# [SOLVED]Beam Deflection

#### karush

##### Well-known member
this is a problem in the topic of Concavity and the second Derivative Test

The deflection $$\displaystyle D$$ of a beam of length $$\displaystyle L$$ is

$$\displaystyle D=2x^4-5Lx^3+3L^2x^2$$,

where $$\displaystyle x$$ is the distance from one end of the beam.

Find the value of $$\displaystyle x$$ that yields the maximum deflection

The answer to this is $$\displaystyle x=\Bigg(\frac{15-\sqrt{33}}{16}\Bigg)L \approx 0.578L$$

well first of all this equation has 2 variables in x and L so not sure what to do perhaps implicit differentiation. also the answer looks it came from a quadratic formula

so not to sure what the first step is.

#### MarkFL

##### Administrator
Staff member
Re: beam deflection

The length of the beam $L$ is a constant.

So what you want to do is compute $D'(x)$ to determine the critical values in the range:

$$\displaystyle [0,L]$$

and then evaluate $D''(x)$ at these critical values, looking for negative values, i.e., concave down, and hence a maximum.

Can you proceed?

#### mathworker

##### Well-known member
Re: beam deflection

L is not a variable its a constant as length is constant and hence you can go for first derivative which is equal to slope.it is maximum when tangent comes parallel to x-axis so = '0' and proceed, you might have mistook L as variable

#### karush

##### Well-known member
Re: beam deflection

$$\displaystyle D'=8x^3-15Lx^2+6L^2x$$
$$\displaystyle =x(8x^2-15Lx+6)=0$$

so $$\displaystyle x=0$$
or $$\displaystyle \frac{15L\pm \sqrt{33}L}{16}$$
$$\displaystyle =\left(\frac{15 \pm \sqrt{33}}{16}\right) L$$
$$\displaystyle =0.5784L$$ or $$\displaystyle 1.2965L$$

so $$\displaystyle D'' = 24x^2-30Lx+6L^2$$

$$\displaystyle D''(0.5784L)>0$$ so max is $$\displaystyle 0.5784L$$

#### MarkFL

##### Administrator
Staff member
Yes, good work! You can discard the critical value that is greater than $L$, since this is outside the relevant domain of $$\displaystyle x\in[0,L]$$.