Center of Gravity and Centroids

Introduction

The centroid is the geometric center of an area. The center of gravity is the point through which the total weight of a body acts. For bodies with uniform density and thickness, the centroid and center of gravity coincide. Centroid calculations are used in engineering to determine equivalent locations of distributed areas, loads, and cross-sectional properties.

Centroid of an Area

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x ‾ = ∫ x d A ∫ d A \bar{x} = \frac{\int_{}^{}xdA}{\int_{}^{}dA} and y ‾ = ∫ y d A ∫ d A \bar{y} = \frac{\int_{}^{}ydA}{\int_{}^{}dA}

where A = ∫ d A A = \int_{}^{}dA

If an object is made of a single, uniform material, center of gravity = center of area.

If the gravitational field around an object is uniform, center of mass = center of gravity.

Centroid for Composite Bodies

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x ‾ = Σ x i A i Σ A i \bar{x} = \frac{\Sigma x_{i}A_{i}}{\Sigma A_{i}} and y ‾ = Σ y i A i Σ A i \bar{y} = \frac{\Sigma y_{i}A_{i}}{\Sigma A_{i}}

where A i A_{i} is the area of each component, and ( x i , y i ) x_{i}\ ,\ y_{i}) is the centroid location of the component.

Differential Area Elements

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Vertical Strip

  • Used when integrating with respect to x

  • d A = y d x dA = ydx

  • Centroid of strip: y = 1 2 ( t o p + b o t t o m ) y = \frac{1}{2}(top + bottom)

Horizontal Strip

  • Used when integrating with respect to y

  • d A = x d y dA = xdy

  • Centroid of strip: y = 1 2 ( l e f t + r i g h t ) y = \frac{1}{2}(left + right)

Area Between Two Curves

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For vertical strips, d A = ( y t o p − y b o t t o m ) d x dA = (y_{top} - y_{bottom})dx

For horizontal strips, d A = ( x r i g h t − x l e f t ) d y dA = (x_{right} - x_{left})dy

Example 1:
Find: Centroid (x, y)

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For a vertical strip, the strip’s centroid is at ( x , y 2 ) . (x,\ \frac{y}{2}).

x ‾ = ∫ x d A ∫ d A = \bar{x} = \frac{\int_{}^{}xdA}{\int_{}^{}dA} = ∫ 0 8 x ( x 2 / 3 ) d x ∫ 0 8 x 2 / 3 d x = 5 i n \frac{\int_{0}^{8}x(x^{2/3})dx}{\int_{0}^{8}x^{2/3}dx} = 5\ in

y ‾ = ∫ y d A ∫ d A = \bar{y} = \frac{\int_{}^{}ydA}{\int_{}^{}dA} = ∫ 0 8 ( y / 2 ) ( x 2 / 3 ) d x ∫ 0 8 x 2 / 3 d x = ∫ 0 8 ( x 2 / 3 2 ) ( x 2 / 3 ) d x ∫ 0 8 x 2 / 3 d x = 1.43 i n \frac{\int_{0}^{8}(y/2)(x^{2/3})dx}{\int_{0}^{8}x^{2/3}dx} = \frac{\int_{0}^{8}(\frac{x^{2/3}}{2})(x^{2/3})dx}{\int_{0}^{8}x^{2/3}dx} = 1.43\ in

Example 2:
Find: Centroid y ‾ \bar{y} of the beam

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Cut where the web meets the flange.

For the top rectangle (flange):

A i = ( 0.30 m ) ( 0.05 m ) = 0.015 m 2 A_{i} = (0.30\ m)(0.05\ m) = 0.015\ m^{2}

y i = 0.3 m + 0.05 m 2 = 0.325 m y_{i} = 0.3\ m + \frac{0.05\ m}{2} = 0.325\ m

For the bottom rectangle (web):

A i = ( 0.05 m ) ( 0.30 m ) = 0.015 m 2 A_{i} = (0.05\ m)(0.30\ m) = 0.015\ m^{2}

y i = 0.3 m 2 = 0.15 m y_{i} = \frac{0.3\ m}{2} = 0.15\ m

y ‾ = Σ y i A i Σ A i = ( 0.015 m 2 ) ( 0.325 m ) + ( 0.015 m 2 ) ( 0.15 m ) 0.015 m 2 + 0.015 m 2 = 0.2375 m ( 237.5 m m ) \bar{y} = \frac{\Sigma y_{i}A_{i}}{\Sigma A_{i}} = \frac{(0.015\ m^{2})(0.325\ m) + (0.015\ m^{2})(0.15\ m)}{0.015\ m^{2} + 0.015\ m^{2}} = 0.2375\ m\ (237.5\ mm)