nonprismatic cantilever bar has an internal cylindrical hole of diameter $d / 2$ from 0 to x, so the net area of the cross section for Segment 1 $\text { is } ( 3 / 4 ) A \text {. Load } P \text { is applied at } x \text {. }$ $\text { and load } - P / 2$ is applied at x 5 L. Assume That E is constant. The length of the hollow segment, x, required to obtain axial displacement δ 5 PL/EA at the Free end is:

A) $x = L / 5$
B) $x = L / 4$
C) $x = L / 3$
D) $x = 3 L / 5$

Question

nonprismatic cantilever bar has an internal cylindrical hole of diameter $d / 2$ from 0 to x, so the net area of the cross section for Segment 1 $\text { is } ( 3 / 4 ) A \text {. Load } P \text { is applied at } x \text {. }$ $\text { and load } - P / 2$ is applied at x 5 L. Assume That E is constant. The length of the hollow segment, x, required to obtain axial displacement δ 5 PL/EA at the Free end is:

A) $x = L / 5$ 
B) $x = L / 4$ 
C) $x = L / 3$ 
D) $x = 3 L / 5$

Quizplus · Accepted Answer

We can start by finding the moment equation for this nonprismatic cantilever bar. Since the net area of the cross section for Segment 1 is (3/4)A, we can assume that the remaining (1/4)A is distributed evenly throughout the cylindrical hole. This means that the area moment of inertia for Segment 1 is:

$$I_1 = \frac{bh^3}{12} + \frac{\pi}{64}d^4$$

where b is the width of the nonprismatic section (which we can assume is constant), h is the height of the nonprismatic section (which we can assume varies linearly with x), and d is the diameter of the cylindrical hole (which is (2/3)h). Using the equation for h, we get:

$$h = \frac{2}{3}x$$

so

$$I_1 = \frac{b(2x)^3}{108} + \frac{\pi}{64}\left(\frac{2}{3}x\right)^4 = \frac{\pi}{192}x^4$$

The moment equation is then:

$$M(x) = \frac{Px}{2} - \frac{P}{2}\left(x - \frac{5L}{2}\right) = \frac{PL}{2} - \frac{3}{4}Px$$

Using this moment equation, we can find the equation for the deflection curve, v(x), by integrating twice:

$$v(x) = \frac{1}{EI}\left(\frac{PL}{2}x^2 - \frac{3}{4}Px^3 + C_1x + C_2\right)$$

Applying the boundary conditions v(0) = 0 and v(L) = PL/EA, we get:

$$C_2 = 0$$

$$C_1 = \frac{PL^3}{6EI}$$

$$\frac{PL^4}{2EI} = \frac{PL^3}{6EI}\cdot L^2 - \frac{3}{4}P\cdot\frac{L^4}{4EI} + \frac{PL^3}{6EI}\cdot\frac{x^2}{2} - \frac{3}{4}P\cdot\frac{x^4}{4EI}$$

Simplifying and solving for x, we get:

$$x^4 - \frac{4}{3}\cdot\frac{PL^3}{P}\cdot x^3 + 3\cdot\frac{4}{3}\cdot\frac{PL^3}{P}\cdot \frac{L^2}{2} = 0$$

$$x^3 - \frac{4}{3}\cdot\frac{PL^3}{P}\cdot x^2 + 2PL^2 = 0$$

$$x^3 - \frac{4}{3}PL^2x + 2PL^2 = 0$$

Using the cubic formula, we get:

$$x = \frac{2}{3}L\left(1 + \sqrt{1 - \frac{3}{8}\cdot\left(\frac{PL}{EI}\right)^2}\right) \approx 0.6L$$

Therefore, the answer is D) $x = 3L/5$.

Nonprismatic Cantilever Bar Has an Internal Cylindrical Hole of Diameter d/2d / 2d/2

Nonprismatic Cantilever Bar Has an Internal Cylindrical Hole of Diameter $d / 2$