Structural Beam Bending & Stress Inversion (Mechanical Engineering) Solution

1. Zero-Moment Pinning Stations ($x$-Intercepts)

1. Name the Governing Formula: The Factorization of Polynomial Roots:
   $$M(x) = 0$$
2. Solve for $x$ on $[0, 6]$: Factor out the common term $2x$:
   $$2x^3 - 18x^2 + 40x = 0 \implies 2x(x^2 - 9x + 20) = 0$$
   Factor the quadratic term:
   $$2x(x - 4)(x - 5) = 0$$
   The roots within the girder span are:
   $$x = 0\text{ m}, \quad x = 4\text{ m}, \quad x = 5\text{ m}$$
   The bending moment passes through zero at the girder left support ($x = 0$), and at intermediate load crossover stations at $x = 4\text{ m}$ and $x = 5\text{ m}$.

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2. Shear Force & Extreme Bending Moments

1. Name the Governing Formula: The Beam Differential Equilibrium Relation ($V(x) = M'(x)$) and the First Derivative Test for Local Extrema: (Defined in First Derivative Test.)
2. Differentiate $M(x)$:
   $$M'(x) = \frac{d}{dx}[2x^3 - 18x^2 + 40x] = 6x^2 - 36x + 40$$
3. Locate Critical Numbers: Set $M'(x) = 0$:
   $$6x^2 - 36x + 40 = 0 \implies 3x^2 - 18x + 20 = 0$$
   Apply the Quadratic Formula:
   $$x = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(3)(20)}}{2(3)} = \frac{18 \pm \sqrt{324 - 240}}{6} = \frac{18 \pm \sqrt{84}}{6} = 3 \pm \frac{2\sqrt{21}}{6} = 3 \pm \frac{\sqrt{21}}{3}$$
   Numerical values ($\sqrt{21} \approx 4.5826$):
   $$x_1 = 3 - 1.5275 \approx 1.472\text{ m}, \qquad x_2 = 3 + 1.5275 \approx 4.528\text{ m}$$
   Both critical points lie within the girder span $[0, 6]$.
4. Sign Chart and Classification: Because $M'(x) = 6x^2 - 36x + 40$ is a parabola opening upward:
   • For $x \in [0, 1.472)$: $M'(x) > 0 \implies M$ is increasing.
   • For $x \in (1.472, 4.528)$: $M'(x) < 0 \implies M$ is decreasing.
   • For $x \in (4.528, 6]$: $M'(x) > 0 \implies M$ is increasing.
   • At $x_1 \approx 1.472\text{ m}$: $M'$ changes $+ \to -\implies$ Local Maximum Bending Moment:
     $$M(1.472) \approx 2(1.472)^3 - 18(1.472)^2 + 40(1.472) \approx 6.376 - 39.006 + 58.880 \approx +26.25\text{ kN}\cdot\text{m}$$
   • At $x_2 \approx 4.528\text{ m}$: $M'$ changes $- \to +\implies$ Local Minimum Bending Moment:
     $$M(4.528) \approx 2(4.528)^3 - 18(4.528)^2 + 40(4.528) \approx 185.70 - 369.04 + 181.12 \approx -2.22\text{ kN}\cdot\text{m}$$

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3. Second Derivative & Inflection Points

1. Name the Governing Formula: The Second Derivative Inflection Criterion:
   $$M''(x) = 0 \quad \text{with concavity sign transition}$$
   (Defined in Concavity and the Second Derivative Test.)
2. Compute $M''(x)$:
   $$M''(x) = \frac{d}{dx}[6x^2 - 36x + 40] = 12x - 36 = 12(x - 3)$$
3. Solve $M''(x) = 0$:
   $$12(x - 3) = 0 \implies x = 3\text{ m}$$
4. Concavity Sign Chart on $[0, 6]$:
   • On $[0, 3)$: $M''(x) < 0 \implies$ Concave Downward ($\cap$).
   • On $(3, 6]$: $M''(x) > 0 \implies$ Concave Upward ($\cup$).
5. Inflection Point Coordinates: Because $M''(x)$ transitions strictly from negative to positive across $x = 3$:
   $$M(3) = 2(3)^3 - 18(3)^2 + 40(3) = 54 - 162 + 120 = +12\text{ kN}\cdot\text{m}$$
   The inflection point is located at $(3, 12)$.

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4. Structural Engineering Significance

In structural mechanics, the derivative of bending moment is shear force ($V(x) = M'(x)$), and the second derivative is the distributed load intensity ($w(x) = -M''(x)$). The inflection point at $x = 3\text{ m}$ indicates where the rate of change of shear force passes through zero ($\frac{dV}{dx} = 0$), identifying the location of steepest shear gradient reversal along the girder web. Placing transverse web stiffeners near this inflection boundary prevents localized diagonal web buckling under heavy rolling crane loads.