Two beams have the same cross-sectional area but different shapes: a solid square and a hollow tube with its material concentrated at a large radius. Which has the higher second moment of area, and why?
AThe solid square, because it has continuous material with no gaps to weaken it
BThey are equal, because second moment of area depends only on total cross-sectional area
CThe hollow tube, because the squared distance term disproportionately amplifies contributions from area far from the neutral axis
DThe hollow tube, but only if the wall thickness exceeds one-quarter of the outer diameter
The second moment of area I = ∫ y² dA weights each bit of area by the square of its distance from the neutral axis. Area at large radius contributes enormously more than the same area near the center. The hollow tube concentrates its material at large radius, so despite having the same total area as the solid square, it achieves a much higher I. This is exactly why structural engineers use hollow sections and I-beams: the same amount of steel resists bending far more effectively when placed at the extremes.
Question 2 Multiple Choice
A rectangular beam's second moment of area about its centroidal axis is I = bh³/12. Doubling the depth h versus doubling the width b produce the same increase in I.
ATrue — both dimensions appear in the formula and changing either doubles I
BFalse — depth h appears cubed, so doubling h increases I by a factor of 8, while doubling b only doubles I
CFalse — doubling b increases I more because width affects the moment arm directly
DTrue, but only for sections where b and h are initially equal
Depth h enters the formula cubed (h³), while width b enters linearly. Doubling h: I becomes b(2h)³/12 = 8bh³/12 — an eightfold increase. Doubling b: I becomes (2b)h³/12 — a doubling. This is why structural design prioritizes beam depth over width for bending resistance. A beam twice as deep is eight times as stiff in bending; a beam twice as wide is only twice as stiff. The cubic relationship is why floor joists are oriented with the long dimension vertical.
Question 3 True / False
For a composite cross-section made of multiple rectangles at different heights, the total second moment of area equals the sum of the individual centroidal moments bh³/12, with no additional correction needed.
TTrue
FFalse
Answer: False
The parallel-axis theorem is required: I = I_c + A·d², where I_c is the centroidal moment of each piece and d is the distance from that piece's centroid to the composite section's overall neutral axis. Simply summing the centroidal moments ignores the offset of each piece — a flange far from the neutral axis has a large A·d² term that dominates its contribution. Omitting this correction dramatically underestimates I and would produce dangerously unconservative structural designs.
Question 4 True / False
The reason I-beams (W-shapes) concentrate material in their top and bottom flanges rather than distributing it uniformly throughout the web is that flanges far from the neutral axis contribute disproportionately more to bending resistance.
TTrue
FFalse
Answer: True
This is the direct engineering application of the squared-distance weighting in I = ∫ y² dA. A unit of area in the flange, far from the neutral axis, contributes y² dA to I, where y is large — so its contribution is much larger than the same unit of area in the web near the neutral axis. The I-beam shape is the efficient solution: put the area where it does the most work (far from the neutral axis) and use just enough web to carry shear forces and connect the flanges. This is form following structural function.
Question 5 Short Answer
A rectangular beam has I = bh³/12 about its centroidal axis. Explain why h appears cubed while b appears only to the first power, and what this means practically for beam design.
Think about your answer, then reveal below.
Model answer: The formula comes from integrating I = ∫ y² dA over the rectangular cross-section. For a rectangle of width b and height h centered at the neutral axis, y ranges from −h/2 to h/2, and dA = b·dy. So I = ∫_{−h/2}^{h/2} y² · b dy = b · [y³/3]_{−h/2}^{h/2} = b · h³/12. Width b enters as a constant multiplier outside the integral — each horizontal strip has the same width. Depth h determines the range of integration AND appears squared inside it (y²), so it contributes cubically. Practically: to maximize bending resistance, increase depth, not width. Doubling depth gives 8× the resistance; doubling width gives only 2×.
This cubed relationship is why wooden floor joists are oriented with the long dimension vertical, why steel beams are much deeper than they are wide, and why the bottom chord of a bridge truss is placed as far below the neutral axis as practical. The structural efficiency gain from adding depth is so large that it almost always dominates the design decision over adding width.