Statics and Dynamics Quiz: Composite Area Moments

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Composite Area MomentsQuestion 1 of 9

A rectangular cross-section of width $b$ and height $h$ is compared to a hollow rectangular section with the same outer dimensions $b \times h$ but with a centered rectangular hole of width $b/2$ and height $h/2$ . Both sections have the same centroidal axis. By what factor does removing the inner rectangle change the second moment of area $I_{xx}$ about the centroidal horizontal axis?

The hollow section retains

\frac{15}{16}

of the solid section's

I_{xx}

, because the removed rectangle contributes

\frac{(b/2)(h/2)^3}{12}

, which is

\frac{1}{16}

of the full section's

I_{xx} = \frac{bh^3}{12}

, and no transfer term is needed since the hole is centered.

The hollow section retains

\frac{3}{4}

of the solid section's

I_{xx}

, because the removed area is

\frac{1}{4}

of the total area and the moment of inertia scales directly with the area when both centroids coincide.

The hollow section retains

\frac{7}{8}

of the solid section's

I_{xx}

, because removing material symmetrically about both axes reduces the depth contribution by half and the width contribution by half, yielding a combined reduction factor of

\frac{1}{8}

The hollow section retains

\frac{1}{2}

of the solid section's

I_{xx}

, because the removed rectangle's dimensions are each half of the outer dimensions, and the moment of inertia scales with the cube of the depth, so the retained fraction is

(1/2)^3 = 1/8

subtracted from a base of

\frac{5}{8}

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Composite Area Moments

Practice Composite Area Moments in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Composite Area Moments, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The hollow section retains $\frac{15}{16}$ of the solid section's $I_{xx}$ , because the removed rectangle contributes $\frac{(b/2)(h/2)^3}{12}$ , which is $\frac{1}{16}$ of the full section's $I_{xx} = \frac{bh^3}{12}$ , and no transfer term is needed since the hole is centered. (correct answer)
The hollow section retains $\frac{3}{4}$ of the solid section's $I_{xx}$ , because the removed area is $\frac{1}{4}$ of the total area and the moment of inertia scales directly with the area when both centroids coincide.
The hollow section retains $\frac{7}{8}$ of the solid section's $I_{xx}$ , because removing material symmetrically about both axes reduces the depth contribution by half and the width contribution by half, yielding a combined reduction factor of $\frac{1}{8}$ .
The hollow section retains $\frac{1}{2}$ of the solid section's $I_{xx}$ , because the removed rectangle's dimensions are each half of the outer dimensions, and the moment of inertia scales with the cube of the depth, so the retained fraction is $(1/2)^3 = 1/8$ subtracted from a base of $\frac{5}{8}$ .

Explanation: When comparing solid and hollow sections, the key principle is superposition of moments of inertia: the hollow section's

I_{xx}

equals the solid outer section's

I_{xx}

minus the

I_{xx}

of the removed region. Critically, since both centroids coincide with the reference axis, no parallel-axis transfer term is needed. For the solid section:

I_{solid} = \frac{bh^3}{12}

. For the removed hole (width

b/2

, height

h/2

, centered):

I_{hole} = \frac{(b/2)(h/2)^3}{12} = \frac{bh^3}{12} \cdot \frac{1}{16}

. Therefore, the hollow section retains

1 - \frac{1}{16} = \frac{15}{16}

of the original

I_{xx}

. Answer A captures this exactly. Answer B confuses moment of inertia with area. While the hole removes

\frac{1}{4}

of the area (since

(b/2)(h/2) = \frac{bh}{4}

I_{xx}

does not scale linearly with area — it depends on the cube of the height and the width independently. These are different quantities. Answer C incorrectly claims that halving both dimensions "halves" each contribution and yields a

\frac{1}{8}

reduction. In reality,

I_{hole}/I_{solid} = \frac{1}{2} \cdot \left(\frac{1}{2}\right)^3 = \frac{1}{16}

, not

\frac{1}{8}

. Answer D presents garbled arithmetic with no consistent logic — the "base of

\frac{5}{8}

" appears from nowhere and the reasoning is internally inconsistent. Your study tip: whenever a hole shares the centroid of the full section, use direct subtraction —

I_{net} = I_{outer} - I_{hole}

— with no parallel-axis correction needed. Always track the

\frac{bh^3}{12}

formula applied separately to each rectangle.

Question 2

An I-beam cross-section is built from three rectangles: two flanges (each $200 \text{ mm} \times 20 \text{ mm}$ ) and one web ( $160 \text{ mm} \times 20 \text{ mm}$ ). The section is symmetric about both the horizontal and vertical centroidal axes. The total height of the section is $200 \text{ mm}$ .

A student claims that the flanges account for more than 80% of the total $I_{xx}$ about the horizontal centroidal axis. Which of the following correctly evaluates this claim?

The claim is correct. Each flange contributes $\frac{200(20)^3}{12} + 200(20)(90)^2 \approx 32.533 \times 10^6 \text{ mm}^4$ , and the web contributes $\frac{20(160)^3}{12} \approx 6.827 \times 10^6 \text{ mm}^4$ , so the two flanges together represent about 90.5% of the total $I_{xx}$ . (correct answer)
The claim is incorrect. Because the flanges and web have equal thickness, the web's larger height gives it a proportionally higher self-inertia that offsets the flanges' area advantage, so each component contributes roughly one-third of the total $I_{xx}$ .
The claim is correct, but only because the flange width (200 mm) exceeds the web height (160 mm); if the dimensions were reversed, the web would dominate and the flanges would contribute less than 20% of the total $I_{xx}$ .
The claim is incorrect. The parallel-axis transfer terms for the flanges are partially cancelled by negative contributions from the web, which reduces the flanges' net percentage below 80% once the full composite calculation is completed.

Explanation: When evaluating moment of inertia contributions in composite sections, remember that the parallel-axis theorem —

I = \frac{bh^3}{12} + Ad^2

— makes distance from the centroidal axis the dominant factor. Components far from the neutral axis accumulate large transfer terms (

Ad^2

), which is precisely why I-beam flanges are so structurally efficient. For this problem, each flange contributes a self-inertia of

\frac{200(20)^3}{12} \approx 0.133 \times 10^6 \text{ mm}^4

plus a transfer term of

200(20)(90)^2 = 32.4 \times 10^6 \text{ mm}^4

, totaling roughly

32.533 \times 10^6 \text{ mm}^4

. The web, centered on the neutral axis, contributes only

\frac{20(160)^3}{12} \approx 6.827 \times 10^6 \text{ mm}^4

with no transfer term. Two flanges together give

\approx 65.07 \times 10^6 \text{ mm}^4

out of a total

\approx 71.89 \times 10^6 \text{ mm}^4

— about 90.5%. Answer A is correct. Answer B is wrong because it ignores the transfer term entirely. Equal thickness does not mean equal contribution; the flanges' distance from the neutral axis overwhelmingly dominates. Answer C introduces a false conditional — reversing the flange width and web height would change magnitudes but would not flip which component dominates, since the flanges' centroidal distance still drives the calculation. Answer D fabricates a concept: there are no "negative contributions" from the web in a standard composite

I_{xx}

calculation. A useful rule of thumb: in I-beam problems, the

Ad^2

transfer terms for flanges typically dwarf every other term. If your answer shows the web dominating, recheck whether you applied the parallel-axis theorem.

Question 3

An engineer doubles the height of a rectangular beam cross-section while halving its width to keep the cross-sectional area constant. The beam is loaded in vertical bending (about the horizontal centroidal axis). Which statement correctly describes the effect on the section modulus $S = I/c$ and the maximum bending stress for the same applied moment?

The moment of inertia $I$ increases by a factor of 8, and since $c$ doubles as well, the section modulus increases by a factor of 4, so the maximum bending stress is reduced to one-quarter of its original value.
The section modulus remains unchanged because the cross-sectional area is constant and the material is the same, so the bending resistance per unit area is conserved regardless of how the dimensions are rearranged.
The moment of inertia $I$ increases by a factor of 4, so the section modulus also increases by a factor of 4, meaning maximum bending stress is reduced to one-quarter of its original value compared to the original section.
The section modulus doubles, so the maximum bending stress is reduced by half. This occurs because $I$ scales with $h^3$ (increasing by factor 8) while $c = h/2$ scales with $h$ (increasing by factor 2), and the simultaneous halving of width reduces $I$ by factor 2, giving a net $S$ increase of $8/(2 \times 2) = 2$ . (correct answer)

Explanation: Bending stress problems like this one test whether you truly understand how cross-sectional geometry affects bending resistance — not just area, but how that area is distributed relative to the neutral axis. Start with the formulas. For a rectangle,

I = \frac{bh^3}{12}

and

c = \frac{h}{2}

, so the section modulus is

S = \frac{I}{c} = \frac{bh^2}{6}

. Maximum bending stress follows from

\sigma = \frac{M}{S}

. Now apply the changes: doubling height means

h \to 2h

, and halving width means

b \to \frac{b}{2}

. The new section modulus becomes

S' = \frac{(b/2)(2h)^2}{6} = \frac{(b/2)(4h^2)}{6} = \frac{2bh^2}{6} = 2S

. Since

S

doubles, the maximum bending stress is cut in half — confirming D. Choice A correctly identifies that

I

increases by 8 and

c

doubles, but it forgets that halving the width also reduces

I

by 2, so the true

I

factor is

8/2 = 4

, not 8. Dividing by the factor-of-2 increase in

c

gives

S

increasing by 2, not 4. Choice B is a fundamental misconception: constant area does not preserve bending resistance. How area is arranged relative to the neutral axis is everything. Choice C correctly identifies a factor-of-4 increase in

I

but then forgets to divide by the factor-of-2 increase in

c

, arriving at the wrong

S

multiplier. Study tip: Always track

I

and

c

separately when dimensions change — combining them into

S = \frac{bh^2}{6}

upfront is the cleanest way to avoid missing a factor.

Question 4

A built-up box section is formed by four rectangular plates. Two horizontal plates (flanges), each $200 \text{ mm} \times 10 \text{ mm}$ , are placed at the top and bottom. Two vertical plates (webs), each $10 \text{ mm} \times 180 \text{ mm}$ , connect them. The overall outer dimensions of the box are $200 \text{ mm}$ wide by $200 \text{ mm}$ tall. The section is symmetric about both centroidal axes.

A junior engineer proposes computing $I_{xx}$ of the box section by calculating the inertia of the outer $200 \times 200 \text{ mm}$ solid rectangle and subtracting the inertia of the inner void. A senior engineer instead sums the contributions of the four individual plates using the parallel-axis theorem. Both methods are applied correctly. Which statement accurately compares the two approaches and their results?

The subtraction method yields a larger $I_{xx}$ because it includes the corner regions where the flanges and webs overlap; these regions are double-counted as missing material when the void is defined, but are actually present as solid plate, causing the void's inertia to be overstated and the net result to be inflated.
The summation method yields a larger $I_{xx}$ than the subtraction method because the parallel-axis transfer terms for the flanges add positive contributions that are not captured when treating the section as a solid rectangle minus a void, so the two approaches account for different portions of the cross-section's area.
Both methods yield the same $I_{xx}$ because they are mathematically equivalent partitions of the same integral $\int y^2 \, dA$ . The subtraction method is generally faster for box sections, while the summation method is more transparent and less prone to errors in defining the void dimensions. (correct answer)
Both methods yield the same $I_{xx}$ , but the subtraction method is only valid when the void's centroid coincides with the composite centroid. For any section where the void is eccentric, the subtraction method breaks down entirely and cannot be used, making the summation method the only reliable approach in general.

Explanation: Whenever you see a question comparing two methods for computing a second moment of area, ask yourself: are both methods partitioning the same integral

\int y^2 \, dA

over the same physical area? If yes, they must agree. The moment of inertia is purely geometric — it doesn't matter how you slice the cross-section mathematically, as long as every bit of material is counted exactly once. The subtraction method treats the section as a

200 \times 200

solid rectangle and removes the inner void (

180 \times 180

mm, since the flanges are 10 mm thick and the webs are 10 mm wide). The summation method adds the four plates individually, using the parallel-axis theorem

I = \bar{I} + Ad^2

to shift each plate's centroidal inertia to the global axis. Both partitions cover identical physical area with no overlaps or gaps, so they yield identical

I_{xx}

. That makes C correct. A is wrong because the void in the subtraction method is defined correctly as the hollow interior — the corner regions are part of the solid plates, not the void, so there is no double-counting issue. The premise of this choice is a fabricated geometric error. B is wrong because it claims the two methods account for different areas, which contradicts the problem statement that both are "applied correctly." The parallel-axis terms in the summation method correspond exactly to material that is already captured in the solid-rectangle calculation. D is wrong because it invents a restriction that doesn't exist. The subtraction method works perfectly for eccentric voids — you simply subtract the void's inertia about the composite centroid, using the parallel-axis theorem if needed. Study tip: Any valid partition of a cross-section — addition, subtraction, or mixed — gives the same result. When two correct methods disagree, look for a setup error, not a method flaw.

Question 5

A composite section consists of two identical rectangles, each $b \times h$ , arranged in two configurations. Configuration 1: The rectangles are stacked vertically (one on top of the other), forming a single $b \times 2h$ section. Configuration 2: The rectangles are placed side by side horizontally, forming a $2b \times h$ section. Both configurations are loaded in vertical bending (bending about the horizontal centroidal axis).

What is the ratio $I_{\text{Config1}} / I_{\text{Config2}}$ of the second moment of area about the horizontal centroidal axis for the two configurations?

$I_{\text{Config1}}/I_{\text{Config2}} = 4$ , but only when the rectangles are rigidly connected (e.g., welded); if simply stacked without bonding, each rectangle acts independently about its own centroid, and the ratio of combined section moduli reverts to 1 since the individual $I$ values are identical.
$I_{\text{Config1}}/I_{\text{Config2}} = 2$ , because each individual rectangle contributes the same self-inertia $bh^3/12$ about its own centroid, and the parallel-axis theorem transfer terms cancel out symmetrically between the two configurations, leaving only a factor-of-2 difference from summing two rectangles.
$I_{\text{Config1}}/I_{\text{Config2}} = 4$ , because Config 1 is a $b \times 2h$ rectangle giving $I_1 = b(2h)^3/12 = 8bh^3/12$ , while Config 2 is a $2b \times h$ rectangle giving $I_2 = 2b(h)^3/12 = 2bh^3/12$ , so the ratio is $8/2 = 4$ . (correct answer)
$I_{\text{Config1}}/I_{\text{Config2}} = 8$ , because doubling the vertical depth in Config 1 scales $I$ by $2^3 = 8$ , while Config 2 keeps the same height $h$ as a single rectangle, so the width doubling in Config 2 has no net effect on the ratio since width does not appear in the bending resistance for horizontal-axis bending.

Explanation: When comparing cross-sections for bending stiffness, your first instinct should be to apply the second moment of area formula directly to the composite shape, treating each configuration as a single unified rectangle. For Configuration 1, stacking two

b \times h

rectangles vertically produces a single

b \times 2h

section. Its second moment of area about the horizontal centroidal axis is

I_1 = \frac{b(2h)^3}{12} = \frac{8bh^3}{12}

. For Configuration 2, placing them side by side creates a

2b \times h

section, giving

I_2 = \frac{2b(h)^3}{12} = \frac{2bh^3}{12}

. The ratio is therefore

I_1/I_2 = 8/2 = 4

, confirming C is correct. A introduces a red herring about bonding conditions. While bonding matters for shear transfer and composite action in some contexts, here both configurations are already treated as solid, uniform cross-sections — connectivity is irrelevant to computing

I

for a monolithic shape. B claims the parallel-axis terms cancel symmetrically, which is simply false. Config 1's depth doubles, and since

I \propto h^3

, the depth dimension has an outsized cubic effect that Config 2's width doubling cannot match. D correctly identifies the

2^3 = 8

scaling of

I_1

relative to a single rectangle, but then wrongly claims the width doubling in Config 2 contributes nothing — width scales

I

linearly, so

I_2 = 2 \cdot \frac{bh^3}{12}

, not

\frac{bh^3}{12}

. A useful tip: remember that

I = \frac{bh^3}{12}

is cubic in the dimension parallel to bending (height for horizontal-axis bending) but only linear in the perpendicular dimension (width). Always identify which dimension is which before comparing configurations.

Question 6

Two beams have identical cross-sectional area $A$ and are made of the same material. Beam 1 is a solid square with side $a$ . Beam 2 is a solid circle with diameter $d$ chosen so that both beams have the same cross-sectional area. When both beams are oriented to resist vertical bending loads (the square oriented with a flat face horizontal), which statement correctly compares their bending resistance?

Beam 2 (circle) has greater bending resistance because the circular shape is more efficient in distributing material uniformly around the centroid, reducing stress concentrations and thus offering a higher effective $I_{xx}$ per unit area than the square.
Beam 1 (square) has greater bending resistance because its $I_{xx} = \frac{a^4}{12}$ exceeds the circle's $I_{xx} = \frac{\pi d^4}{64}$ when $d$ is set by equal area, since the square concentrates more area at extreme fiber distances from the neutral axis than the circle does. (correct answer)
Both beams have identical bending resistance because they have equal cross-sectional areas and are made of the same material, and bending stiffness depends only on the amount of material present, not on how it is distributed across the section.
Beam 1 (square) has greater bending resistance, but only when loaded along a principal axis; when loaded at an arbitrary angle, the circle's rotational symmetry makes it superior in every bending direction simultaneously, so neither section is universally preferable.

Explanation: Bending resistance questions test your understanding that the second moment of area (

I

), not just the amount of material, determines how well a beam resists bending. The key formula is

\sigma = \frac{Mc}{I}

: for a given moment

M

, a larger

I

means lower stress and greater resistance. So whenever cross-sections are compared at equal area, you must compute and compare

I

directly. Start by linking the areas. Equal area means

a^2 = \frac{\pi d^2}{4}

, giving

d = \frac{2a}{\sqrt{\pi}}

. Now compute each moment of area about the horizontal neutral axis:

I_{\text{square}} = \frac{a^4}{12}, \qquad I_{\text{circle}} = \frac{\pi d^4}{64} = \frac{\pi}{64}\left(\frac{4a^2}{\pi}\right)^2 = \frac{a^4}{4\pi}

Numerically,

\frac{1}{12} \approx 0.0833

while

\frac{1}{4\pi} \approx 0.0796

. The square wins — confirming B is correct. The square pushes more area toward its top and bottom faces (the extreme fibers), maximizing the lever arm contribution to

I

. A is wrong because "uniform distribution around the centroid" actually hurts bending resistance — material near the neutral axis contributes very little to

I

. The circle's symmetry is irrelevant to bending efficiency. C is a fundamental misconception. Bending resistance depends critically on how material is distributed, not merely how much exists — this is why I-beams outperform solid rectangles at the same weight. D correctly notes the circle's rotational symmetry but wrongly implies this makes it superior overall. For the specific case of vertical loading, the square still outperforms. Study tip: Always compute

I

explicitly when comparing cross-sections at equal area — intuition about "shape efficiency" often misleads. Remember that material far from the neutral axis contributes proportionally to

y^2

, so spreading material to the extremes always wins.

Question 7

A structural engineer is designing a T-section beam. The flange has width $b_f = 120 \text{ mm}$ , thickness $t_f = 20 \text{ mm}$ , and the web has height $h_w = 80 \text{ mm}$ and thickness $t_w = 20 \text{ mm}$ . The section sits with the flange on top. The centroid of the composite section is located at $\bar{y} = 73.33 \text{ mm}$ measured from the bottom of the web.

Using the parallel-axis theorem, the engineer computes the second moment of area of the flange about the composite centroidal axis. The flange's own centroidal axis is at $y_{flange} = 90 \text{ mm}$ from the bottom of the section. Which expression correctly gives the flange's contribution $I_{flange}$ to the total $I_{xx}$ about the composite centroid?

$I_{flange} = \frac{120(20)^3}{12} + (120 \times 20)(90 - 73.33)^2$ , giving the flange's own centroidal inertia plus the transfer term using the correct distance from the composite centroid to the flange centroid. (correct answer)
$I_{flange} = \frac{20(120)^3}{12} + (120 \times 20)(90 - 73.33)^2$ , giving the flange's inertia about its strong axis plus the transfer term, treating the flange as if bending occurs about the vertical axis.
$I_{flange} = \frac{120(20)^3}{12} + (120 \times 20)(90)^2$ , giving the flange's own centroidal inertia plus the transfer term measured from the bottom of the section rather than from the composite centroid.
$I_{flange} = \frac{120(20)^3}{12} + (120 \times 20)(20 - 73.33)^2$ , giving the flange's own centroidal inertia plus the transfer term using the flange thickness rather than its centroidal distance from the base.

Explanation: Whenever you apply the parallel-axis theorem to a composite section, you need two things for each sub-area: its own centroidal moment of inertia

I_{\bar{x}}

, and the squared perpendicular distance

d^2

between that sub-area's centroid and the composite centroidal axis. The formula is

I = I_{\bar{x}} + Ad^2

. For the flange, the relevant bending axis is horizontal (the x-axis), so you compute

I_{\bar{x}}

by rotating the rectangle with its width (120 mm) in the horizontal direction and thickness (20 mm) in the vertical direction. That gives

\frac{b_f t_f^3}{12} = \frac{120(20)^3}{12}

. The transfer distance is the vertical gap between the flange's centroid (at 90 mm from the bottom) and the composite centroid (at 73.33 mm from the bottom):

d = 90 - 73.33 = 16.67 \text{ mm}

. Answer A captures both pieces correctly. Answer B flips the flange dimensions in the first term, using

\frac{20(120)^3}{12}

, which would be the moment of inertia about the vertical axis — a completely wrong axis for a beam bending problem. Answer C uses

90^2

as the transfer distance squared instead of

(90 - 73.33)^2

. This is the classic trap of measuring the sub-area centroid from the bottom of the section rather than from the composite centroid, which is what the parallel-axis theorem requires. Answer D uses the flange thickness (20 mm) instead of the flange centroid location (90 mm) in the transfer term, confusing a geometric dimension with a positional coordinate. Study tip: Always write down the composite centroid

\bar{y}

first and compute every transfer distance as

d = y_{sub} - \bar{y}

. This single habit eliminates the most common errors on parallel-axis theorem problems.

Question 8

A composite cross-section is formed by welding a solid circular rod of diameter $D = 40 \text{ mm}$ on top of a rectangular bar of width $b = 80 \text{ mm}$ and height $h = 60 \text{ mm}$ . The bottom of the rectangle is the reference level ( $y = 0$ ). The circle sits on top of the rectangle so that its lowest point is tangent to the top of the rectangle at $y = 60 \text{ mm}$ .

What is the $y$ -coordinate of the composite centroid measured from the bottom of the rectangle, and which component experiences the larger parallel-axis transfer term in the computation of $I_{xx}$ about that centroid?

$\bar{y} \approx 40.4 \text{ mm}$ ; the rectangle has the larger transfer term because its area $(80 \times 60 = 4800 \text{ mm}^2)$ greatly exceeds the circle's area $(\pi(20)^2 \approx 1257 \text{ mm}^2)$ , and area is the dominant factor in the $Ad^2$ transfer term.
$\bar{y} \approx 40.4 \text{ mm}$ ; the circle has the larger transfer term because, although its area is smaller, its centroid is at $y = 80 \text{ mm}$ — farther from $\bar{y}$ than the rectangle's centroid at $y = 30 \text{ mm}$ — and the distance is squared in $Ad^2$ , making the circle's contribution dominate. (correct answer)
$\bar{y} \approx 46.5 \text{ mm}$ ; the circle has the larger transfer term because its centroid lies above the composite centroid while the rectangle's centroid lies below, and transfer terms are always larger for components above the neutral axis due to the direction of the bending moment.
$\bar{y} \approx 40.4 \text{ mm}$ ; neither component dominates the transfer term because the parallel-axis contributions are equal and opposite by the definition of the centroid, which requires $\sum A_i d_i = 0$ , so both $Ad^2$ terms must be numerically equal.

Explanation: When finding the centroid and moment of inertia of a composite section, you must track both area and distance carefully — especially in the parallel-axis theorem, where distance is squared and can easily outweigh a larger area. Start by locating each centroid. The rectangle's centroid sits at

y_1 = 30 \text{ mm}

, and the circle's centroid sits at

y_2 = 60 + 20 = 80 \text{ mm}

. With areas

A_1 = 4800 \text{ mm}^2

and

A_2 = \pi(20)^2 \approx 1257 \text{ mm}^2

, the composite centroid is:

\bar{y} = \frac{4800(30) + 1257(80)}{4800 + 1257} = \frac{144000 + 100531}{6057} \approx 40.4 \text{ mm}

Now compare the parallel-axis transfer terms

Ad^2

. The rectangle's distance is

d_1 = 40.4 - 30 = 10.4 \text{ mm}

, giving

4800(10.4)^2 \approx 518{,}000 \text{ mm}^4

. The circle's distance is

d_2 = 80 - 40.4 = 39.6 \text{ mm}

, giving

1257(39.6)^2 \approx 1{,}971{,}000 \text{ mm}^4

. The circle's transfer term is nearly four times larger, entirely because

d

is squared — B is correct. Choice A gets

\bar{y}

right but incorrectly concludes that area alone drives

Ad^2

. Choice C states an incorrect centroid location and introduces a false rule that components "above" the neutral axis always dominate. Choice D misreads the centroid condition:

\sum A_i d_i = 0

means the signed first moments cancel, but the squared terms

Ad^2

are always positive and need not be equal. Study tip: When comparing

Ad^2

terms, always check the distance first — because it's squared, a component with a small area but large offset will often dominate over a bulky component sitting close to the centroid.

Question 9

A composite cross-section consists of a large rectangle $100 \text{ mm} \times 200 \text{ mm}$ (width × height) with a circular hole of diameter $60 \text{ mm}$ centered at a point $50 \text{ mm}$ above the section's centroidal axis. The composite centroid remains on the horizontal axis of symmetry of the outer rectangle.

Which expression correctly gives the second moment of area $I_{xx}$ of the composite section about its centroidal horizontal axis?

$I_{xx} = \frac{100(200)^3}{12} - \left[\frac{\pi(60)^4}{64} + \frac{\pi(60)^2}{4}(100)^2\right]$ , subtracting the circle's self-inertia and a transfer term measured from the bottom of the section rather than from the composite centroidal axis.
$I_{xx} = \frac{100(200)^3}{12} - \frac{\pi(60)^4}{64}$ , subtracting only the circle's centroidal self-inertia because the hole is located within the rectangle and its area is already accounted for in the rectangle's moment of inertia.
$I_{xx} = \frac{100(200)^3}{12} - \frac{\pi(60)^4}{64} + \frac{\pi(60)^2}{4}(50)^2$ , subtracting the circle's self-inertia but adding the transfer term because removing off-center material reduces the total area but increases the remaining section's effective arm.
$I_{xx} = \frac{100(200)^3}{12} - \left[\frac{\pi(60)^4}{64} + \frac{\pi(60)^2}{4}(50)^2\right]$ , subtracting both the circle's self-inertia and the parallel-axis transfer term because the hole's centroid is offset from the composite centroidal axis. (correct answer)

Explanation: Whenever you see a composite section with a hole, reach for the subtraction method: treat the hole as a negative area and subtract its full contribution to

I_{xx}

, including both its own centroidal inertia and any parallel-axis transfer term. The composite centroid sits at the center of the outer rectangle. The rectangle's moment of inertia about that axis is simply

\frac{bh^3}{12} = \frac{100(200)^3}{12}

. The circular hole's centroid is offset 50 mm from the composite centroidal axis, so by the parallel-axis theorem, the hole's total contribution to

I_{xx}

— had it been solid material — would be

\frac{\pi(60)^4}{64} + \frac{\pi(60)^2}{4}(50)^2

. Because the hole removes that material, you subtract the entire bracketed quantity. That gives exactly answer D, the correct expression. Answer A is wrong because it uses 100 mm as the transfer distance instead of 50 mm (the actual offset from the centroidal axis to the hole's center). Using the distance from the bottom of the section is a classic parallel-axis theorem misapplication. Answer B omits the transfer term entirely, which would only be valid if the hole's centroid coincided with the composite centroidal axis (i.e., zero offset). Since the hole is 50 mm off-center, its positional contribution cannot be ignored. Answer C adds the transfer term instead of subtracting it. Removing off-center material always reduces

I_{xx}

, so both the self-inertia and the transfer term must be subtracted together. Study tip: For any cutout, always ask two questions: (1) What is the shape's own centroidal inertia? (2) Is its centroid offset from the composite centroidal axis? If yes to both, subtract the full parallel-axis expression — never just one piece.