This question asks us to find the product of the deflection and the flexural rigidity (EI) at the free end of a cantilever beam subjected to a specific loading condition.

Understanding the Cantilever Beam Setup

We have a cantilever beam of length L. A cantilever beam is fixed at one end and free at the other. The load is distributed linearly along the beam, starting from zero at the free end and increasing to a maximum value of w per unit length at the fixed end. This type of loading is often referred to as a triangular load.

Coordinate System and Load Intensity

Let's set up a coordinate system. We'll measure the distance x from the free end of the cantilever. So, x=0 corresponds to the free end, and x=L corresponds to the fixed end.

The load intensity, w'(x), varies linearly from 0 at x=0 to w at x=L. We can express the load intensity function as:

$$ w'(x) = \frac{w}{L}x $$

Calculating the Bending Moment

To find the deflection, we first need the bending moment equation, M(x). Consider a section of the beam at a distance x from the free end. The bending moment at this section is caused by the loads acting on the portion of the beam between the free end (0) and the section (x).

Consider a small segment of the beam of length ds at a distance s from the free end. The load on this segment is dw = w'(s) ds = $\frac{w}{L}s ds$. This load acts at distance s.

The moment caused by this small load element dw at the section x is dM = dw × (distance) = dw × (x - s).

The total bending moment M(x) at section x is the integral of these moments from the free end (s=0) to the section (s=x):

$$ M(x) = \int_{0}^{x} (x-s) dw = \int_{0}^{x} (x-s) \frac{w}{L}s ds $$

$$ M(x) = \frac{w}{L} \int_{0}^{x} (sx - s^2) ds $$

$$ M(x) = \frac{w}{L} \left[ \frac{s^2x}{2} - \frac{s^3}{3} \right]_{0}^{x} $$

$$ M(x) = \frac{w}{L} \left( \frac{x^2 \cdot x}{2} - \frac{x^3}{3} \right) = \frac{w}{L} \left( \frac{x^3}{2} - \frac{x^3}{3} \right) $$

$$ M(x) = \frac{w}{L} \left( \frac{3x^3 - 2x^3}{6} \right) = \frac{w}{L} \frac{x^3}{6} = \frac{wx^3}{6L} $$

Assuming downward load causes negative bending moment, the equation is $M(x) = -\frac{wx^3}{6L}$.

Integrating to Find Deflection

The relationship between bending moment and the elastic curve (deflection y) is given by the Euler-Bernoulli beam equation:

$$ EI \frac{d^2y}{dx^2} = M(x) $$

Substituting the bending moment equation:

$$ EI \frac{d^2y}{dx^2} = -\frac{wx^3}{6L} $$

Now, we integrate this equation twice with respect to x to find the deflection y(x).

First integration (for slope, $\frac{dy}{dx}$):

$$ EI \frac{dy}{dx} = \int -\frac{wx^3}{6L} dx = -\frac{w}{6L} \frac{x^4}{4} + C_1 = -\frac{wx^4}{24L} + C_1 $$

Second integration (for deflection, y):

$$ EI y = \int \left(-\frac{wx^4}{24L} + C_1\right) dx = -\frac{w}{24L} \frac{x^5}{5} + C_1 x + C_2 $$

$$ EI y = -\frac{wx^5}{120L} + C_1 x + C_2 $$

Applying Boundary Conditions

We need to determine the constants of integration, C₁ and C₂, using the boundary conditions for a cantilever beam:

1. At the fixed end (x=L), the slope is zero: $\frac{dy}{dx} = 0$.
2. At the fixed end (x=L), the deflection is zero: $y = 0$.

Using the first boundary condition ($EI \frac{dy}{dx} = 0$ at x=L):

$$ 0 = -\frac{wL^4}{24L} + C_1 $$

$$ 0 = -\frac{wL^3}{24} + C_1 \implies C_1 = \frac{wL^3}{24} $$

Now, substitute C₁ back into the deflection equation:

$$ EI y = -\frac{wx^5}{120L} + \frac{wL^3}{24} x + C_2 $$

Using the second boundary condition ($EI y = 0$ at x=L):

$$ 0 = -\frac{wL^5}{120L} + \frac{wL^3}{24} (L) + C_2 $$

$$ 0 = -\frac{wL^4}{120} + \frac{wL^4}{24} + C_2 $$

$$ 0 = \frac{-wL^4 + 5wL^4}{120} + C_2 = \frac{4wL^4}{120} + C_2 $$

$$ 0 = \frac{wL^4}{30} + C_2 \implies C_2 = -\frac{wL^4}{30} $$

So, the final deflection equation is:

$$ EI y = -\frac{wx^5}{120L} + \frac{wL^3}{24} x - \frac{wL^4}{30} $$

Calculating the Product at the Free End

We need the product of deflection and flexural rigidity at the free end, which corresponds to x=0. This is the value of EI y at x=0.

Substitute x=0 into the deflection equation:

$$ EI y_{x=0} = -\frac{w(0)^5}{120L} + \frac{wL^3}{24} (0) - \frac{wL^4}{30} $$

$$ EI y_{x=0} = 0 + 0 - \frac{wL^4}{30} $$

$$ EI y_{x=0} = -\frac{wL^4}{30} $$

The deflection itself at the free end is $y_{free\_end} = \frac{wL^4}{30EI}$. The product of deflection and flexural rigidity is $EI \times y_{free\_end}$. Since the deflection is downward, it's often represented as a positive magnitude in formulas. Therefore, the product is:

$$ EI \times \delta_{max} = EI \times \frac{wL^4}{30EI} = \frac{wL^4}{30} $$

Conclusion

The product of the deflection and flexural rigidity at the free end of the cantilever beam is $\frac{wL^4}{30}$.