CELE Engineering Mechanics — Centroids and Moments of InertiaRevision Notes

Formulas

Exam Tips

• Draw a neat sketch, label every part, and assign coordinates before computing — this eliminates sign errors.
• For a shape with one axis of symmetry, you immediately halve your work (only one coordinate to compute).
• Tabulate: Part | Area | x̄ | Ax̄ | ȳ | Aȳ — this format is fast and reduces arithmetic mistakes.
• Board problems on centroids are typically 3–5 minute items; if your solution takes longer, recheck the datum assignment.
• Memorize 4r/(3π) for semicircle and quarter-circle — it appears in roughly one of every three centroid problems.

Key Points

• The centroid is the area-weighted geometric center of a cross-section — it is NOT necessarily on the boundary.
• For composite areas, divide the shape into simple parts (rectangles, triangles, circles, semicircles), then apply the weighted-average formula.
• A cut-out (hole) is treated as a NEGATIVE area — subtract both its area and its area-moment contribution.
• The centroid lies on every axis of symmetry. For doubly symmetric sections (square, circle, I-section), the centroid is at the intersection of the two axes.
• Semicircle of radius r: centroid is at 4r/(3π) ≈ 0.424r from the diameter.
• Triangle: centroid is at h/3 above the base (one-third rule).
• Quarter circle: centroid at 4r/(3π) from each straight edge.
• Always define a clear, consistent reference origin (usually bottom-left corner or base) before computing.
• Board exams frequently combine two or three standard shapes with one cut-out — practice this exact pattern.

Definitions

Section Title

Centroid of a Composite Area

Common Mistakes

• Forgetting to subtract the hole: treating a hollow section as solid overcounts area and shifts the centroid toward the hole.
• Measuring centroid distances from the wrong reference — always state your datum clearly and be consistent.
• Using the base formula (h/3 from base for a triangle) instead of the correct distance from the chosen datum.
• For a semicircle sitting on a flat edge: confusing whether 4r/(3π) is measured upward (centroid above diameter) or downward.
• Not identifying the axis of symmetry first — if the shape is symmetric about y, then x̄ is at the axis and only ȳ needs computation.

Formulas

Exam Tips

• When the problem asks for I 'about the base', use the base formula (bh³/3 for rectangle) directly — no parallel-axis needed.
• For a hollow section where both rectangles share the same centroid, simply compute I_outer − I_inner. No Ad² term needed.
• Always check units: if b and h are in mm, I comes out in mm⁴. For stress calculations convert to consistent units.
• The board exam often gives a T or I section and asks for I about the centroidal axis — this always requires the parallel-axis theorem because the parts do NOT share a centroid.
• Memorize: πd⁴/64 for area MoI; πd⁴/32 for polar MoI — the 64 and 32 relate as 2:1, same as J = 2I for circles.

Key Points

• The moment of inertia I = ∫y² dA quantifies resistance to bending — the larger I, the stiffer the section.
• It has units of length⁴ (mm⁴ or m⁴). Never mix units.
• Standard formulas must be memorized: rectangle, triangle, circle about centroidal axes.
• I about the base of a rectangle = bh³/3; I about the centroidal axis = bh³/12. These two are frequently confused.
• For triangles: I about base = bh³/12; I about centroidal axis (at h/3) = bh³/36.
• For a circle: Iₓ = Iy = πd⁴/64 = πr⁴/4.
• I is always positive — there is no negative moment of inertia even for holes (the negative sign comes in through the parallel-axis transfer).
• The parallel-axis theorem is the primary tool for composite sections: I = Ī + Ad².
• Polar moment J = Iₓ + Iy (perpendicular-axis theorem); for a circle J = πd⁴/32.

Definitions

Section Title

Moment of Inertia (Second Moment of Area)

Common Mistakes

• Using bh³/3 (base formula) instead of bh³/12 (centroidal formula) when applying the parallel-axis theorem — this double-counts the Ad² term.
• Applying I = πd⁴/64 to a hollow circle without subtracting the inner circle: I_hollow = π(d_o⁴ − d_i⁴)/64.
• Confusing b and h — in bh³/12, h is the dimension measured in the direction perpendicular to the bending axis (the 'depth' direction).
• Forgetting that the triangle formula bh³/36 is for the centroidal axis; about the base it is bh³/12.
• Computing J = Iₓ + Iy but using different axes for Iₓ and Iy — both must be about the same point (centroid).

Formulas

Exam Tips

• Set up a table: Part | A | ȳ_part | Aȳ | Ī | d = ȳ_part − ȳ_composite | Ad² | I_transferred. Sum the last column.
• Find the composite centroid FIRST — you cannot compute d without knowing where the composite centroid is.
• Check your answer for reasonableness: I of a composite section should always be larger than the I of its largest single part computed about the SAME axis.
• If the section has a horizontal axis of symmetry, the composite centroid is on it — this eliminates the centroid-finding step.
• Board exams love the T-section: flange wide but thin, web narrow but deep. Practice it until the table takes under 4 minutes.

Key Points

• Statement: I = Ī + Ad² — where Ī is the centroidal MoI of the part, A is the part's area, and d is the perpendicular distance from the PART's centroid to the NEW (composite) centroidal axis.
• This theorem is the most-tested MoI concept in PRC board exams — it appears in nearly every composite-section problem.
• Steps: (1) Find composite centroid. (2) For each part, compute Ī (centroidal MoI). (3) Compute d = |part centroid − composite centroid|. (4) Add Ad² to Ī. (5) Sum all transferred I values.
• For a cut-out (hole): subtract both the centroidal Ī and the Ad² term — holes reduce total I.
• The transfer distance d is always from the part's OWN centroid to the TARGET axis — never between two arbitrary axes.
• You cannot transfer from a non-centroidal axis to another non-centroidal axis in one step. You must pass through the centroid.
• The minimum I for any set of parallel axes is always Ī — at the centroidal axis.
• Skipping the Ad² transfer when parts are offset from the composite centroid is the most common and most costly board exam error.

Definitions

Section Title

Parallel-Axis Theorem

Common Mistakes

• Using d = distance from base to part centroid instead of d = distance from COMPOSITE centroid to part centroid.
• Applying the parallel-axis theorem when the part's centroid already coincides with the composite centroid — in this case d=0 and Ad²=0 (no transfer needed, as in a centered hollow rectangle).
• Transferring from the base directly to the composite centroid using I_base = bh³/3 + Ad² — this is WRONG. Always use the CENTROIDAL I (bh³/12) as Ī.
• Forgetting to square d — using Ad instead of Ad².
• For a hole: subtracting only the centroidal Ī but forgetting to also subtract the Ad² term.

Formulas

Exam Tips

• For a solid circular section, r = d/4 (since r = √(πd⁴/64 ÷ πd²/4) = d/4) — memorize this shortcut.
• For a rectangular section, r_weak = b_small/√12 — the weaker axis governs column buckling.
• Board exams often give I and A and ask for r — one calculation. Or give r and A and ask for I = r²A. Practice both directions.
• Polar moment problems usually involve circular cross-sections (shaft design). For rectangles and T-sections, J = Iₓ + Iy by the perpendicular-axis theorem.
• When comparing two sections of equal area, the one with larger r is more efficient as a column — this is why hollow sections (HSS) are preferred in steel column design (AISC 360).

Key Points

• Radius of gyration: r = √(I/A). It is the distance from the axis at which the entire area could be concentrated to produce the same I.
• Units of r: mm or m — same as length, NOT mm².
• r is used in column design: slenderness ratio KL/r (NSCP 2015 Section 505 / AISC 360-16 Chapter E). Higher r means less slender = more buckling resistance.
• For a rectangular section b×h: r_x = h/√12 = 0.289h (about centroidal x-axis); r_y = b/√12 = 0.289b (about centroidal y-axis).
• Composite sections: compute I and A of the composite first, then r = √(I/A). You cannot average the individual r values.
• Polar moment J = Iₓ + Iy (perpendicular-axis theorem). For any shape, J ≥ I about any single axis.
• For a circle: J = πd⁴/32 = 2I (since Iₓ = Iy = πd⁴/64).
• Shear flow in thin-walled sections and torsional constant in design (AISC 360) both involve J.

Definitions

Section Title

Radius of Gyration and Polar Moment

Common Mistakes

• Computing r = I/A instead of r = √(I/A) — forgetting the square root.
• Using the wrong I: for a column buckling about the weak axis, use I_min (about the axis with smaller I), not I_max.
• Averaging radii of gyration of parts to get composite r — WRONG. Always compute composite I and A first.
• Confusing J (polar, area) with the torsional constant J used in Saint-Venant torsion for non-circular sections — these are different for open sections.
• For the perpendicular-axis theorem, using axes that are not both centroidal and not both in the same plane.

Exam Tips

• If the problem gives a standard steel section (W, S, C, L), look up I and r in AISC tables — no computation needed for those properties.
• For PRC board problems, the composite section is almost always a T, I, L, or hollow box — these four cover 90% of exam items.
• Check answers against bounding values: I_composite must be between the smallest part I (all area at centroid, d=0) and the sum of I_part computed about the centroid of the largest part.
• When the answer choices are widely spaced (e.g., ×10⁶ vs ×10⁷), a quick estimate of the dominant Ad² term can narrow the choice before full computation.
• For the L-shaped (angle) section: neither axis is an axis of symmetry, so both x̄ and ȳ must be computed. Do NOT assume either centroid is at mid-dimension.

Key Points

• Step 1 — Sketch and Label: Draw the section with all dimensions. Mark the reference origin (usually bottom-left or bottom center). Label each part.
• Step 2 — Decompose: Break the section into rectangles, triangles, circles, or semicircles. Identify any holes (negative areas).
• Step 3 — Tabulate Centroids: For each part, record Aᵢ, xᵢ, yᵢ, Aᵢxᵢ, Aᵢyᵢ. Sum columns. Divide to get x̄ and ȳ.
• Step 4 — Compute Centroidal I: For each part, compute Ī using the standard formula (bh³/12, πd⁴/64, etc.).
• Step 5 — Compute Transfer Distances: d = |part centroid − composite centroid|. Use ȳ from Step 3.
• Step 6 — Apply Parallel-Axis: For each part, I_part = Ī + Ad². Subtract for holes.
• Step 7 — Sum: I_composite = Σ I_part. Check units (mm⁴).
• Step 8 — Derived Quantities: r = √(I/A); J = Iₓ + Iy; σ_max = Mc/I where c = distance from centroid to extreme fiber.
• Time budget: centroid step should take ≤2 min for a 3-part section; full I computation ≤5 min.

Section Title

Board-Exam Problem-Solving Strategy

Common Mistakes

• Skipping the centroid step and assuming the centroid is at mid-depth — valid only for doubly symmetric sections.
• Using c = total depth instead of c = distance from centroid to the extreme fiber in σ = Mc/I.
• Mixing mm and cm in the same computation — work entirely in mm and mm⁴.
• Not identifying which axis (x or y, strong or weak) the problem is asking about.
• Using I about the base in the flexure formula — the formula σ = Mc/I requires I about the CENTROIDAL axis.