Geometric properties of sections

Geometric properties of a cross-section — its centroid, moment of inertia and radius of gyration — decide how a beam bends and how a column buckles. They appear in almost every structural formula, from the flexure equation $\sigma = My/I$ to Euler's buckling load. For NEC MCQs you should be able to locate centroids of built-up shapes, apply the parallel-axis theorem and recall standard values for rectangles, circles and triangles.

Key concepts

Axes of symmetry

• If a section has an axis of symmetry, its centroid lies on that axis. With two axes of symmetry (rectangle, I-section, circle), the centroid is at their intersection.
• For symmetric sections the product of inertia about symmetry axes is zero, so these are principal axes.

Centre of gravity and centroid

• Centre of gravity: point through which the resultant weight acts. For plane areas we use the centroid (geometric centre of area).
• For a composite (built-up) section of parts with areas $A_i$ and centroid distances $x_i$, $y_i$: $\bar x = \frac{\sum A_i x_i}{\sum A_i},\qquad \bar y = \frac{\sum A_i y_i}{\sum A_i}$
• Holes/cut-outs are entered with negative area.

Shape	Centroid location
Rectangle $b\times d$	$d/2$ from base
Triangle (height $h$)	$h/3$ from base ($2h/3$ from apex)
Semicircle (radius $r$)	$4r/3\pi$ from diameter
Quarter circle	$4r/3\pi$ from each straight edge
Circle	Centre

Standard steel sections

• Rolled sections such as I-beams (ISMB, ISJB), channels (ISMC), angles (ISA), T-sections. Properties ($A$, $I_{xx}$, $I_{yy}$, $r$, centroid distance $c_y$ for channels/angles) are read from steel tables (IS Handbook SP 6).
• Built-up sections: two channels back-to-back or face-to-face, plates added to I-beams, laced/battened columns — centroid found by the composite formula and $I$ by parallel-axis theorem.

Moment of inertia (second moment of area)

$I_x = \int y^2\,dA,\qquad I_y = \int x^2\,dA$

• Unit: mm⁴ or m⁴. Measures resistance to bending about an axis.
• Parallel axis theorem: $I = I_G + Ah^2$ ($h$ = distance between centroidal axis and the parallel axis).
• Perpendicular axis theorem (plane laminae): $I_z = I_x + I_y$ (polar moment $J$).

Section	$I$ about centroidal axis	$I$ about base
Rectangle	$bd^3/12$	$bd^3/3$
Triangle	$bh^3/36$	$bh^3/12$
Circle (dia $D$)	$\pi D^4/64$	— ($J = \pi D^4/32$)
Hollow circle	$\pi(D^4-d^4)/64$	—
Semicircle	≈ $0.11r^4$ (about own centroid)	$\pi D^4/128$ (about diameter)

Section modulus and radius of gyration

• Section modulus $Z = I/y_{max}$ (mm³); rectangle $Z = bd^2/6$; circle $Z = \pi D^3/32$. Bending stress $\sigma = M/Z$.
• Radius of gyration $k = \sqrt{I/A}$ (mm): distance at which the whole area could be concentrated to give the same $I$. Rectangle $k = d/\sqrt{12} = 0.289d$; solid circle $k = D/4$.
• Columns buckle about the axis with the least radius of gyration; slenderness ratio = $L_e/k_{min}$.

High-yield points

• Rectangle: $I = bd^3/12$ (centroid), $bd^3/3$ (base); triangle: $bh^3/36$ (centroid), $bh^3/12$ (base).
• Solid circle: $I = \pi D^4/64$, $J = \pi D^4/32$, $Z = \pi D^3/32$, $k = D/4$.
• Triangle centroid at $h/3$ from base; semicircle at $4r/3\pi$ from the diameter.
• Rectangle: $Z = bd^2/6$, $k = d/\sqrt{12} = 0.289d$.
• About an axis of symmetry the product of inertia is zero (principal axis).

Formulas

• Centroid: $\bar y = \sum A_iy_i/\sum A_i$ (holes as negative areas).
• Parallel axis: $I = I_G + Ah^2$; perpendicular axis: $I_z = I_x + I_y$.
• Section modulus $Z = I/y_{max}$; radius of gyration $k = \sqrt{I/A}$.
• Slenderness ratio $= L_e/k_{min}$.

Worked example

A T-section has a flange 100 mm × 20 mm and a web 20 mm × 80 mm (web below flange, total depth 100 mm). Find the centroid from the bottom of the web and $I_{xx}$ about the centroidal axis.

• Flange: $A_1 = 2000$ mm², $y_1 = 80 + 10 = 90$ mm
• Web: $A_2 = 1600$ mm², $y_2 = 40$ mm
• $\bar y = (2000\times90 + 1600\times40)/3600 = (180\,000 + 64\,000)/3600 = 67.78$ mm
• $I_1 = 100\times20^3/12 + 2000(90-67.78)^2 = 66\,667 + 2000\times493.8 = 1\,054\,321$ mm⁴
• $I_2 = 20\times80^3/12 + 1600(67.78-40)^2 = 853\,333 + 1600\times771.6 = 2\,087\,901$ mm⁴
• $I_{xx} \approx 3.14\times10^6$ mm⁴; $k = \sqrt{3.14\times10^6/3600} = 29.5$ mm.

Exam tips

• Rectangle about its base: $bd^3/3$ (four times $bd^3/12$); triangle about base: $bh^3/12$, about centroid $bh^3/36$.
• Radius of gyration of a solid circle = $D/4$; of a rectangle = $d/\sqrt{12}$.
• Parallel axis theorem works only from the centroidal axis — never shift between two non-centroidal axes directly.
• For a given area, hollow sections and I-sections give higher $I$ — material placed far from the neutral axis.
• Columns fail about the axis of minimum $k$ (minimum $I$).

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. For the T-section with a 100 mm × 20 mm flange on a 20 mm × 100 mm web (total depth 120 mm), the moment of inertia about the horizontal centroidal axis is about

• A. $4.27\times10^{6}$ mm⁴
• B. $3.47\times10^{6}$ mm⁴
• C. $5.33\times10^{6}$ mm⁴
• D. $6.40\times10^{6}$ mm⁴

Answer: C. $\bar y = 80$ mm. Flange: $\frac{100\times20^3}{12} + 2000\times30^2 = 1.867\times10^6$; web: $\frac{20\times100^3}{12} + 2000\times30^2 = 3.467\times10^6$; total $5.33\times10^6$ mm⁴.

2. The moment of inertia of a solid circular section of diameter d about its diameter is

• A. $\frac{\pi d^4}{16}$
• B. $\frac{\pi d^4}{64}$
• C. $\frac{\pi d^3}{32}$
• D. $\frac{\pi d^4}{32}$

Answer: B. $I = \frac{\pi d^4}{64}$ about a diameter; $\frac{\pi d^4}{32}$ is the polar moment of inertia.

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