This problem asks us to find the slope at the free end of a cantilever beam that has a varying flexural rigidity along its length and is subjected to a moment at the free end. The beam is fixed at one end and free at the other. Its flexural rigidity is different in the first half compared to the second half.

Understanding the Varying Rigidity Cantilever Beam

We have a cantilever beam of length \(L\). It has two distinct sections based on flexural rigidity:

• Section 1: From the fixed end (where \(x=0\)) to the midpoint (\(x=L/2\)), the flexural rigidity is \(EI\).
• Section 2: From the midpoint (\(x=L/2\)) to the free end (\(x=L\)), the flexural rigidity is \(EI/2\).

A moment \(M\) is applied at the free end (\(x=L\)). For a cantilever beam with a moment applied at the free end, the bending moment is constant throughout the beam. Let's assume the moment \(M\) creates a positive bending moment (causing tension on the bottom fibers). So, the bending moment at any section \(x\) from the fixed end is \(M(x) = M\).

Calculating Slope Using the Moment-Area Method

The moment-area method is a convenient way to determine slopes and deflections in beams. The first moment-area theorem states that the change in slope between two points on the elastic curve is equal to the area of the \(\frac{M}{EI}\) diagram between those two points.

For a cantilever beam, the slope at the fixed end (\(x=0\)) is zero. Therefore, the slope at the free end (\(\theta_L\)) is equal to the total area of the \(\frac{M}{EI}\) diagram from the fixed end to the free end.

First, let's construct the \(\frac{M}{EI}\) diagram. Since the bending moment \(M(x) = M\) is constant, we need to divide it by the respective flexural rigidity for each section.

• For Section 1 (\(0 \le x \le L/2\)): \(\frac{M}{EI(x)} = \frac{M}{EI}\).
• For Section 2 (\(L/2 \le x \le L\)): \(\frac{M}{EI(x)} = \frac{M}{EI/2} = \frac{2M}{EI}\).

The \(\frac{M}{EI}\) diagram consists of two rectangular areas:

• Area \(A_1\): From \(x=0\) to \(x=L/2\), height \(\frac{M}{EI}\), width \(\frac{L}{2}\).
• Area \(A_2\): From \(x=L/2\) to \(x=L\), height \(\frac{2M}{EI}\), width \(\frac{L}{2}\).

The total slope at the free end is the sum of these areas:

\(\theta_L = A_1 + A_2\)

\(A_1 = \left(\frac{M}{EI}\right) \times \left(\frac{L}{2}\right) = \frac{ML}{2EI}\)

\(A_2 = \left(\frac{2M}{EI}\right) \times \left(\frac{L}{2}\right) = \frac{ML}{EI}\)

\(\theta_L = \frac{ML}{2EI} + \frac{ML}{EI} = \frac{ML}{2EI} + \frac{2ML}{2EI} = \frac{3ML}{2EI}\)

Thus, the slope at the free end is \(\frac{3ML}{2EI}\).

Calculating Slope Using the Double Integration Method

We can also solve this problem using the differential equation of the elastic curve, \(EI \frac{d^2y}{dx^2} = M(x)\). Since \(EI\) varies, we need to write the equation for each section.

For Section 1 (\(0 \le x \le L/2\)):

\(EI \frac{d^2y_1}{dx^2} = M\)

\(\frac{d^2y_1}{dx^2} = \frac{M}{EI}\)

Integrating once gives the slope:

\(\frac{dy_1}{dx} = \int \frac{M}{EI} dx = \frac{Mx}{EI} + C_1\)

Integrating again gives the deflection:

\(y_1(x) = \int \left(\frac{Mx}{EI} + C_1\right) dx = \frac{Mx^2}{2EI} + C_1 x + C_2\)

For Section 2 (\(L/2 \le x \le L\)):

\(\frac{EI}{2} \frac{d^2y_2}{dx^2} = M\)

\(\frac{d^2y_2}{dx^2} = \frac{2M}{EI}\)

Integrating once gives the slope:

\(\frac{dy_2}{dx} = \int \frac{2M}{EI} dx = \frac{2Mx}{EI} + C_3\)

Integrating again gives the deflection:

\(y_2(x) = \int \left(\frac{2Mx}{EI} + C_3\right) dx = \frac{Mx^2}{EI} + C_3 x + C_4\)

We need to find the constants \(C_1, C_2, C_3, C_4\) using boundary and continuity conditions.

Boundary conditions at the fixed end (\(x=0\)):

• Slope at fixed end is zero: \(\frac{dy_1}{dx}(0) = 0\).
• Deflection at fixed end is zero: \(y_1(0) = 0\).

\(\frac{dy_1}{dx}(0) = \frac{M(0)}{EI} + C_1 = 0 \implies C_1 = 0\)

\(y_1(0) = \frac{M(0)^2}{2EI} + C_1(0) + C_2 = 0 \implies C_2 = 0\)

So, for Section 1: \(\frac{dy_1}{dx} = \frac{Mx}{EI}\) and \(y_1(x) = \frac{Mx^2}{2EI}\).

Continuity conditions at \(x=L/2\):

• Slope must be continuous: \(\frac{dy_1}{dx}(L/2) = \frac{dy_2}{dx}(L/2)\).
• Deflection must be continuous: \(y_1(L/2) = y_2(L/2)\).

\(\frac{M(L/2)}{EI} = \frac{2M(L/2)}{EI} + C_3\)

\(\frac{ML}{2EI} = \frac{ML}{EI} + C_3 \implies C_3 = \frac{ML}{2EI} - \frac{ML}{EI} = -\frac{ML}{2EI}\)

\(\frac{M(L/2)^2}{2EI} = \frac{M(L/2)^2}{EI} + C_3(L/2) + C_4\)

\(\frac{ML^2}{8EI} = \frac{ML^2}{4EI} + \left(-\frac{ML}{2EI}\right)\left(\frac{L}{2}\right) + C_4\)

\(\frac{ML^2}{8EI} = \frac{ML^2}{4EI} - \frac{ML^2}{4EI} + C_4\)

\(\frac{ML^2}{8EI} = C_4\)

So, for Section 2, the slope is:

\(\frac{dy_2}{dx} = \frac{2Mx}{EI} + C_3 = \frac{2Mx}{EI} - \frac{ML}{2EI}\)

We want the slope at the free end, which is at \(x=L\). Using the equation for Section 2:

\(\theta_L = \frac{dy_2}{dx}(L) = \frac{2M(L)}{EI} - \frac{ML}{2EI} = \frac{2ML}{EI} - \frac{ML}{2EI} = \frac{4ML}{2EI} - \frac{ML}{2EI} = \frac{3ML}{2EI}\)

Both methods yield the same result for the slope at the free end.

Revision Table: Key Concepts

Additional Information on Beam Deflection

Understanding beam deflection and slope is crucial in structural engineering to ensure that structures are safe and perform as intended without excessive deformation. The methods used here, Moment-Area and Double Integration, are fundamental techniques for indeterminate beams or beams with varying properties or complex loading.

The elastic curve is the shape of the deformed beam under load. The slope at any point on the elastic curve is \(\frac{dy}{dx}\), and the deflection is \(y\). The relationship between the bending moment \(M\), flexural rigidity \(EI\), and the curvature of the elastic curve is given by \(\frac{d^2y}{dx^2} = \frac{M}{EI}\) (for small deflections).

For beams with varying flexural rigidity, the term \(EI\) inside the integration or in the \(\frac{M}{EI}\) diagram calculation becomes a function of \(x\), as seen in this problem where it changes value at \(L/2\).

Other methods for finding beam deflections include the Conjugate Beam Method and energy methods like Castigliano's Theorem, especially useful for more complex problems or when deflections at specific points are needed.

The final answer is \(\frac{3ML}{2EI}\).