In this example, it is impossible to determine the point at which the maximum deflection will occur, but we will address this issue later.

To calculate it, we will use the Clebsch's Method.

\begin{aligned} &\sum M_B=0\\ &10+V_A\cdot 5-10\cdot 2,5-5\cdot 5\cdot 2,5=0\\ &V_A=15.5\ kN\\ \\ &I_{yc}=\frac{a\cdot (3a)^3}{12}=2.25a^4\\ \end{aligned}

\begin{aligned} &EIw''=-Mg\\ &Mg=V_{A}x+10\cdot x^0-\frac{1}{2}\cdot 5x^2-10\cdot (x-2.5)\\ &EIw''=-15.5x-10\cdot x^0+2.5x^2+10\cdot (x-2.5)\\ &EIw'=-\frac{15.5}{2}x^2-10x+\frac{2.5}{3}x^3+\frac{10}{2}(x-2.5)^2+C\\ &EIw=-\frac{15.5}{6}x^3-\frac{10}{2}x^2+\frac{2.5}{12}x^4+\frac{10}{6}(x-2.5)^3+Cx+D\\ \end{aligned}

Boundary conditions:

\begin{aligned} &w(0)=0\\ &D=0\\ &w(5)=0\\ &-\frac{15.5}{6}\cdot 5^3-\frac{10}{2}\cdot 5^2+\frac{2.5}{12}\cdot 5^4+\frac{10}{6}(5-2.5)^3+5C=0\\ &5C=\frac{875}{3}\\ &C=\frac{175}{3}\\ \end{aligned}

The maximum (extremal) deflection will occur at the point in the interval where the first derivative of the deflection function equals zero, i.e., where the angle of deflection equals zero. In the analyzed example, this will be:

\begin{aligned} &x_1=-6.22\ m\\ &x_2=2.44\ m\\ &x_3=7.09\ m\\ \\ &x_2\in (0; 5)\\ \\ &w_{max}=\frac{w(2.44)}{EI}=-\frac{15.5}{6}\cdot 2.44^3-\frac{10}{2}\cdot 2.44^2+\frac{2.5}{12}\cdot 2.44^4+\frac{175}{3}\cdot 2.44=\frac{82.42}{EI}\\ \end{aligned}

Condition from the task:

\begin{aligned} &w_{max}\le w_{dop}\\ \\ &\frac{82.42\cdot 10^3}{200\cdot 10^9\cdot 2.25a^4}\le 0.01\\ &\sqrt[4]{1.83\cdot 10^{-5}}\\ \\ &a\ge 0.0654\ m\\ &a=7\ cm\\ \end{aligned}

Deflections

Strength of materials - Deflections - Example 4

Example 4