Question

A cantilever beam of length$\mathit{L}$is embedded at its right end, and a horizontal tensile force of$\mathit{P}$pounds is applied to its free left end. When the origin is taken at its free end, as shown in Figure 5.2.8, the deflection$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$of the beam can be shown to satisfy the differential equation

$\mathit{E}\mathit{I}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathit{P}\mathit{y}\mathbf{-}\mathit{w}\mathbf{\left(}\mathit{x}\mathbf{\right)}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{.}$

Find the deflection of the cantilever beam if $\mathit{w}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{w}}_{\mathbf{0}}\mathit{x}\mathbf{,}\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathit{L}\mathbf{,}$and

Short answer

Therefore, the solution is

$y\left(x\right)=C_1\cosh \sqrt{P/\left(EI\right)}x+C_2\sinh \sqrt{P/\left(EI\right)}x+\frac{w_0}{2P}{x}^2+\frac{w_{0}EI}{P^2}$

Where

$C_1=-\frac{w_{0}EI}{P^2}$and$C_2=\left[-\frac{w_{0}L}{P}+\frac{w_{0}EI}{P^2}\sqrt{\frac{P}{EI}}\sinh \sqrt{\frac{PL}{EI}}\right]/\sqrt{\frac{P}{EI}}\cosh \sqrt{\frac{PL}{EI}}$

Step-by-step solution

 Step 1: Given Information

The given equation is:

$EI{y}^{\text{'}\text{'}}=Py-w\left(x\right)\frac{x}{2}$

To calculate the cantilever beam's deflection

We have

$EI{y}^{\text{'}\text{'}}-Py=-w\left(x\right)\frac{x}{2}$

$=-w_0\frac{x^2}{2}$Where 0 < X < L

Our aim is to find the deflection $\text{y}\left(\text{x}\right)$of the cantilever beam, which has the boundary conditions

$\text{y}\left(\text{x}\right)y\left(0\right)=y^{\text{'}}\left(L\right)=0$

We, first, find the complementary solution, which represents the solution of the homogeneous form of the given differential equation. The auxiliary equation is

$$\begin{gathered} E{Im}^2-P=0 \\ m=\pm \sqrt{\frac{P}{EI}} \end{gathered}$$

Which has two distinct and real roots, yields

$y_c\left(x\right)=C_1\cosh \sqrt{P/\left(EI\right)}x+C_2\sinh \sqrt{P/\left(EI\right)}x$

To find the non-homogeneous term's specific solution

We apply the method of indeterminate coefficients to get the specific solution of the non-homogeneous term, which produces a solution of the type

$y_p\left(x\right)=A{x}^2+Bx+C$

We find the second derivative of the particular solution, yields

$[2EIA-PC]-\left[PA\right]x^2-Bx=-\frac{w_0}{2}{x}^2$

We substitute into the given differential equation, yields

$EI\left[2A\right]-P\left[A{x}^2+Bx+C\right]=-\frac{w_0}{2}{x}^2$

$[2EIA-PC]-\left[PA\right]x^2-Bx=-\frac{w_0}{2}{x}^2$

The yields are calculated by comparing the coefficients.

$A=\frac{w_0}{2P},B=0$and$C=\frac{w_{0}EI}{P^2}$

yields the particular solution

$y_p\left(x\right)=\frac{w_0}{2P}{x}^2+\frac{w_{0}EI}{P^2}$

And the general solution to have the form

$=C_1\cosh \sqrt{P/\left(EI\right)}x+C_2\sinh \sqrt{P/\left(EI\right)}x+\frac{w_0}{2P}{x}^2+\frac{w_{0}EI}{P^2}\dots \left(1\right)$

Now, we find the values of constants, subjected to the boundary conditions, using (1), yields

$C_1+\frac{w_{0}EI}{P^2}=0$for$y\left(0\right)=0$

$C_1\sqrt{\frac{P}{EI}}\sinh \sqrt{\frac{PL}{EI}}+C_2\sqrt{\frac{P}{EI}}\cosh \sqrt{\frac{PL}{EI}}+\frac{w_{0}L}{P}=0$for$y^{\text{'}}\left(L\right)=0$

Figuring out how to solve the two equations

Solving the two equations for $c_1$and $c_2$yields

$C_1=-\frac{w_{0}EI}{P^2}$and

$C_2=\left[-\frac{w_{0}L}{P}+\frac{w_{0}EI}{P^2}\sqrt{\frac{P}{EI}}\sinh \sqrt{\frac{PL}{EI}}\right]/\sqrt{\frac{P}{EI}}\cosh \sqrt{\frac{PL}{EI}}$

Substitute the value of$c_1$ and$c_2$ into (1), yields the general solution subjected to the boundary conditions.

$y\left(x\right)=C_1\cosh \sqrt{P/\left(EI\right)}x+C_2\sinh \sqrt{P/\left(EI\right)}x+\frac{w_0}{2P}{x}^2+\frac{w_{0}EI}{P^2}$

Where

$C_1=-\frac{w_{0}EI}{P^2}$and$C_2=\left[-\frac{w_{0}L}{P}+\frac{w_{0}EI}{P^2}\sqrt{\frac{P}{EI}}\sinh \sqrt{\frac{PL}{EI}}\right]/\sqrt{\frac{P}{EI}}\cosh \sqrt{\frac{PL}{EI}}$