Cantilever Beam Elastic Deflection & Tangent Stiffness (Civil Engineering) Solution

Problem Overview

We calculate the slope $\theta(x) = \frac{dy}{dx}$ and mid-span tangent line equation for an engineered structural steel cantilever beam governed by the fourth-order elastic deflection equation:

y(x) = \frac{w_0}{24EI}\left(x^4 - 4Lx^3 + 6L^2x^2\right)

over the domain

0 \le x \le L

, where

C = \frac{w_0}{24EI}

is the constant flexural rigidity scaling coefficient and

L = 6.0\text{ m}

Step-by-Step Mathematical Solution

Step 1: Compute the First Derivative $\frac{dy}{dx}$

The governing mathematical rules are the Constant Multiple Rule and Power Rule:

\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x), \qquad \frac{d}{dx}[x^n] = n x^{n-1}

(These rules are defined in Power Rule.)

Treating $C = \frac{w_0}{24EI}$ and $L$ as constants, differentiate $y(x)$ term-by-term with respect to $x$ :

\theta(x) = \frac{dy}{dx} = C \cdot \frac{d}{dx}\left[x^4 - 4Lx^3 + 6L^2x^2\right]

\frac{dy}{dx} = C \cdot \left[4x^3 - 4L(3x^2) + 6L^2(2x)\right]

\frac{dy}{dx} = C \cdot \left[4x^3 - 12Lx^2 + 12L^2x\right] = 4Cx\left(x^2 - 3Lx + 3L^2\right)

Substituting back $C = \frac{w_0}{24EI}$ :

\theta(x) = \frac{w_0}{6EI}\left(x^3 - 3Lx^2 + 3L^2x\right)

Step 2: Evaluate Boundary Slopes at Support ( $x = 0$ ) and Tip ( $x = L$ )

At the Wall Support ( $x = 0$ ): $\theta(0) = \frac{w_0}{6EI}\left(0^3 - 3L(0)^2 + 3L^2(0)\right) = 0$ This rigorously confirms the structural boundary condition for a rigid clamp: zero slope at the fixed root.
At the Free Cantilever Tip ( $x = L$ ): $\theta(L) = \frac{w_0}{6EI}\left(L^3 - 3L(L^2) + 3L^2(L)\right) = \frac{w_0}{6EI}\left(L^3 - 3L^3 + 3L^3\right) = \frac{w_0 L^3}{6EI}$ This is the classical structural engineering formula for the maximum tip rotation of a cantilever beam under uniform load.

Step 3: Formulate the Tangent Line at Mid-Span ( $x = L/2 = 3.0\text{ m}$ )

The governing formula is the Point-Slope Equation of a Tangent Line:

y - y_1 = m(x - x_1), \qquad \text{where } m = \left.\frac{dy}{dx}\right|_{x_1}

(This formula is defined in Definition of the Derivative.)

Evaluate the deflection height $y_1$ at $x_1 = L/2$ : $y(L/2) = C \left[\left(\frac{L}{2}\right)^4 - 4L\left(\frac{L}{2}\right)^3 + 6L^2\left(\frac{L}{2}\right)^2\right]$ $y(L/2) = C \left[\frac{L^4}{16} - 4L\left(\frac{L^3}{8}\right) + 6L^2\left(\frac{L^2}{4}\right)\right] = C L^4 \left[\frac{1}{16} - \frac{1}{2} + \frac{3}{2}\right] = C L^4 \left[\frac{1}{16} + 1\right] = \frac{17}{16} C L^4$
Evaluate the tangent slope $m$ at $x_1 = L/2$ : $m = \theta(L/2) = C \left[4\left(\frac{L}{2}\right)^3 - 12L\left(\frac{L}{2}\right)^2 + 12L^2\left(\frac{L}{2}\right)\right]$ $m = C \left[4\left(\frac{L^3}{8}\right) - 12L\left(\frac{L^2}{4}\right) + 6L^3\right] = C L^3 \left[\frac{1}{2} - 3 + 6\right] = \frac{7}{2} C L^3$
Write the Tangent Line Equation:
$y - \frac{17}{16} C L^4 = \frac{7}{2} C L^3 \left(x - \frac{L}{2}\right)$
For $L = 6.0\text{ m}$ :
- $L^3 = 216$ , $L^4 = 1296$ .
- $y_1 = \frac{17}{16} C (1296) = 1377 C$ .
- $m = \frac{7}{2} C (216) = 756 C$ . $y - 1377 C = 756 C (x - 3) \implies y = 756 C x - 891 C$

Engineering Conclusion

In structural civil engineering, computing the instantaneous slope $\frac{dy}{dx}$ is just as critical as computing total vertical displacement. When building multi-story cantilever decks, architectural facade glazing (such as floor-to-ceiling glass curtain walls) attaches directly to cantilever floor slabs. While the slab may safely withstand vertical bending stress, excessive rotational slope $\theta(x)$ imparts destructive in-plane racking and shear strain onto the rigid glass framing. By formulating tangent line equations, structural engineers evaluate the angular clearance envelopes at mid-span and tip connections, ensuring rubber expansion gaskets and slotted slip-joints absorb beam rotation without shattering architectural glass.