Statics  ·  Chapter 9
Engineering Statics · Chapter 9

Center of Gravity and Centroid

Learning objectives

  • Distinguish centroid, center of mass, and center of gravity.
  • Select line, area, volume, or weight moments.
  • Calculate composite centroids and resultant load locations.
  • Verify coordinates using symmetry, units, and bounds.

Topics: center of gravity · center of mass · centroid by integration · composite bodies · Pappus–Guldinus theorems · distributed loading · fluid pressure

9.1  Center of Gravity, Center of Mass, and the Centroid of a Body

Center of Gravity (G)

The center of gravity is the single point through which the resultant weight of a body acts. Every particle of mass \(dm\) within a body contributes a differential weight \(dW\). These parallel weight forces sum to the total weight \(W\), and a moment balance about each coordinate axis locates G.

Center of Gravity — General Body (Eq. 9–1)
\[\bar{x} = \frac{\int \tilde{x}\,dW}{\int dW}, \qquad \bar{y} = \frac{\int \tilde{y}\,dW}{\int dW}, \qquad \bar{z} = \frac{\int \tilde{z}\,dW}{\int dW}\]
where \(\tilde{x},\tilde{y},\tilde{z}\) are the coordinates of an arbitrary differential weight element \(dW\).

Center of Mass (Cm)

Substituting \(dW = g\,dm\) and cancelling the constant \(g\) yields the center-of-mass equations — identical in form to the center of gravity but based on mass distribution. Needed for dynamic analysis where acceleration matters.

Center of Mass (Eq. 9–2)
\[\bar{x} = \frac{\int \tilde{x}\,dm}{\int dm}, \qquad \bar{y} = \frac{\int \tilde{y}\,dm}{\int dm}, \qquad \bar{z} = \frac{\int \tilde{z}\,dm}{\int dm}\]

Centroid of a Volume

For a homogeneous body with constant density \(\rho\), set \(dm = \rho\,dV\). The density cancels and the result is the pure geometric centroid C, dependent only on shape.

Centroid of a Volume (Eq. 9–3)
\[\bar{x} = \frac{\int_V \tilde{x}\,dV}{\int_V dV}, \qquad \bar{y} = \frac{\int_V \tilde{y}\,dV}{\int_V dV}, \qquad \bar{z} = \frac{\int_V \tilde{z}\,dV}{\int_V dV}\]

Centroid of an Area

For a planar area bounded by a curve \(y = f(x)\), a single integration using rectangular strip elements locates the centroid. A vertical strip has \(d A = y\,dx\), \(\tilde{x}=x\), \(\tilde{y}=y/2\). A horizontal strip has \(dA = x\,dy\), \(\tilde{x}=x/2\), \(\tilde{y}=y\).

Centroid of an Area (Eq. 9–4)
\[\bar{x} = \frac{\int_A \tilde{x}\,dA}{\int_A dA}, \qquad \bar{y} = \frac{\int_A \tilde{y}\,dA}{\int_A dA}\]

Centroid of a Line (Curve)

For a curved rod or wire described by \(y = f(x)\), the arc-length element is \(dL = \sqrt{1+(dy/dx)^2}\,dx\) (or equivalently in terms of \(dy\)).

Centroid of a Line (Eq. 9–5)
\[\bar{x} = \frac{\int_L \tilde{x}\,dL}{\int_L dL}, \qquad \bar{y} = \frac{\int_L \tilde{y}\,dL}{\int_L dL}\] \[\text{where}\quad dL = \sqrt{\,1+\!\left(\frac{dy}{dx}\right)^{\!2}}\,dx \quad\text{or}\quad dL = \sqrt{\left(\frac{dx}{dy}\right)^{\!2}+1}\,dy\]

Important Points

The centroid is the pure geometric center of a shape — it coincides with the center of mass or center of gravity only when the material is homogeneous. Both center of gravity and centroid equations express a moment balance: the resultant's moment equals the sum of the moments of all parts. A centroid may lie outside the physical body (e.g., a ring). The centroid always lies on every axis of symmetry.

Procedure for Analysis — Integration

Choose an appropriate coordinate system and select a differential element (line segment \(dL\), rectangular area strip \(dA\), or disk \(dV\)).
Position the element so it touches the arbitrary boundary point \((x,y,z)\).
Express the element size (\(dL\), \(dA\), or \(dV\)) and moment arms \(\tilde{x}, \tilde{y}, \tilde{z}\) in terms of the curve's coordinates.
Substitute into Eqs. 9–1 through 9–5, express everything in terms of one variable, and integrate over the full extent of the body.

9.2  Composite Bodies

Composite Body Method

A composite body is divided into simpler parts whose individual centroids or centers of gravity are already known. A weighted sum over all parts replaces the integration. For a hole or cutout, treat that region as a part with negative area, volume, or weight.

Center of Gravity — Composite Body (Eq. 9–6)
\[\bar{x} = \frac{\sum \tilde{x}\,W}{\sum W}, \qquad \bar{y} = \frac{\sum \tilde{y}\,W}{\sum W}, \qquad \bar{z} = \frac{\sum \tilde{z}\,W}{\sum W}\]
For constant density, replace \(W\) by \(A\), \(V\), or \(L\) as appropriate.

Procedure for Composite Bodies

Sketch: Divide the shape into simple composite parts (rectangles, triangles, semicircles, standard volumes).
Holes: Represent holes as additional parts with negative sign.
Moment Arms: Establish coordinate axes; identify \(\tilde{x}, \tilde{y}, \tilde{z}\) for each part from standard tables or prior integration results.
Summation: Apply Eq. 9–6. If the shape is symmetric about an axis, the centroid lies on that axis — use this to reduce unknowns.
Tabulate each part's area/volume/length and moment arm for clarity.

9.3  Theorems of Pappus and Guldinus

Bodies of Revolution

When a plane curve or plane area is revolved about an external axis, it sweeps out a surface of revolution or a solid of revolution. The Pappus–Guldinus theorems provide a shortcut: knowing the centroid of the generating curve or area is enough to compute the surface area or volume.

First Theorem — Surface Area of Revolution (Eq. 9–7)

The surface area generated by revolving a curve of length \(L\) through angle \(\theta\) about an external axis equals the product of \(\theta\), the perpendicular distance \(\bar{r}\) from the axis to the centroid of the curve, and the curve length \(L\).

\[A = \theta\,\bar{r}\,L\]

For a full revolution, \(\theta = 2\pi\). \(\bar{r}\) = distance from axis to centroid of the generating curve.

Second Theorem — Volume of Revolution (Eq. 9–8)

The volume generated by revolving a plane area \(A\) through angle \(\theta\) about an external axis equals the product of \(\theta\), the perpendicular distance \(\bar{r}\) from the axis to the centroid of the area, and the generating area \(A\).

\[V = \theta\,\bar{r}\,A\]
Composite Shapes — Pappus–Guldinus (Eqs. 9–9 / 9–10)
\[A = \theta\,\sum(\tilde{r}\,L), \qquad V = \theta\,\sum(\tilde{r}\,A)\]
Sum over all composite line segments or area segments, each with its own centroidal distance \(\tilde{r}\) from the axis.

9.4  Resultant of a General Distributed Loading

Distributed Load on a Flat Surface

A general distributed load on a flat plate is described by the pressure function \(p = p(x,y)\) in Pa (N/m²). Each differential area \(dA\) carries force \(dF = p\,dA\). The pressure-height surface forms a 3D loading diagram above the plate.

Resultant Force — General Distributed Load (Eq. 9–11)
\[F_R = \int_A p(x,y)\,dA = \int_V dV = V\]
The magnitude of \(F_R\) equals the total volume under the distributed-loading diagram.
Location of Resultant Force (Eq. 9–12)
\[\bar{x} = \frac{\int_V x\,dV}{\int_V dV}, \qquad \bar{y} = \frac{\int_V y\,dV}{\int_V dV}\]
The resultant force acts through the centroid of the volume under the loading diagram.

Key Result

The line of action of \(F_R\) passes through the geometric center (centroid) of the 3D volume formed by the loading surface. This is a direct extension of the 2D beam loading concept to general plate surfaces.

9.5  Fluid Pressure

Pascal's Law

A fluid at rest creates pressure at a point that is equal in all directions. For an incompressible liquid, the pressure at depth \(z\) below the free surface depends only on the fluid's specific weight and depth. Gases are compressible and this law does not apply to them.

Hydrostatic Pressure (Pascal's Law) (Eq. 9–13)
\[p = \gamma z = \rho g z\]
\(\gamma = \rho g\) = specific weight of fluid; \(z\) = depth below free surface. For water: \(\gamma = 62.4\,\text{lb/ft}^3 = 9810\,\text{N/m}^3\).

Flat Plate of Constant Width

Trapezoidal Pressure Distribution

Since pressure varies linearly with depth, the load intensity on a constant-width plate forms a trapezoid (force per unit length). The resultant force equals the trapezoidal area; it acts at the centroid of that trapezoid — the center of pressure P, which is below the plate's geometric centroid.

Curved Plate of Constant Width

Horizontal and Vertical Components

For a curved submerged surface, resolve the resultant into horizontal (\(F_h\)) and vertical (\(F_v\)) components. \(F_h\) equals the resultant hydrostatic force on the vertical projection of the curved surface. \(F_v\) equals the weight of the real or imaginary fluid column directly above the curved surface.

Flat Plate of Variable Width

Resultant on Any Flat Submerged Plate (Eq. 9–14)
\[F_R = \gamma\,\bar{z}\,A\]
\(\bar{z}\) = depth to the centroid of the plate area \(A\). Valid for any flat plate shape — \(F_R\) acts at the center of pressure P, not at the centroid of the area.

Physical meaning

A centroid is a weighted average location. Area, volume, mass, weight, or pressure supplies the weight; its first moment is the amount multiplied by its coordinate.

Problem-solving strategy

  1. Choose coordinates and the correct measure: \(L,A,V,m,W\), or pressure load.
  2. Exploit symmetry before tabulating parts or selecting a differential element.
  3. Sum signed first moments, treating holes as negative measures.
  4. Divide by total measure and check that the location is physically credible.
EXAMPLE 1Offset hole in a bracket plate
A uniform 200 mm by 120 mm rectangular plate has a 40 mm diameter circular hole centered at \((150,60)\) mm from its lower-left corner. Locate the remaining area centroid.

Recognition strategy

This is a composite-area problem: rectangle minus circular hole, with a horizontal symmetry line.

Given information

  • Rectangle centroid \((100,60)\) mm; hole area is negative and centered at \((150,60)\) mm.

Model and interpretation

Uniform thickness and density allow area weighting. Coordinates refer to the same lower-left datum.

Step-by-step solution
1. Areas. \(A_R=200(120)=24000\ \text{mm}^2\), while \(A_H=-\pi20^2=-1256.6\ \text{mm}^2\).
2. Total. The remaining area is \(A=24000-1256.6=22743.4\ \text{mm}^2\).
3. Moment. \(\sum xA=24000(100)-1256.6(150)=2.2115\times10^6\ \text{mm}^3\).
4. Location. \(\bar{x}=\sum xA/A=97.24\ \text{mm}\), and symmetry gives \(\bar y=60.0\ \text{mm}\).

Verification

Units: First moments are mm³ divided by mm², correctly leaving millimetres for centroid coordinates.

Signs and direction: The hole has negative area, so removing material on the right shifts \(\bar x\) left of 100 mm.

Plausibility: The result lies within the plate and only 2.76 mm left of the rectangle center because the hole is small.

Final result: \(\boxed{(\bar x,\bar y)=(97.2,\ 60.0)\ \text{mm}}\).

Engineering interpretation

The hole shifts the balance point away from itself; fixture or lifting calculations should use the finished geometry, including all cutouts.

EXAMPLE 2Graded-density cylinder
A 0.20 m radius cylinder is 0.60 m tall, with density \(\rho=800+400z\) kg/m³ where \(z\) is metres from its base. Find its center-of-mass height.

Recognition strategy

Density varies only axially, so disk area cancels in the mass-moment ratio.

Given information

  • \(0\le z\le0.60\) m and \(dm=\rho\pi(0.20)^2dz\).

Model and interpretation

Axisymmetry gives \(\bar x=\bar y=0\); density is supplied and varies linearly with height.

Step-by-step solution
1. Ratio. Write \(\bar z=\int z\rho dz/\int\rho dz\), after cancelling the constant disk area.
2. Mass integral. \(\int_0^{.6}(800+400z)dz=552\ \text{kg/m}^2\).
3. Moment integral. \(\int_0^{.6}z(800+400z)dz=172.8\ \text{kg/m}\).
4. Divide. \(\bar z=172.8/552=0.313\ \text{m}\) above the base.

Verification

Units: The numerator has one more metre than the denominator, so their ratio is metres.

Signs and direction: Positive \(z\) is upward and positive density makes the computed height physically above the base.

Plausibility: Increasing density upward moves the mass center above the uniform-cylinder value of 0.300 m, as 0.313 m does.

Final result: \(\boxed{\bar z=0.313\ \text{m above base}}\).

Engineering interpretation

Material grading changes balance even with unchanged exterior geometry; area-centroid tables alone cannot represent this part.

EXAMPLE 3Vertical gate loading
A 1.20 m wide vertical rectangular gate is 2.0 m tall with its top 1.0 m below fresh water. Find the hydrostatic resultant and its depth below the free surface.

Recognition strategy

Hydrostatic pressure is linear with depth, so use the area centroid for force and the pressure-center relation for location.

Given information

  • \(\gamma=9810\ \text{N/m}^3\), \(A=2.40\ \text{m}^2\), and \(\bar z=2.00\ \text{m}\).

Model and interpretation

Water is static, incompressible, and open to atmosphere; pressure is gauge pressure measured downward from its free surface.

Step-by-step solution
1. Force. \(F_R=\gamma\bar zA=9810(2.00)(2.40)=47.1\ \text{kN}\).
2. Inertia. About the horizontal centroidal axis, \(I_G=bh^3/12=1.20(2.0)^3/12=0.800\ \text{m}^4\).
3. Center pressure. Use \(z_P=\bar z+I_G/(\bar zA)\) for the line of action.
4. Evaluate. \(z_P=2.00+0.800/[2.00(2.40)]=2.167\ \text{m}\) below the surface.

Verification

Units: Specific weight times depth times area gives N; \(I_G/(\bar zA)\) gives metres as required for depth.

Signs and direction: Gauge pressure acts normal from water toward the gate, and increasing downward pressure places the resultant below the area centroid.

Plausibility: The 2.167 m center lies between the 1 m and 3 m gate edges and below its 2.0 m centroid.

Final result: \(\boxed{F_R=47.1\ \text{kN},\ z_P=2.167\ \text{m below free surface}}\).

Engineering interpretation

Gate hinges and anchors must resist both this force and its moment; water level variation, seals, impact, code load combinations, and structural strength require separate design review.

Chapter recap

  • Centroids are first moments divided by the relevant total measure.
  • Use signed components for holes and mass weights for nonuniform materials.
  • Distributed loads act through their loading centroid, often not the geometric center.
  • Before moving on: choose a datum, state assumptions, tabulate moments, and check symmetry and bounds.