Centroid Notes FE Sem 1 Engineering Mechanics | Mumbai University
On this page
The centroid is the geometric centre of an area, the point where the whole area may be assumed to be concentrated. This module covers the centroids of primitive shapes and the method of composite areas for laminas made by joining or cutting such shapes. Locating centroids is needed later for placing resultant loads, for moments of inertia in strength of materials, and for designing beam sections such as T, L and I sections. The module carries a guaranteed numerical in the examination, usually a composite lamina with a cut-out.
Summary
Centroid and centre of gravity: the centre of gravity of a body is the point through which the resultant of the weights of all its particles acts. For a thin plane lamina of uniform thickness and density, the weight of each part is proportional to its area, so the centre of gravity coincides with the centroid, the point where the whole area may be considered concentrated. The centroid depends only on geometry.
Method of moments of area: by Varignon's principle applied to areas, the moment of the total area about an axis equals the sum of the moments of its elemental areas. Hence x-bar = Sum(A x)/Sum A and y-bar = Sum(A y)/Sum A, where x and y are the centroid coordinates of each part. For continuous shapes the sums become integrals of x dA and y dA divided by A.
Axes of symmetry: if a lamina has an axis of symmetry, the centroid lies on that axis; if it has two axes of symmetry, the centroid is at their intersection. This halves the work in sections like T and I sections, where only y-bar needs calculation.
Centroids of primitive geometrical shapes: rectangle at the intersection of diagonals (b/2, h/2); triangle at h/3 from the base, the intersection of medians; circle at its centre; semicircle at 4r/(3 pi) from the diameter on the axis of symmetry; quarter circle at 4r/(3 pi) from each straight edge; circular sector of half angle alpha at 2r sin alpha/(3 alpha) from the centre along its axis of symmetry. These standard results are used directly in composite problems and some are derived by integration.
Composite laminas: a plane lamina consisting of primitive shapes is split into those shapes. A table is prepared with each part's area A, centroid coordinates x and y, and products Ax and Ay. Added parts take positive area; holes or cut-outs are treated as negative areas. Then x-bar = Sum Ax/Sum A and y-bar = Sum Ay/Sum A, all measured from chosen reference axes.
Choice of reference axes and checks: take the reference axes along the lowest and leftmost edges so that all coordinates are positive, or along an axis of symmetry. The answer must lie within the overall extent of the figure and should shift towards the larger area and away from cut-outs, which is a quick check of the result.
Mind map
Centroid of Plane Laminas
- Concepts
- Centre of gravity
- Centroid
- Moment of area
- Axis of symmetry
- Standard shapes
- Rectangle b/2, h/2
- Triangle h/3
- Circle at centre
- Semicircle 4r/3pi
- Quarter circle 4r/3pi
- Sector 2r sin(alpha)/3alpha
- Derivations
- Triangle by strip integration
- Semicircle by polar element
- Quarter circle
- Composite areas
- Split into primitives
- Table A, x, y, Ax, Ay
- Cut-outs as negative area
- Typical sections
- T-section
- L-section
- I-section
- Plate with hole
Procedure to locate the centroid of a composite lamina
- Sketch the lamina with all dimensions and fix reference axes (usually bottom edge as x axis and left edge as y axis).
- Note any axis of symmetry; the centroid lies on it, so one coordinate is known directly.
- Divide the figure into primitive shapes: rectangles, triangles, circles, semicircles, quarter circles.
- For each part write the area A (negative for holes and cut-outs) and its centroid coordinates x and y from the reference axes using standard results.
- Tabulate A, x, y, Ax and Ay and find Sum A, Sum Ax and Sum Ay.
- Compute x-bar = Sum Ax / Sum A and y-bar = Sum Ay / Sum A.
- Check that the centroid lies within the figure's extent and shifts sensibly, then mark it on the sketch.
Important definitions
| Term | Meaning |
|---|---|
| Centre of gravity | The point through which the resultant of the gravitational forces on all particles of a body acts, whatever the orientation of the body. |
| Centroid | The geometric centre of a plane area or line; the point at which the total area may be assumed to be concentrated for taking first moments. |
| Plane lamina | A thin flat plate of uniform thickness whose centre of gravity coincides with the centroid of its area. |
| First moment of area | The product of an area and the perpendicular distance of its centroid from a reference axis, units mm^3 or m^3. |
| Axis of symmetry | A line dividing an area into two mirror-image halves; the centroid always lies on it. |
| Composite area | An area made up of a number of simple shapes, added or removed, whose centroid is found by the method of moments. |
| Reference axes | The x and y axes from which centroid distances of all parts are measured. |
Formulas
| Formula | Meaning |
|---|---|
| \( \bar{x} = \frac{\sum A_i x_i}{\sum A_i},\quad \bar{y} = \frac{\sum A_i y_i}{\sum A_i} \) | Centroid of a composite lamina; A_i is the area of each part (negative for holes), x_i, y_i its centroid from the reference axes. |
| \( \bar{x} = \frac{\int x\,dA}{\int dA},\quad \bar{y} = \frac{\int y\,dA}{\int dA} \) | Centroid of a continuous area by integration; used to derive standard results. |
| \( A = bh,\quad (\bar{x}, \bar{y}) = \left(\frac{b}{2}, \frac{h}{2}\right) \) | Rectangle of width b and height h, measured from a corner. |
| \( A = \frac{1}{2}bh,\quad \bar{y} = \frac{h}{3} \) | Triangle: centroid at one third of the height from the base (two thirds from the apex). For a right triangle it is b/3 from the vertical side too. |
| \( A = \pi r^2 \) | Circle: centroid at the centre. |
| \( A = \frac{\pi r^2}{2},\quad \bar{y} = \frac{4r}{3\pi} \) | Semicircle: centroid on the axis of symmetry at 4r/(3 pi) = 0.4244 r from the diameter. |
| \( A = \frac{\pi r^2}{4},\quad \bar{x} = \bar{y} = \frac{4r}{3\pi} \) | Quarter circle: centroid at 4r/(3 pi) from each of the two straight edges. |
| \( A = \alpha r^2,\quad \bar{x} = \frac{2r\sin\alpha}{3\alpha} \) | Circular sector of radius r and half angle alpha (radians), measured from the centre along the axis of symmetry. |
Key points to remember
- For a homogeneous lamina of uniform thickness the centre of gravity and the centroid coincide.
- If there is one axis of symmetry, calculate only the other coordinate; two axes of symmetry fix the centroid directly.
- Holes and cut-outs are entered with negative area, but their centroid coordinates keep their actual signs.
- The triangle centroid is h/3 from the base, not h/3 from the apex; check which edge you are measuring from.
- The semicircle centroid is 4r/(3 pi) from the diameter, so when a semicircle sits on top of a rectangle its y is the rectangle height plus 4r/(3 pi), and when it is cut from the top it is the height minus 4r/(3 pi).
- Angles in the sector formula must be in radians.
- Use a neat table; examiners award marks for the tabulation even if arithmetic slips.
- The centroid can lie outside the material, for example in an L-section or a ring, but it is always within the overall bounding extent.
Solved examples
Example 1. Locate the centroid of a T-section with flange 120 mm x 20 mm and web 20 mm x 100 mm (total depth 120 mm).
The section is symmetric about the vertical axis, so x-bar lies on it. Take the x axis at the bottom of the web. Flange: A1 = 120 x 20 = 2400 mm^2, y1 = 100 + 10 = 110 mm. Web: A2 = 20 x 100 = 2000 mm^2, y2 = 50 mm. Sum A = 4400 mm^2. Sum Ay = 2400(110) + 2000(50) = 264000 + 100000 = 364000 mm^3. y-bar = 364000/4400 = 82.73 mm from the bottom of the web, on the axis of symmetry.
Example 2. Find the centroid of an L-section (unequal angle) 100 mm x 80 mm x 10 mm, the 100 mm leg vertical and the 80 mm leg horizontal.
Take the reference axes along the outer edges at the corner. Part 1, vertical leg: 10 x 100 = 1000 mm^2, x1 = 5 mm, y1 = 50 mm. Part 2, remaining horizontal leg: 70 x 10 = 700 mm^2, x2 = 10 + 35 = 45 mm, y2 = 5 mm. Sum A = 1700 mm^2. Sum Ax = 1000(5) + 700(45) = 5000 + 31500 = 36500, x-bar = 36500/1700 = 21.47 mm. Sum Ay = 1000(50) + 700(5) = 50000 + 3500 = 53500, y-bar = 53500/1700 = 31.47 mm. Centroid is at (21.47 mm, 31.47 mm) from the outer corner; note it lies outside the material.
Example 3. An unsymmetrical I-section has top flange 100 mm x 20 mm, web 20 mm x 120 mm and bottom flange 160 mm x 20 mm, total depth 160 mm. Find its centroid.
Symmetric about the vertical axis, so find y-bar from the bottom. Bottom flange: A1 = 3200 mm^2, y1 = 10 mm. Web: A2 = 2400 mm^2, y2 = 20 + 60 = 80 mm. Top flange: A3 = 2000 mm^2, y3 = 140 + 10 = 150 mm. Sum A = 7600 mm^2. Sum Ay = 3200(10) + 2400(80) + 2000(150) = 32000 + 192000 + 300000 = 524000 mm^3. y-bar = 524000/7600 = 68.95 mm from the bottom, on the axis of symmetry.
Example 4. A lamina consists of a rectangle 40 mm wide and 30 mm high with a semicircle of diameter 40 mm placed on its top edge. Find the centroid.
Axis of symmetry is vertical through the middle, so x-bar = 20 mm from the left edge. Rectangle: A1 = 1200 mm^2, y1 = 15 mm. Semicircle: r = 20 mm, A2 = pi(20^2)/2 = 628.32 mm^2, y2 = 30 + 4(20)/(3 pi) = 30 + 8.49 = 38.49 mm. Sum A = 1828.32 mm^2. Sum Ay = 1200(15) + 628.32(38.49) = 18000 + 24183 = 42183 mm^3. y-bar = 42183/1828.32 = 23.07 mm. Centroid at (20 mm, 23.07 mm) from the bottom-left corner.
Example 5. A quarter circle of radius 50 mm is cut from one corner of a 100 mm x 100 mm square plate, the centre of the quarter circle being at that corner. Find the centroid of the remaining plate.
Take the axes along the two edges meeting at the cut corner. Square: A1 = 10000 mm^2, x1 = y1 = 50 mm. Quarter circle (removed): A2 = -pi(50^2)/4 = -1963.50 mm^2, x2 = y2 = 4(50)/(3 pi) = 21.22 mm. Sum A = 8036.50 mm^2. Sum Ax = 10000(50) - 1963.50(21.22) = 500000 - 41667 = 458333 mm^3. x-bar = 458333/8036.50 = 57.03 mm. By symmetry about the diagonal, y-bar = 57.03 mm. Centroid at (57.03 mm, 57.03 mm), shifted away from the cut corner as expected.
Example 6. A lamina is made of a rectangle 60 mm long and 40 mm high, with a right-angled triangle attached to its right side; the triangle has its vertical side of 40 mm along the right edge of the rectangle and its base of 30 mm along the x axis. Find the centroid.
Origin at the bottom-left corner. Rectangle: A1 = 2400 mm^2, x1 = 30, y1 = 20 mm. Triangle: A2 = 0.5(30)(40) = 600 mm^2, its right angle is at (60, 0), so x2 = 60 + 30/3 = 70 mm and y2 = 40/3 = 13.33 mm. Sum A = 3000 mm^2. Sum Ax = 2400(30) + 600(70) = 72000 + 42000 = 114000, x-bar = 38.0 mm. Sum Ay = 2400(20) + 600(13.33) = 48000 + 8000 = 56000, y-bar = 18.67 mm. Centroid at (38.0 mm, 18.67 mm).
Example 7. A rectangular plate 200 mm x 150 mm has a circular hole of 60 mm diameter whose centre is 150 mm from the left edge and 100 mm from the bottom edge. Find the centroid of the plate.
Plate: A1 = 30000 mm^2 at (100, 75). Hole: A2 = -pi(30^2) = -2827.43 mm^2 at (150, 100). Sum A = 27172.57 mm^2. Sum Ax = 30000(100) - 2827.43(150) = 3000000 - 424115 = 2575885, x-bar = 94.80 mm. Sum Ay = 30000(75) - 2827.43(100) = 2250000 - 282743 = 1967257, y-bar = 72.40 mm. Centroid at (94.80 mm, 72.40 mm), moved away from the hole.
Important questions and answers
Q1. Differentiate between centroid and centre of gravity. (3 marks)
Centre of gravity is the point through which the resultant weight of a body acts; it applies to bodies having mass and depends on the distribution of weight. Centroid is the geometric centre of an area or volume and depends only on shape. For a homogeneous lamina of uniform thickness in a uniform gravity field the two coincide. Centroid is used for areas and lines (no mass), centre of gravity for physical bodies.
Q2. Derive the centroid of a triangle from its base using integration. (4 marks)
Take a triangle of base b and height h with the apex at the top. Consider a horizontal strip of thickness dy at height y from the base; by similar triangles its width is b(h - y)/h, so dA = b(h - y)/h dy. Total area A = integral from 0 to h of b(h - y)/h dy = bh/2. First moment about the base = integral from 0 to h of y b(h - y)/h dy = (b/h)(h^3/2 - h^3/3) = bh^2/6. y-bar = (bh^2/6)/(bh/2) = h/3 from the base.
Q3. Derive the centroid of a semicircular lamina of radius r. (5 marks)
Place the diameter on the x axis and the centre at O; by symmetry x-bar = 0. Take an elemental triangular sector of angle d theta at angle theta; its area is dA = (1/2) r^2 d theta and its centroid is at 2r/3 from O, so its height above the diameter is (2r/3) sin theta. First moment about the diameter = integral from 0 to pi of (2r/3) sin theta (1/2) r^2 d theta = (r^3/3)[-cos theta] from 0 to pi = 2r^3/3. Area = pi r^2/2. y-bar = (2r^3/3)/(pi r^2/2) = 4r/(3 pi).
Q4. State the centroids of standard shapes: rectangle, triangle, semicircle, quarter circle and circular sector. (3 marks)
Rectangle b x h: (b/2, h/2). Triangle: h/3 from the base on the median. Semicircle: 4r/(3 pi) from the diameter on the axis of symmetry. Quarter circle: 4r/(3 pi) from each straight edge. Circular sector of half angle alpha: 2r sin alpha/(3 alpha) from the centre along the axis of symmetry.
Q5. Find the centroid of a channel section with flanges 80 mm x 10 mm each and web 10 mm x 150 mm (overall depth 170 mm, flanges projecting to the right from the web). (5 marks)
Symmetric about the horizontal mid axis, so y-bar = 85 mm from the bottom. Find x-bar from the back of the web. Web: 10 x 150 = 1500 mm^2, x = 5 mm. Top flange: 80 x 10 = 800 mm^2, x = 40 mm. Bottom flange: 800 mm^2, x = 40 mm. Sum A = 3100 mm^2. Sum Ax = 1500(5) + 800(40) + 800(40) = 7500 + 64000 = 71500. x-bar = 71500/3100 = 23.06 mm. Centroid at (23.06 mm, 85 mm).
Q6. Explain why a hole is treated as a negative area in centroid calculations. (3 marks)
The composite area equals the full shape minus the removed part. Since first moments are additive, the moment of the remaining area equals the moment of the full shape minus the moment of the removed part. Writing the hole's area as negative in Sum A and Sum Ax, Sum Ay does exactly this subtraction, while its centroid coordinates are entered with their actual values.
Q7. Find the centroid of a semicircular plate of radius 60 mm from which a circle of 40 mm diameter is removed, the circle centre lying on the axis of symmetry 20 mm above the diameter. (5 marks)
x-bar = 0 on the axis of symmetry. Semicircle: A1 = pi(60^2)/2 = 5654.87 mm^2, y1 = 4(60)/(3 pi) = 25.46 mm. Circle removed: A2 = -pi(20^2) = -1256.64 mm^2, y2 = 20 mm. Sum A = 4398.23 mm^2. Sum Ay = 5654.87(25.46) - 1256.64(20) = 144000 - 25133 = 118867 mm^3. y-bar = 118867/4398.23 = 27.03 mm from the diameter.
Q8. What is the use of an axis of symmetry in finding centroids? (3 marks)
Every elemental area on one side of the axis has a mirror element on the other side at equal and opposite distance, so the first moment of the area about the axis is zero. Hence the centroid lies on the axis of symmetry. With one axis only the other coordinate needs computation; with two axes the centroid is their intersection, e.g. rectangle, circle, symmetric I-section.
Q9. Find the centroid of a quarter circle of radius r using integration. (4 marks)
Use the polar element of area dA = (1/2) r^2 d theta with centroid at 2r/3 from the centre. Area = pi r^2/4. Moment about the x axis = integral from 0 to pi/2 of (2r/3) sin theta (1/2) r^2 d theta = (r^3/3)[-cos theta] from 0 to pi/2 = r^3/3. y-bar = (r^3/3)/(pi r^2/4) = 4r/(3 pi). By symmetry about the 45 degree line, x-bar = 4r/(3 pi).
Exam tips
- Always draw the figure, mark the reference axes and number the parts before making the table.
- Show the standard formula used for each part's centroid (for example 4r/3pi) in the table or next to it.
- Use the axis of symmetry and say so explicitly; it saves time and earns a mark.
- Recheck signs: hole area negative, coordinates as measured.
- Give the final answer as coordinates with units and mark the centroid on the sketch.
FAQs
Are centroids of lines or volumes in this module?
No. The MU syllabus limits this module to centroids of plane laminas made of primitive geometrical shapes.
Do I need to memorise derivations?
Standard results are used directly in numericals, but derivations of the triangle, semicircle and quarter-circle centroids are commonly asked as short theory questions, so learn them.
What if the centroid falls outside the figure?
That is possible for shapes like L-sections, channels and rings. It must still lie within the overall extent of the figure.
These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Engineering Mechanics) and were prepared with AI help and checked.
More in Engineering Mechanics
See all →- System of Forces Notes FE Sem 1 Engineering Mechanics | MU MU FE Sem 1 Engineering Mechanics System of Forces notes: resultants of concurrent, parallel and general forces, moments, couples, Varignon,...
- Equilibrium and Friction Notes FE Sem 1 Mechanics | Mumbai Univ MU FE Sem 1 Engineering Mechanics Module 3: equilibrium conditions, free body diagrams, beam support reactions, laws of friction, inclined p...
- Kinematics of Particles and Rigid Bodies Notes FE Sem 1 | MU MU FE Sem 1 Engineering Mechanics Module 4: variable acceleration, curvilinear motion, tangential and normal acceleration, ICR for plane mot...
- Kinetics of Particle Notes FE Sem 1 Engineering Mechanics | MU MU FE Sem 1 Engineering Mechanics Module 5: D'Alembert's principle, inertia force, impulse-momentum, restitution, direct and oblique impact.
- Robot Kinematics Notes FE Sem 1 Engineering Mechanics | MU MU FE Sem 1 Engineering Mechanics Module 6: robot basics, degrees of freedom, D-H parameters, homogeneous transformation, 2-DOF forward kine...