Start » Exercises » Engineering Mechanics I » Center of area, with bore

Center of area, with bore

The common center of area has to be calculated for the shown circular cross-section with a bore.

Solution

The individual areas are numbered in the first step.

$\require{cancel}$

Since the component is mirror-symmetrical about the x-axis, it is sufficient to consider the x-coordinate. The general equation for the x-coordinate of the center of area is:

$\tag{1} x_S = \frac{\sum A_i \cdot x_i}{\sum A_i}$

And for the area of ​​the circle

$\tag{2} A = \frac{\pi}{4} \cdot d^2$

The individual areas and their centre point coordinates are listed in a table. Empty areas, such as e.g. the bore with the diameter of 22mm, are negative.

$i$$A_i$$x_i \, in \, mm$$A_i \cdot x_i\, in \,mm^3$
13.84800
2-38011-4.180
$\sum$3.468-4.180

$\tag{3} x_S = \frac{-4180mm^3}{3468mm^2}$

$\tag{4} x_S = -1,21 mm$

And for the sake of completeness:

$\tag{5} y_S = 0 mm$