Determine the percent reduction in the polar moment of inertia of the square plate due to the introduction of the circular nole. Assumea - 2.36. Awwer: 8 ◻ x eTextbook and Media Altempts Got 5 used Using multiple attempts will inpact your score.
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Assume a = 2.3R. Answer: n = eTextbook and Media Save for Later % Using multiple attempts will impact your score. Attempts: 0 of 5 Submit Ans Show more…
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Determine the polar moment of inertia of the circular area without and with the central square hole. Answers: Without the square hole Iz With the square hole
Frank D.
'For the shape shown, determine (a) the polar moment of inertia of the circular area alone (without the square hole), (b) the polar moment of inertia of the area shown (with the square hole), (c) the percent change in moment of inertia in the shape by having added the square hole, (d) the percent change in area by having added the square hole: (The square is centered on the circle )'
Sri K.
If $M$ mass of the square plate before cutting the holes, then mass of portion of each hole. $$ m=\frac{M}{16 R^{2}} \times \pi R^{2}=\frac{\pi}{16} M $$ $\therefore$ Moment of inertia of remaining portion $$ \begin{aligned} I &=I_{\text {square }}-4 l_{\text {hole }} \\ &=\frac{M}{12}\left(16 R^{2}+16 R^{2}\right)-4\left[\frac{m R^{2}}{2}+m(\sqrt{2} R)^{2}\right] \\ &=\frac{M}{12} \times 32 R^{2}-10 \mathrm{mR}^{2} \\ &=\frac{8}{3} M R^{2}-\frac{10 \pi}{16} M R^{2}=\left(\frac{8}{3}-\frac{10 \pi}{16}\right) M R^{2} \end{aligned} $$
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