CELE Engineering Mechanics — Centroids and Moments of InertiaDetailed Explanation
If the summary was not enough, this is the deep dive. Detailed explanations for Centroids and Moments of Inertia in the CELE Engineering Mechanics context, written to turn surface familiarity into genuine understanding. Professional Regulation Commission (PRC) — Board of Civil Engineering's toughest CELE questions on this chapter are answered by the reasoning built here.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Engineering Mechanics section sits under a "Core" weighting, and Centroids and Moments of Inertia is the 6th chapter in the 8-chapter CELE Engineering Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Engineering Mechanics.
Centroids and Moments of Inertia - Detailed Explanation
Centroids and Moments of Inertia are among the most heavily tested topics in the PRC Civil Engineer Licensure Examination. Every structural design calculation — from computing bending stresses using σ = Mc/I, to checking column slenderness via r = √(I/A), to determining beam deflections — depends directly on these two quantities. This chapter builds from the geometric centroid of composite areas, through the standard moment-of-inertia formulas for common shapes, to the essential parallel-axis theorem that makes analysis of built-up sections possible. Mastery here is non-negotiable: board problems regularly disguise centroid-and-MoI calculations inside structural analysis, reinforced concrete design (ACI 318), and steel design (AISC 360 / NSCP 2015 Section 5) problems. Study each concept in the correct sequence — centroid first, then moment of inertia, then the parallel-axis transfer — because each step depends on the one before it.
Concepts
Centroid of a Composite Area
The centroid (geometric center) of an area is the point at which the entire area may be assumed to be concentrated for purposes of computing first moments. For a composite area — one built up from simple shapes — the centroid coordinates are the area-weighted averages of the component centroids: x̄ = (ΣAᵢxᵢ) / ΣAᵢ ȳ = (ΣAᵢyᵢ) / ΣAᵢ where Aᵢ is the area of part i and (xᵢ, yᵢ) is the centroid of part i measured from the chosen reference origin. A cut-out or hole is treated as a NEGATIVE area — both its area and its moment (Aᵢxᵢ, Aᵢyᵢ) are subtracted. Symmetry is your best friend: the centroid lies on every axis of symmetry, so a doubly-symmetric section (rectangle, circle, I-beam with equal flanges) has its centroid at the geometric center. Centroidal locations for common shapes (all measured from the base or reference edge): • Rectangle (b × h): centroid at h/2 from base, b/2 from side. • Right triangle: centroid at h/3 from base (the right-angle corner side). • Semicircle (radius r): centroid at 4r/(3π) ≈ 0.4244r from the flat (diameter) edge. • Quarter-circle: centroid at 4r/(3π) from each straight edge. Procedure for composite centroid problems: 1. Choose a convenient reference origin (usually bottom-left corner). 2. Break the composite into simple shapes; assign + for solid parts, − for holes. 3. Tabulate: Shape | Aᵢ | xᵢ | Aᵢxᵢ | yᵢ | Aᵢyᵢ 4. Sum columns: x̄ = ΣAᵢxᵢ / ΣAᵢ, ȳ = ΣAᵢyᵢ / ΣAᵢ. 5. Verify using symmetry whenever possible.
Examples
Notice that the centroid shifted away from (50, 50) toward the lower-left because material was removed from the upper-right. The diagonal symmetry axis (y = x) confirms x̄ = ȳ, serving as a built-in check.
Scenario
A 100 × 100 mm square has a 40 × 40 mm square cut-out at its top-right corner (spanning from x = 60 mm to x = 100 mm, y = 60 mm to y = 100 mm). Origin is at the bottom-left corner. Find the centroid.
Solution
Step 1 — Full square: A₁ = 100 × 100 = 10,000 mm²; centroid at (50, 50) mm. Step 2 — Cut-out (negative): A₂ = −40 × 40 = −1,600 mm²; centroid at (80, 80) mm. Step 3 — Net area: A = 10,000 − 1,600 = 8,400 mm². Step 4 — x̄: [10,000(50) − 1,600(80)] / 8,400 = (500,000 − 128,000) / 8,400 = 372,000 / 8,400 = 44.29 mm. Step 5 — ȳ: By the 45° diagonal symmetry of this shape, ȳ = 44.29 mm (same calculation).
This L-section reappears in structural steel tables. Note that the flange at the top pulls the centroid upward significantly compared to the web alone.
Scenario
An L-shaped section consists of: Piece 1 — a 200 × 50 mm horizontal plate (flange), and Piece 2 — a 50 × 150 mm vertical plate (leg) below it on the left side. Origin at the bottom-left corner. Find ȳ (from the bottom).
Solution
Piece 1 (flange, 200 × 50 mm, top): A₁ = 10,000 mm², centroid ȳ₁ = 150 + 25 = 175 mm. Piece 2 (leg, 50 × 150 mm): A₂ = 7,500 mm², centroid ȳ₂ = 75 mm. ΣA = 17,500 mm². ΣAᵢyᵢ = 10,000(175) + 7,500(75) = 1,750,000 + 562,500 = 2,312,500 mm³. ȳ = 2,312,500 / 17,500 = 132.14 mm from the bottom.
Applications
- Locating the neutral axis of a reinforced concrete or steel beam cross-section (required for bending stress calculation: σ = Mc/I).
- Determining eccentricity in column design — if load does not act through the centroid, bending moments develop (ACI 318 Section 6.6, NSCP 2015 Section 4).
- Finding the shear center of thin-walled open sections to prevent torsion.
- Computing hydraulic pressure resultants on submerged surfaces (fluid mechanics).
Misconceptions
- WRONG: Triangle centroid at h/2 from base. CORRECT: h/3 from base (2h/3 from apex).
- WRONG: Semicircle centroid at r/2. CORRECT: 4r/(3π) ≈ 0.424r from the diameter.
- WRONG: Adding hole area as positive. CORRECT: Holes are NEGATIVE areas.
- WRONG: Using the wrong reference axis mid-problem. CORRECT: Fix ONE reference origin at the start and never change it.
- WRONG: Assuming centroid is inside the material for concave shapes. CORRECT: The centroid can fall outside the physical material (e.g., a C-channel).
Related Concepts
- Moment of Inertia (second moment of area)
- Parallel-Axis Theorem
- Neutral Axis in Beam Bending
- First Moment of Area (used in shear stress calculation: τ = VQ/Ib)
Common Exam Questions
Example
Find ȳ of a T-section with a 150 × 20 mm flange on top of a 20 × 100 mm web. Answer: tabulate two rectangles, compute ȳ from bottom.
Approach
Set up the tabular method. Watch for holes (negative area). Use symmetry as a check. Common traps: wrong centroid location for triangle (h/3 from base), or placing semicircle centroid at r/2 instead of 4r/(3π).
Question Type
Multiple-choice / Computation
Example
A composite section consists of a rectangle with a circular hole. Locate the centroid of the remaining cross-section.
Approach
Full tabular solution. Show all sub-areas, centroids, and moments. State reference axis. Double-check arithmetic.
Question Type
Problem-solving (5–10 pts)
Key Points To Remember
- Centroid = area-weighted average of component centroids.
- Holes/cut-outs are NEGATIVE areas — subtract both area and moment.
- Centroid lies on every axis of symmetry.
- Triangle centroid is at h/3 from BASE, NOT from apex.
- Semicircle centroid: 4r/(3π) from the flat side (≈ 0.4244r).
- Always define your reference origin clearly before computing moments.
- Board exam tip: tabulate neatly — arithmetic errors are the #1 source of wrong answers.
Moment of Inertia (Second Moment of Area)
The moment of inertia of an area (also called the second moment of area) is defined as: Iₓ = ∫y² dA (about the x-axis) Iᵧ = ∫x² dA (about the y-axis) It quantifies how the area is distributed relative to an axis — the farther the area from the axis, the larger I. This is NOT the mass moment of inertia (which uses density and volume); in structural analysis, 'moment of inertia' always means the area moment of inertia unless stated otherwise. Standard centroidal formulas to MEMORIZE (axis through centroid): • Rectangle (b wide, h tall): - About centroidal x-axis: Ī = bh³/12 - About base (NOT centroidal): I_base = bh³/3 • Triangle (base b, height h): - About centroidal axis (parallel to base): Ī = bh³/36 - About the base: I_base = bh³/12 • Circle (diameter d, radius r): - Iₓ = Iᵧ = πd⁴/64 = πr⁴/4 • Hollow circle (outer D, inner d): - I = π(D⁴ − d⁴)/64 • Semicircle (radius r, flat side down): - About centroidal axis: Ī = 0.1098r⁴ (= [π/8 − 8/(9π)]r⁴) - About the flat (diameter) axis: I_diameter = πr⁴/8 Key insight: bh³/12 vs bh³/3. The factor of 3 difference arises because the centroidal axis is at the mid-height, while the base is farther from most of the area. This distinction causes the most board-exam errors.
Examples
Because the hole is concentric with the outer rectangle, no parallel-axis transfer is needed — both shapes share the centroid. This is the simplest MoI subtraction case.
Scenario
A hollow rectangular section is 100 mm wide × 150 mm deep (outside), with a centered rectangular hole 60 mm wide × 110 mm deep. Find Iₓ about the centroidal horizontal axis.
Solution
Both rectangles share the same centroid (doubly symmetric). Subtract directly: I_outer = (100)(150)³/12 = 100 × 3,375,000 / 12 = 28,125,000 mm⁴ I_hole = (60)(110)³/12 = 60 × 1,331,000 / 12 = 6,655,000 mm⁴ Iₓ = 28,125,000 − 6,655,000 = 21,470,000 mm⁴ ≈ 2.147 × 10⁷ mm⁴
For a solid circle, all diameters are axes of symmetry, so Iₓ = Iᵧ. This value is used in computing the bending stress in a circular rod under transverse loading.
Scenario
A circular steel rod has a diameter of 50 mm. Compute its moment of inertia about a diameter.
Solution
I = πd⁴/64 = π(50)⁴/64 = π × 6,250,000 / 64 = 306,796 mm⁴ ≈ 3.068 × 10⁵ mm⁴
Applications
- Bending stress in beams: σ = Mc/I (NSCP 2015; ACI 318 for cracked/uncracked sections).
- Beam deflection: δ = f(P, L, E, I) — larger I means less deflection.
- Steel section selection: AISC 360 / NSCP 2015 Section 5 tabulates Iₓ, Iᵧ for standard W, S, C, and L shapes.
- Euler's critical buckling load: Pcr = π²EI/(KL)² — I directly controls column capacity.
- Cracked-section moment of inertia Icr in ACI 318-19 Section 24.2 for deflection control.
Misconceptions
- WRONG: Using bh³/3 for centroidal axis. CORRECT: bh³/3 is about the BASE; bh³/12 is about the CENTROID.
- WRONG: Treating b and h interchangeably. CORRECT: h is ALWAYS perpendicular to the axis.
- WRONG: Using mass moment of inertia formulas (mr²/2 etc.) in structural problems. CORRECT: Area MoI (mm⁴) is used in structural/mechanics problems.
- WRONG: Forgetting to use the parallel-axis theorem when the hole/part centroid does not coincide with the composite centroid.
- WRONG: MoI can be negative. CORRECT: Area MoI is always positive (it involves y² or x²).
Related Concepts
- Centroid of area (must be located first)
- Parallel-Axis Theorem (transfer MoI between axes)
- Polar Moment of Inertia
- Radius of Gyration
- Section Modulus S = I/c
Common Exam Questions
Example
Find Iₓ of a 200 × 400 mm rectangle about its centroidal x-axis: I = (200)(400)³/12 = 1.067 × 10⁹ mm⁴.
Approach
Identify the shape, confirm which axis (centroidal or base), substitute. Watch the 'h' dimension — it is always perpendicular to the axis.
Question Type
Direct formula application
Example
Iₓ of a 200 mm diameter circle with a 100 mm diameter concentric circular hole: I = π(200⁴ − 100⁴)/64.
Approach
If both shapes share a centroid, subtract I directly. If not, use the parallel-axis theorem first.
Question Type
Hollow section (subtraction)
Key Points To Remember
- Rectangle centroidal MoI: bh³/12 (most-used formula in the exam).
- Rectangle about base: bh³/3 (note h is always the dimension perpendicular to the axis).
- Triangle centroidal MoI: bh³/36; about base: bh³/12.
- Circle MoI: πd⁴/64 or πr⁴/4.
- h is ALWAYS the dimension PERPENDICULAR to the axis of interest.
- Units: mm⁴ or m⁴ (area × length², since I = ∫y² dA).
- MoI is always POSITIVE (it's a squared quantity).
Parallel-Axis Theorem
The parallel-axis theorem (also called Steiner's theorem) allows you to compute the moment of inertia about ANY axis given the centroidal moment of inertia and the perpendicular distance between the two parallel axes: I = Ī + Ad² where: Ī = moment of inertia about the CENTROIDAL axis of the part A = area of the part d = perpendicular distance from the PART's centroid to the TARGET axis (the composite centroid) This is the workhorse formula for composite/built-up sections (T-beams, I-beams, channel sections, built-up columns). The procedure for a composite section: 1. Locate the composite centroid (using the centroid formula from Concept 1). 2. For each part: compute Ī (centroidal MoI of that part about its own centroid). 3. Compute d = distance from part centroid to the composite centroid. 4. Add the transfer term: I_part = Ī + Ad². 5. Sum all parts: I_total = Σ(Ī_i + Aᵢdᵢ²). For HOLES: subtract both Ī and Ad² (use negative area for holes). Critical insight: The Ad² term dominates for parts that are far from the centroid. A flange at the top of a T-beam contributes enormous inertia NOT because of its own Ī (which is small, being thin) but because of its large Ad² term. This is why wide-flange steel sections (W-shapes) are so efficient — the flanges are far from the neutral axis.
Examples
Notice that for the flange, Ī₁ = 800,000 mm⁴ is tiny compared to Ad² = 15,967,000 mm⁴ — the transfer term is 20× bigger! This demonstrates why wide flanges at large distances from the neutral axis are structurally efficient. Omitting Ad² would give a catastrophically wrong answer — the board exam classic mistake.
Scenario
A T-section: flange = 150 mm wide × 40 mm thick (on top); web = 40 mm wide × 160 mm deep (below flange). Total depth = 200 mm. Origin at the bottom. Find Iₓ about the composite centroidal axis.
Solution
STEP 1 — Locate composite centroid (ȳ from bottom): Flange: A₁ = 150 × 40 = 6,000 mm², ȳ₁ = 160 + 20 = 180 mm Web: A₂ = 40 × 160 = 6,400 mm², ȳ₂ = 80 mm ΣA = 12,400 mm² ΣAᵢyᵢ = 6,000(180) + 6,400(80) = 1,080,000 + 512,000 = 1,592,000 mm³ ȳ = 1,592,000 / 12,400 = 128.4 mm from bottom STEP 2 — Transfer each part: d₁ (flange) = 180 − 128.4 = 51.6 mm Ī₁ = (150)(40)³/12 = 800,000 mm⁴ I₁ = 800,000 + 6,000(51.6)² = 800,000 + 15,966,960 = 16,766,960 mm⁴ d₂ (web) = 128.4 − 80 = 48.4 mm Ī₂ = (40)(160)³/12 = 13,653,333 mm⁴ I₂ = 13,653,333 + 6,400(48.4)² = 13,653,333 + 14,990,464 = 28,643,797 mm⁴ STEP 3 — Total: Iₓ = 16,766,960 + 28,643,797 ≈ 4.54 × 10⁷ mm⁴
The two flanges together contribute 205 × 10⁶ mm⁴ while the web contributes only 45 × 10⁶ mm⁴ — flanges at large d dominate. This explains the efficiency of wide-flange (W-shape) steel sections.
Scenario
An I-section is made of three plates: top flange 200 × 20 mm, web 20 × 300 mm, bottom flange 200 × 20 mm. The section is doubly symmetric. Find Iₓ.
Solution
By symmetry, ȳ = 170 mm from the bottom (mid-total-depth). Top flange: d = 170 − 10 = 160 mm (wait — let us set origin at bottom): Total depth = 20 + 300 + 20 = 340 mm. Centroid at 170 mm from bottom. Top flange centroid: 340 − 10 = 330 mm → d = 330 − 170 = 160 mm Bottom flange centroid: 10 mm → d = 170 − 10 = 160 mm (symmetric) Web centroid: 170 mm → d = 0 I_top flange = (200)(20)³/12 + 4000(160)² = 133,333 + 102,400,000 = 102,533,333 mm⁴ I_bottom flange = same = 102,533,333 mm⁴ I_web = (20)(300)³/12 + 6000(0)² = 45,000,000 + 0 = 45,000,000 mm⁴ Iₓ = 102,533,333 + 102,533,333 + 45,000,000 = 250,066,667 mm⁴ ≈ 2.501 × 10⁸ mm⁴
Applications
- Computing Iₓ and Iᵧ of built-up steel sections (AISC 360; NSCP 2015 Section 5) for bending and buckling.
- Transformed section method for composite beams (reinforced concrete, T-beams): ACI 318 transforms steel area to equivalent concrete area using n = Eₛ/Ec.
- Evaluating effective moment of inertia Iₑ for deflection in ACI 318-19 Eq. 24.2.3.5.
- Transfer formula used in computing product of inertia for unsymmetric bending.
- Column radius of gyration calculation for slenderness ratio KL/r.
Misconceptions
- WRONG: d = distance from the reference axis (bottom) to the part centroid. CORRECT: d = distance from PART centroid to COMPOSITE centroid.
- WRONG: Applying the parallel-axis theorem between two non-centroidal axes. CORRECT: Always start from the CENTROIDAL axis of the part.
- WRONG: Forgetting to locate the composite centroid first. CORRECT: You CANNOT compute d without the composite centroid.
- WRONG: Using positive area for holes when applying the parallel-axis theorem. CORRECT: Holes use negative area; subtract both Ī and Ad².
- WRONG: Squaring the wrong dimension. If d = 51.6 mm, then d² = 2,662.6 mm², not 51.6 mm².
Related Concepts
- Centroid of composite area
- Moment of inertia of basic shapes
- Polar moment of inertia
- Radius of gyration
- Section modulus S = I/c
Common Exam Questions
Example
A T-section with given flange and web dimensions — find centroid, then Iₓ, then section modulus S = Iₓ/c.
Approach
Always locate centroid first. Then tabulate: Part | Ā | d | Ad² | Ī | I_part. Sum. Do not skip the centroid step.
Question Type
Multi-part computation (T-beam, I-beam, channel)
Example
A beam must have Iₓ ≥ 5 × 10⁷ mm⁴. Does the given T-section satisfy this requirement?
Approach
Given a required Iₓ (from deflection or bending stress limit), check if the composite section satisfies it.
Question Type
Structural design application
Key Points To Remember
- I = Ī + Ad²: Ī is centroidal MoI of the PART, d is from PART centroid to COMPOSITE centroid.
- d is ALWAYS measured from the PART's centroid to the reference axis (usually composite centroid).
- STEP ORDER: Find composite centroid FIRST, then apply parallel-axis theorem.
- For holes: use negative area — subtract both Ī and Ad².
- The Ad² term is usually larger than Ī for parts far from the centroid — never ignore it.
- You can only transfer TO a centroidal axis using I = Ī + Ad² (not between two non-centroidal axes directly).
- Board exam tip: The most common error is using the wrong d — double-check that d is measured from the PART centroid, not from an edge.
Polar Moment of Inertia and Radius of Gyration
POLAR MOMENT OF INERTIA (J) The polar moment of inertia is the moment of inertia about an axis perpendicular to the plane of the area (the z-axis, passing through the reference point): J = ∫r² dA = Iₓ + Iᵧ This is the perpendicular-axis theorem: J equals the sum of the two in-plane moments of inertia. For a circle: J = πd⁴/32 = πr⁴/2 J is used primarily in TORSION problems: the shear stress in a circular shaft under torque T is: τ = Tr/J (or τ = Tc/J, where c = outer radius) For non-circular sections, torsion is more complex, but J still appears in the perpendicular-axis context. RADIUS OF GYRATION (r or k) The radius of gyration is the distance from the axis at which the entire area could be concentrated to produce the same moment of inertia: r = √(I/A) So: I = Ar² For both principal axes: rₓ = √(Iₓ/A) rᵧ = √(Iᵧ/A) The radius of gyration governs column buckling in the Euler formula: Pcr = π²EA/(KL/r)² where KL/r is the slenderness ratio. A larger r → smaller slenderness ratio → more buckling resistance. For doubly-symmetric sections (I-beams, square tubes), rₓ ≠ rᵧ — buckling occurs first about the axis with the SMALLER r (usually the weak axis rᵧ for wide-flange sections). NSCP 2015 Section 5 (AISC 360 Chapter E) uses KL/r to select the correct column curve.
Examples
This r value would be used in the slenderness ratio KL/r for column design. The hollow section has a higher r than a solid bar of the same area — material is distributed far from the centroid.
Scenario
For the hollow rectangular section in the MoI example: outer 100 × 150 mm, inner hole 60 × 110 mm. Compute rₓ.
Solution
A = 100(150) − 60(110) = 15,000 − 6,600 = 8,400 mm² Iₓ = 2.147 × 10⁷ mm⁴ (from the MoI example) rₓ = √(Iₓ/A) = √(21,470,000 / 8,400) = √(2,556) = 50.6 mm
For a circle, Iₓ = Iᵧ (complete symmetry), so J = 2Iₓ = πd⁴/32. This checks out perfectly. J is used in computing torsional shear stress in shafts.
Scenario
A solid circular steel shaft has a diameter of 80 mm. Find J and verify: J = Iₓ + Iᵧ.
Solution
J = πd⁴/32 = π(80)⁴/32 = π × 40,960,000 / 32 = 4,021,239 mm⁴ ≈ 4.02 × 10⁶ mm⁴ By perpendicular-axis theorem: Iₓ = Iᵧ = πd⁴/64 = 2,010,619 mm⁴ J = Iₓ + Iᵧ = 2,010,619 + 2,010,619 = 4,021,238 mm⁴ ✓
Applications
- Torsion in circular shafts: τ_max = Tc/J; angle of twist: φ = TL/GJ.
- Column design: slenderness ratio KL/r (NSCP 2015 Section 503.3 / AISC 360 Chapter E).
- Weld group analysis: polar moment of inertia of weld patterns under eccentric shear.
- Bolt group analysis: torsional shear on bolt groups (NSCP 2015 Section 510).
- Perpendicular-axis theorem: useful shortcut when Iₓ is known to find Iᵧ (or vice versa) for symmetric shapes.
Misconceptions
- WRONG: J = Iₓ × Iᵧ. CORRECT: J = Iₓ + Iᵧ (sum, not product).
- WRONG: Radius of gyration r is the radius of the section. CORRECT: r = √(I/A) — a derived quantity, not a physical radius.
- WRONG: Using the larger r for column slenderness. CORRECT: Use the SMALLER r (weak axis) — it gives the LARGER slenderness ratio and controls design.
- WRONG: J = πd⁴/64. CORRECT: πd⁴/64 is Iₓ (or Iᵧ) for a circle; J = πd⁴/32.
- WRONG: Radius of gyration has units of mm². CORRECT: r = √(mm⁴/mm²) = mm — it has units of length.
Related Concepts
- Moment of Inertia (Iₓ, Iᵧ)
- Torsion in circular shafts
- Column buckling (Euler formula)
- NSCP 2015 Section 5 / AISC 360 Chapter E (column design)
- Weld and bolt group eccentric loading
Common Exam Questions
Example
A 60 mm diameter shaft carries 800 N·m torque. Find max shear stress: τ = T(d/2)/J.
Approach
Compute J = πd⁴/32 for solid circle or π(D⁴−d⁴)/32 for hollow circle. Apply τ = Tc/J.
Question Type
Torsion computation
Example
A W200×52 column (rₓ = 89.9 mm, rᵧ = 51.7 mm), K = 1.0, L = 4 m. Governing KL/r = 4000/51.7 = 77.4 (about weak axis).
Approach
Compute I and A for the section. Find r = √(I/A). Compute KL/r. Compare to λc or use NSCP column tables.
Question Type
Column slenderness
Key Points To Remember
- Polar MoI: J = Iₓ + Iᵧ (perpendicular-axis theorem).
- Circle: J = πd⁴/32 (torsion formula: τ = Tc/J).
- Radius of gyration: r = √(I/A); units are mm or m.
- r governs the slenderness ratio KL/r in column design (NSCP 2015 Section 5).
- Larger r = less slender = more buckling resistance.
- For columns, always check BOTH rₓ and rᵧ — use the SMALLER one (critical axis).
- Radius of gyration is NOT the centroidal distance; it is a derived quantity combining I and A.
Practice Problems
This is a key insight tested on the board exam: when a hole is centered at the composite centroid, the centroid of the remaining area is unchanged. Symmetry confirms this.
Problem
PROBLEM 1 — Composite Centroid with Hole A plate 240 mm wide × 180 mm tall has a circular hole of diameter 60 mm centered at (120, 90) mm from the bottom-left corner. Find the centroid of the remaining area.
Solution
Rectangle: A₁ = 240 × 180 = 43,200 mm², centroid at (120, 90). Circular hole (negative): A₂ = −π(60)²/4 = −2,827.4 mm², centroid at (120, 90). Net area: A = 43,200 − 2,827.4 = 40,372.6 mm². x̄ = [43,200(120) − 2,827.4(120)] / 40,372.6 = (5,184,000 − 339,288) / 40,372.6 = 4,844,712 / 40,372.6 = 120.0 mm. ȳ = similarly = 90.0 mm. The centroid remains at (120, 90) — the center of the rectangle — because the hole is concentric with the rectangle's centroid. Removing a concentric hole does not shift the centroid.
Note that d₁ = d₂ = 55 mm in this particular section — a coincidence. The smaller section modulus governs bending strength: S_bottom < S_top, so the bottom fiber reaches the allowable stress first. In an unsymmetric T-beam, always compute both section moduli.
Problem
PROBLEM 2 — MoI of T-Section (Standard Board Format) A T-section has: flange = 200 mm wide × 20 mm thick; web = 20 mm wide × 200 mm deep. The flange is on top. Total depth = 220 mm. Find: (a) ȳ from the bottom, (b) Iₓ about the centroidal axis, (c) section modulus S_top and S_bottom.
Solution
(a) CENTROID: Flange: A₁ = 200 × 20 = 4,000 mm², ȳ₁ = 200 + 10 = 210 mm Web: A₂ = 20 × 200 = 4,000 mm², ȳ₂ = 100 mm ΣA = 8,000 mm² ΣAy = 4,000(210) + 4,000(100) = 840,000 + 400,000 = 1,240,000 mm³ ȳ = 1,240,000 / 8,000 = 155 mm from bottom. (b) MOMENT OF INERTIA (parallel-axis theorem): d₁ = 210 − 155 = 55 mm (flange to composite centroid) Ī₁ = (200)(20)³/12 = 133,333 mm⁴ I₁ = 133,333 + 4,000(55)² = 133,333 + 12,100,000 = 12,233,333 mm⁴ d₂ = 155 − 100 = 55 mm (web to composite centroid) Ī₂ = (20)(200)³/12 = 13,333,333 mm⁴ I₂ = 13,333,333 + 4,000(55)² = 13,333,333 + 12,100,000 = 25,433,333 mm⁴ Iₓ = 12,233,333 + 25,433,333 = 37,666,667 mm⁴ ≈ 3.767 × 10⁷ mm⁴ (c) SECTION MODULUS: c_top = 220 − 155 = 65 mm (top fiber to centroid) c_bottom = 155 mm (bottom fiber to centroid) S_top = Iₓ/c_top = 37,666,667 / 65 = 579,487 mm³ ≈ 5.795 × 10⁵ mm³ S_bottom = Iₓ/c_bottom = 37,666,667 / 155 = 242,882 mm³ ≈ 2.429 × 10⁵ mm³
The weak axis (smaller rᵧ) always governs column buckling when KL is the same about both axes. For rectangular solid sections, the narrow dimension controls — the axis about which the section is least stiff is the critical one.
Problem
PROBLEM 3 — Radius of Gyration and Column Application A built-up column section is formed by two 150 × 150 × 12 mm equal-leg angles placed back-to-back to form a T-shape (simplified: treat as two rectangles 150 × 12 mm each, arranged to form a + cross). For simplicity, model it as a solid rectangle 162 × 12 mm (web) plus two 69 × 12 mm flanges extending symmetrically (total width 150 mm, total depth 162 mm). Find rₓ and rᵧ for a more practical I-shaped stub: outer 150 × 162 mm minus two cutouts 63 × 138 mm each side. Actual simplified problem: For a 150 mm × 162 mm solid rectangle column stub, compute rₓ and rᵧ. Then determine the governing slenderness ratio if K = 1.0 and L = 3.5 m.
Solution
A = 150 × 162 = 24,300 mm² Iₓ = (150)(162)³/12 = 150 × 4,251,528 / 12 = 53,144,100 mm⁴ Iᵧ = (162)(150)³/12 = 162 × 3,375,000 / 12 = 45,562,500 mm⁴ rₓ = √(Iₓ/A) = √(53,144,100/24,300) = √(2,187) = 46.8 mm rᵧ = √(Iᵧ/A) = √(45,562,500/24,300) = √(1,875) = 43.3 mm Governing KL/r (use smaller r = rᵧ = 43.3 mm): KL/r = 1.0 × 3,500 / 43.3 = 80.8 With KL/r = 80.8, check NSCP 2015 Section 503.3: λc = (KL/πr)√(Fy/E). For A36 steel (Fy = 248 MPa, E = 200,000 MPa): λc = (80.8/π)√(248/200,000) = 25.72 × 0.03521 = 0.906. Since λc < 1.5, use inelastic buckling formula.
The torsional shear stress varies linearly from zero at the center to a maximum at the outer surface. For hollow shafts, the inner material near the axis contributes little to J while adding weight — hence hollow shafts are more efficient for torsion.
Problem
PROBLEM 4 — Polar Moment and Torsional Shear Stress A hollow circular steel shaft has outer diameter D = 100 mm and inner diameter d = 60 mm. It transmits a torque T = 5 kN·m. Find: (a) polar moment of inertia J, (b) maximum shear stress τ_max, (c) shear stress at the inner surface.
Solution
(a) J = π(D⁴ − d⁴)/32 = π(100⁴ − 60⁴)/32 = π(100,000,000 − 12,960,000)/32 = π(87,040,000)/32 = 8,545,132 mm⁴ ≈ 8.545 × 10⁶ mm⁴ (b) τ_max occurs at outer surface, c = D/2 = 50 mm: τ_max = Tc/J = (5 × 10⁶ N·mm)(50 mm) / (8,545,132 mm⁴) = 250,000,000 / 8,545,132 = 29.3 MPa (c) Shear stress at inner surface, r = d/2 = 30 mm: τ_inner = Tr/J = (5 × 10⁶)(30) / 8,545,132 = 150,000,000 / 8,545,132 = 17.6 MPa Check: τ_inner/τ_max = 30/50 = 0.6 → τ_inner = 0.6 × 29.3 = 17.6 MPa ✓ (shear stress varies linearly with r)
Semicircle centroid location (4r/3π from the flat edge) is frequently forgotten. Always add the flat-edge elevation to get the centroid above the origin. The semicircle pulls the composite centroid upward compared to the L-shape alone.
Problem
PROBLEM 5 — Centroid of L-Shape with Semicircle An L-shaped section is formed by: Rectangle 1 = 200 mm wide × 40 mm thick (horizontal leg); Rectangle 2 = 40 mm wide × 200 mm tall (vertical leg, sharing the lower-left corner with Rectangle 1). A semicircle of radius 40 mm is added on top of Rectangle 2 (its flat edge aligns with the top of Rectangle 2). Find ȳ from the bottom.
Solution
Set origin at bottom-left. Rect 1 (horizontal): A₁ = 200 × 40 = 8,000 mm², ȳ₁ = 20 mm Rect 2 (vertical): A₂ = 40 × 200 = 8,000 mm², ȳ₂ = 40 + 100 = 140 mm (above Rect 1) Semicircle (on top of Rect 2): A₃ = π(40)²/2 = 2,513.3 mm² Centroid of semicircle above its flat edge: 4r/(3π) = 4(40)/(3π) = 16.98 mm ȳ₃ = 40 + 200 + 16.98 = 256.98 mm ΣA = 8,000 + 8,000 + 2,513.3 = 18,513.3 mm² ΣAy = 8,000(20) + 8,000(140) + 2,513.3(256.98) = 160,000 + 1,120,000 + 645,865 = 1,925,865 mm³ ȳ = 1,925,865 / 18,513.3 = 104.0 mm from the bottom
Exam Preparation Tips
- MEMORIZE the six essential formulas before exam day: x̄ = ΣAᵢxᵢ/ΣAᵢ; Ī_rect = bh³/12; Ī_tri = bh³/36; I_circle = πd⁴/64; I = Ī + Ad² (parallel axis); r = √(I/A).
- ALWAYS set up a tabular solution for composite centroid problems: columns for Shape | Aᵢ | xᵢ | Aᵢxᵢ | yᵢ | Aᵢyᵢ. This prevents arithmetic errors and earns partial credit.
- STEP ORDER is non-negotiable: (1) Find centroid first, (2) compute d for each part, (3) apply parallel-axis theorem. Doing these out of order causes systematic errors.
- SIGN DISCIPLINE: holes are NEGATIVE. If you accidentally add a hole instead of subtracting, your answer will be way off — and the error is hard to spot. Label each row as '+' or '−' in your table.
- WATCH THE h DIMENSION: in bh³/12, 'h' is ALWAYS perpendicular to the axis. If you want Iᵧ of a rectangle (about the vertical axis), use 'b' as h: Iᵧ = hb³/12.
- SEMICIRCLE TRAP: centroid is at 4r/(3π) ≈ 0.424r from the flat diameter, NOT at r/2. This comes up frequently in centroid problems involving arched sections.
- TRIANGLE TRAP: centroid is at h/3 from the BASE (the shorter distance), not h/3 from the apex. Many examinees reverse this.
- For column problems (NSCP 2015 / AISC 360): always compute BOTH rₓ and rᵧ and use the SMALLER one — the larger KL/r governs buckling.
- UNIT CHECKS: centroidal distances are in mm; areas in mm²; first moments (Ay) in mm³; second moments (I) in mm⁴; radius of gyration in mm.
- BOARD EXAM STRATEGY: If a T-section or I-section problem appears, automatically locate the centroid, then immediately set up the parallel-axis table. Do not attempt to 'estimate' I without the transfer term — it is always wrong by a factor of 2–10x.
- PRACTICE with actual NSCP/PRC board exam problems from past years. The T-section MoI and the composite centroid with a hole are the two most recurring question types.
- DOUBLE-CHECK with symmetry: if the section is doubly symmetric, x̄ and ȳ should both be at the geometric center. Use this to verify your tabular calculation.
In summary
Centroids and Moments of Inertia are foundational to every branch of structural analysis and design that appears in the PRC Civil Engineer Licensure Examination. The centroid tells you WHERE the neutral axis is; the moment of inertia tells you HOW STIFF the section is against bending or buckling. These two quantities unlock the flexure formula (σ = Mc/I), Euler's buckling equation (Pcr = π²EI/(KL)²), and the torsion formula (τ = Tc/J) — all examined regularly in both the engineering mechanics and structural engineering portions of the board exam. The non-negotiable skills to master are: (1) the tabular method for composite centroids, with strict negative-area treatment of holes; (2) the six standard MoI formulas, especially distinguishing bh³/12 (centroidal) from bh³/3 (about base); and (3) the parallel-axis theorem I = Ī + Ad², applied only AFTER locating the composite centroid. Skipping or misapplying any one of these three steps produces answers that are typically off by 50–200%, making the error obvious and costly on the board exam. As a PRC Civil Engineer licensee practicing under RA 544 and the National Structural Code of the Philippines (NSCP 2015), you will apply these principles every time you compute section properties for beams and columns — making this not just a licensure exam topic, but a career-long skill. Invest the time to memorize the formulas, practice the tabular procedure until it is automatic, and always verify your answers using symmetry checks and unit analysis.
Ready to practise for the CELE 2026?
Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.