Study Companion: S25 — Products of Inertia and Rotation of Axes

S24's moments of inertia were taken about axes parallel to an area's edges. For a symmetric section that is enough: the axis of symmetry and the one at right angles to it are where the moment of inertia is largest and smallest. An unsymmetrical section, such as a steel angle, is different: its strongest and weakest axes are tilted. Finding them needs one more integral, the product of inertia \int xy\,dA, and a rule for how moments of inertia change as the axes rotate. The result is the principal axes and Mohr's circle.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Appendix A, article A/4 ("Products of Inertia and Rotation of Axes"), with Sample Problems A/9–A/12. Axes: x to the right, y up; rotation angles \theta and \alpha counter-clockwise from x.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

Meriam's angle section (Sample Problem A/12): a 40 \times 10\,\mathrm{mm} leg and a 10 \times 40\,\mathrm{mm} leg, 50\,\mathrm{mm} tall overall. About the horizontal axis through its centroid, I_x = 18.17(10^4)\,\mathrm{mm^4}; about the vertical one, I_y = 10.17(10^4)\,\mathrm{mm^4}.

Is 18.17(10^4)\,\mathrm{mm^4} the largest moment of inertia this section has about any axis through C, and which way is it weakest? The plausible guess is "yes: the axes along the legs are the strong and weak ones, as for a rectangle". The answer is in Nugget 05.

Nugget 01: The Product of Inertia Measures Lopsidedness About a Pair of Axes

Claim. The product of inertia I_{xy} = \int xy\,dA weighs each element by the product of its two coordinates. Unlike a moment of inertia it can be positive, negative or zero, and it is zero whenever either axis is an axis of symmetry of the area.

The picture. Meriam's Fig. A/5, and the sign of xy in each quadrant.

Symmetric about the x-axis, every element at (x, y) has a twin at (x, -y); their contributions xy\,dA and x(-y)\,dA cancel. Area in the first and third quadrants adds positively, area in the second and fourth negatively.

Check with numbers. A 3 \times 2 rectangle with a corner at the origin, in the first quadrant: I_{xy} = \int_0^3\int_0^2xy\,dy\,dx = (\frac92)(2) = 9 = b^2h^2/4, positive. The same rectangle centred on the x-axis (-1 \le y \le 1): \int_0^3\int_{-1}^{1}xy\,dy\,dx = 0, by symmetry.

The idea. Meriam (A/4): in unsymmetrical sections, and in moments of inertia about rotated axes, the expression dI_{xy} = xy\,dA occurs; integrated,

the product of inertia of the area about the x–y axes. Moments of inertia of a positive area are always positive; the product may be positive, negative or zero. It is zero whenever either reference axis is an axis of symmetry (Fig. A/5), because the area is made of pairs of symmetric elements whose terms cancel.

The equation (definition: product of inertia):

Worked number. Any rectangle, circle, I-beam or channel about centroidal axes along its symmetry lines has \bar I_{xy} = 0; an angle section, which has no axis of symmetry parallel to its legs, does not: Meriam's angle has \bar I_{xy} = -7.5(10^4)\,\mathrm{mm^4} (Nugget 02).

Real example. Section tables list I_{xy} only for unsymmetrical shapes, angles and Z-sections; for every symmetrical rolled section it is zero and omitted.

Plausible wrong answer. "Like I_x, I_{xy} is never negative." It weighs by xy, which changes sign between quadrants.

Check. An area lies wholly in the second quadrant. What is the sign of I_{xy}? (Negative: x < 0, y > 0.)

Narration spine. An oval sits on the x-axis; two small squares light up mirror-wise above and below, with + and - products that cancel, and the whole oval's I_{xy} reads zero. The axes then show four quadrants coloured by the sign of xy.

Bridge. Like moments of inertia, products are usually known about centroidal axes and needed about others.

Nugget 02: Products Transfer With Signed Offsets

Claim. The product of inertia about axes parallel to the centroidal ones is the centroidal product plus the area times the two offsets: I_{xy} = \bar I_{xy} + d_xd_yA. The offsets carry signs, so the transfer term can be negative.

The picture. Meriam's Sample Problems A/9 and A/11.

Check with numbers. SP A/9: a rectangle b \times h whose sides are parallel to the axes has \bar I_{xy} = 0 by symmetry, so I_{xy} = d_xd_ybh. For b = 3, h = 2 with a corner at the origin, the centroid offsets are 1 and 1.5: I_{xy} = (1)(1.5)(6) = 9, as integrated in Nugget 01. SP A/11: a semicircle hanging below the x-axis, its diameter from x = 0 to 2r; its centroid is r to the right and 4r/3\pi below, and its own y_0 is an axis of symmetry, so

For r = 30\,\mathrm{mm}: -0.54(10^6)\,\mathrm{mm^4}.

The idea. Meriam (A/4), transfer of axes: with the centroidal coordinates x_0, y_0 and the offsets of Fig. A/4, I_{xy} = \int(x_0 + d_y)(y_0 + d_x)\,dA = \int x_0y_0\,dA + d_x\int x_0\,dA + d_y\int y_0\,dA + d_xd_y\int dA. The first term is \bar I_{xy}; the middle two vanish because the first moment of an area about its own centroid is zero; the last is d_xd_yA:

Meriam names each offset by the axes it separates: d_x is the distance between the x- and x_0-axes (measured along y), d_y that between the y- and y_0-axes (measured along x). Their signs must follow the positive directions of the axes (SP A/9's caution); SP A/11 shows that the transfer saves much labour.

The equation (mathematical identity: transfer of axes for the product of inertia):

Worked number. A 60 \times 40\,\mathrm{mm} rectangle whose centroid is at (50, 30)\,\mathrm{mm}: I_{xy} = (30)(50)(2400) = 3.6(10^6)\,\mathrm{mm^4}. Moved to (-50, 30)\,\mathrm{mm}, the same rectangle has -3.6(10^6)\,\mathrm{mm^4}.

Real example. For a built-up section of plates and angles, engineers tabulate each part's \bar I_{xy} + d_xd_yA exactly as for I (S24), signs included, before finding the section's principal axes.

Plausible wrong answer. "Add Ad^2 as for a moment of inertia." The transfer term is Ad_xd_y, a product of two signed offsets, not a square.

Check. A part's centroid lies on the x-axis. What is its transfer term? (Zero: d_x = 0.)

Narration spine. A rectangle slides away from the origin; two dashed offsets grow and a bar Ad_xd_y grows with them. It crosses into the second quadrant and the bar flips red, negative. Then the semicircle drops below the axis, and its product reads -2r^4/3.

Bridge. When no part has a known product, integrate, choosing the element with care.

Nugget 03: Integrate With Elements Whose Own Product Is Zero

Claim. A thin rectangular strip parallel to an axis has zero product about its own centroidal axes, so by the transfer theorem its product is just its area times its centroid's two coordinates. One integration then covers the area.

The picture. Meriam's Sample Problem A/10: the area between x = ay^2/b^2 and the line x = a.

Check with numbers. Solution I, second-order element: I_{xy} = \int_0^b\int_{ay^2/b^2}^{a}xy\,dx\,dy = \int_0^b\frac12(a^2 - \frac{a^2y^4}{b^4})y\,dy = \frac16a^2b^2. Solution II, vertical strip dA = y\,dx of height y = b\sqrt{x/a}, centroid at (x, y/2): dI_{xy} = 0 + (\frac12y)(x)(y\,dx), so I_{xy} = \int_0^a\frac{y^2}{2}x\,dx = \int_0^a\frac{xb^2}{2a}x\,dx = \frac{b^2}{6a}a^3 = \frac16a^2b^2. With a = 4, b = 3: I_{xy} = 24 (units)⁴.

The idea. Meriam (A/4, SP A/10): starting with dA = dx\,dy needs two integrations; a first-order strip saves one by using SP A/9's result for its own product, with the strip's centroidal offsets d_x = y/2 and d_y = x. A horizontal strip works too: dI_{xy} = y\cdot\frac12(a + x)[(a - x)\,dy], which integrates to the same.

The equation (mathematical identity: A/8 applied to a thin rectangular strip):

Worked number. A right triangle with legs b along x and h along y: vertical strips of height h(1 - x/b) give I_{xy} = \int_0^b x\cdot\frac h2(1 - x/b)\cdot h(1 - x/b)\,dx = b^2h^2/24; for b = 60, h = 30\,\mathrm{mm}, 13.5(10^4)\,\mathrm{mm^4}. Transferred to its centroid (b/3, h/3): \bar I_{xy} = b^2h^2/24 - (b/3)(h/3)(bh/2) = -b^2h^2/72.

Real example. Finite-element and CAD programs compute section properties exactly so, summing over small elements each with its own centroid; the product of inertia is one of the standard outputs.

Plausible wrong answer. "For a vertical strip, dI_{xy} = xy\,dA with y the strip's height." Its centroid is at y/2; using y doubles the answer.

Check. Why is the strip's own product of inertia zero? (Its centroidal axes are axes of symmetry of the thin rectangle.)

Narration spine. The parabolic area; a vertical strip sweeps across with a dot at its mid-height, the product x\cdot y/2\cdot dA accumulating to a^2b^2/6, and a horizontal strip does the same with its dot at the middle of its length.

Bridge. With I_x, I_y and I_{xy} known, the moments of inertia about any rotated axes follow.

Nugget 04: Rotating the Axes Mixes I and the Product Through the Double Angle

Claim. About axes x', y' rotated by \theta, the moments and product of inertia are combinations of I_x, I_y and I_{xy} with \cos2\theta and \sin2\theta. Their sum I_{x'} + I_{y'} does not change.

The picture. Meriam's Fig. A/6.

An element's coordinates in the rotated axes are x' = y\sin\theta + x\cos\theta and y' = y\cos\theta - x\sin\theta.

Check with numbers. A 120 \times 40\,\mathrm{mm} rectangle about its centroid, I_x = 0.640(10^6), I_y = 5.76(10^6), I_{xy} = 0 (all mm⁴), rotated 30^\circ:

I_{y'} = 3.20 + 1.28 = 4.48(10^6), I_{x'y'} = -2.56\sin60^\circ = -2.217(10^6)\,\mathrm{mm^4}. A direct double integral of (y\cos\theta - x\sin\theta)^2 over the rectangle gives the same 1.92(10^6), and 1.92 + 4.48 = 6.40(10^6) = I_x + I_y.

The idea. Meriam (A/4): I_{x'} = \int y'^2\,dA = \int(y\cos\theta - x\sin\theta)^2\,dA and I_{y'} = \int(y\sin\theta + x\cos\theta)^2\,dA. Expanding, with \sin^2\theta = (1 - \cos2\theta)/2, \cos^2\theta = (1 + \cos2\theta)/2 and the definitions of I_x, I_y, I_{xy}:

and similarly, with \sin\theta\cos\theta = \frac12\sin2\theta and \cos^2\theta - \sin^2\theta = \cos2\theta,

Adding Eqs. A/9: I_{x'} + I_{y'} = I_x + I_y = I_z, the polar moment about O, which checks Eq. A/3.

The equation (mathematical identity: transformation of second moments under rotation of axes):

Worked number. The ruler of S24 on edge (I_x = 6750, I_y = 67.5\,\mathrm{mm^4} about its centroid) tilted 10^\circ: I_{x'} = 3408.75 + 3341.25\cos20^\circ = 6548\,\mathrm{mm^4}, already 3\,\% down; at 45^\circ it is halfway, 3409\,\mathrm{mm^4}.

Real example. A joist installed slightly out of plumb loses stiffness about the vertical load direction by exactly this formula, and gains a product term that makes it deflect sideways too.

Plausible wrong answer. "Rotating by \theta changes I like \cos\theta." The dependence is on 2\theta: a 90^\circ turn swaps I_x and I_y, a 180^\circ turn returns to the start.

Check. What is I_{x'} + I_{y'} for the rectangle at \theta = 70^\circ? (6.40(10^6)\,\mathrm{mm^4}, the same as at any angle.)

Narration spine. The axes turn slowly about O; I_{x'} and I_{y'} bars rise and fall in opposite directions while a third bar for I_{x'y'} swings through zero; a counter for their sum never moves.

Bridge. As the axes turn, I_{x'} passes through a largest and a smallest value. Those directions matter most.

Nugget 05: Principal Axes: Largest and Smallest I, Zero Product

Claim. The moment of inertia is largest and smallest about two perpendicular axes, the principal axes, at the angle \alpha given by \tan2\alpha = 2I_{xy}/(I_y - I_x). About them the product of inertia is zero, and the principal moments are \frac{I_x + I_y}{2} \pm \sqrt{(\frac{I_x - I_y}{2})^2 + I_{xy}^2}.

The picture. Meriam's angle section with its principal axes.

Check with numbers (SP A/12). Origin at the outer corner, the centroid is at (12.5, 17.5)\,\mathrm{mm}. Each leg's own product is zero by symmetry; transferring each to C: part I (the 40 \times 10 leg, centroid 7.5\,\mathrm{mm} right and 12.5\,\mathrm{mm} below C): I_{xy} = 0 + (-12.5)(7.5)(400) = -3.75(10^4); part II (the 10 \times 40 leg, 7.5\,\mathrm{mm} left and 12.5\,\mathrm{mm} above): (12.5)(-7.5)(400) = -3.75(10^4); total -7.5(10^4)\,\mathrm{mm^4}. Moments of inertia by S24: I_x = 6.58 + 11.58 = 18.17(10^4), I_y = 7.58 + 2.58 = 10.17(10^4)\,\mathrm{mm^4}. Then

and Eq. A/9 at \alpha (\cos2\alpha = 0.471, \sin2\alpha = 0.882): I_{\max} = [14.17 + 4.00(0.471) + 7.50(0.882)](10^4) = 22.7(10^4)\,\mathrm{mm^4}, I_{\min} = 5.67(10^4)\,\mathrm{mm^4}.

The puzzle. No: about an axis tilted 31.0^\circ counter-clockwise from x the angle's moment of inertia is 22.7(10^4)\,\mathrm{mm^4}, 25\,\% above I_x, and about the axis at right angles (121^\circ) it is only 5.67(10^4), 56\,\% of I_y. Neither leg direction is a principal axis, because the section has no symmetry about either.

The idea. Meriam (A/4): set dI_{x'}/d\theta = (I_y - I_x)\sin2\theta - 2I_{xy}\cos2\theta = 0; the critical angle \alpha satisfies

Its two solutions for 2\alpha differ by \pi, so the two axes differ by \pi/2: one is the axis of maximum moment of inertia, the other of minimum, the principal axes of inertia. Substituting into Eq. A/9a shows the product is zero on them. Substituting \sin2\alpha, \cos2\alpha into Eqs. A/9:

Which root is the maximum is settled by evaluating Eq. A/9 at it, as SP A/12 does.

The equation (mathematical identity: the extremum of Eq. A/9):

Worked number. An equal-leg angle, legs 50 \times 10\,\mathrm{mm} and 10 \times 40\,\mathrm{mm} (each 50\,\mathrm{mm} long overall): I_x = I_y = 19.64(10^4), I_{xy} = -11.11(10^4)\,\mathrm{mm^4}. Then I_y - I_x = 0, \tan2\alpha is infinite, \alpha = 45^\circ: the principal axes are the angle's diagonals, with I_{\max} = 30.75(10^4) and I_{\min} = 8.53(10^4)\,\mathrm{mm^4}.

Real example (one you can see). A steel angle used as a shelf bracket or lintel and loaded straight down sags partly sideways: its principal axes are tilted, so a vertical load is not along either of them. Steel tables list each angle's principal axes and the angle \alpha for this reason.

Plausible wrong answer. "The strong axis of any section is parallel to its longest edge." Only when that edge is parallel to an axis of symmetry.

Check. What are the principal axes of a section with an axis of symmetry? (The symmetry axis and the one perpendicular to it through the centroid: I_{xy} = 0 there already.)

Narration spine. The angle's axes rotate; a gauge for I_{x'} rises past 18.17 to a peak of 22.7 at 31^\circ while the product gauge crosses zero at the same moment. The axes freeze there, a bold strong axis and a thin weak one at right angles.

Bridge. All of Eqs. A/9–A/11 are the geometry of one circle.

Nugget 06: Mohr's Circle Draws Every Rotated Value at Once

Claim. Plot A = (I_x, I_{xy}) and B = (I_y, -I_{xy}) on axes of moment of inertia (horizontal) and product (vertical), and draw the circle with AB as diameter. Rotating the axes of the area by \theta moves the diameter by 2\theta in the same sense; the circle's ends on the horizontal axis are I_{\max} and I_{\min}.

The picture. Mohr's circle for the angle section (Meriam's Fig. A/7 and SP A/12).

Check with numbers. A = (18.17, -7.50), B = (10.17, 7.50), in 10^4\,\mathrm{mm^4}. The centre is midway, (18.17 + 10.17)/2 = 14.17; the radius is \sqrt{4.00^2 + 7.50^2} = 8.50. So I_{\max} = 14.17 + 8.50 = 22.67 and I_{\min} = 14.17 - 8.50 = 5.67. The radius to A makes \tan^{-1}(7.50/4.00) = 61.9^\circ with the horizontal: 2\alpha.

The idea. Meriam (A/4): Eqs. A/9, A/9a, A/10 and A/11 are represented by Mohr's circle. Choose a horizontal axis for moments of inertia and a vertical one for products; locate A at (I_x, I_{xy}) and B at (I_y, -I_{xy}); draw the circle on AB as diameter. The angle from the radius OA to the horizontal axis is 2\alpha, twice the angle from the area's x-axis to the axis of maximum moment of inertia, measured in the same sense. Any point C at angle 2\theta from A gives (I_{x'}, I_{x'y'}), and the opposite point D gives (I_{y'}, -I_{x'y'}). The trigonometry of the circle reproduces Eqs. A/9, A/9a and A/10. SP A/12's hint: the circle is an alternative to Eqs. A/11.

The equation (geometric relationship: Mohr's circle of inertia):

Worked number. The 120 \times 40\,\mathrm{mm} rectangle: centre 3.20, radius 2.56 (10^6\,\mathrm{mm^4}); rotating its axes 30^\circ moves the diameter 60^\circ, to I_{x'} = 3.20 - 2.56\cos60^\circ = 1.92, as in Nugget 04. A circle's section has I_x = I_y and I_{xy} = 0: its Mohr's circle shrinks to a point, and every axis is principal.

Real example. Mohr's circle is the same construction engineers use for stresses and strains: the second moments of an area, a stress tensor and a strain tensor all transform by the double angle, and one picture serves all three.

Plausible wrong answer. "Turning the area's axes by \theta turns the circle's diameter by \theta." By 2\theta: a 90^\circ turn of the axes swaps A and B, a half-turn round the circle.

Check. Where on Mohr's circle are the principal moments, and what is the product there? (At the ends of the horizontal diameter; the product is zero.)

Narration spine. Points A and B drop onto the plane, a diameter joins them and a circle grows round it. A radius sweeps from A down to the horizontal axis through 61.9^\circ while, beside it, the angle section's axes turn 31^\circ, and the readout lands on 22.67.

Synthesis

The angle section's moment of inertia is not largest about the horizontal axis: it reaches 22.7(10^4)\,\mathrm{mm^4} about an axis 31^\circ round, and only 5.67(10^4) about the axis at right angles. The leg directions, natural for a rectangle, are wrong for a section without symmetry, and the product of inertia, -7.5(10^4)\,\mathrm{mm^4}, is what measures that.

The second moments of an area form a set that transforms together. Moments and product about rotated axes mix through the double angle (Eqs. A/9, A/9a); their sum stays fixed; the largest and smallest occur on perpendicular principal axes where the product vanishes; and Mohr's circle draws it all. For any area there are two perpendicular principal axes through each point, of greatest and least moment of inertia and zero product of inertia; an axis of symmetry is always one of them.

• Central idea: I_{xy} = \int xy\,dA completes the second moments; with it, I can be found about any axis.
• Key relationship: \tan2\alpha = 2I_{xy}/(I_y - I_x); I_{\max,\min} = \frac{I_x + I_y}{2} \pm \sqrt{(\frac{I_x - I_y}{2})^2 + I_{xy}^2}.
• Most important assumption: signs of the offsets and of I_{xy} follow the axes' positive directions; angles in one consistent sense.
• Open question: Part 5 replaced distributed forces by resultants. What happens when surfaces in contact resist sliding? (Part 6, friction, S26.)

Computed views

Neither MATLAB reference covers products of inertia. These views are our own, built on the lesson's Meriam examples, in both languages so you can work in whichever you prefer. Both print the same numbers.

The computer's route to the principal axes is worth knowing. Put the moments into the inertia tensor:

Its eigenvalues are I_{\max} and I_{\min}, and its eigenvectors are the principal directions. Mohr's circle is a drawing of the same eigenvalue problem.

Python (numpy, scipy). S25_products_of_inertia.py builds the puzzle's angle from two rectangles with the shared sections.py. It then rotates the axes, finds the principal axes with numpy.linalg.eigh, and repeats this for the equal-leg angle.

Part 1: the angle section, legs 40 x 10 and 10 x 40 mm (10^4 mm^4)
  A = 800 mm^2, C = (12.5, 17.5) mm
  I_x = 18.17, I_y = 10.17, I_xy = -7.50
  right triangle 60 x 30 about its legs: I_xy = 13.50
Part 2: rotating the axes
  theta   0 deg: I_x' = 18.17
  theta  31 deg: I_x' = 22.67
  theta  45 deg: I_x' = 21.67
  theta  90 deg: I_x' = 10.17
  theta 121 deg: I_x' = 5.67
  eigenvalues: I_max = 22.67 at 31.0 deg, I_min = 5.67
  Mohr's circle: centre 14.17, radius 8.50
  ruler on edge tilted 10 deg: I_x' = 6548 mm^4
Part 3: the equal-leg angle, legs 50 x 10 and 10 x 40 mm
  I_x = I_y = 19.64, I_xy = -11.11; I_max = 30.75 at 45.0 deg, I_min = 8.53

MATLAB (Octave-compatible). S25_products_of_inertia.m uses section_props.m and eig.

angle: A = 800, C = (12.5, 17.5), I_x = 18.17, I_y = 10.17, I_xy = -7.50 (10^4)
theta   0 deg: I_x' = 18.17
theta  31 deg: I_x' = 22.67
theta  45 deg: I_x' = 21.67
theta  90 deg: I_x' = 10.17
theta 121 deg: I_x' = 5.67
I_max = 22.67 at 31.0 deg, I_min = 5.67 (10^4 mm^4)
equal legs: I_xy = -11.11, I_max = 30.75 at 45.0 deg, I_min = 8.53

Traps

• Assuming I_{xy} \ge 0. It takes the sign of xy. (From Nugget 01.)
• Forgetting that one symmetry axis is enough. Either axis of symmetry makes I_{xy} = 0. (From Nugget 01.)
• Using Ad^2 for a product. The transfer term is Ad_xd_y, with signs. (From Nugget 02.)
• Unsigned offsets. A centroid left of or below the axes has negative offsets. (From Nugget 02.)
• Using the strip's top as its arm. A vertical strip's centroid is at y/2. (From Nugget 03.)
• Rotating with \theta instead of 2\theta. Eqs. A/9 use the double angle. (From Nugget 04.)
• Assuming the leg directions are principal. Only symmetry guarantees it. (From Nugget 05.)
• Taking either root of Eq. A/10 as the maximum. Check with Eq. A/9 which one it is. (From Nugget 05.)
• Turning the Mohr diameter by \theta. It turns by 2\theta, in the same sense. (From Nugget 06.)

Self-check

1. A small area lies in the fourth quadrant. What is the sign of its product of inertia about the x–y axes?
2. A 60 \times 40\,\mathrm{mm} rectangle has its sides parallel to the axes and its centroid at (50, 30)\,\mathrm{mm}. Find I_{xy}.
3. Meriam's Sample Problem A/11 with r = 30\,\mathrm{mm}: find I_{xy}.
4. Meriam's Sample Problem A/10 with a = 4, b = 3: find I_{xy} by horizontal strips.
5. A right triangle has legs 60\,\mathrm{mm} along x and 30\,\mathrm{mm} along y, its right angle at the origin. Find I_{xy} about the legs and about its centroidal axes parallel to them.
6. A 120 \times 40\,\mathrm{mm} rectangle (\bar I_x = 0.640, \bar I_y = 5.76, \times10^6\,\mathrm{mm^4}) has its axes rotated 30^\circ. Find I_{x'}, I_{y'} and I_{x'y'}, and check the sum.
7. Find SP A/12's principal moments from Eqs. A/11.
8. Find the centre and radius of SP A/12's Mohr's circle, and read off I_{\max}, I_{\min} and 2\alpha.
9. What are the principal axes of an I-section, and why?
10. Why is every centroidal axis of a circle a principal axis?
11. An equal-leg angle (legs 50 \times 10 and 10 \times 40\,\mathrm{mm}, each 50\,\mathrm{mm} overall) has I_x = I_y = 19.64(10^4) and I_{xy} = -11.11(10^4)\,\mathrm{mm^4} about its centroid. Find its principal axes and moments.
12. Show that I_{x'} + I_{y'} is the same for every \theta, and say what it equals.

Fully worked solutions

1. Negative: x > 0, y < 0, so xy < 0 for every element.

2. \bar I_{xy} = 0; I_{xy} = d_xd_yA = (30)(50)(2400) = 3.6(10^6)\,\mathrm{mm^4}.

3. I_{xy} = -2r^4/3 = -2(30)^4/3 = -0.54(10^6)\,\mathrm{mm^4}.

4. Horizontal strip at y, from x = ay^2/b^2 to a, centroid at ((a + x)/2, y), own product zero: I_{xy} = \int_0^b y\frac{a + x}{2}(a - x)\,dy = \int_0^b\frac y2(a^2 - x^2)\,dy = \frac{a^2}{2}\int_0^b(y - y^5/b^4)\,dy = \frac{a^2}2(\frac{b^2}2 - \frac{b^2}6) = \frac{a^2b^2}{6} = 24.

5. Vertical strips: I_{xy} = \int_0^b x\cdot\frac h2(1 - \frac xb)h(1 - \frac xb)\,dx = b^2h^2/24 = 13.5(10^4)\,\mathrm{mm^4}. Centroid at (20, 10)\,\mathrm{mm}, A = 900\,\mathrm{mm^2}: \bar I_{xy} = 13.5(10^4) - (10)(20)(900) = -4.5(10^4)\,\mathrm{mm^4} (= -b^2h^2/72).

6. I_{x'} = 3.20 - 2.56\cos60^\circ = 1.92, I_{y'} = 3.20 + 1.28 = 4.48, I_{x'y'} = -2.56\sin60^\circ = -2.217 (10^6\,\mathrm{mm^4}). Sum 6.40(10^6), equal to 0.640 + 5.76.

7. \frac{18.17 + 10.17}{2} \pm \frac12\sqrt{(18.17 - 10.17)^2 + 4(7.5)^2} = 14.17 \pm \frac12\sqrt{64 + 225} = 14.17 \pm 8.50: 22.67 and 5.67(10^4)\,\mathrm{mm^4}.

8. Centre 14.17, radius \sqrt{4.00^2 + 7.50^2} = 8.50; I_{\max} = 22.67, I_{\min} = 5.67 (10^4\,\mathrm{mm^4}); 2\alpha = \tan^{-1}(7.5/4.0) = 61.9^\circ, so \alpha = 31.0^\circ.

9. Its two axes of symmetry through the centroid: each makes I_{xy} = 0, so they are principal; the axis along the web carries I_{\min}, the one across it I_{\max}.

10. Every diameter is an axis of symmetry, so I_{xy} = 0 about any centroidal pair, and I_x = I_y: the Mohr's circle is a single point.

11. I_y - I_x = 0 makes \tan2\alpha infinite: 2\alpha = 90^\circ, \alpha = 45^\circ, the diagonals. I_{\max,\min} = 19.64 \pm 11.11: 30.75 and 8.53(10^4)\,\mathrm{mm^4}; Eq. A/9 at 45^\circ gives 19.64 - (-11.11)(1) = 30.75, so the maximum is along +45^\circ.

12. Adding Eqs. A/9, the \cos2\theta and \sin2\theta terms cancel: I_{x'} + I_{y'} = I_x + I_y for all \theta, which is the polar moment I_z about the origin (Eq. A/3), unchanged by any rotation about it.