Centroid & second moment of area: practice problems with step-by-step solutions
Also called: moment of inertia · parallel axis theorem
Free Centroid & second moment of area practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: centroid, second moment of area, bending stress.
Free practice problemEasy
Find the height ȳ of the centroid of the T-section above its bottom edge.
Learn Centroid & second moment of area from scratch
1Why the shape of a section matters
Cut a beam straight across and look at the cut face. That shape is the cross-section, or just the section.
Take a wooden plank of 100 × 20 mm. Lay it flat across a gap and it sags a lot. Stand the same plank on its edge and it is far stiffer, although it has exactly the same area.
What matters is how far the material sits from the axis the beam bends about. Material far from that axis works much harder.
Two properties describe this. The centroid is the geometric centre of the section: the beam bends about an axis through it. The second moment of area I measures how far the area is spread from that axis.
Self-check The same plank is used flat (a) and on edge (b), bent about a horizontal axis. Which is stiffer?
Both are the same, since the area is the same
Flat (a)
It depends only on the material
On edge (b)
Answer: On edge (b). On edge, the material lies further from the bending axis. The area is the same, but its distance from the axis is what counts.
2The centroid of a composite section
For a rectangle, the centroid is in the middle. Many beam sections, such as a T, are made of rectangles.
Split the section into rectangles. For each one, find its area Ai and the height yi of its own centroid, measured from one reference line, usually the bottom edge.
The centroid of the whole section is the area-weighted average of those heights:
ȳ = Σ Ai·yi / Σ Ai
A large part pulls the centroid towards itself. In a T-section with the flange on top, the centroid sits above mid-height.
Self-check Part 1 has A = 3000 mm² at y = 20 mm. Part 2 has A = 1000 mm² at y = 100 mm. Where is the centroid?
ȳ = 160 mm
ȳ = 40 mm
ȳ = 60 mm
ȳ = 80 mm
Answer: ȳ = 40 mm. ȳ = (3000·20 + 1000·100)/4000 = 160 000/4000 = 40 mm. 60 mm is the plain average of the heights, which ignores the areas. 80 mm swaps the areas.
3The second moment of area
For a rectangle of width b and height h, bent about a horizontal axis through its own centroid:
Ic = b · h³ / 12
The height is cubed, which is why the plank on edge is 25 times stiffer: 20·100³ / (100·20³) = 25.
For a composite section, each part sits some distance d from the axis through the overall centroid. The parallel-axis theorem adds a term for that offset:
I = Σ ( Ic,i + Ai · di² )
The unit is mm⁴. The numbers get large, so answers are often given as “× 10⁶ mm⁴”.
Once you have I, the bending stress at a distance y from the centroidal axis follows from the bending moment M:
σ = M · y / I
Self-check A rectangle 100 mm wide and 20 mm high: what is I about its own horizontal centroidal axis?
1 666 667 mm⁴
400 000 mm⁴
66 667 mm⁴
800 000 mm⁴
Answer: 66 667 mm⁴. I = bh³/12 = 100·20³/12 = 66 667 mm⁴. 1 666 667 swaps b and h (that is the plank on edge); 800 000 forgets to divide by 12.
4The method step by step
Split the section into rectangles and number them.
For each part: area Ai and centroid height yi, all measured from the bottom edge.
Centroid: ȳ = ΣAiyi / ΣAi.
For each part: own Ic = bh³/12 and distance di = yi − ȳ.
Add up: I = Σ(Ic,i + Aidi²).
If a stress is asked: σ = M·y/I, with y the distance from the axis to the edge you need.
A small table with columns A, y, A·y, Ic, d and A·d² keeps everything in order.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A T-section has a flange of 100 × 20 mm on top of a web of 20 × 100 mm. Find the centroid height ȳ, the second moment of area Ix, and the bending stresses at the top and bottom edges for a sagging moment of 4 kN·m.
1
Split into two rectangles
Flange (1): A1 = 100·20 = 2000 mm², centroid at y1 = 100 + 10 = 110 mm.
Web (2): A2 = 20·100 = 2000 mm², centroid at y2 = 50 mm.
2
Centroid
ȳ = (2000·110 + 2000·50) / 4000 = 80 mm
The two areas are equal, so ȳ is halfway between 110 and 50 mm.
3
Own second moments
Ic,1 = 100·20³/12 = 66 667 mm⁴
Ic,2 = 20·100³/12 = 1 666 667 mm⁴
4
Parallel-axis terms
Each centroid is 30 mm from the axis x: d1 = 110 − 80 = 30 mm and d2 = 50 − 80 = −30 mm.
M = 4 kN·m = 4 × 10⁶ N·mm. The bottom edge is 80 mm from the axis, the top edge 120 − 80 = 40 mm.
σbottom = 4 × 10⁶ · 80 / 5 333 333 = 60 MPa
σtop = 4 × 10⁶ · 40 / 5 333 333 = 30 MPa
The bottom edge is in tension and the top edge in compression, as in any sagging beam.
Answer
ȳ = 80 mm, Ix = 5.33 × 10⁶ mm⁴, σbottom = 60 MPa and σtop = 30 MPa.
6Common mistakes
Forgetting the A·d² term. Without it you only add the parts’ own I values, which is far too small.
Measuring the heights yi from different reference lines. Pick the bottom edge and stick to it.
Using d from the bottom edge instead of from the overall centroid.
Swapping b and h in bh³/12. The dimension at right angles to the bending axis is cubed.
Using the wrong edge distance in σ = M·y/I. The edge furthest from the axis has the largest stress.
Self-check A 100 × 20 mm rectangle (width × height) has its centroid 50 mm from an axis x. What is its contribution to Ix?
0.067 × 10⁶ mm⁴
5.00 × 10⁶ mm⁴
5.07 × 10⁶ mm⁴
6.67 × 10⁶ mm⁴
Answer: 5.07 × 10⁶ mm⁴. I = bh³/12 + A·d² = 66 667 + 2000·50² = 5 066 667 mm⁴. 0.067 × 10⁶ forgets A·d²; 5.00 × 10⁶ forgets the own term; 6.67 × 10⁶ uses 1 666 667 (b and h swapped) plus A·d².
7How exam questions look
Easy: centroid. Find ȳ of a T-section above its bottom edge. Variants: an angle section, or I of a single rectangle, circle (πd⁴/64) or box.
Medium: second moment of area. Find Ix of the T-section about its centroidal axis. Variants: I-sections, channels, and the radius of gyration r = √(I/A).
Hard: bending stress. The T-section carries a moment M; find the stress at the top or bottom edge with σ = M·y/I.
The given dimensions are bf and tf for the flange, and tw and hw for the web.
Find the height ȳ of the centroid of the T-section above its bottom edge.
bf = 260 mmtf = 10 mmtw = 8 mmhw = 280 mm
Find ȳ(mm)
Show hints
Split the T into a flange and a web, and find the area and centroid height of each.
ȳ = ΣA·y / ΣA, with every y measured from the bottom edge.
Af = 2600 mm², Aw = 2240 mm².
Show the answer and the step-by-step solution
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Split into two rectangles
Measure every y from the bottom edge.
flange: Af = 260·10 = 2600 mm², yf = 280 + 10/2 = 285 mm
A 100 mm × 240 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 156 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 240 mmd = 30 mmhole centre: y = 156 mm
Find Ix(× 10⁶ mm⁴)
Show hints
Find the centroid first: the plate minus the hole.
ȳ = 118.9 mm.
Then I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²) about ȳ.
Show the answer and the step-by-step solution
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Centroid
ȳ = (24000·120 − 706.9·156)/(24000 − 706.9) = 118.9 mm
Full plate about ȳ
100·240³/12 + 24000·1.092² = 115.2 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·37.09² = 1.012 × 10⁶ mm⁴
Subtract
Ix = 114.2 × 10⁶ mm⁴
Answer
Ix = 114.2 × 10⁶ mm⁴
Example 3Hard
A 100 mm × 200 mm plate has a rectangular cut-out 30 mm wide and 20 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 200 mmcut-out: 30 × 20 mmhole centre: y = 120 mm
Find Ix(× 10⁶ mm⁴)
Show hints
Find the centroid first: the plate minus the hole.
ȳ = 99.38 mm.
Then I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²) about ȳ.
Show the answer and the step-by-step solution
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Centroid
ȳ = (20000·100 − 600·120)/(20000 − 600) = 99.38 mm
Full plate about ȳ
100·200³/12 + 20000·0.6186² = 66.67 × 10⁶ mm⁴
Hole about ȳ
30·20³/12 + 600·20.62² = 0.2751 × 10⁶ mm⁴
Subtract
Ix = 66.4 × 10⁶ mm⁴
Answer
Ix = 66.4 × 10⁶ mm⁴
Questions and answers
What is Centroid & second moment of area?
Cut a beam straight across and look at the cut face. That shape is the cross-section, or just the section.
Take a wooden plank of 100 × 20 mm. Lay it flat across a gap and it sags a lot. Stand the same plank on its edge and it is far stiffer, although it has exactly the same area.
How do I solve Centroid & second moment of area problems step by step?
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
What do the Centroid & second moment of area exercises cover?
Easy: centroid
Medium: second moment of area
Hard: bending stress
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.