Problem $2$

Calculate the centroid, $I_{},I_{xy},$ and $I_{yy}($all w$.$r$.$t$.$ the centroid$)$ for the c$-$s below. Use any

symmetry to your advantage. The cross$-$section is of constant radius $r$ and constant thickness $t.$

Apply thin$-$walled approximations. Please start your $$ dimension at the positive $x-$axis, with $+$

being counterclockwise. See the following page for some very good advice.

Notes on circular$-$arc cross$-$sections

When your cross$-$section has a circular component, you've really only got one good option $-$ you

have to set up your integrals for that component with respect to the angle $,$ the vertex of which

is at the center of the circular arc.

For calculating the centroid, we must revert to the general form of the equation: $\stackrel{}{x}=\frac{{\displaystyle \underset{A}{\overset{\phantom{}}{}}}\text{x}\text{d}\text{A}}{{\displaystyle \underset{A}{\overset{\phantom{}}{}}}dA}.$

Generally, dA would be replaced with dxdy$,$ and this would become a double integral. In the

case of the thin$-$walled cross$-$sections we use in aerostructures, $dA$ is replaced with $t^{**}ds,$ and the

area integral becomes a line integral. In the case of a circular component, we must further

replace $ds$ with $r**d.$ In the numerator, $x$ is replaced with an expression for the $x-$coordinate,

with respect to some convenient point $($temporary origin$),$ of the "infinitesimal element" as a

function of $.$ For calculating the centroid, you may as well put the temporary origin at the

center of the arc. Due to the symmetry of this particular shape, we know the centroid is on the x$-$

axis, so there is no need to calculate $\frac{?}{\text{b}\text{a}\text{r}}(y).$

For calculating $I_{},I_{xy},$ and $I_{yy},($referred to generically as $I_{mn})$ it's time to expand your thought

process. In the past, you have probably always calculated $I_{mn}$ for a segment or c$-$s with respect to

segment's or cross$-$section$\text{'}$s own centroid. Then, if you wanted to determine $I_{mn}$ about some

other point $P,$ you used the parallel axis theorem: $I_{mn,P}\frac{?}{b}=ar(I{)}_{mn}+A{}_1{}_2,$ where $A$ is the area of the

segment or cross section and ${}_1$ and ${}_2$ are the appropriate distances. You did this above for

each of the segments, where $$ was the distance from the segment centroid to the total c$-$s

centroid. Well, the parallel axis theorem can be used in reverse, so long as you are moving TO

the centroid: $\frac{?}{\text{b}\text{a}\text{r}}(I{)}_{mn}=I_{mn,P}-A{}_1{}_2.$ So$,$ using the standard equation, $I_{}={\displaystyle \underset{A}{\overset{\phantom{}}{}}}{y}^{2}dA,$ you can

calculate $I_{mn}$ about the center of the arc by simply expressing $x$ and $y$ w$.$r$.$t$.$ the center of the arc.

Then, use the parallel axis theorem in reverse to get $I_{mn}$ about the centroid.