Geometrical Properties of Sections

Principal axes

For a plane area, the principal axes are the pair of mutually perpendicular axes through a point (normally the centroid) about which the product of inertia is zero. They are the axes about which the moments of inertia take extreme (maximum and minimum) values. For a section with an axis of symmetry, the axis of symmetry and the axis perpendicular to it are principal axes.

Principal moments of inertia

The moments of inertia about the principal axes are called the principal moments of inertia: the maximum $I_{max}$ about one principal axis and the minimum $I_{min}$ about the other. For a section with $I_x$, $I_y$ and $I_{xy}$ about the centroidal axes:

$$I_{max,min} = \frac{I_x + I_y}{2} \pm \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2}$$

The inclination of the principal axes to the x-axis is given by

$$\tan 2\theta_p = \frac{-2I_{xy}}{I_x - I_y}$$

Two directions, $\theta_p$ and $\theta_p + 90^\circ$, satisfy this equation. Note that $I_{max} + I_{min} = I_x + I_y$ (the polar moment of inertia is constant) and $I_{max}I_{min} = I_xI_y - I_{xy}^2$.

The product of inertia of an area about a pair of perpendicular axes $x$ and $y$ is the sum, over the whole area, of the products of each elemental area and its two coordinates:

$$I_{xy} = \int xy\,dA$$

Its unit is length$^4$ (for example mm$^4$). Unlike $I_x$ and $I_y$, it may be positive, negative or zero: it is zero if either axis is an axis of symmetry, and it is positive when most of the area lies in the first and third quadrants. By the parallel-axis theorem $I_{xy} = \bar I_{xy} + A\bar x\bar y$.

Dimensions and assumptions (read from the figure):

• Overall outline (mm): vertices (0,−100), (150,−100), (150,−50), (200,−50), (200,50), (0,100); the top edge slopes from 100 mm to 50 mm above the x-axis.
• Semicircular cut-out of radius 50 mm with its centre on the left edge at (0, −50), removed.
• Circular hole of diameter 60 mm with centre (100, −25), removed.

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 25746\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.530\times10^{6}}{25746} = 98.26\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{37869}{25746} = 1.471\ \text{mm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 7.246\times10^{7}\pm1.437\times10^{7} \\ I_{max} &= 8.682\times10^{7}\ \text{mm}^4,\qquad I_{min} = 5.809\times10^{7}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{2.768\times10^{7}}{-7.709\times10^{6}}$, so $2\theta_p = 105.56^\circ$ (and $-74.44^\circ$).

• The axis of $I_{max}$ makes 52.78° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -37.22° (perpendicular), both passing through the centroid (98.26, 1.471) measured from the reference origin.

Answer: $I_{max} = 8.682\times10^{7}\ \text{mm}^4$, $I_{min} = 5.809\times10^{7}\ \text{mm}^4$; principal axes through the centroid at 52.78° (for $I_{max}$) and -37.22° (for $I_{min}$) from the +x axis.

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Strip: 1000 × 200 (x 0 to 1000, y 0 to 200). Upper-left right triangle: vertices (0,200), (600,200), (0,800). Lower-right right triangle: vertices (1000,0), (400,0), (1000,−600).

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 5.600\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.800\times10^{8}}{5.600\times10^{5}} = 500.0\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.600\times10^{7}}{5.600\times10^{5}} = 100.0\ \text{mm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 4.827\times10^{10}\pm3.688\times10^{10} \\ I_{max} &= 8.514\times10^{10}\ \text{mm}^4,\qquad I_{min} = 1.139\times10^{10}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{7.200\times10^{10}}{-1.600\times10^{10}}$, so $2\theta_p = 102.53^\circ$ (and $-77.47^\circ$).

• The axis of $I_{max}$ makes 51.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -38.74° (perpendicular), both passing through the centroid (500.0, 100.0) measured from the reference origin.

Answer: $I_{max} = 8.514\times10^{10}\ \text{mm}^4$, $I_{min} = 1.139\times10^{10}\ \text{mm}^4$; principal axes through the centroid at 51.26° (for $I_{max}$) and -38.74° (for $I_{min}$) from the +x axis.

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Rectangle 1200 × 200 (x 0 to 1200, y 0 to 200); triangle above its left end with base 600 and height 500 (vertices (0,200), (600,200), (0,700)); quarter circle of radius 200 below the right end, centre at (1200, 0), occupying x 1000 to 1200, y −200 to 0.

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 4.214\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{2.090\times10^{8}}{4.214\times10^{5}} = 496.0\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{7.633\times10^{7}}{4.214\times10^{5}} = 181.1\ \text{mm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 3.580\times10^{10}\pm2.914\times10^{10} \\ I_{max} &= 6.494\times10^{10}\ \text{mm}^4,\qquad I_{min} = 6.667\times10^{9}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.343\times10^{10}}{-4.773\times10^{10}}$, so $2\theta_p = 145.00^\circ$ (and $-35.00^\circ$).

• The axis of $I_{max}$ makes 72.50° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -17.50° (perpendicular), both passing through the centroid (496.0, 181.1) measured from the reference origin.

Answer: $I_{max} = 6.494\times10^{10}\ \text{mm}^4$, $I_{min} = 6.667\times10^{9}\ \text{mm}^4$; principal axes through the centroid at 72.50° (for $I_{max}$) and -17.50° (for $I_{min}$) from the +x axis.

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Bottom rectangle 25 × 5; middle rectangle 25 × 10 above it; top part is a trapezoid with a vertical left side of 15 cm, top width 5 cm and a sloping right side down to the full width of 25 cm (vertices (0,15), (25,15), (5,30), (0,30)).

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $cm^2$, I in $cm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 600.0\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{6625}{600.0} = 11.04\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{7500}{600.0} = 12.50\ \text{cm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 32487\pm10183 \\ I_{max} &= 42670\ \text{cm}^4,\qquad I_{min} = 22304\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{19375}{6276}$, so $2\theta_p = 72.05^\circ$ (and $-107.95^\circ$).

• The axis of $I_{max}$ makes 36.03° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -53.97° (perpendicular), both passing through the centroid (11.04, 12.50) measured from the reference origin.

Answer: $I_{max} = 42670\ \text{cm}^4$, $I_{min} = 22304\ \text{cm}^4$; principal axes through the centroid at 36.03° (for $I_{max}$) and -53.97° (for $I_{min}$) from the +x axis.

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Base rectangle 600 × 200; triangle above its left end (vertices (0,200), (400,200), (0,700)), so the left edge is 700 mm high; quarter circle of radius 200 added at the right of the triangle with centre (400, 200), occupying x 400 to 600, y 200 to 400.

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 2.514\times10^{5}\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{6.457\times10^{7}}{2.514\times10^{5}} = 256.8\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{5.762\times10^{7}}{2.514\times10^{5}} = 229.2\ \text{mm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 6.913\times10^{9}\pm2.756\times10^{9} \\ I_{max} &= 9.669\times10^{9}\ \text{mm}^4,\qquad I_{min} = 4.157\times10^{9}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{5.100\times10^{9}}{-2.092\times10^{9}}$, so $2\theta_p = 112.31^\circ$ (and $-67.69^\circ$).

• The axis of $I_{max}$ makes 56.15° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -33.85° (perpendicular), both passing through the centroid (256.8, 229.2) measured from the reference origin.

Mohr's circle check

• Centre $C = \left(\dfrac{I_x+I_y}{2},\,0\right) = (6.913\times10^{9},\ 0)$; radius $R = 2.756\times10^{9}$.
• Plot $X(I_x,\ I_{xy}) = (5.867\times10^{9},\ -2.550\times10^{9})$ and $Y(I_y,\ -I_{xy}) = (7.959\times10^{9},\ 2.550\times10^{9})$; the line XY is a diameter.
• The circle cuts the horizontal axis at $I_{max} = 9.669\times10^{9}$ and $I_{min} = 4.157\times10^{9}$, equal to the values found analytically. The angle from CX to the $I_{max}$ end of the horizontal diameter is $2\theta_p = 112.31^\circ$ (turned from the x axis towards the principal axis in the same sense as the physical rotation).

Answer: $I_{max} = 9.669\times10^{9}\ \text{mm}^4$, $I_{min} = 4.157\times10^{9}\ \text{mm}^4$; principal axes through the centroid at 56.15° (for $I_{max}$) and -33.85° (for $I_{min}$) from the +x axis.

Dimensions and assumptions (read from the figure):

• Dimensions in mm. Rectangle 65 × 80 (x 0 to 65, y 0 to 80); quarter circle of radius 65 on top with centre at (0, 80); right-angled triangle on the right of the rectangle: vertices (65,0), (145,0), (65,145) (base 80, height 145).

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $mm^2$, I in $mm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 14318\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{7.922\times10^{5}}{14318} = 55.33\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{8.453\times10^{5}}{14318} = 59.04\ \text{mm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 1.935\times10^{7}\pm6.807\times10^{6} \\ I_{max} &= 2.615\times10^{7}\ \text{mm}^4,\qquad I_{min} = 1.254\times10^{7}\ \text{mm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{1.326\times10^{7}}{3.103\times10^{6}}$, so $2\theta_p = 76.82^\circ$ (and $-103.18^\circ$).

• The axis of $I_{max}$ makes 38.41° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -51.59° (perpendicular), both passing through the centroid (55.33, 59.04) measured from the reference origin.

Answer: $I_{max} = 2.615\times10^{7}\ \text{mm}^4$, $I_{min} = 1.254\times10^{7}\ \text{mm}^4$; principal axes through the centroid at 38.41° (for $I_{max}$) and -51.59° (for $I_{min}$) from the +x axis.

Dimensions and assumptions (read from the figure):

• Dimensions in cm. The section is the polygon (0,0), (50,0), (50,50), (30,50), (30,90), (10,90), (0,30): base 50, left vertical edge 30, sloping edge of horizontal 10 and rise 60, top width 20, right side 50 + 40 with a 20 cm step.

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $cm^2$, I in $cm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 3400\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{79500}{3400} = 23.38\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{1.255\times10^{5}}{3400} = 36.91\ \text{cm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 1.270\times10^{6}\pm7.156\times10^{5} \\ I_{max} &= 1.986\times10^{6}\ \text{cm}^4,\qquad I_{min} = 5.546\times10^{5}\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.740\times10^{5}}{1.381\times10^{6}}$, so $2\theta_p = 15.15^\circ$ (and $-164.85^\circ$).

• The axis of $I_{max}$ makes 7.57° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -82.43° (perpendicular), both passing through the centroid (23.38, 36.91) measured from the reference origin.

Answer: $I_{max} = 1.986\times10^{6}\ \text{cm}^4$, $I_{min} = 5.546\times10^{5}\ \text{cm}^4$; principal axes through the centroid at 7.57° (for $I_{max}$) and -82.43° (for $I_{min}$) from the +x axis.

The radius of gyration $k$ of an area about an axis is the distance from that axis at which the whole area could be assumed concentrated at a point so that the moment of inertia remains the same:

$$I = Ak^2 \quad\Longrightarrow\quad k = \sqrt{\frac{I}{A}}$$

Unit: length (mm, m). For example, $k_x = \sqrt{I_x/A}$ and $k_y = \sqrt{I_y/A}$; for a rectangle $b\times d$, $k_x = d/\sqrt{12}$ about the centroidal axis parallel to $b$. It measures how far the area is spread from the axis and is used in column design through the slenderness ratio $L_e/k_{min}$.

Dimensions and assumptions (read from the figure):

• Dimensions in cm. Rectangle 15 × 15 (x 0 to 15, y 0 to 15); quarter circle of radius 15 above it with centre at (0, 15); right-angled triangle on the right: vertices (15,0), (45,0), (15,15).

Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $cm^2$, I in $cm^4$)

Centroid

$$\begin{aligned} A &= \sum A_i = 626.7\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{8438}{626.7} = 13.46\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{6588}{626.7} = 10.51\ \text{cm} \end{aligned}$$

Moments and product of inertia about the centroidal axes

With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$:

Principal moments of inertia and principal axes

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 52301\pm30354 \\ I_{max} &= 82655\ \text{cm}^4,\qquad I_{min} = 21947\ \text{cm}^4 \end{aligned}$$

Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{55052}{-25590}$, so $2\theta_p = 114.93^\circ$ (and $-65.07^\circ$).

• The axis of $I_{max}$ makes 57.47° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -32.53° (perpendicular), both passing through the centroid (13.46, 10.51) measured from the reference origin.

Answer: $I_{max} = 82655\ \text{cm}^4$, $I_{min} = 21947\ \text{cm}^4$; principal axes through the centroid at 57.47° (for $I_{max}$) and -32.53° (for $I_{min}$) from the +x axis.