Question

(a) Find the maximum deflection of the cantilever beam in Problem 1.

(b) How does the maximum deflection of a beam that is half as long compare with the value in part (a)?

(c) Find the maximum deflection of the simply supported beam in Problem 2.

(d) How does the maximum deflection of the simply supported beam in part (c) compare with the value of maximum deflection of the embedded beam in Example 1?

Short answer

Therefore, the solution is

$$\begin{gathered} \text{(a)}{y}_{\max }=\frac{w_0}{8EI}{L}^4\text{.} \\ \text{(b)}1/16. \\ \text{(c)}{y}_{\max }=\frac{5w_0}{384EI}{L}^4\text{.} \end{gathered}$$

(d) It means that for two identical beams; the the beam with simply ends gives maximum deflection give five times that from the embedded ends beam.

Step-by-step solution

(a) Find the maximum deflection of the cantilever beam

Recall Problem 1,

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

Since the beam is free the right end. Hence, the maximum deflection occurs at its right end; that is at $x=L.$So that

$y_{\max }=\frac{w_0}{8EI}{L}^4$

(b) Find the maximum deflection of a beam

From the result in (a) the defecation of the beam proportional to$L^4.$ So that the maximum deflection of a beam that is half' is $\frac{1}{16}$form that in (a).

(c) Find the maximum deflection of the supported beam

According to Problem 2,

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

Since the beam is simply in the both sides. Hence, the maximum deflection occurs at the middle; that is at $x=\frac{L}{2}.$So that

$y_{\max }=\frac{5w_0}{384EI}{L}^4$width="118">ymax=5w0384EIL4

(d) Compare the maximum deflection of the simply supported beam and deflection of the embedded beam

In Example 1, the maximum value occurs also at the middle; that is

$y_{\max }=\frac{w_0}{384EI}{L}^4$

It means that for two identical beams; the beam with simply ends gives maximum deflection give five times that from the embedded ends beam.