Question 9.5: Using numerical integration, calculate the moments of inerti...
Using numerical integration, calculate the moments of inertia about the x– and y-axes for the half parabolic complement. Use Simpson’s rule with four panels. Compare your results with the exact values in Table 9.2.
\bar{I}_{x}=\frac{bh^3}{36} I _{x}=\frac{bh^3}{12} \bar{I}_{y}=\frac{bh}{36}(a^2-ab+b^2) I_{y}=\frac{bh}{12}(a^2+ab+b^2) \bar{I}_{xy}=\frac{bh^2}{72}(2a-b) I_{xy}=\frac{bh^2}{24}(2a+b) |
\bar{I}_{x}=\frac{37bh^3}{2100} I _{x}=\frac{bh^3}{21} \bar{I}_{y}=\frac{b^3h}{80} I_{y}=\frac{b^3h}{5} \bar{I}_{xy}=\frac{b^2h^2}{120} I_{xy}=\frac{b^2h^2}{12} |
\bar{I}_{x}=\bar{I}_{y}=0.05488R^4 I_{x}=I_{y=}\frac{πR^4}{16} \bar{I}_{xy}=-0.01647R^4 I _{xy}=\frac{R^4}{8} |
\bar{I}_{x}=\frac{8bh^3}{175} I _{x}=\frac{2bh^3}{7} \bar{I}_{y}=\frac{19b^3h}{480} I_{y}=\frac{2b^3h}{15} \bar{I}_{xy}=\frac{b^2h^2}{60} I_{xy}=\frac{b^2h^2}{6} |
\bar{I}_{x}=0.05488ab^3 I _{x}=\frac{πab^3}{16} \bar{I}_{y}=0.05488a^3b I_{y}=\frac{πa^3b}{16} \bar{I}_{xy}=-0.01647a^2b^2 I_{xy}=\frac{a^2b^2}{8} |
I _{x}=\frac{R^4}{8}(2α-\sin2α) I_{y}=\frac{R^4}{8}(2α+\sin2α) I_{xy}=0 |
Table 9.2 Inertial Properties of Plane Areas: Part 2
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