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

$\mathit{w}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathit{w}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathit{L}$

(b) Use a graphing utility to graph the deflection curve when${\mathit{w}}_{\mathbf{0}}\mathbf{=}\mathbf{48}\mathit{E}\mathit{I}$and$\mathit{L}\mathbf{=}\mathbf{1}$

Short answer

Therefore, the solution is:

$\left(a\right)y\left(x\right)=\frac{w_0}{48EI}\left(3L^2{x}^2-5L{x}^3+2x^4\right)$

(b) See graph for$y\left(x\right)=3x^2-5x^3+2x^4$

Step-by-step solution

Given Information.

The given value is:

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

(a) Determining the y(x):

The following non-homogeneous fourth order differential equation describes the deflection.

$EI\frac{d^{4}y}{d{x}^4}=w\left(x\right)=w_0$ where $0<x<L$

w(x) is the weight per unit length. The goal is to solve the above differential equation while keeping the four boundary conditions in mind. Since the beam is immersed at x=0 and supported at x=L, we can derive the four boundary conditions from Table 5.2.1.

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

We start by finding the complementary solution, which is the homogeneous form of the given differential equation's solution. The auxiliary formula is

$El{m}^4=0$

$m=0$

It has four equal roots and produces

$y_c\left(x\right)=c_1+c_{2}x+c_2{x}^2+c_3{x}^3$

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

$y_p\left(x\right)=c{x}^4$

We calculate the fourth derivative and plug it into the differential equation, which gives us

$\frac{d^4{y}_p}{d{x}^4}=4!C$

$$\begin{gathered} EI.24C=w_0 \\ C=\frac{w_0}{24EI} \end{gathered}$$

Consequently, we have

$y_p\left(x\right)=\frac{w_0}{24EI}{x}^4$

as well as a general solution for having the form

$y\left(x\right)=y_c\left(x\right)+y_p\left(x\right)$

$=c_1+c_{2}x+c_3{x}^2+c_4{x}^3+\frac{w_0}{24EI}{x}^4$

Now, using (1), we can obtain the values of constants that are subject to the boundary conditions.

$$\begin{gathered} c_1=0fory\left(0\right)=0 \\ {c}_2=0for{y}^{\text{'}}\left(0\right)=0 \end{gathered}$$

localid="1664347881017" $c_1+c_{2}L+c_3{L}^2+c_4{L}^3+\frac{w_0{L}^4}{24EI}=0fory\left(L\right)=0$

$2c_3+6c_{4}L+\frac{12w_0{L}^2}{24EI}=0fory\text{'}\text{'}\left(L\right)=0$

Using Mathematica

Using Mathematica to solve yields

$c_1=0,$$c_2=0$ , $c_3=\frac{L^2{w}_0}{16EI}$ and $c_4=\frac{-5L{w}_0}{48EI}$

Yields, which is the general solution of the given differential equation when the boundary conditions are met.

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

$=\frac{L^2{w}_0}{16EI}{x}^2-\frac{5L{w}_0}{48EI}{x}^3+\frac{w_0}{24EI}{x}^4$

$=\frac{w_0}{48EI}\left[3L^2{x}^2-5L{x}^3+2x^4\right]$

Mapping Graph

When $w_0=48EI$and $L=1$, we get

$y\left(x\right)=\left[3x^2-5x^3+2x^4\right]$

$=3x^2-5x^3+2x^4$

We get the following results using Mathematica.

This indicates that the deflection is positive when measured from the axis of symmetry y=0, implying that the deflection curve is downward the axis of symmetry.