Solving for y in A(total)*y(horz centroidal axis): Where Did I Go Wrong?

[smr101]

Hi, having problems with (a) here, I'll show my attempt:

A1 = (0.025 * 0.05) - ((pi*0.01^2)/ 2)
1.093 x10^-3 m^2

A2 = (0.075 * 0.05) - ((pi*0.01^2)/2)
= 3.593 x10^-3 m^2

A(total)*y(horz centroidal axis) = A1y1 + A2y2

y = 1.093 x10^-3 * 0.0875 + 3.593x10^-3 * 0.0375 /(4.686x10^-3)
= 49.19 mm

Correct answer is 48.32 mm, any idea where I've gone wrong?

Thanks.

 

[SteamKing]

Hi, having problems with (a) here, I'll show my attempt:

A1 = (0.025 * 0.05) - ((pi*0.01^2)/ 2)
1.093 x10^-3 m^2

A2 = (0.075 * 0.05) - ((pi*0.01^2)/2)
= 3.593 x10^-3 m^2

A(total)*y(horz centroidal axis) = A1y1 + A2y2

y = 1.093 x10^-3 * 0.0875 + 3.593x10^-3 * 0.0375 /(4.686x10^-3)
= 49.19 mm

Correct answer is 48.32 mm, any idea where I've gone wrong?

Thanks.

You would be better off calculating the area and centroid of the entire rectangular piece and subtracting from that the area and centroid of the circular hole.
The centroid of a circle is easy: it's the center.

The way you did the moments originally, you need to know the centroid of a semicircle, which is not given in your formula list.

 

[smr101]

You would be better off calculating the area and centroid of the entire rectangular piece and subtracting from that the area and centroid of the circular hole.
The centroid of a circle is easy: it's the center.

The way you did the moments originally, you need to know the centroid of a semicircle, which is not given in your formula list.

Right, so the centroid, y, of the circle is just 75mm?

 

[SteamKing]

Right, so the centroid, y, of the circle is just 75mm?

Yes. The dashed lines on the figure are just there to locate the center of the circle relative to other parts of the cross section.

The centroids of simple figures like circles and rectangles should be learned, not least because they are pretty obvious.