Question

In Problems 1-5 solve equation (4) subject to the appropriate boundary conditions. The beam is of length L, and ${\mathit{w}}_{\mathbf{0}}$is a constant.

1. (a) The beam is embedded at its left end and free at its right end, and$\mathbf{w}\left(x\right)\mathbf{=}\mathbf{w}\mathbf{\_}\left\{0\right\}\mathbf{,}\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}$

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

Short answer

$$\begin{gathered} \left(a\right)y\left(x\right)=\frac{w_0}{24EI}{x}^2\left(6L^2-4Lx+x^2\right) \\ \left(b\right)y\left(x\right)=x^2\left(6-4x+x^2\right) \end{gathered}$$

Step-by-step solution

Definition of Embedded or clamped

$\text{EI}\frac{d^{4}y}{d{x}^4}=w\left(x\right)$

Boundary conditions associated with the above equation depend on how the ends of the beam are supported. A cantilever beam is embedded or clamped at one end and free at the other.

Using the embedded system

(a) Consider a beam embedded at its left end and free at its right end with

$w\left(x\right)=w_0,0⩽x⩽L$

where L is the length of the beam and$w_0$is a constant. According to the given conditions, we have

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

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

Integrating the obtained differential equation four times, we get

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

Using the boundary conditions

Using the boundary conditions$y\left(0\right)=y^{\text{'}}\left(0\right)=0$, we get

$c_1=c_2=0$

and So,

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

The condition$y^{\text{'}\text{'}}\left(L\right)=0$gives

$2c_3+6c_{4}L+\frac{w_0}{2EI}{L}^2=0$

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

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

By solving them, we have

$c_4=-\frac{w_0}{6EI}L$

and

$c_3=\frac{w_0}{4EI}{L}^2=0$

Substitution

Thus

$y\left(x\right)=\frac{w_0}{4EI}{L}^2{X}^2-\frac{W_0}{6EI}L{x}^3+\frac{W_0}{24EI}{X}^4$

$=\frac{W_0}{24EI}(6x^2{L}^2-4L{x}^3+x^{4)}$

If $w_0=24EI$and $L=1$, the equation goes to be

$y\left(x\right)=x^2\left(6-4x+x^2\right)$

which represented by the following graph

Result

$$\begin{gathered} \left(a\right)y\left(x\right)=\frac{W_0}{24EI}{x}^2\left(6L^2-4Lx+x^2\right) \\ \left(b\right)y\left(x\right)=x^2\left(6-4x+x^2\right) \end{gathered}$$