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Structural Beam Bending Moment Optimization (Civil & Construction Engineering) Solution

Problem Overview

We analyze the internal bending moment distribution M(x)=−2x3+15x2−24x+50M(x) = -2x^3 + 15x^2 - 24x + 50 (kN⋅m\text{kN}\cdot\text{m}) across a 6-meter span (x∈[0,6]x \in [0, 6]). We verify the Extreme Value Theorem, locate critical points, execute the Closed Interval Method, and determine structural steel rebar placement to maximize safety while minimizing material costs.


Step-by-Step Mathematical Solution

Step 1: Verify Extreme Value Theorem Hypotheses

The governing theorem is the Extreme Value Theorem (EVT):

A continuous function on a closed interval [a,b][a, b] is guaranteed to achieve an absolute maximum and an absolute minimum. (This theorem is defined in Extreme Value Theorem.)

  1. Continuity: M(x)M(x) is a cubic polynomial. Polynomials are continuous for all real numbers R\mathbb{R}; hence M(x)M(x) is continuous on [0,6][0, 6].
  2. Closed Domain: The domain [0,6][0, 6] explicitly includes both structural support wall boundaries (x=0x = 0 and x=6x = 6).
  3. Conclusion: Both hypotheses are satisfied. Absolute extrema are guaranteed to exist.

Step 2: Differentiate and Locate Interior Critical Numbers

The governing concept is Critical Numbers:

M′(x)=0orM′(x) does not existM'(x) = 0 \quad \text{or} \quad M'(x) \text{ does not exist}
(This concept is defined in Critical Numbers and Critical Points.)

  1. Differentiate using the Power Rule:
    M′(x)=ddx[−2x3+15x2−24x+50]=−6x2+30x−24M'(x) = \frac{d}{dx}[-2x^3 + 15x^2 - 24x + 50] = -6x^2 + 30x - 24
    (In beam mechanics, the derivative of the bending moment is the internal shear force: V(x)=dMdxV(x) = \frac{dM}{dx}. Thus, critical points of the bending moment occur precisely where shear force is zero).
  2. Set M′(x)=0M'(x) = 0 to find stationary points:
    −6(x2−5x+4)=0-6(x^2 - 5x + 4) = 0
    Factoring the quadratic:
    −6(x−1)(x−4)=0  ⟹  x=1 mandx=4 m-6(x - 1)(x - 4) = 0 \implies x = 1\text{ m} \quad \text{and} \quad x = 4\text{ m}
  3. Check singular points: M′(x)M'(x) is a polynomial, defined everywhere.
  4. Candidate Filtering: Both x=1 mx = 1\text{ m} and x=4 mx = 4\text{ m} lie strictly within the interior interval (0,6)(0, 6). Both are retained.

Step 3: Execute Candidate Height Evaluation (Closed Interval Method)

The governing algorithm is the Closed Interval Method: Evaluate the original bending moment function M(x)M(x) at all surviving interior critical numbers and boundary endpoints:

  1. Left Support Wall (x=0 mx = 0\text{ m}):
    M(0)=−2(0)3+15(0)2−24(0)+50=50 kN⋅mM(0) = -2(0)^3 + 15(0)^2 - 24(0) + 50 = 50\text{ kN}\cdot\text{m}
  2. First Critical Point (x=1 mx = 1\text{ m}):
    M(1)=−2(1)3+15(1)2−24(1)+50=−2+15−24+50=39 kN⋅mM(1) = -2(1)^3 + 15(1)^2 - 24(1) + 50 = -2 + 15 - 24 + 50 = 39\text{ kN}\cdot\text{m}
  3. Second Critical Point (x=4 mx = 4\text{ m}):
    M(4)=−2(4)3+15(4)2−24(4)+50=−2(64)+15(16)−96+50M(4) = -2(4)^3 + 15(4)^2 - 24(4) + 50 = -2(64) + 15(16) - 96 + 50
    M(4)=−128+240−96+50=66 kN⋅mM(4) = -128 + 240 - 96 + 50 = 66\text{ kN}\cdot\text{m}
  4. Right Support Wall (x=6 mx = 6\text{ m}):
    M(6)=−2(6)3+15(6)2−24(6)+50=−2(216)+15(36)−144+50M(6) = -2(6)^3 + 15(6)^2 - 24(6) + 50 = -2(216) + 15(36) - 144 + 50
    M(6)=−432+540−144+50=14 kN⋅mM(6) = -432 + 540 - 144 + 50 = 14\text{ kN}\cdot\text{m}

Step 4: Compare Heights and Identify Extrema

Comparing the four candidate values {50,39,66,14}\{50, 39, 66, 14\}:

  • Absolute Maximum Bending Moment Value: 66 kN⋅m66\text{ kN}\cdot\text{m} at location x=4 metersx = 4\text{ meters}.
  • Absolute Minimum Bending Moment Value: 14 kN⋅m14\text{ kN}\cdot\text{m} at location x=6 metersx = 6\text{ meters}.

Step 5: Civil Engineering Design Optimization

  1. Structural Safety and Rebar Clustering: The absolute peak bending moment of 66 kN⋅m66\text{ kN}\cdot\text{m} represents the point of maximum tensile flexural stress in the concrete beam fibers. If structural failure occurs under dynamic loading, cracking will initiate at the bottom tensile face at x=4 metersx = 4\text{ meters}. The civil engineer uses this calculus calculation to specify dense groupings of high-yield steel rebar stirrups and longitudinal tension bars concentrated in the zone around x=4 metersx = 4\text{ meters}.
  2. Material Cost Savings vs. Uniform Sizing: At the right support wall (x=6 metersx = 6\text{ meters}), the internal bending moment drops to its minimum of 14 kN⋅m14\text{ kN}\cdot\text{m}—nearly 80% lower than the maximum. If an engineer lazily specified uniform maximum rebar reinforcement across the entire 6-meter span, the project would waste tons of expensive steel reinforcement in low-stress zones. By tapering rebar density according to the moment profile, construction costs are minimized while satisfying structural building codes.