How unit 5 is examined
This unit covers centroid, moment and product of inertia, radius of gyration, and shear force and bending moment diagrams; the beam diagrams and the moment of inertia numericals carry most of the marks.
Centroid and Centre of Gravity
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Definition. <mark>The centroid is the geometric centre of a line, area or volume, whereas the centre of gravity is the point through which the whole weight of a body acts.</mark>
Key points.
- The centroid depends only on shape, so it is defined for lines and plane areas, which have no weight.
- The centre of gravity depends on weight distribution, so it is used for three-dimensional bodies.
- In a uniform gravity field and for a homogeneous body, the centroid and the centre of gravity coincide.
- For a symmetrical figure the centroid lies on the axis of symmetry, and at the intersection of two such axes.
- For a composite area, $\bar{x} = \frac{\sum A_i x_i}{\sum A_i}$ and $\bar{y} = \frac{\sum A_i y_i}{\sum A_i}$, with a hole taken as a negative area.
- Rectangle: centroid at $d/2$ from the base; triangle: at $h/3$ from the base; semicircle: at $\frac{4r}{3\pi}$ from the diameter.
Also asked with this term (Nov 2022). Profile cross-sectioning is levelling along the centre line of a proposed road, canal or railway (longitudinal section) and at right angles to it (cross-sections) to get the ground profile and the earthwork quantities. A plane table is a drawing board on a tripod, used to plot a map directly in the field. Its accessories are the alidade (sighting rule), trough compass, plumbing fork with plumb bob, and spirit level.
Answer frame. Open with the definition of each term; give the difference in one line; then write the composite-area formula; close with the plane-table accessories list and the profile-levelling purpose.
Asked: [14 marks] (Nov 2022) Explain: a) Centroid and Centre of Gravity; b) Profile Cross-sectioning; c) Plane table and its used instruments
Moment Inertia of Area and Mass
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Definition. ==The moment of inertia of an area about an axis is the sum of the products of each elemental area and the square of its distance from the axis, $I = \int y^2\,dA$.==
Key points.
- It is the second moment of area, and it measures the resistance of a section to bending; its unit is $\text{mm}^4$ or $\text{m}^4$.
- For mass the moment of inertia is $I = \int r^2\,dm$ or $\sum m r^2$, with unit $\text{kg m}^2$; it measures resistance to angular acceleration.
- It is always positive, because the distance is squared.
- Parallel axis theorem: $I_{AB} = I_G + Ah^2$, valid only when one axis passes through the centroid.
- Perpendicular axis theorem (plane areas): $I_z = I_x + I_y$.
- Rectangle: $I_{xx} = \frac{bd^3}{12}$ about the centroidal axis, $\frac{bd^3}{3}$ about the base. Triangle about base: $\frac{bh^3}{12}$. Circle: $\frac{\pi d^4}{64}$. Semicircle about diameter: $\frac{\pi r^4}{8}$.
Derivation of parallel axis theorem. Let $G$ be the centroidal axis, $AB$ an axis parallel to it at distance $h$, and $dA$ an element at distance $y$ from $G$.
$$I_{AB} = \int (y+h)^2 dA = \int y^2 dA + 2h\int y\,dA + h^2\int dA$$
Since $\int y\,dA = 0$ (moment about the centroid), $\int y^2 dA = I_G$ and $\int dA = A$, we get $I_{AB} = I_G + Ah^2$.
Example (Nov 2022, Jun 2023, Dec 2023). Given: triangle base 50, height 100 (above X-axis, base on the axis); semicircle radius 50 below it; circular hole $d = 50$ centred on the axis.
| Part | Formula | $I_x$ ($\text{mm}^4$) |
|---|---|---|
| Triangle | $50 \times 100^3/12$ | $4.167\times10^6$ |
| Semicircle | $\pi \times 50^4/8$ | $2.454\times10^6$ |
| Hole (subtract) | $\pi \times 50^4/64$ | $0.307\times10^6$ |
$I_{xx} = 4.167 + 2.454 - 0.307 \approx 6.31\times10^6\ \text{mm}^4$.
Answer frame. Definition question: define $I$ and $k$, give both formulas and units, then a one-line significance. Theorem question: state, sketch $A$, $G$, $AB$, $h$, $dA$, expand as above. Numerical: split into standard shapes, tabulate $I$ of each about the axis (add $Ah^2$ where needed), add, subtract holes, write the boxed value.
Pitfall: Applying $Ah^2$ from a non-centroidal axis, or adding the hole instead of subtracting it.
Asked: [8 marks] (Nov 2022, Jun 2023, Dec 2023) Calculate the Moment of Inertia about the X-axis of the given composite lamina Asked: [7 marks] (Jun 2022, Dec 2023) Explain about the Moment of inertia and Radius of Gyration Asked: [7 marks] (Dec 2023) Define and derive an expression for parallel axis theorem
Radius of Gyration
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Definition. <mark>The radius of gyration is the distance from an axis at which the whole area (or mass) could be concentrated and still give the same moment of inertia.</mark>
Key points.
- For an area, $k = \sqrt{I/A}$; for a mass, $k = \sqrt{I/M}$.
- Its unit is mm or m, the same as length.
- Equivalently $I = Ak^2$, so a section with a larger $k$ resists bending or buckling better.
- For a rectangle, $k_{xx} = d/\sqrt{12}$; for a circle, $k = d/4$.
Introduction to product of Inertia and Principle Axes
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Definition. ==The product of inertia of an area about a pair of perpendicular axes is $I_{xy} = \int xy\,dA$; principal axes are the pair of axes about which $I_{xy} = 0$.==
Key points.
- $I_{xy}$ may be positive, negative or zero, because $x$ and $y$ carry signs.
- If either axis is an axis of symmetry, $I_{xy} = 0$, so the axes of a symmetrical section are principal.
- Parallel axis theorem: $I_{xy} = I_{\bar{x}\bar{y}} + A\bar{x}\bar{y}$, with signed coordinates of the centroid.
- Principal axes are inclined at $\tan 2\theta = -\frac{2I_{xy}}{I_x - I_y}$ to the x-axis.
- About principal axes, $I$ is maximum about one and minimum about the other: $I_{max,min} = \frac{I_x+I_y}{2} \pm \sqrt{\left(\frac{I_x-I_y}{2}\right)^2 + I_{xy}^2}$.
- Example: a rectangle about its centroidal axes has $I_{xy} = 0$; an unsymmetrical angle section has $I_{xy} \neq 0$ and its principal axes are tilted.
- Rectangle: $I_{xx} = \int_{-d/2}^{d/2} b\,y^2\,dy = \frac{bd^3}{12}$ (strip $b\,dy$ at distance $y$), and about the base $\frac{bd^3}{3}$.
| Moment of inertia | Product of inertia |
|---|---|
| $I_{xx} = \int y^2 dA$ | $I_{xy} = \int xy\,dA$ |
| Always positive | Positive, negative or zero |
| Uses one axis | Uses a pair of perpendicular axes |
| Measures resistance to bending | Measures asymmetry of the section |
| Not zero for any area | Zero if either axis is a symmetry axis |
| Maximum and minimum on principal axes | Zero on principal axes |
Answer frame. Comparison: open with both definitions, then the table, close with the principal-axes link. Short note (any two): define, give the formula, one example, and for the rectangle the strip derivation.
Asked: [6 marks] (Nov 2022) Differentiate the difference between Moment of inertia and Product of inertia Asked: [14 marks] (Dec 2024) Explain in brief any two: i) Principle Axes with the example; ii) Product of inertia with the example; iii) Moment of inertia of Rectangular Shape
Support Reactions, Shear force and bending moment Diagram for Cantilever & simply supported beam with concentrated, distributed load and Couple
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Definition. <mark>Shear force at a section is the algebraic sum of the vertical forces on one side of it, and bending moment is the algebraic sum of the moments of those forces about the section.</mark>
Key points.
- Support reactions are the forces and moments a support exerts to keep the beam in equilibrium: roller (one vertical reaction), hinge (vertical and horizontal), fixed (vertical, horizontal and moment).
- A cantilever is fixed at one end and free at the other; a simply supported beam rests on a hinge and a roller.
- Sign convention: shear is positive when the left part tends to move up; bending moment is positive when sagging.
- Find reactions first from $\sum F_y = 0$, $\sum M = 0$.
- Point load: SFD jumps by the load, BMD is linear. UDL: SFD is sloping straight, BMD is parabolic. Couple: SFD unchanged, BMD jumps by the couple.
- Bending moment is maximum where shear is zero; contraflexure is where the moment changes sign.
- Cantilever with load $W$ at the free end: shear $= W$ throughout, moment $= -Wx$, maximum $-WL$ at the fixed end (hogging). With UDL $w$: shear $= wx$, moment $= -wx^2/2$, maximum $-wL^2/2$.
| Beam (span $L$) | Reactions | SFD | Maximum BM |
|---|---|---|---|
| UDL $W$ | $WL/2$ each | linear, $+WL/2$ to $-WL/2$, zero at midspan | $WL^2/8$ at midspan |
| Point load $W$, $a$ from A, $b$ from B | $R_A = Wb/L$, $R_B = Wa/L$ | $+Wb/L$, drop $W$, then $-Wa/L$ | $Wab/L$ under load |
For UDL: $V_x = \frac{WL}{2} - Wx$, $M_x = \frac{WL}{2}x - \frac{Wx^2}{2}$.
Example (Jun 2023, Dec 2023). UVL 0 at A to 15 kN/m at B, 6 m span, 25 kN at 3 m. Resultant of UVL is $\frac12 \times 15 \times 6 = 45$ kN at 4 m from A. $\sum M_A = 0$: $6R_B = 25\times3 + 45\times4 = 255$, so $R_B = 42.5$ kN and $R_A = 70 - 42.5 = 27.5$ kN.
| Section | SF (kN) | BM (kNm) |
|---|---|---|
| A | $+27.5$ | 0 |
| C just left ($x=3$) | $27.5 - 11.25 = +16.25$ | $27.5\times3 - 11.25 = 71.25$ |
| C just right | $-8.75$ | $71.25$ |
| B | $-42.5$ | 0 |
Shear is $V = 27.5 - 1.25x^2$ (curved), BM $= 27.5x - 0.417x^3$. Shear changes sign at the load, so $M_{max} = 71.25$ kNm at midspan.
Variant (10 kN/m, 20 kN at centre, 5 m): $R_A = R_B = 35$ kN; SF $+35, +10 | -10, -35$; $M_{max} = 35\times2.5 - 10\times2.5^2/2 = 56.25$ kNm.
Answer frame. Open by naming the beam and load; draw the beam with reactions, SFD below it and BMD below that, all labelled at the key points; write section equations, then values; close with $V=0$ giving $M_{max}$. For the procedure question, list the steps: reactions, sign convention, effect of point load, UDL and couple, then zero shear and contraflexure.
Pitfall: Placing the UVL resultant at $L/2$ instead of $2L/3$ from the zero end.
Asked: [8 marks] (Jun 2022) Draw SFD and BMD for a simply supported beam with a point load W at distance a from the left and b from the right, span L Asked: [8 marks] (Nov 2022, Dec 2024) Draw SFD and BMD for a simply supported beam carrying UDL W kN/m over span L, and find maximum bending moment and shear force Asked: [7 marks] (Dec 2023, Jun 2023) Draw SFD and BMD for a simply supported beam of 6 m carrying a UVL of 15 kN/m and 25 kN at centre; or 5 m with UDL 10 kN/m and 20 kN at centre Asked: [7 marks] (Dec 2023) What do you understand by support reactions and its type? Explain shear force and bending moment for a cantilever beam Asked: [6 marks] (Dec 2024) Detailed procedure to draw the SFD and BMD under point load, U.D.L. and Couple
Last-minute revision
- Centroid is geometric; centre of gravity is where the weight acts; both coincide for a homogeneous body.
- $I = \int y^2 dA$; parallel axis $I = I_G + Ah^2$; perpendicular axis $I_z = I_x + I_y$.
- Rectangle $\frac{bd^3}{12}$ (centroid), $\frac{bd^3}{3}$ (base); triangle $\frac{bh^3}{12}$; circle $\frac{\pi d^4}{64}$; semicircle $\frac{\pi r^4}{8}$.
- Radius of gyration $k = \sqrt{I/A}$.
- $I_{xy} = \int xy\,dA$, zero for a symmetry axis; $\tan 2\theta = -\frac{2I_{xy}}{I_x-I_y}$.
- Simply supported UDL: reactions $WL/2$, $M_{max} = WL^2/8$.
- Point load: $R_A = Wb/L$, $M_{max} = Wab/L$.
- UVL resultant $\frac12 wL$ acts at $2L/3$ from the zero end.
- Fixed support gives 3 reactions, hinge 2, roller 1.
- Bending moment is maximum where shear is zero.
- Composite area: $I = \sum(I_G + Ah^2)$, holes subtracted.
Memory hooks
- "Centroid = shape, CG = weight."
- "$I$ squares the distance, $I_{xy}$ multiplies two signs."
- "UDL: shear line, moment curve; couple: shear ignores it."
- "Triangle one-third, UVL two-thirds."
- "Roller 1, hinge 2, fixed 3."
Coverage checklist
- Centroid and Centre of Gravity: Nov 2022 (14 marks).
- Moment Inertia of Area and Mass: composite MI numerical (Nov 2022, Jun 2023, Dec 2023), MI and radius of gyration (Jun 2022, Dec 2023), parallel axis theorem (Dec 2023).
- Radius of Gyration: no past questions; covered above.
- Introduction to product of Inertia and Principle Axes: Nov 2022 comparison, Dec 2024 short note.
- Support Reactions, Shear force and bending moment Diagram for Cantilever & simply supported beam with concentrated, distributed load and Couple: point load (Jun 2022), UDL (Nov 2022, Dec 2024), UVL with point load (Jun 2023, Dec 2023), supports and cantilever (Dec 2023), procedure (Dec 2024).