34 Engineering integral applications

Centroids, second moments, and RMS values

A steel I-beam holds up a floor. How much can it carry before it bends too far? The answer depends on a number called the second moment of area — an integral you compute once from the beam’s cross-section. Every structural engineering table of beam properties lists this number.

An AC power supply delivers 240V. But 240V of what? The voltage oscillates between +340V and −340V. The RMS value — root mean square — is the equivalent DC voltage that delivers the same power. It is an integral.

These are not abstract exercises. They are the specific results that structural, mechanical, and electrical engineering courses use in every problem set.

34.1 What this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

I_x = y^2 , dA: second moment of area about the x-axis
V_{} = : RMS voltage over one period T
{x} = : x-coordinate of the centroid

Definitions to keep handy

These are the words we keep coming back to. If one feels slippery, come back here and steady it before you push on.

centroid: The geometric centre of a shape — the point where it would balance if made from uniform material.
first moment of area: The integral of position times area element — used to locate the centroid.
second moment of area: The integral of (distance from axis)² times area element — measures resistance to bending.
radius of gyration: The distance from an axis at which the total area could be concentrated to give the same second moment.
RMS value: Root mean square — the square root of the mean of the squared values over one cycle. Gives the effective value of an alternating quantity.

Here is the main move this chapter is making, in plain terms. You do not need to be fast. You just need to keep the thread.

Coming in: You know how to evaluate a definite integral and compute areas and volumes. Integration is accumulation: adding up infinitely many infinitesimally thin slices of something.
Leaving with: The same accumulation idea extends to more complex quantities. The centroid of a shape is the average position of its area — computed by integration. The second moment of area measures how area is distributed around an axis — the key to predicting beam deflection and shaft stress. The RMS value of a waveform is the effective value of an alternating quantity — what makes a 240V AC supply deliver the same power as 240V DC. These are the integrals that engineering courses assume you know.

34.2 Section A: Mean values and RMS values

34.2.1 The mean value of a function

You already know the mean of a list of numbers: add them up, divide by how many there are. The mean value of a function over an interval is the same idea, extended to a continuous quantity.

The mean value of f(x) over [a, b] is:

\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx

Read that as: “the integral (the total accumulation) divided by the width of the interval.” It is the height of a rectangle with the same area as the region under the curve.

This matters directly for electrical engineering. The mean value of a sinusoidal voltage over one full cycle is zero — the positive half cancels the negative half. That is why a voltmeter measuring DC cannot read AC: it reads zero. But a half-wave rectified signal (only the positive half, the negative half clipped to zero) has a nonzero mean.

Worked example. Find the mean value of v = 10 \sin \theta over \theta = 0 to \pi (one positive half-cycle).

\bar{v} = \frac{1}{\pi - 0} \int_0^\pi 10 \sin \theta \, d\theta

= \frac{10}{\pi} \left[ -\cos \theta \right]_0^\pi

= \frac{10}{\pi} \left( -\cos \pi + \cos 0 \right)

= \frac{10}{\pi} \left( -(-1) + 1 \right) = \frac{10}{\pi} \times 2 = \frac{20}{\pi}

\bar{v} \approx 6.37 \text{ V}

Interpretation: if you build a half-wave rectifier circuit and connect a DC voltmeter across the output, it reads 20/\pi \approx 6.37 V. That is the mean value of the half-sine waveform.

34.2.2 Root mean square (RMS)

The mean value answers “what is the average height of the waveform?” But for power calculations you need something different.

Electrical power dissipated in a resistor is P = V^2 / R. It depends on the square of the voltage. To find the equivalent DC voltage that delivers the same average power, you take the mean of V^2 and then take the square root. That is: root, mean, square — in that order, applied in reverse.

The RMS value of f(x) over [a, b] is:

f_{\text{rms}} = \sqrt{\frac{1}{b-a} \int_a^b \left[f(x)\right]^2 dx}

For a sinusoidal voltage v = V_0 \sin(\omega t) over one complete period T = 2\pi/\omega:

\int_0^T V_0^2 \sin^2(\omega t) \, dt

Use the identity \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}:

= V_0^2 \int_0^T \frac{1 - \cos(2\omega t)}{2} \, dt = V_0^2 \cdot \frac{T}{2}

(The cosine term integrates to zero over a full period.)

So: V_{\text{rms}} = \sqrt{\frac{1}{T} \cdot V_0^2 \cdot \frac{T}{2}} = \frac{V_0}{\sqrt{2}} \approx 0.707 \, V_0

This is why UK mains is described as “240V”: the peak is 240\sqrt{2} \approx 340 V, but the RMS — the power-equivalent DC value — is 240 V. Your kettle consumes the same power as it would from a 240 V DC supply.

Worked example 1. Find the RMS value of v = 150 \sin(100\pi t) volts.

Here V_0 = 150 V. Therefore:

V_{\text{rms}} = \frac{150}{\sqrt{2}} = \frac{150\sqrt{2}}{2} \approx 106.1 \text{ V}

Worked example 2. Find the RMS value of a square wave that alternates between +10 V and -10 V.

The square of \pm 10 is always 100, so [f(t)]^2 = 100 at every point in time. The mean of a constant is that constant, and its square root is 10. Therefore V_{\text{rms}} = 10 V. For a square wave, the RMS equals the peak — there is no reduction from 1/\sqrt{2} because the waveform never dips toward zero between peaks.

Why squaring first?

Taking the mean of v directly would give zero for any symmetric AC waveform — positive and negative halves cancel. Squaring makes everything positive, so the mean measures the actual “size” of the oscillations. Taking the square root at the end brings the units back from V² to V. The ordering is deliberate: square first, then mean, then root.

34.3 Section B: Centroids and first moments of area

34.3.1 What a centroid is

Hold a flat shape at its centroid and it balances. Let it go and it stays level. The centroid is the average position of all the area in the shape — the point where gravity acts on a uniform lamina.

In structural engineering, the centroid of a beam’s cross-section is the neutral axis — the line along which the cross-section neither stretches nor compresses under bending. Everything above it is in compression; everything below it is in tension. Locating the centroid correctly is the first step in any bending calculation.

34.3.2 The first moment of area

The notation \bar{x}, read as “x-bar”, gives the x-coordinate of the centroid:

\bar{x} = \frac{\int x \, dA}{A}

And \bar{y}, the y-coordinate:

\bar{y} = \frac{\int y \, dA}{A}

Both expressions follow from the definition of a mean: integrate (accumulate) the position weighted by area, then divide by the total area. The integral \int x \, dA is called the first moment of area about the y-axis; \int y \, dA is the first moment about the x-axis.

For a shape defined by a function y(x) from x = a to x = b, taking vertical strips of width dx and height y(x):

\bar{x} = \frac{\int_a^b x \cdot y(x) \, dx}{A} \qquad \bar{y} = \frac{\int_a^b \frac{y(x)}{2} \cdot y(x) \, dx}{A} = \frac{\frac{1}{2}\int_a^b [y(x)]^2 \, dx}{A}

The factor of \frac{1}{2} in the \bar{y} formula: each vertical strip has its centre of mass at its midpoint, which is at height y(x)/2 above the x-axis.

34.3.3 Standard results derived

Rectangle (width b, height h, base along x-axis):

A = bh

By symmetry, \bar{x} = b/2 and:

\bar{y} = \frac{\frac{1}{2}\int_0^b h^2 \, dx}{bh} = \frac{\frac{1}{2} h^2 \cdot b}{bh} = \frac{h}{2}

The centroid is at the geometric centre. Not surprising — but useful to confirm, because it validates the method on a shape whose answer you already know.

Triangle (base b, height h, apex at top, base along x-axis):

The height at position x from the left is y(x) = h\left(1 - x/b\right) for the case where the apex is above the left corner. For a general triangle, integration gives \bar{y} = h/3 from the base. The centroid sits one-third of the height up from the base — closer to the base than the apex.

Setting up the integral:

A = \frac{1}{2}bh

\int_0^b \frac{1}{2}[y(x)]^2 \, dx = \frac{1}{2}\int_0^b h^2\left(1 - \frac{x}{b}\right)^2 dx

Let u = 1 - x/b, du = -dx/b:

= \frac{1}{2} h^2 b \int_0^1 u^2 \, du = \frac{1}{2} h^2 b \cdot \frac{1}{3} = \frac{h^2 b}{6}

Therefore:

\bar{y} = \frac{h^2 b / 6}{bh/2} = \frac{h}{3}

Semicircle (radius r, flat edge along x-axis):

The centroid lies on the axis of symmetry. By integration (or a result you will derive in Vol 7 using polar coordinates):

\bar{y} = \frac{4r}{3\pi} \approx 0.424 \, r

This is the standard result. The centroid is slightly less than halfway up from the flat edge — pulled down because there is more area near the flat edge than near the curved top.

34.3.4 Composite shapes

Real cross-sections are rarely simple shapes. An L-section is two rectangles; a T-section is three. The key fact: the centroid of a composite shape is the area-weighted mean of the individual centroids.

\bar{y} = \frac{\sum_i A_i \bar{y}_i}{\sum_i A_i}

Worked example. Find the centroid of an L-section: a vertical rectangle 80 mm tall × 20 mm wide, sitting on a horizontal rectangle 60 mm wide × 20 mm thick (the foot).

Take y = 0 at the base of the foot.

Part	Width (mm)	Height (mm)	Area A_i (mm²)	\bar{y}_i (mm)	A_i \bar{y}_i (mm³)
Foot	60	20	1200	10	12 000
Web	20	80	1600	20 + 40 = 60	96 000

\bar{y} = \frac{12\,000 + 96\,000}{1200 + 1600} = \frac{108\,000}{2800} \approx 38.6 \text{ mm from the base}

34.3.5 Pappus’s theorem

Rotate a plane figure of area A about an external axis. The volume of the solid of revolution is:

V = 2\pi \bar{y} \cdot A

where \bar{y} is the distance from the centroid to the axis of rotation.

Verification. Rotate a rectangle of width w and height h about a parallel axis at distance d from its near edge. The rectangle generates a hollow cylinder with outer radius d + w and inner radius d.

The centroid of the rectangle is at distance \bar{y} = d + w/2 from the axis. Pappus gives:

V = 2\pi \left(d + \frac{w}{2}\right) \cdot wh

The actual volume of the hollow cylinder:

V = \pi(d+w)^2 h - \pi d^2 h = \pi h \left[(d+w)^2 - d^2\right] = \pi h \cdot w(2d + w)

These are the same — confirmed.

34.4 Section C: Second moments of area

This is the material that every structural and mechanical engineering course assumes. If a chapter in this series earns the label “gating content,” this is it.

34.4.1 Why the second moment appears

A beam bends under load. The bending stress at a point in the cross-section is:

\sigma = \frac{M \cdot y}{I}

where M is the bending moment (the internal moment at that cross-section), y is the distance from the neutral axis to the point in question, and I is the second moment of area about the neutral axis.

The second moment of area, written I_x for the axis x, is:

I_x = \int y^2 \, dA

Read that as: “the integral of y squared times area element, accumulated over the whole cross-section.” It is not a moment of force — it is purely a geometric property of the cross-section shape.

A larger I means less stress for the same bending moment. This is why I-beams have that shape: most of the material (the flanges) sits far from the neutral axis (the web), maximising I without maximising weight.

The units of I are length to the fourth power: mm⁴ or m⁴. The numbers get large. An I-beam with 150 mm total depth typically has I in the range 10^6–10^7 mm⁴.

34.4.2 Standard results derived by integration

Rectangle (width b, height h, axis through the centroid):

Take the x-axis through the centroid. Each strip at height y has area dA = b \, dy:

I_x = \int_{-h/2}^{h/2} y^2 \cdot b \, dy = b \left[\frac{y^3}{3}\right]_{-h/2}^{h/2} = b \cdot \frac{2(h/2)^3}{3} = \frac{bh^3}{12}

\boxed{I_x = \frac{bh^3}{12}}

Rectangle (axis along the base, i.e., y = 0 at the base):

I_{\text{base}} = \int_0^h y^2 \cdot b \, dy = b \cdot \frac{h^3}{3} = \frac{bh^3}{3}

\boxed{I_{\text{base}} = \frac{bh^3}{3}}

Triangle (base b, height h, axis through centroid at h/3 from base):

The strip width at height y from the apex is b(h - y)/h (the cross-section narrows linearly to zero at the apex). Integrating y^2 times this width from 0 to h gives I about the base as bh^3/12. Shifting to the centroidal axis (at h/3) using the parallel axis theorem:

I_{\text{centroid}} = \frac{bh^3}{12} - \frac{bh}{2} \cdot \left(\frac{h}{3}\right)^2 = \frac{bh^3}{12} - \frac{bh^3}{18} = \frac{bh^3}{36}

\boxed{I_{\text{centroid}} = \frac{bh^3}{36}}

Solid circle (radius r, axis through centre):

Using circular strips (or converting to polar): each strip at radius \rho has dA = 2\pi\rho \, d\rho, and its contribution to I_x involves both x^2 and y^2 components. By symmetry I_x = I_y, and I_x + I_y = I_z where I_z = \int \rho^2 \, dA:

I_z = \int_0^r \rho^2 \cdot 2\pi\rho \, d\rho = 2\pi \cdot \frac{r^4}{4} = \frac{\pi r^4}{2}

So I_x = I_y = I_z / 2:

\boxed{I_x = \frac{\pi r^4}{4}}

Hollow circle (outer radius R, inner radius r):

Subtract the inner circle’s contribution:

\boxed{I_x = \frac{\pi(R^4 - r^4)}{4}}

Semicircle (radius r, axis along the diameter):

\boxed{I_x = \frac{\pi r^4}{8}}

Reference table:

Shape	Axis	I
Rectangle b \times h	Centroidal (x)	bh^3/12
Rectangle b \times h	Base	bh^3/3
Triangle base b, height h	Centroidal	bh^3/36
Solid circle radius r	Centre	\pi r^4/4
Hollow circle R, r	Centre	\pi(R^4 - r^4)/4
Semicircle radius r	Diameter	\pi r^4/8

34.4.3 The parallel axis theorem

You now know I for simple shapes about their own centroidal axes. But a composite section has parts whose centroids are not at the overall centroid. To shift I from one axis to a parallel one:

I = I_{\text{centroid}} + A d^2

where d is the perpendicular distance between the two parallel axes.

Why it works. Let the centroidal axis be at y = 0. Any parallel axis is at y = -d, so distances from it are y + d. Then:

I = \int (y + d)^2 \, dA = \int y^2 \, dA + 2d \int y \, dA + d^2 \int dA

The middle integral is \int y \, dA — which equals zero by definition of the centroid (the first moment about the centroidal axis is zero). The last integral is A. So:

I = I_{\text{centroid}} + Ad^2

The 2d\int y \, dA term always drops out when the first axis IS the centroidal axis. This is the key step — it only works if you start from the centroidal axis.

Worked example 1 — verification. Rectangle b \times h, axis along the base. Using the parallel axis theorem (centroid is at h/2 from the base):

I_{\text{base}} = \frac{bh^3}{12} + bh \cdot \left(\frac{h}{2}\right)^2 = \frac{bh^3}{12} + \frac{bh^3}{4} = \frac{bh^3 + 3bh^3}{12} = \frac{bh^3}{3} \checkmark

This confirms the direct result derived above.

Worked example 2 — numerical. A solid circle of radius 50 mm, axis 100 mm from its centre.

I_{\text{centroid}} = \frac{\pi (50)^4}{4} = \frac{\pi \times 6\,250\,000}{4} \approx 4.91 \times 10^6 \text{ mm}^4

A = \pi (50)^2 \approx 7854 \text{ mm}^2

I = 4.91 \times 10^6 + 7854 \times (100)^2 = 4.91 \times 10^6 + 78.54 \times 10^6 \approx 83.4 \times 10^6 \text{ mm}^4

The Ad^2 term dominates — moving the axis 100 mm away contributes 78.5 \times 10^6 mm⁴ versus the intrinsic 4.9 \times 10^6 mm⁴. Area far from the axis contributes disproportionately.

34.4.4 The perpendicular axis theorem

For a flat lamina (a thin plate lying in the xy-plane):

I_z = I_x + I_y

where I_z is the second moment about the axis perpendicular to the lamina through the same point. This follows directly from r^2 = x^2 + y^2 in the definition I_z = \int r^2 \, dA = \int (x^2 + y^2) \, dA.

For a rectangle b \times h: I_x = bh^3/12, I_y = hb^3/12, so I_z = bh(h^2 + b^2)/12. The theorem holds.

34.4.5 Radius of gyration

The radius of gyration k is defined by:

k = \sqrt{\frac{I}{A}}

Read it as: the distance from the axis at which all the area could be concentrated to produce the same I. Two cross-sections with the same A and the same I have the same k, regardless of their shape.

For a solid circle of radius r: k = \sqrt{\pi r^4/4 \div \pi r^2} = r/2.

For a thin hollow cylinder of radius r (all area at radius r): k = r.

The hollow cylinder has a larger radius of gyration for the same area — this is why thin-walled tubes are efficient structural members.

34.4.6 Full worked example: I-beam composite section

An I-beam has:

Top flange: 120 mm wide × 15 mm thick
Web: 10 mm wide × 150 mm tall
Bottom flange: 120 mm wide × 15 mm thick

Total depth: 15 + 150 + 15 = 180 mm. The section is symmetric, so the centroid is at the mid-depth: \bar{y} = 90 mm from the base.

Step 1 — I of each part about its own centroidal axis.

Flanges (each): I_{\text{own}} = \frac{120 \times 15^3}{12} = \frac{120 \times 3375}{12} = 33\,750 \text{ mm}^4

Web: I_{\text{own}} = \frac{10 \times 150^3}{12} = \frac{10 \times 3\,375\,000}{12} = 2\,812\,500 \text{ mm}^4

Step 2 — distance from each part’s centroid to the overall centroid.

Bottom flange centroid: 7.5 mm from base → d = 90 - 7.5 = 82.5 mm

Web centroid: 15 + 75 = 90 mm from base → d = 0 mm (it IS the centroid)

Top flange centroid: 15 + 150 + 7.5 = 172.5 mm from base → d = 172.5 - 90 = 82.5 mm

Step 3 — apply parallel axis theorem to each part.

Bottom flange: I = 33\,750 + (120 \times 15) \times 82.5^2 = 33\,750 + 1800 \times 6806.25 = 33\,750 + 12\,251\,250 = 12\,285\,000 \text{ mm}^4

Web: I = 2\,812\,500 + 0 = 2\,812\,500 \text{ mm}^4

Top flange: same as bottom flange by symmetry: 12\,285\,000 \text{ mm}^4

Step 4 — total I.

I = 12\,285\,000 + 2\,812\,500 + 12\,285\,000 = 27\,382\,500 \text{ mm}^4 \approx 2.74 \times 10^7 \text{ mm}^4

Interpretation. If this beam carries a bending moment M = 50 kN·m = 50 \times 10^6 N·mm, the maximum bending stress (at the outermost fibre, y_{\max} = 90 mm from the neutral axis) is:

\sigma_{\max} = \frac{M \cdot y_{\max}}{I} = \frac{50 \times 10^6 \times 90}{2.74 \times 10^7} \approx 164 \text{ N/mm}^2 = 164 \text{ MPa}

Structural steel yields at around 250 MPa, so this beam has a reasonable safety margin under this load.

34.4.7 Ship stability — a note

The righting moment of a ship after heeling by a small angle involves the metacentric height GM, which depends on I_{WP}/V — the second moment of the waterplane area about the ship’s centreline, divided by the underwater volume. A large waterplane I gives a large restoring force and a stiff (but sometimes too stiff) ride. Naval architects optimise I_{WP} just as structural engineers optimise beam I. The mathematics is the same; the application is a hull form rather than a cross-section.

Why the second moment of area matters in bending

The bending stress formula \sigma = My/I tells you that stress is proportional to M (how hard you’re bending it) and y (how far from the neutral axis), and inversely proportional to I (how area is distributed). For a given loading, you reduce stress by increasing I. The cheapest way to increase I is to move material further from the neutral axis — which is exactly what the flanges of an I-beam do.

34.5 Section D: Volumes of revolution — brief recap

You met volumes of revolution in Vol 5 Ch 3. This section places them in the engineering applications cluster and adds the shell method.

Disk method (rotation about the x-axis):

V = \pi \int_a^b \left[y(x)\right]^2 dx

Each slice perpendicular to the x-axis is a disk of radius y(x) and thickness dx.

Shell method (rotation about the y-axis):

V = 2\pi \int_a^b x \cdot y(x) \, dx

Each shell is a cylindrical surface at radius x, height y(x), and thickness dx.

Worked example — disk method. Find the volume of a cone with base radius r and height h, formed by rotating the line y = (r/h)x about the x-axis from x = 0 to x = h.

V = \pi \int_0^h \left(\frac{r}{h} x\right)^2 dx = \pi \frac{r^2}{h^2} \cdot \frac{h^3}{3} = \frac{\pi r^2 h}{3}

This is the standard formula for the volume of a cone.

Worked example — shell method. Find the volume of the solid formed by rotating y = x^2 between x = 0 and x = 2 about the y-axis.

V = 2\pi \int_0^2 x \cdot x^2 \, dx = 2\pi \int_0^2 x^3 \, dx = 2\pi \left[\frac{x^4}{4}\right]_0^2 = 2\pi \times 4 = 8\pi \approx 25.1

34.6 Where this goes

Second moments of area connect directly to the beam bending equation in Vol 7 (structural analysis). There, the bending moment M, shear force V, and deflection \delta all depend on EI — the product of Young’s modulus and I. Integrating the bending moment diagram twice, with I as a geometric factor, gives beam deflections. The composite section calculation you did above is exactly what engineers run before writing those integrals.

Centroids and first moments extend into Vol 7’s vector integral calculus. In three dimensions, the centre of mass of a solid body involves a triple integral, and the moment of inertia tensor — a 3×3 matrix whose entries are second moments and products of inertia — governs 3D rotation. The flat cross-section results from this chapter are the I_{xx}, I_{yy}, and I_{xy} components of that tensor restricted to two dimensions. RMS connects to Vol 7’s Fourier series: Parseval’s theorem states that the RMS of a periodic function equals the RMS computed from its Fourier coefficients. That is, the RMS is preserved when you decompose a waveform into sinusoids — a statement about energy that uses exactly the integral definition you have here.

Where this shows up

Structural engineers compute I for every beam and column in a building frame — the number determines both stress and deflection.
Mechanical engineers use centroids to find the balance point of rotating components, and second moments (as moments of inertia) to predict angular acceleration under applied torque.
Electrical engineers use RMS values for every AC power calculation — the rating on any transformer, motor, or cable is an RMS quantity.
Naval architects compute the second moment of the waterplane area to assess ship stability and design hull forms.
Aerospace engineers compute moments of inertia of aircraft and spacecraft for stability and control analysis.

34.7 Exercises

These are puzzles, not drills. Each has one clean numerical answer — the interesting part is setting up the integral correctly and tracking the units.

Code

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1. Find the mean value of i = 15 \sin \theta over 0 \leq \theta \leq \pi/2 (the first quarter of a sine cycle, as in the positive quarter-wave of a rectified current).

Code

makeStepperHTML(1, [
  { op: "Write the mean value formula",
    eq: "\\bar{i} = \\frac{1}{\\pi/2 - 0}\\int_0^{\\pi/2} 15\\sin\\theta\\,d\\theta" },
  { op: "Factor out constants",
    eq: "= \\frac{2}{\\pi} \\cdot 15 \\int_0^{\\pi/2} \\sin\\theta\\,d\\theta" },
  { op: "Integrate",
    eq: "= \\frac{30}{\\pi}\\left[-\\cos\\theta\\right]_0^{\\pi/2}" },
  { op: "Evaluate limits",
    eq: "= \\frac{30}{\\pi}(-\\cos(\\pi/2) + \\cos 0) = \\frac{30}{\\pi}(0 + 1) = \\frac{30}{\\pi}",
    note: "cos(π/2) = 0 and cos(0) = 1." },
  { op: "Check",
    eq: "\\bar{i} = 30/\\pi \\approx 9.55\\text{ A}. \\text{ Reasonable: the peak is 15 A, the mean over a quarter-sine is between 0 and 15.}" }
])

2. Find the RMS value of v = 100 \sin(314t) volts, where t is time in seconds. This represents a 50 Hz mains waveform with 100 V peak.

Code

makeStepperHTML(2, [
  { op: "Identify V₀",
    eq: "V_0 = 100\\text{ V}" },
  { op: "Apply the sine RMS formula",
    eq: "V_{\\text{rms}} = \\frac{V_0}{\\sqrt{2}}",
    note: "For any sinusoidal v = V₀ sin(ωt), the RMS is V₀/√2 regardless of frequency." },
  { op: "Compute",
    eq: "V_{\\text{rms}} = \\frac{100}{\\sqrt{2}} = \\frac{100\\sqrt{2}}{2} = 50\\sqrt{2}" },
  { op: "Check",
    eq: "50\\sqrt{2} \\approx 70.7\\text{ V}. \\text{ This is the effective voltage — a 70.7 V DC supply delivers the same power.}" }
])

3. A triangular lamina has base b = 6 cm and height h = 8 cm. Find the y-coordinate of its centroid, measured from the base.

Code

makeStepperHTML(3, [
  { op: "Write the centroid formula for a triangle",
    eq: "\\bar{y} = \\frac{h}{3}" },
  { op: "Substitute h = 8 cm",
    eq: "\\bar{y} = \\frac{8}{3} \\approx 2.67\\text{ cm from the base}" },
  { op: "Check",
    eq: "\\text{Centroid at } h/3 \\text{ from base, i.e., } 2h/3 \\text{ from apex. For h=8: } 8/3 \\approx 2.67\\text{ cm}. \\checkmark" }
])

4. Find I_x for a rectangle 40 mm wide × 60 mm tall: (a) about its centroidal axis, and (b) about its base.

Code

makeStepperHTML(4, [
  { op: "Write centroidal formula",
    eq: "I_{\\text{centroid}} = \\frac{bh^3}{12}" },
  { op: "Substitute b = 40, h = 60",
    eq: "I_{\\text{centroid}} = \\frac{40 \\times 60^3}{12} = \\frac{40 \\times 216\\,000}{12} = 720\\,000\\text{ mm}^4" },
  { op: "Write base formula",
    eq: "I_{\\text{base}} = \\frac{bh^3}{3}" },
  { op: "Substitute",
    eq: "I_{\\text{base}} = \\frac{40 \\times 60^3}{3} = \\frac{40 \\times 216\\,000}{3} = 2\\,880\\,000\\text{ mm}^4" },
  { op: "Verify with parallel axis theorem",
    eq: "I_{\\text{base}} = I_{\\text{centroid}} + Ad^2 = 720\\,000 + (40\\times60)\\times30^2 = 720\\,000 + 2\\,160\\,000 = 2\\,880\\,000\\text{ mm}^4\\checkmark",
    note: "d = h/2 = 30 mm is the distance from centroid to base." }
])

5. An I-beam has a web 100 mm × 8 mm, and equal top and bottom flanges each 60 mm × 10 mm. The section is symmetric. Find I about the centroidal axis (parallel to the flanges).

Code

makeStepperHTML(5, [
  { op: "Find total depth and centroid",
    eq: "\\text{Total depth} = 10 + 100 + 10 = 120\\text{ mm};\\quad \\bar{y} = 60\\text{ mm from base (by symmetry)}" },
  { op: "I of each flange about its own centroidal axis",
    eq: "I_{\\text{flange,own}} = \\frac{60 \\times 10^3}{12} = 5000\\text{ mm}^4" },
  { op: "Distance from each flange centroid to overall centroid",
    eq: "d_{\\text{bottom flange}} = 60 - 5 = 55\\text{ mm};\\quad d_{\\text{top flange}} = 55\\text{ mm (symmetric)}" },
  { op: "I of each flange about overall centroid (parallel axis)",
    eq: "I_{\\text{flange}} = 5000 + (60\\times10)\\times55^2 = 5000 + 1\\,815\\,000 = 1\\,820\\,000\\text{ mm}^4" },
  { op: "I of web about overall centroid (centroid of web coincides with overall centroid)",
    eq: "I_{\\text{web}} = \\frac{8 \\times 100^3}{12} = \\frac{8\\,000\\,000}{12} \\approx 666\\,667\\text{ mm}^4" },
  { op: "Total I",
    eq: "I = 2\\times 1\\,820\\,000 + 666\\,667 = 3\\,640\\,000 + 666\\,667 \\approx 4\\,307\\,000\\text{ mm}^4 \\approx 4.31\\times10^6\\text{ mm}^4" },
  { op: "Check",
    eq: "\\text{Flanges contribute } 3.64\\times10^6\\text{ mm}^4 \\text{ (84\\%), web } 0.67\\times10^6\\text{ mm}^4 \\text{ (16\\%)}. \\checkmark \\text{ Flanges dominate, as expected.}" }
])

6. A solid circle has diameter 50 mm (radius 25 mm). Find (a) I about its centroidal axis, and (b) I about a parallel axis 30 mm from the centre.

Code

makeStepperHTML(6, [
  { op: "Write I for solid circle",
    eq: "I_{\\text{centroid}} = \\frac{\\pi r^4}{4}" },
  { op: "Substitute r = 25 mm",
    eq: "I_{\\text{centroid}} = \\frac{\\pi \\times 25^4}{4} = \\frac{\\pi \\times 390\\,625}{4} \\approx 306\\,796\\text{ mm}^4" },
  { op: "Apply parallel axis theorem, d = 30 mm",
    eq: "I = I_{\\text{centroid}} + Ad^2" },
  { op: "Compute A",
    eq: "A = \\pi r^2 = \\pi \\times 625 \\approx 1963\\text{ mm}^2" },
  { op: "Compute Ad²",
    eq: "Ad^2 = 1963 \\times 900 \\approx 1\\,766\\,858\\text{ mm}^4" },
  { op: "Total",
    eq: "I = 306\\,796 + 1\\,766\\,858 \\approx 2\\,073\\,654\\text{ mm}^4 \\approx 2.07\\times10^6\\text{ mm}^4" },
  { op: "Check",
    eq: "\\text{The } Ad^2 \\text{ term (1.77}\\times10^6\\text{) is about 5.8× the centroidal I (0.31}\\times10^6\\text{).} \\checkmark \\text{ Moving the axis far out dominates.}" }
])