Chapter 8, Problem 8.1P

Calculate the moment of intertia with respect to the X-X centroidal axes for the areas shown.

To determine

Moment of inertia with respect to X-X centroidal axes.

Answer to Problem 8.1P

  $I_{\overline{X}}=13623in{.}^4$ .

Explanation of Solution

Given:

Concept Used: Since the shape is symmetrical about X and Y axis, so the centroidal axis lies in the center of the cross-section. The moment of inertia of the ring will be equal to the moment of inertia of the outer rectangle minus the moment of inertia on the inner circle.

Calculation: Moment of inertia of the Square about the centroid is

  $\begin{array}{l}{I}_{{\overline{X}}_1}=\frac{b{h}^3}{12}\\ I_{{\overline{X}}_1}=\frac{12\times {24}^3}{12}in{.}^4\\ I_{{\overline{X}}_1}=13824in{.}^4\end{array}$

Moment of inertia of the Circle about the centroid is

  $\begin{array}{l}{I}_{{\overline{X}}_2}=\frac{\pi d^4}{64}in{.}^4\\ I_{{\overline{X}}_2}=\frac{\pi \times 8^4}{64}in{.}^4\\ I_{{\overline{X}}_2}=201.062in{.}^4\end{array}$

Total moment of inertia of the shape is

  $\begin{array}{l}{I}_{\overline{X}}=I_{{\overline{X}}_1}-I_{{\overline{X}}_2}\\ I_{\overline{X}}=13824in{.}^4-201.062in{.}^4\\ I_{\overline{X}}=13623in{.}^4\end{array}$

Conclusion: Moment of inertia with respect to X-X centroidal axes $I_{\overline{X}}=13623in{.}^4$

To determine

Moment of inertia with respect to X-X centroidal axes.

Answer to Problem 8.1P

  $I_{\overline{X}}\text{= 1017}\text{.82 }i{n}^4$ .

Explanation of Solution

Given:

Concept Used: Since the shape is symmetrical about X and Y axis, so the centroidal axis lies in the center of the cross-section. The moment of inertia of the ring will be equal to the moment of inertia of the outer circle $\text{1}2\text{ in}.$ minus the moment of inertia on the inner circle of diameter $\text{1}0\text{ in}.$

Calculation: Total moment of inertia of the circular ring is

  $\begin{array}{l}{I}_{\overline{X}}=I_{{\overline{X}}_1}-I_{{\overline{X}}_2}\\ I_{\overline{X}}=\frac{\pi d_1{}^4}{64}-\frac{\pi d_2{}^4}{64}i{n}^4\\ I_{\overline{X}}=\frac{\pi \times {12}^4}{64}-\frac{\pi \times 1^4}{64}i{n}^4\\ I_{\overline{X}}\text{= 1017}\text{.82 }i{n}^4\end{array}$

Conclusion:

Moment of inertia of the circular ring is $I_{\overline{X}}\text{= 1017}\text{.82 }i{n}^4$

To determine

Moment of inertia with respect to X-X centroidal axes.

Answer to Problem 8.1P

  $I_{\overline{X}}=7190000c{m}^4$ .

Explanation of Solution

Given:

Concept Used: Since the shape is symmetrical about X and Y axis so the centroidal axis lies in the center of the cross-section. The moment of inertia of the ring will be equal to the moment of inertia of the outer square minus the moment of inertia on the inner rectangle.

Calculation: Moment of inertia of the Outer square is

  $\begin{array}{l}{\text{I}}_{{\text{<mover accent="true"><mo>X</mo><mo>¯</mo></mover>}}_{\text{1}}}\text{=}\frac{{\text{bh}}^{\text{3}}}{\text{12}}\\ {\text{I}}_{{\text{<mover accent="true"><mo>X</mo><mo>¯</mo></mover>}}_{\text{1}}}\text{=}\frac{{\text{100×100}}^{\text{3}}}{\text{12}}{\text{mm}}^{\text{4}}\\ {\text{I}}_{{\text{<mover accent="true"><mo>X</mo><mo>¯</mo></mover>}}_{\text{1}}}\text{=8333333}{\text{.33 mm}}^{\text{4}}.\end{array}$

Moment of inertia of the Inner rectangle is

  $\begin{array}{l}{\text{I}}_{{\text{<mover accent="true"><mo>X</mo><mo>¯</mo></mover>}}_{\text{2}}}\text{=}\frac{{\text{bh}}^{\text{3}}}{\text{12}}\\ {\text{I}}_{{\text{<mover accent="true"><mo>X</mo><mo>¯</mo></mover>}}_{\text{2}}}\text{=}\frac{{\text{40×70}}^{\text{3}}}{\text{12}}{\text{mm}}^{\text{4}}\\ {\text{I}}_{{\text{<mover accent="true"><mo>X</mo><mo>¯</mo></mover>}}_{\text{2}}}\text{=1143333}{\text{.33 mm}}^{\text{4}}.\end{array}$

Total moment of inertia of the cross-section is

  $\begin{array}{l}{I}_{\overline{X}}=I_{{\overline{X}}_1}-I_{{\overline{X}}_2}\\ I_{\overline{X}}=8333333.33m{m}^4-1143333.33m{m}^4.\\ I_{\overline{X}}=7190000m{m}^4.\end{array}$

Conclusion: Moment of inertia with respect to X-X centroidal axes ${\text{I}}_{\text{<mover accent="true"><mo>X<mo>¯</mo></mo></mover>}}{\text{=7190000 mm}}^{\text{4}}$ .