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Benjamin_harsh

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- Homework Statement
- Find the Moment of inertia of the shaded region w.r.t the line AB.

- Relevant Equations
- ##\large \frac {4R}{3π} \normalsize = 1.7 cm##

Sol: ##I_{AB} = I_{AB1} + I_{AB2} - I_{AB3}##

##I_{AB1} = I_{G1x} + A_{1}.Y_{2}##

moment of inertia of a triangle,##I_{XX} = \large \frac{bh^3}{36}##

moment of inertia of a triangle,##I_{YY} = \large\frac{hb^3}{36}##

##I_{AB1} = \large \frac {8*8^{3}}{36} + (\frac {1}{2})\normalsize*8*8 (\large\frac {8}{3})^{2} \normalsize = 341.33 cm^{4}##

##I_{AB2} = I_{G2x} + A_{1}.Y_{2}##

##\large \frac {4R}{3π} \normalsize = 1.7 cm##

Moment of inertia of semi circle section parallel to the base, ##I_{AB2} = 0.11*R^{4}##

##I_{AB2} = 0.11*44 + (\large \frac{π*4^4}{2})^{2}\normalsize (1.7)^{2}##

##I_{AB2} = 100.79 cm^{4}##

moment of inertia of a circle w.r.t centroidal X or centroidal Y: ##\frac {π}{64}d^{4}##

##I_{AB3} = I_{G2x} + A_{1}.Y^{2}##

##I_{AB3} = \large\frac{π}{64}\normalsize*4^{4}+(π*2^{2}).0##

##I_{AB3} = 12.56 cm^{4}##

##I_{AB} = I_{AB1} + I_{AB2} - I_{AB3} ##

##I_{AB} = 341.33 + 100.79 – 12.56 = 429.55 cm^{4}##

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