A prismatic, straight elastic, cantilever beam is subjected to a linearly distributed transverse load as shown below. If the beam length is L, Young’s modulus E, and area moment of inertia I, the magnitude of the maximum deflection is

The correct answer is qL⁴/30EI.

Problem Description & Coordinate System:
As shown in the diagram, we have a cantilever beam of length $L$ fixed at the left end and free at the right end. The coordinate system is set up with:
1. The origin at the free end (right end).
2. The horizontal axis $x$ directed to the left.
3. The vertical axis $y$ pointing upwards.
4. A linearly distributed load (triangular load) acting downwards, with intensity $q$ at the fixed support ($x=L$) and $0$ at the free end ($x=0$).

Thus, the load intensity at any point $x$ from the free end is given by:
$w\left(x\right)=\frac{qx}{L}$

Bending Moment Equation:
Consider a section at a distance $x$ from the free end. The total load acting on the segment of length $x$ to the right of the section is:
$W\left(x\right)=\frac{1}{2}·x·w\left(x\right)=\frac{q{x}^2}{2L}$

This load acts at a distance of $\frac{x}{3}$ from the section. Therefore, the bending moment $M\left(x\right)$ at this section is:
$M\left(x\right)=-W\left(x\right)·\frac{x}{3}=-\frac{q{x}^3}{6L}$

Euler-Bernoulli Beam Equation:
According to the Euler-Bernoulli beam deflection theory:
$EI\frac{d^{2}y}{d{x}^2}=M\left(x\right)=-\frac{q{x}^3}{6L}$

Integrating the equation once with respect to $x$ to find the slope:
$EI\frac{dy}{dx}=-\frac{q{x}^4}{24L}+C_1$

Applying the boundary condition for slope at the fixed end ($x=L$), where $\frac{dy}{dx}=0$:
$0=-\frac{q{L}^4}{24L}+C_1\Rightarrow C_1=\frac{q{L}^3}{24}$

Thus, the slope equation becomes:
$EI\frac{dy}{dx}=-\frac{q{x}^4}{24L}+\frac{q{L}^3}{24}$

Integrating a second time to find the deflection $y$:
$EIy=-\frac{q{x}^5}{120L}+\frac{q{L}^{3}x}{24}+C_2$

Applying the boundary condition for deflection at the fixed end ($x=L$), where $y=0$:
$0=-\frac{q{L}^5}{120L}+\frac{q{L}^4}{24}+C_2$
$0=-\frac{q{L}^4}{120}+\frac{5q{L}^4}{120}+C_2\Rightarrow C_2=-\frac{q{L}^4}{30}$

Thus, the deflection curve equation is:
$EIy\left(x\right)=-\frac{q{x}^5}{120L}+\frac{q{L}^{3}x}{24}-\frac{q{L}^4}{30}$

Maximum Deflection:
The maximum deflection occurs at the free end ($x=0$):
$EIy\left(0\right)=-\frac{q{L}^4}{30}\Rightarrow y\left(0\right)=-\frac{q{L}^4}{30EI}$

Taking the magnitude of the deflection:
$\left|{\delta }_{\text{max}}\right|=\frac{q{L}^4}{30EI}$