Determine the moments of inertia Ix and Iy for the parabolic semiseg -ment OAB shown in Fig. 12-12. The equation of the parabolic boundary is y=f(x)=h(1-x^2/b^2)(a)(This same area was considered previously in Example 12-1.)

Question

Determine the moments of inertia [latex]I_{x}[/latex] and [latex]I_{y}[/latex] for the parabolic semiseg -ment OAB shown in Fig. 12-12. The equation of the parabolic boundary is

[latex]y=f(x)=h\left(1-\frac{x^{2}}{b^{2}}\right)[/latex] (a)

(This same area was considered previously in Example 12-1.)

Accepted Answer

To determine the moments of inertia by integration, we will use Eqs. (12-9a) [latex]I_{x}=\int y^{2} d A[/latex] and (12-9b)[latex]I_{y}=\int x^{2} d A[/latex]. The differential element of area dA is selected as a vertical strip of width dx and height y, as shown in Fig. 12-12. The area of this element is

[latex]d A=y d x=h\left(1-\frac{x^{2}}{b^{2}} ight) d x[/latex] (b)

Since every point in this element is at the same distance from the y axis, the moment of inertia of the element with respect to the y axis is [latex]x^{2}[/latex] dA. Therefore, the moment of inertia of the entire area with respect to the y axis is obtained as follows:

[latex]I_{y}=\int x^{2} d A=\int_{0}^{b} x^{2} h\left(1-\frac{x^{2}}{b^{2}} ight) d x=\frac{2 h b^{3}}{15}[/latex] (c)

To obtain the moment of inertia with respect to the x axis, we note that the differential element of area dA has a moment of inertia [latex]dI_{x}[/latex] with respect to the x axis equal to

[latex]d l_{x}=\frac{1}{3}(d x) y^{3}=\frac{y^{3}}{3} d x[/latex]

as obtained from Eq. (12-12) [latex]I_{B B}=\int y^{2} d A=\int_{0}^{h} y^{2} b d y=\frac{b h^{3}}{3}[/latex]. Hence, the moment of inertia of the entire area with respect to the x axis is

[latex]I_{x}=\int_{0}^{b} \frac{y^{3}}{3} d x=\int_{0}^{b} \frac{h^{3}}{3}\left(1-\frac{x^{2}}{b^{2}} ight)^{3} d x=\frac{16 b h^{3}}{105}[/latex] (d)

These same results for [latex]I_{x}[/latex] and [latex]I_{y}[/latex] can be obtained by using an element in the form of a horizontal strip of area dA = xdy or by using a rectangular element of area dA = dxdy and performing a double integration. Also, note that the preceding formulas for [latex]I_{x}[/latex] and [latex]I_{y}[/latex] agree with those given in Case 17 of Appendix D.

Appendix D ( case 17)

Parabolic semisegment (Origin of axes at corner)
[latex]\begin{aligned}&y=f(x)=h\left(1-\frac{x^{2}}{b^{2}} ight) \&A=\frac{2 b h}{3} \quad \bar{x}=\frac{3 b}{8} \quad \bar{y}=\frac{2 h}{5} \&I_{x}=\frac{16 b h^{3}}{105} \quad I_{y}=\frac{2 h b^{3}}{15} \quad I_{x y}=\frac{b^{2} h^{2}}{12}\end{aligned}[/latex]

Question 12.3: Determine the moments of inertia Ix and Iy for the parabolic...

Related Answered Questions

Determine the product of inertia Ixy of the Z-section shown in Fig. 12-23. The section has width b, height h, and constant thickness t. ...

Verified Answer:

Verified Answer:

The cross section of a steel beam is constructed of a HE 450A wide-flange section with 25 cm × 1.5 cm a cover plate welded to the top flange and a UPN 320 channel section welded to the bottom flange (Fig. 12-8). Locate the centroid C of the cross-sectional area. ...

Verified Answer:

The parabolic semisegment OAB shown in Fig. 12-15 has base b and height h. Using the parallel-axis theorem, determine the moments of inertia and with respect to the centroidal axes xc and yc I . ...

Verified Answer:

A parabolic semisegment OAB is bounded by the x axis, the y axis, and a par-abolic curve having its vertex at A (Fig. 12-5). The equation of the curve is y=f(x)=h (1-x²/b²) (a) in which b is the base and h is the height of the semisegment. Locate the centroid C of the semisegment. ...

Verified Answer:

Determine the orientations of the principal centroidal axes and the magnitudes of the principal centroidal moments of inertia for the cross-sectional area of the Z-section shown in Fig. 12-28. Use the following numerical data: height h = 200 mm, width b = 90 mm, and constant thickness t = 15 mm. ...

Verified Answer: