Statics Quiz: Area Moment Of Inertia Standard Shapes

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Area Moment Of Inertia Standard ShapesQuestion 1 of 4

A student uses the formula I = πd⁴/64 to calculate the area moment of inertia for a circular cross-section and obtains 201 in⁴. Later, the student discovers that the given dimension was actually the radius, not the diameter. What should the correct moment of inertia be?

I = 804 \text{ in}^4

I = 3216 \text{ in}^4

I = 1608 \text{ in}^4

I = 50.25 \text{ in}^4

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Statics Quiz

Statics Quiz: Area Moment Of Inertia Standard Shapes

Practice Area Moment Of Inertia Standard Shapes in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Area Moment Of Inertia Standard Shapes, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$I = 804 \text{ in}^4$
$I = 3216 \text{ in}^4$ (correct answer)
$I = 1608 \text{ in}^4$
$I = 50.25 \text{ in}^4$

Explanation: The student used I = πd⁴/64 = 201, so d⁴ = 201 × 64/π = 4104. If d was actually the radius r, then the true diameter is 2r = 2d. The correct moment of inertia is I = π(2d)⁴/64 = π(16d⁴)/64 = 16πd⁴/64 = 16 × 201 = 3216 in⁴. Alternatively, since I ∝ d⁴ and the diameter is actually twice what was used, the moment of inertia increases by a factor of 2⁴ = 16.

Question 2

A thin rectangular plate with dimensions $a = 12$ in by $b = 8$ in has a mass uniformly distributed. If the plate is oriented with the longer side horizontal, what is the ratio of the moment of inertia about the horizontal centroidal axis to the moment of inertia about the vertical centroidal axis?

$\frac{I_x}{I_y} = 0.44$ (correct answer)
$\frac{I_x}{I_y} = 2.25$
$\frac{I_x}{I_y} = 1.50$
$\frac{I_x}{I_y} = 0.67$
$\frac{I_x}{I_y} = 1.78$

Explanation: When analyzing moments of inertia for rectangular plates, remember that the moment of inertia depends on how the mass is distributed relative to the axis of rotation. For a thin rectangular plate, mass farther from the axis contributes more significantly to the moment of inertia. For a rectangular plate with uniform mass distribution, the moments of inertia about the centroidal axes are:

$I_x = \frac{mb^2}{12}$ (about horizontal axis)
$I_y = \frac{ma^2}{12}$ (about vertical axis)

Since the plate is oriented with the longer side (

a = 12

in) horizontal and the shorter side (

b = 8

in) vertical, you can calculate:

\frac{I_x}{I_y} = \frac{\frac{mb^2}{12}}{\frac{ma^2}{12}} = \frac{b^2}{a^2} = \frac{8^2}{12^2} = \frac{64}{144} = 0.44

This confirms answer A is correct. Answer B (2.25) represents the inverse ratio

\frac{a^2}{b^2}

, which would occur if you confused which dimension goes with which axis. Answer C (1.50) equals

\frac{a}{b} = \frac{12}{8}

, suggesting you used the dimensions directly instead of their squares. Answer D (0.67) equals

\frac{b}{a} = \frac{8}{12}

, also using dimensions rather than their squares. Study tip: Always remember that moment of inertia formulas involve the square of the perpendicular distance. For rectangular plates, the dimension that matters is perpendicular to the axis of rotation—horizontal axis uses the vertical dimension, and vice versa.

Question 3

A right triangle has base $b = 6$ in and height $h = 8$ in, with the right angle at the origin and legs along the positive x and y axes. What is the moment of inertia about an axis parallel to the base passing through the centroid?

$I = 85.3$ in⁴ (correct answer)
$I = 96.0$ in⁴
$I = 64.0$ in⁴
$I = 128.0$ in⁴
$I = 42.7$ in⁴

Explanation: When calculating the moment of inertia for composite or basic shapes, you need to find the centroid location first, then apply the appropriate formula about the desired axis. For this right triangle with base

b = 6

in and height

h = 8

in, the centroid is located at

(\bar{x}, \bar{y}) = (b/3, h/3) = (2, 8/3)

in from the origin. The moment of inertia of a triangle about its base is

I_{base} = \frac{bh^3}{12}

. Since we want the moment of inertia about an axis parallel to the base passing through the centroid, we use the base formula:

I = \frac{bh^3}{12} = \frac{6 \times 8^3}{12} = \frac{6 \times 512}{12} = \frac{3072}{12} = 256

in⁴. Wait - this seems too high. The key insight is that this formula gives the moment about the base edge, but we need it about the centroidal axis parallel to the base. For a triangle, the moment of inertia about a centroidal axis parallel to the base is

I = \frac{bh^3}{36} = \frac{6 \times 8^3}{36} = \frac{3072}{36} = 85.3

in⁴. Answer A (

85.3

in⁴) is correct. Answer B (

96.0

in⁴) likely uses an incorrect formula or factor. Answer C (

64.0

in⁴) might result from using

h^2

instead of

h^3

. Answer D (

128.0

in⁴) could come from doubling the wrong intermediate calculation. Study tip: Memorize that for triangles, the centroidal moment of inertia uses the denominator 36, while the base moment uses 12. Always verify which axis the problem asks for.

Question 4

A thin-walled circular tube with outer radius R = 4 in and wall thickness t = 0.25 in is compared to a solid circular shaft with the same cross-sectional area. What is the ratio of the area moment of inertia of the tube to that of the solid shaft?

$2.85$
$3.21$ (correct answer)
$2.47$
$3.68$

Explanation: Tube: Outer radius R = 4 in, inner radius r = 3.75 in. Area = π(4² - 3.75²) = π(16 - 14.06) = 6.09 in². I_tube = π(R⁴ - r⁴)/4 = π(256 - 197.75)/4 = 45.7 in⁴. Solid shaft with same area: πr² = 6.09, so r = 1.39 in. I_solid = πr⁴/4 = π(1.39⁴)/4 = 14.2 in⁴. Ratio = 45.7/14.2 = 3.21.