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 simply supported at both ends, and$\mathit{w}\left(x\right)\mathbf{=}{\mathit{w}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathit{L}$

(b) Use a graphing utility to graph the deflection curve when${\mathit{w}}_{\mathbf{0}}\mathbf{=}\mathbf{24}\mathit{E}\mathit{I}$ and$\mathit{L}\mathbf{=}\mathbf{1}$.

Short answer

Therefore, the solution is

$\left(a\right)y\left(x\right)=\frac{w_0}{24EI}x\left(L^3-2L{x}^2+x^3\right)$

$\left(b\right)y\left(x\right)=2x^2{\left(x-1\right)}^2$

Step-by-step solution

Given Information.

The given value is:

$w\left(x\right)=w_0,0<x<L=1$

(a) Determining the  :

Consider a beam that is only supported on both ends.

$w\left(x\right)=w_0,0⩽x⩽L$

When L is the beam's length and$w_0$is a constant determined by the conditions, we get

$$\begin{gathered} EI\frac{d^{4}y}{d{x}^4}=w_0,y\left(0\right) \\ =y^{\text{'}\text{'}}\left(0\right)=y\left(L\right)=y^{\text{'}\text{'}}\left(L\right) \\ =0 \end{gathered}$$

where E represents the Young's modules of elasticity of the beam's material and I represents the moment of interia of a cross section of the beam.

Integrating  

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

$y\left(x\right)=c_1+c_{2}x+c_3{x}^2+c_4{x}^3+\frac{w_0}{24EI}{x}^4$

Using$y\left(0\right)=y^{\text{'}\text{'}}\left(0\right)=0$as a boundary condition, we get

$c_1=c_3=0$

as well as

$y\left(x\right)=c_{2}x+c_4{x}^3+\frac{w_0}{24EI}{x}^4$

If$y\left(L\right)=0$is true, the result is

$c_2+c_4{L}^2+\frac{w_0}{24EI}{L}^3=0$

and the$y^{\text{'}\text{'}}\left(L\right)=0$condition yields

$6c_4+\frac{w_0}{2EI}L=0$

We have accomplished this by resolving them.

$c_4=-\frac{w_0}{12EI}L$

Then

$c_2=\frac{w_0}{24EI}{L}^3$

And

$y\left(x\right)=\frac{w_0}{24EI}{L}^{3}x-\frac{w_0}{12EI}L{x}^3+\frac{w_0}{24EI}{x}^4$

$=\frac{w_0}{24EI}\left(L^{3}x-2L{x}^3+x^4\right)$

$=\frac{w_0}{24EI}x\left(L^3-2L{x}^2+x^3\right)$

(b) Mapping Graph

When $w_0=24EI$and L=1, the equation is

$y\left(x\right)=x\left(x^3-2x^2+1\right)$

This as depicted in the graph below