CELE Engineering Mechanics — Centroids and Moments of InertiaExam Answer Templates

Marks

1

Topic

Centroid of an Area

Difficulty

easy

Template Id

T1

Examiner Tip

One crisp, complete sentence with the key phrase 'first moment of area equals zero' secures the mark immediately. Avoid vague answers like 'center point of the shape.'

Model Answer

The centroid of a plane area is the geometric center of that area — the point at which the entire area may be considered concentrated such that the first moment of area about any axis through that point is zero.

Question Type

very_short_answer

Answer Structure

• One sentence: state what the centroid is (geometric center / point where first moment of area equals zero) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing the centroid only as the 'center of gravity' without qualifying it as a purely geometric property of area.
• Omitting the 'first moment of area equals zero' condition — this is the defining mathematical property.

Key Phrases To Include

• geometric center
• first moment of area
• equals zero
• plane area

Marks

2

Topic

Parallel-Axis Theorem

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners specifically look for the overbar notation Ī to distinguish centroidal from non-centroidal inertia. Using the same symbol for both loses the definition mark.

Model Answer

The parallel-axis theorem states that the moment of inertia I of a plane area about any axis is equal to its centroidal moment of inertia Ī about a parallel axis plus the product of the area A and the square of the perpendicular distance d between the two axes: I = Ī + A·d² Where: I = moment of inertia about the new (parallel) axis (mm⁴) Ī = moment of inertia about the centroidal axis parallel to the new axis (mm⁴) A = total area of the section (mm²) d = perpendicular distance from the centroid to the new axis (mm)

Question Type

short_answer

Answer Structure

• Line 1: State the theorem in words — I equals centroidal I plus Ad² [1 mark]
• Line 2: Write the formula I = Ī + Ad² and define I, Ī, A, and d with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing I = I_bar + Ad instead of Ad² — squaring d is essential.
• Failing to define d clearly as the distance from the part's own centroid to the transfer axis.
• Omitting units from the formula definition.

Key Phrases To Include

• I = Ī + Ad²
• centroidal moment of inertia
• parallel axis
• perpendicular distance
• mm⁴

Marks

2

Topic

Moment of Inertia

Difficulty

easy

Template Id

T3

Examiner Tip

The examiners know that reviewees confuse bh³/12 and bh³/3. Stating clearly 'about the centroidal axis' and 'about the base' respectively shows you understand the distinction.

Model Answer

(a) Rectangle (b × h) about its centroidal axis parallel to b: Ī_x = bh³/12 (b) Solid circle of diameter d about any centroidal axis: Ī = πd⁴/64

Question Type

very_short_answer

Answer Structure

• Part (a): Correct formula bh³/12 with correct orientation note (axis parallel to b, i.e., h is the dimension perpendicular to the axis) [1 mark]
• Part (b): Correct formula πd⁴/64 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing bh³/12 (centroidal) with bh³/3 (about the base) — this is the single most common formula error in board exams.
• Writing πd⁴/32 (which is the polar moment J, not I_x).
• Reversing b and h so that the wrong dimension is cubed.

Key Phrases To Include

• bh³/12
• πd⁴/64
• centroidal axis
• h perpendicular to axis

Marks

2

Topic

Moment of Inertia

Difficulty

easy

Template Id

T4

Examiner Tip

Including the parallel-axis verification at the end demonstrates mastery and protects against arithmetic errors — examiners reward it even when not explicitly required.

Model Answer

Given: b = 80 mm, h = 120 mm (a) Centroidal x-axis: Ī_x = bh³/12 = 80(120)³/12 = 80(1,728,000)/12 = 11,520,000 mm⁴ = 1.152 × 10⁷ mm⁴ (b) About the base: I_base = bh³/3 = 80(120)³/3 = 80(1,728,000)/3 = 46,080,000 mm⁴ = 4.608 × 10⁷ mm⁴ Check (parallel-axis theorem): I_base = Ī_x + A·d² = 1.152×10⁷ + 80(120)(60)² = 1.152×10⁷ + 3.456×10⁷ = 4.608×10⁷ mm⁴ ✓

Question Type

numerical

Answer Structure

• State formula for each case before substituting [implicit in marks]
• Part (a): Ī_x = bh³/12 → substitute → numerical answer in mm⁴ [1 mark]
• Part (b): I_base = bh³/3 → substitute → numerical answer in mm⁴ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using bh³/12 for both — failing to distinguish centroidal from base formula.
• Arithmetic error in cubing 120 (correct: 1,728,000).
• Omitting mm⁴ units from the final answer.

Key Phrases To Include

• bh³/12
• bh³/3
• 1.152 × 10⁷ mm⁴
• 4.608 × 10⁷ mm⁴

Marks

3

Topic

Radius of Gyration

Difficulty

medium

Template Id

T5

Examiner Tip

The structural significance mark is often left blank by students who only know the formula. One sentence connecting r to the slenderness ratio in column design earns that third mark reliably.

Model Answer

The radius of gyration r of a cross-section about an axis is the distance from that axis at which the entire cross-sectional area A would have to be concentrated to produce the same moment of inertia I as the actual section: r = √(I / A) Where: r = radius of gyration (mm) I = moment of inertia about the axis of interest (mm⁴) A = cross-sectional area (mm²) Structural significance: In column design, slenderness ratio SR = L_eff / r governs the buckling behavior of a compression member (NSCP 2015 / AISC 360). A larger r (i.e., a section whose area is distributed farther from the axis) resists buckling more effectively. Designers maximise r by choosing hollow sections (HSS/pipes) or wide-flange sections, which concentrate area away from the centroidal axis.

Question Type

short_answer

Answer Structure

• Line 1–2: Define radius of gyration in words (distance at which area is concentrated to give same I) [1 mark]
• Line 3: Write formula r = √(I/A) and define each symbol with units [1 mark]
• Line 4–5: State structural significance — governs column slenderness SR = L_eff/r; larger r reduces buckling tendency [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing r = I/A instead of r = √(I/A) — forgetting the square root.
• Stating significance as 'used in beam design' — radius of gyration is primarily relevant to column/compression member design.
• Not citing any code reference (NSCP 2015 or AISC 360) when significance is discussed at review level.

Key Phrases To Include

• r = √(I/A)
• slenderness ratio
• SR = L_eff / r
• buckling
• compression member
• mm

Marks

3

Topic

Centroid of an Area

Difficulty

medium

Template Id

T6

Examiner Tip

Clearly labeling ȳ₁ and ȳ₂ from the same reference axis is the single most important organisational step. Examiners trace your reference axis first; if it is ambiguous, every subsequent value is suspect.

Model Answer

Reference axis: bottom of the rectangle (y = 0). Part 1 — Rectangle (A₁): A₁ = 200 × 100 = 20,000 mm² ȳ₁ = 100/2 = 50 mm (centroid at mid-height of rectangle) Part 2 — Triangle (A₂) on top: A₂ = ½ × 200 × 60 = 6,000 mm² ȳ₂ = 100 + 60/3 = 100 + 20 = 120 mm (triangle centroid at h/3 above base) Composite centroid: ȳ = (A₁ȳ₁ + A₂ȳ₂) / (A₁ + A₂) = (20,000 × 50 + 6,000 × 120) / (20,000 + 6,000) = (1,000,000 + 720,000) / 26,000 = 1,720,000 / 26,000 = 66.15 mm from the bottom

Question Type

numerical

Answer Structure

• Identify reference axis explicitly [sets up correct d values later] [0.5 mark — part of method mark]
• Compute A₁ and ȳ₁ for the rectangle [0.5 mark]
• Compute A₂ and ȳ₂ for the triangle, correctly using h/3 above base and adding rectangle height [1 mark]
• Apply ȳ = ΣAᵢȳᵢ / ΣAᵢ and compute correct numerical answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Placing triangle centroid at h/3 from the top of the triangle instead of h/3 from its base.
• Forgetting to add the rectangle height (100 mm) to the triangle centroid coordinate.
• Using h/2 for the triangle centroid — this is the rectangle formula, not the triangle.

Key Phrases To Include

• ȳ = ΣAᵢȳᵢ / ΣAᵢ
• h/3 from base
• reference axis
• 66.15 mm

Marks

3

Topic

Polar Moment of Inertia

Difficulty

medium

Template Id

T7

Examiner Tip

Always verify using the direct formula J = πd⁴/32. If your summed answer does not equal this, you have a computational error — find it before writing the final answer.

Model Answer

Perpendicular-Axis Theorem: For a plane area, the polar moment of inertia J about an axis perpendicular to the plane and passing through a point equals the sum of the moments of inertia about any two mutually perpendicular axes in the plane passing through the same point: J = I_x + I_y Numerical computation for a solid circle (d = 100 mm): For a circle: I_x = I_y = πd⁴/64 (by symmetry) I_x = I_y = π(100)⁴/64 = π(100,000,000)/64 = 4,908,739 mm⁴ ≈ 4.909 × 10⁶ mm⁴ J = I_x + I_y = 2 × 4,908,739 = 9,817,477 mm⁴ ≈ 9.817 × 10⁶ mm⁴ Alternatively, directly: J = πd⁴/32 = π(100)⁴/32 = 9,817,477 mm⁴ ✓

Question Type

numerical

Answer Structure

• State perpendicular-axis theorem: J = I_x + I_y [1 mark]
• Identify I_x = I_y = πd⁴/64 for a circle and substitute d = 100 mm [1 mark]
• Compute J correctly and state J = πd⁴/32 as a check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using J = πd⁴/64 (confusing I and J formulas for the circle).
• Not recognising that for a circle I_x = I_y, leading to incorrect summing.
• Arithmetic error in computing (100)⁴ = 10⁸ (not 10⁶ or 10⁷).

Key Phrases To Include

• J = I_x + I_y
• perpendicular-axis theorem
• πd⁴/32
• πd⁴/64
• 9.817 × 10⁶ mm⁴

Marks

3

Topic

Moment of Inertia

Difficulty

medium

Template Id

T8

Examiner Tip

State explicitly 'hole is concentric — no transfer needed' before computing. This one sentence shows the examiner you understand the parallel-axis theorem condition and protects you from a setup error.

Model Answer

Since both the outer rectangle and the hole are centered at the same centroid (the hole is concentric), no parallel-axis transfer is needed. Outer rectangle (120 mm wide × 180 mm deep): Ī_outer = (120)(180)³/12 = 120 × 5,832,000 / 12 = 58,320,000 mm⁴ = 5.832 × 10⁷ mm⁴ Inner hole (60 mm wide × 90 mm deep): Ī_hole = (60)(90)³/12 = 60 × 729,000 / 12 = 3,645,000 mm⁴ = 3.645 × 10⁶ mm⁴ Net section (subtract hole): I_x = Ī_outer − Ī_hole = 5.832 × 10⁷ − 3.645 × 10⁶ = 5.832 × 10⁷ − 0.3645 × 10⁷ = 5.4675 × 10⁷ mm⁴ ≈ 5.47 × 10⁷ mm⁴

Question Type

numerical

Answer Structure

• Recognise that hole is concentric → no parallel-axis transfer required [1 mark — method]
• Compute Ī_outer = 5.832 × 10⁷ mm⁴ [1 mark]
• Compute Ī_hole = 3.645 × 10⁶ mm⁴ and subtract: I_x = 5.47 × 10⁷ mm⁴ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying parallel-axis theorem to the hole even though it is centred — this adds an unnecessary (and incorrect) Ad² term.
• Using the hole's outer dimensions incorrectly or mixing up which dimension is h.
• Forgetting to subtract the hole — computing only the gross section inertia.

Key Phrases To Include

• concentric
• no parallel-axis transfer
• subtract hole
• 5.47 × 10⁷ mm⁴
• bh³/12

Marks

3

Topic

Centroid of an Area

Difficulty

medium

Template Id

T9

Examiner Tip

Draw the L-section to scale with coordinates marked before computing. The y-coordinate of the top plate (y₁ = 140 mm) is the most error-prone value — verify it against your sketch.

Model Answer

Reference: origin at bottom-left corner of the section. Part 1 — Horizontal plate (150 mm × 20 mm): A₁ = 150 × 20 = 3,000 mm² x₁ = 150/2 = 75 mm y₁ = 130 + 20/2 = 140 mm Part 2 — Vertical web (20 mm × 130 mm): A₂ = 20 × 130 = 2,600 mm² x₂ = 20/2 = 10 mm y₂ = 130/2 = 65 mm Total area: A = 3,000 + 2,600 = 5,600 mm² Centroid: x̄ = (A₁x₁ + A₂x₂) / A = (3,000 × 75 + 2,600 × 10) / 5,600 = (225,000 + 26,000) / 5,600 = 251,000 / 5,600 = 44.82 mm from left ȳ = (A₁y₁ + A₂y₂) / A = (3,000 × 140 + 2,600 × 65) / 5,600 = (420,000 + 169,000) / 5,600 = 589,000 / 5,600 = 105.18 mm from bottom

Question Type

numerical

Answer Structure

• Clearly define reference origin [setup mark included in method]
• Compute A₁, x₁, y₁ and A₂, x₂, y₂ correctly (note y₁ = 130 + 10 = 140 mm) [1 mark]
• Apply x̄ = ΣAᵢxᵢ/ΣAᵢ → 44.82 mm [1 mark]
• Apply ȳ = ΣAᵢyᵢ/ΣAᵢ → 105.18 mm [1 mark]

Scoring Breakdown

Common Mark Deductions

• Placing y₁ at 20/2 = 10 mm (measuring from the wrong reference — the top of the plate) instead of 130 + 10 = 140 mm from the bottom.
• Not accounting for the flush-left alignment when computing x-coordinates.
• Arithmetic errors in the moment sums ΣAᵢxᵢ and ΣAᵢyᵢ.

Key Phrases To Include

• reference origin
• x̄ = ΣAᵢxᵢ / ΣAᵢ
• ȳ = ΣAᵢyᵢ / ΣAᵢ
• 44.82 mm
• 105.18 mm

Marks

2

Topic

Parallel-Axis Theorem

Difficulty

medium

Template Id

T10

Examiner Tip

The question 'why' demands a mathematical justification, not just a statement. One equation (I = Ī + Ad²) followed by 'since Ad² ≥ 0' is a complete and mark-winning answer.

Model Answer

Centroidal moment of inertia Ī: the second moment of area computed about the axis passing through the centroid of the section. It is the minimum value among all parallel axes. Non-centroidal moment of inertia I: the second moment of area about any parallel axis at a perpendicular distance d from the centroid, given by I = Ī + Ad². Why Ī is minimum: The transfer term Ad² is always non-negative (since A > 0 and d² ≥ 0). Therefore I = Ī + Ad² ≥ Ī for any d ≥ 0, with equality only when d = 0 (i.e., the axis passes through the centroid). The centroidal axis thus yields the smallest possible I among all parallel axes.

Question Type

short_answer

Answer Structure

• Define Ī (centroidal) vs I (non-centroidal) with the connecting formula I = Ī + Ad² [1 mark]
• Explain why Ī is minimum: Ad² ≥ 0 always; minimum occurs at d = 0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating Ī is minimum without explaining the mathematical reason (non-negativity of Ad²).
• Confusing the statement — some students write 'I is always greater than Ī' without the qualifying condition d ≠ 0.

Key Phrases To Include

• I = Ī + Ad²
• Ad² is always non-negative
• minimum when d = 0
• centroidal axis

Marks

3

Topic

Centroid of an Area

Difficulty

medium

Template Id

T11

Examiner Tip

Write '(−) Cut-out' explicitly beside A₂ in your table. This one label tells the examiner you are applying the negative-area technique correctly and prevents sign errors throughout.

Model Answer

Treat as: Full square (positive area) MINUS the cut-out square (negative area). Full square: A₁ = 100 × 100 = 10,000 mm² (+) x̄₁ = 50 mm, ȳ₁ = 50 mm Cut-out square (40 × 40 mm, centred at (80, 80)): A₂ = 40 × 40 = 1,600 mm² (−) x̄₂ = 60 + 40/2 = 80 mm ȳ₂ = 60 + 40/2 = 80 mm Net area: A = 10,000 − 1,600 = 8,400 mm² x̄ = (A₁x̄₁ − A₂x̄₂) / (A₁ − A₂) = (10,000 × 50 − 1,600 × 80) / 8,400 = (500,000 − 128,000) / 8,400 = 372,000 / 8,400 = 44.29 mm ȳ = (A₁ȳ₁ − A₂ȳ₂) / (A₁ − A₂) = (10,000 × 50 − 1,600 × 80) / 8,400 = 372,000 / 8,400 = 44.29 mm ∴ Centroid: x̄ = ȳ = 44.29 mm from the bottom-left corner.

Question Type

numerical

Answer Structure

• Identify composite approach: full square minus cut-out (negative area) — label cut-out as negative [1 mark]
• Correctly locate centroid of cut-out at (80, 80) mm [1 mark]
• Apply ȳ = ΣAᵢyᵢ/ΣAᵢ with correct signs → x̄ = ȳ = 44.29 mm [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding the cut-out area instead of subtracting it (treating it as positive).
• Placing the cut-out centroid at (40, 40) — the centroid of a 40 × 40 square, not at the correct position (80, 80) relative to the origin.
• Forgetting to subtract A₂ from the total area in the denominator.

Key Phrases To Include

• negative area
• cut-out
• 80 mm centroid of hole
• 44.29 mm
• ΣAᵢxᵢ / ΣAᵢ

Marks

5

Topic

Parallel-Axis Theorem

Difficulty

hard

Template Id

T12

Examiner Tip

In a 5-mark T-section problem, the parallel-axis transfer step in Part (b) is where most marks are won or lost. Write each part's calculation as its own block with Ī, d, Ad², and sum — this makes partial credit easy for the examiner to award even if your final total has an arithmetic error.

Model Answer

Reference: y measured downward from the top of the flange. --- PART (a): Locate centroid ȳ --- Flange (F): A_F = 200 × 30 = 6,000 mm²; ȳ_F = 30/2 = 15 mm Web (W): A_W = 30 × 170 = 5,100 mm²; ȳ_W = 30 + 170/2 = 115 mm Total area: A = 6,000 + 5,100 = 11,100 mm² ȳ = (A_F·ȳ_F + A_W·ȳ_W) / A = (6,000 × 15 + 5,100 × 115) / 11,100 = (90,000 + 586,500) / 11,100 = 676,500 / 11,100 = 60.95 mm from the top of the flange --- PART (b): Moment of Inertia I_x (parallel-axis theorem) --- For each part: I_part = Ī_part + A_part·d² where d = distance from part's own centroid to composite centroid ȳ = 60.95 mm Flange: d_F = |ȳ_F − ȳ| = |15 − 60.95| = 45.95 mm Ī_F = (200)(30)³/12 = 200 × 27,000 / 12 = 450,000 mm⁴ I_F = 450,000 + 6,000 × (45.95)² = 450,000 + 6,000 × 2,111.4 = 450,000 + 12,668,400 = 13,118,400 mm⁴ Web: d_W = |ȳ_W − ȳ| = |115 − 60.95| = 54.05 mm Ī_W = (30)(170)³/12 = 30 × 4,913,000 / 12 = 12,282,500 mm⁴ I_W = 12,282,500 + 5,100 × (54.05)² = 12,282,500 + 5,100 × 2,921.4 = 12,282,500 + 14,899,140 = 27,181,640 mm⁴ Total: I_x = I_F + I_W = 13,118,400 + 27,181,640 = 40,300,040 mm⁴ ≈ 4.03 × 10⁷ mm⁴ --- PART (c): Radius of Gyration r_x --- r_x = √(I_x / A) = √(4.03 × 10⁷ / 11,100) = √(3,630.6) = 60.25 mm

Question Type

numerical

Answer Structure

• Part (a) — 2 marks: Set up table with A, ȳ for flange and web; apply ȳ = ΣAᵢyᵢ/ΣAᵢ; correct answer ȳ = 60.95 mm [2 marks]
• Part (b) — 2 marks: For each part compute Ī + Ad² with correct d values; sum correctly to I_x ≈ 4.03 × 10⁷ mm⁴ [2 marks]
• Part (c) — 1 mark: Apply r_x = √(I_x/A) = 60.25 mm [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not applying the parallel-axis transfer (Ad²) — using only centroidal Ī for each part underestimates I_x by over 50%.
• Measuring d from the base of the section rather than from each part's own centroid to the composite centroid.
• Using wrong formula for Ī of web: using 30³×170/12 instead of 30×170³/12 (mixing up b and h).
• Forgetting units (mm⁴ for I, mm for r) in any answer line.
• Rounding ȳ prematurely in Part (a) and carrying forward a rounded value into Part (b), causing cascading errors.

Key Phrases To Include

• ȳ = ΣAᵢyᵢ / ΣAᵢ
• I = Ī + Ad²
• d = distance from part centroid to composite centroid
• bh³/12
• 60.95 mm
• 4.03 × 10⁷ mm⁴
• 60.25 mm
• r_x = √(I/A)

Marks

2

Topic

Centroid of an Area

Difficulty

medium

Template Id

T13

Examiner Tip

Connect the answer directly to the derivation: 'the flexure formula is derived assuming bending about the neutral axis, which IS the centroidal axis.' This shows code-level understanding, not just formula recall.

Model Answer

In beam bending theory (Euler-Bernoulli), the neutral axis of the cross-section — the axis along which bending stress is zero — coincides with the centroidal axis of the cross-section. The flexure formula σ = Mc/I gives the bending stress at distance c from the neutral axis, where I is the moment of inertia about that neutral (centroidal) axis. I must be computed about the centroidal axis because the derivation of σ = Mc/I assumes that the internal bending moment is resisted by stresses distributed symmetrically about the centroid. Using I about any other axis would misrepresent the section's bending stiffness and yield incorrect stress values — potentially unconservative and unsafe design.

Question Type

short_answer

Answer Structure

• Line 1–2: State that the neutral axis coincides with the centroidal axis; bending stress is zero at the centroid [1 mark]
• Line 3–4: Explain that I in the flexure formula must be the centroidal I because the derivation assumes moments are computed about the neutral (centroidal) axis; incorrect axis → incorrect stresses [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 'I must be about the centroid' without explaining why (i.e., not linking to the neutral axis / derivation assumption).
• Confusing the centroidal axis with the axis of maximum stress (the extreme fiber at distance c).

Key Phrases To Include

• neutral axis
• centroidal axis
• bending stress zero at centroid
• σ = Mc/I
• centroidal moment of inertia

Marks

2

Topic

Centroid of an Area

Difficulty

easy

Template Id

T14

Examiner Tip

Memorise ȳ = 4r/(3π) for the semicircle and 4r/(3π) from each straight edge for the quarter circle — these two appear frequently in board exams and must be recalled instantly without derivation.

Model Answer

For a semicircular area of radius r, the centroid is located at a distance ȳ from the diameter: ȳ = 4r / (3π) ← standard result from integration Substituting r = 80 mm: ȳ = 4 × 80 / (3π) = 320 / (3 × 3.14159) = 320 / 9.4248 = 33.95 mm ∴ The centroid of the semicircular area is 33.95 mm from the flat (diameter) edge.

Question Type

numerical

Answer Structure

• State standard formula ȳ = 4r/(3π) for a semicircle [1 mark]
• Substitute r = 80 mm and compute ȳ = 33.95 mm with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using ȳ = r/2 (the formula for the centroid of a half-ring/arc, not a semicircular area).
• Using ȳ = 4r/(3π) correctly but computing 3π incorrectly (e.g., using π = 3.14 and getting 9.42 instead of 9.425).
• Measuring the centroid from the curved edge rather than the flat diameter edge.

Key Phrases To Include

• ȳ = 4r/(3π)
• 33.95 mm
• from the diameter
• flat edge

Marks

5

Topic

Radius of Gyration

Difficulty

hard

Template Id

T15

Examiner Tip

The design-relevance mark in Part (b) is often missed by students who stop at the number. One sentence — 'r_y < r_x, so the weak axis governs column buckling per AISC 360/NSCP 2015' — secures that third mark and demonstrates professional-level understanding.

Model Answer

Given: A = 11,400 mm², I_x = 142 × 10⁶ mm⁴, I_y = 48.3 × 10⁶ mm⁴ --- PART (a): Radius of gyration r_x --- r_x = √(I_x / A) = √(142 × 10⁶ / 11,400) = √(12,456) = 111.6 mm --- PART (b): Radius of gyration r_y --- r_y = √(I_y / A) = √(48.3 × 10⁶ / 11,400) = √(4,237) = 65.1 mm Note: r_y < r_x because I_y < I_x; the y-axis (weak axis) governs column buckling. In AISC 360 / NSCP 2015, slenderness about the weak axis SR_y = KL/r_y controls compression capacity for most W-sections. --- PART (c): Polar moment of inertia J --- J = I_x + I_y (perpendicular-axis theorem) = 142 × 10⁶ + 48.3 × 10⁶ = 190.3 × 10⁶ mm⁴ = 1.903 × 10⁸ mm⁴

Question Type

numerical

Answer Structure

• Part (a): Apply r_x = √(I_x/A) → 111.6 mm [1 mark]
• Part (b): Apply r_y = √(I_y/A) → 65.1 mm [1 mark]
• Part (b) note: Identify r_y as weak-axis governs column buckling (design relevance) [1 mark]
• Part (c): Apply J = I_x + I_y → 1.903 × 10⁸ mm⁴ [2 marks: 1 formula + 1 answer]

Scoring Breakdown

Common Mark Deductions

• Computing r = I/A without the square root — a formula recall error.
• Not identifying which r governs column design (the smaller r_y).
• Using J = I_x − I_y or J = I_x only instead of J = I_x + I_y.
• Errors in unit conversion when computing √(mm⁴/mm²) = mm.

Key Phrases To Include

• r = √(I/A)
• 111.6 mm
• 65.1 mm
• weak axis governs
• SR = KL/r_y
• J = I_x + I_y
• 1.903 × 10⁸ mm⁴
• AISC 360
• NSCP 2015