\frac{20}{7}x)\cdot (-\frac{1}{7}\cdot x) dx+\int_{0}^{4}(\frac{120}{7}x-2.5x^2)\cdot (\frac{1}{7}\cdot x) dx=\\ &-\frac{20}{49}\cdot \frac{1}{3}\cdot 3^3+\int_{0}^{4}(\frac{120}{49}x^2-\frac{5}{14}x^3) dx=\\ &-\frac{180}{49}+\frac{120}{49}\cdot \frac{1}{3}\cdot 4^3-\frac{5}{14}\cdot \frac{1}{4}\cdot 4^4=\frac{720}{7}\\ &\varphi_B=\frac{720}{7EI}\\ \end{aligned}

Energy Methods - Displacements in Statically Determinate Systems

Strength of materials - Energy Methods - Displacements in Statically Determinate Systems - Example 3

Example 3