Question

In Problems 1–5 solve equation (4) subject to the appropriate

boundary conditions. The beam is of length L, and w0 is a constant.

a) The beam is embedded at its left end and simply supported at its right end, and

$\mathit{w}\left(x\right)\mathbf{=}{\mathit{w}}_{\mathbf{0}}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\pi }\mathbf{x}}{\mathbf{L}}\mathbf{,}\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathit{L}$

(b) Use a graphing utility to graph the deflection curve when${\mathit{w}}_{\mathbf{0}}\mathbf{=}\mathbf{2}{\mathit{\pi }}^{\mathbf{3}}\mathit{E}\mathit{I}$ and$\mathit{L}\mathbf{=}\mathbf{1}$

c) Use a root-_finnding application of a CAS (or a graphic calculator) to approximate the point in the graph in part (b) at which the maximum deflection occurs. What is the maximum deflection?

Short answer

Therefore, the solution is

$$\begin{gathered} \left(a\right)y\left(x\right)=\frac{w_{0}L}{2EI{\pi }^3}\left(-2L^{2}x+3L{x}^2-x^3+\frac{2L^3}{\pi }sin\left(\frac{\pi x}{L}\right)\right) \\ \left(b\right)y\left(x\right)=-2x+3x^2-x^3+\frac{2}{\pi }sin\left(\pi x\right) \\ \left(c\right)y_{\max }=0.2708 \end{gathered}$$

Step-by-step solution

Step 1: Given Information.

$w\left(x\right)=w_{0}sin\frac{\pi x}{L},0<x<L$

(a) Determining the y(x) :

Consider a beam with an embedded left end and a support at the right end.

$w\left(x\right)=w_{0}sin\left(\frac{\pi x}{L}\right),0⩽x⩽L$

where L denotes the beam's length and w denotes a constant. We have made decisions based on the circumstances.

$EI\frac{d^{4}y}{d{x}^4}=w_{0}sin\left(\frac{\pi x}{L}\right),y\left(0\right)=y\text{'}\left(0\right)=y\left(L\right)=y\left(L\right)=0$

where E is the moment of inertia of a cross section of the beam and I is the young's modules of elasticity of the beam's material.

Using Integration

Taking the derived differential equation and integrating it four times gives us

$y\left(x\right)=c_1+c_{2}x+c_3{x}^2+c_4{x}^3+\frac{w_0{L}^4}{EI{\pi }^4}sin\left(\frac{\pi x}{L}\right)$

Using the $y\left(0\right)=y\text{'}\left(0\right)=0$border condition, we get

localid="1664275233208" $c_1=0$ , $c_2=\frac{w_0{L}^3}{EI{\pi }^3}$

as well as

$y\left(x\right)=-\frac{W_0{L}^3}{EI{\mathrm{\pi }}^3}X+C_3{X}^2+C_4{X}^3+\frac{W_0{L}^4}{EI{\mathrm{\pi }}^4}\sin \left(\frac{\mathrm{\pi x}}{L}\right)$

If y(L)=0, the result is

$c_3+c_{4}L=\frac{w_0{L}^2}{EI{\pi }^3}$

If y”(L)=0, the result is

$c_3+3c_{4}L=0$

We have accomplished this by resolving them.

$c_4=\frac{w_{0}L}{2EI{\pi }^3}$

Then,

$c_3=\frac{3w_0{L}^2}{2EI{\pi }^3}$

$y\left(x\right)=-\frac{w_0{L}^3}{EI{\pi }^3}x+\frac{3w_0{L}^2}{2EI{\pi }^3}{x}^2-\frac{w_{0}L}{2EI{\pi }^3}{x}^3+\frac{w_0{L}^4}{EI{\pi }^4}sin\left(\frac{\pi x}{L}\right)$

$y\left(x\right)=\frac{w_{0}L}{2EI{\pi }^3}\left(-2L^{2}x+3L{x}^2-x^3+\frac{2L^3}{\pi }sin\left(\frac{\pi x}{L}\right)\right)$

(b) Mapping Graph

When$w_0=2{\pi }^{3}EI$and $L=1$, we get

$y\left(x\right)=-2x+3x^2-x^3+\frac{2}{\pi }sin\left(\pi x\right)$

This as depicted in the graph below