Strongest Rectangular Beam from a Circular Log A rectangular beam is to be cut from a circular log. For maximum flexural strength (maximum section modulus), what should be the ratio of width to depth (width/depth)?

Difficulty: Medium

Correct Answer: 0.707

Explanation:


Introduction / Context:
When a rectangle is inscribed in a given circle (e.g., a beam cut from a circular log), its dimensions are constrained by the circle's diameter. To maximize flexural strength, we maximize the section modulus Z of the rectangle subject to the circular constraint.


Given Data / Assumptions:

  • Rectangle of width b and depth d fits inside a circle of fixed diameter.
  • For a rectangular section, Z ∝ b * d^2 about the centroidal axis parallel to the width.
  • Geometric constraint: b^2 + d^2 = constant^2 (the circle).


Concept / Approach:
We maximize f = b * d^2 subject to b^2 + d^2 = C^2. Using optimization (e.g., Lagrange multiplier) gives a fixed ratio between b and d for the optimum.


Step-by-Step Solution:
Let f = b * d^2, g = b^2 + d^2 - C^2 = 0∂f/∂b = d^2, ∂f/∂d = 2 b d∂g/∂b = 2 b, ∂g/∂d = 2 dSet gradients proportional: d^2 = 2 λ b and 2 b d = 2 λ dFrom the second: b = λ; from the first: d^2 = 2 b^2 ⇒ d = √2 * bTherefore width/depth = b/d = 1/√2 ≈ 0.707


Verification / Alternative check:
The result is classical: the depth should be √2 times the width to maximize Z, giving b/d = 1/√2 ≈ 0.707.


Why Other Options Are Wrong:

  • 0.303, 0.404, 0.505, 0.606: do not satisfy the optimization condition b/d = 1/√2.


Common Pitfalls:
Maximizing area b * d instead of section modulus b * d^2 leads to a different ratio and an incorrect answer for strength design.


Final Answer:
0.707

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