Centroids and second moment of area

Engineering · First-year engineering · about 10 min

The centroid locates the geometric centre of an area. The second moment of area measures how that area is spread about an axis. Both are essential when analysing beam bending and deflection, especially for L-shaped, T-shaped and other built-up sections.

The key idea

Split a section into simple, non-overlapping areas. Find its centroid, then shift each part's second moment to the required centroidal axis:

\bar y=\frac{\sum A_i y_i}{\sum A_i},\qquad I_{\bar x}=\sum_i\left[I_{x,i}^{(c)}+A_i(y_i-\bar y)^2\right]

Here $A_i$ and $y_i$ are each part's area and centroid height, and $I_{x,i}^{(c)}$ is its second moment about its own horizontal centroidal axis.

Draw one upright rectangle of width b and height h. Mark a small dot at its centre and a thin horizontal line crossing that dot, extended beyond both sides, to identify the centroidal x axis. Draw a short vertical axis y through the dot with its arrow upwards. Draw external dimension lines with arrowheads for full width b below the rectangle and full height h to the left. Label the horizontal centroidal axis x, the centre C, and the two dimensions. Do not add any numerical values or other axes. — For a rectangle, vertical depth is the cubed dimension in the horizontal-axis second moment.

Try it yourself

Change the shapes and check how the centroid and second moment move.

Locate the centroid

Place an origin and divide the shape into rectangles, triangles or other areas with known centroids. For a composite area, use $\bar x=\sum A_i x_i/\sum A_i$ and $\bar y=\sum A_i y_i/\sum A_i$ . Distances $x_i$ and $y_i$ must be measured from the same origin. The centroid of a rectangle lies halfway across its width and height. Symmetry can save work: a section symmetric about a vertical centreline has its centroid on that line. Report centroid coordinates in a length unit, such as $\mathrm{mm}$ .

Measure the area spread

The second moment of area about a horizontal axis is $I_x=\int y^2\,\mathrm{d}A$ . For a rectangle of width $b$ and vertical height $h$ , $I_{x}^{(c)}=bh^3/12$ about its horizontal centroidal axis. The cube makes depth particularly influential: doubling $h$ at fixed $b$ multiplies this value by $8$ . Units are length to the fourth power, commonly $\mathrm{mm^4}$ . Do not confuse this geometric quantity with the mass moment of inertia, which measures resistance to angular acceleration.

Shift axes and handle openings

For an axis parallel to a component's centroidal axis, use $I_x=I_x^{(c)}+Ad^2$ , where $d$ is the perpendicular distance between the axes. Calculate the whole-section centroid first and use the distance from each component to that axis. A circular or rectangular opening can be represented with negative area, provided its centroid and second moment are also subtracted consistently. The $Ad^2$ term is always non-negative for a solid component. As a reasonableness check, the composite centroid should lie within the range of the component centroid coordinates for an all-solid section.

Worked example: Centroid and horizontal second moment of an L-section

An L-shaped area consists of a vertical rectangle $20\,\mathrm{mm}$ wide and $120\,\mathrm{mm}$ tall, plus a bottom rectangle extending a further $80\,\mathrm{mm}$ to the right and measuring $20\,\mathrm{mm}$ tall. The two rectangles do not overlap. The origin is at the outside lower-left corner. Find $\bar x$ , $\bar y$ , and the second moment $I_{\bar x}$ about the horizontal axis through the composite centroid.

Draw an exact orthogonal L-shaped area as the union of a left vertical bar from x=0 to x=20 mm and y=0 to y=120 mm, plus a bottom bar from x=20 to x=100 mm and y=0 to y=20 mm. The full outside width is 100 mm, outside height is 120 mm, and thickness is 20 mm. Put the origin O at the outside lower-left corner; x points right along the lower edge, y points up along the left edge. Add a thin dashed vertical partition at x=20 mm from y=0 to y=20 mm, separating a full-height vertical rectangle labelled 1 and the non-overlapping right bottom rectangle labelled 2. Show dimension arrows for full width 100 mm, full height 120 mm, top width of the vertical bar 20 mm, and thickness of the bottom bar 20 mm. Do not mark the composite centroid or a numerical second moment. — Divide the L into two rectangles without counting the corner twice.

Show the solution

Call the vertical rectangle part 1 and the extra bottom rectangle part 2. All coordinates are measured from the lower-left origin.

Areas and part centroids are

A_1=20(120)=2400\,\mathrm{mm^2},\quad (x_1,y_1)=(10,60)\,\mathrm{mm}

A_2=80(20)=1600\,\mathrm{mm^2},\quad (x_2,y_2)=(60,10)\,\mathrm{mm}.

Total area and centroid coordinates are

A=4000\,\mathrm{mm^2},\quad \bar x=\frac{2400(10)+1600(60)}{4000}=30\,\mathrm{mm}

\bar y=\frac{2400(60)+1600(10)}{4000}=40\,\mathrm{mm}.

Find each rectangle's own horizontal centroidal second moment:

I_{x,1}^{(c)}=\frac{20(120)^3}{12}=2{,}880{,}000\,\mathrm{mm^4}

I_{x,2}^{(c)}=\frac{80(20)^3}{12}=53{,}333.33\,\mathrm{mm^4}.

Shift both to $y=\bar y=40\,\mathrm{mm}$ :

I_{\bar x}=2{,}880{,}000+2400(60-40)^2+53{,}333.33+1600(10-40)^2

I_{\bar x}\approx5.3333\times10^6\,\mathrm{mm^4}.

Answer: $\bar x=30.0\,\mathrm{mm}$ , $\bar y=40.0\,\mathrm{mm}$ and $I_{\bar x}=5.33\times10^6\,\mathrm{mm^4}$ .

Common mistakes

Counting an overlapping corner twice

Use disjoint rectangles, or subtract the overlap once from area and moments.

Using a different origin for each centroid

Measure every component coordinate from the same reference axes.

Forgetting the parallel-axis term

Add $Ad^2$ to each part when its centroidal axis differs from the target axis.

Using $bh^3/12$ with the wrong height

The cubed dimension is perpendicular to the axis about which you calculate $I$ .

Check your understanding

1. What is $I_x^{(c)}$ for a rectangle $60\,\mathrm{mm}$ wide and $20\,\mathrm{mm}$ tall, about its horizontal centroidal axis?

2. Two non-overlapping areas of $1000\,\mathrm{mm^2}$ each have centroid heights $10\,\mathrm{mm}$ and $50\,\mathrm{mm}$ . What is their combined centroid height?

3. A solid part has area $500\,\mathrm{mm^2}$ . Its centroidal axis is $20\,\mathrm{mm}$ from a parallel target axis. What term must be added to its centroidal second moment?

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