Beam Load and Reaction Analysis

The problem describes a simply supported beam of span L. The load is distributed such that the first half of the span (0 to L/2) has a total weight W, and the second half (L/2 to L) has a total weight 2W. We need to find the maximum shear force.

First, let's determine the intensity of the distributed loads:

• Intensity for the first half ($w_1$): $w_1 \times \frac{L}{2} = W \implies w_1 = \frac{2W}{L}$
• Intensity for the second half ($w_2$): $w_2 \times \frac{L}{2} = 2W \implies w_2 = \frac{4W}{L}$

The total load on the beam is $W_{total} = W + 2W = 3W$. Let the reactions at the supports A (left) and B (right) be $R_A$ and $R_B$, respectively.

Calculating Support Reactions

Using the principle of moments about support A:

$ R_B \times L = \left(W \times \frac{L}{4}\right) + \left(2W \times \frac{3L}{4}\right) $

The term $W \times \frac{L}{4}$ represents the moment due to the load W (whose centroid is at L/4). The term $2W \times \frac{3L}{4}$ represents the moment due to the load 2W (whose centroid is at L/2 + L/4 = 3L/4).

$ R_B \times L = \frac{WL}{4} + \frac{6WL}{4} = \frac{7WL}{4} $

$ R_B = \frac{7W}{4} = 1.75W $

Now, using the equilibrium equation for vertical forces ($\sum F_y = 0$):

$ R_A + R_B = W_{total} $

$ R_A + 1.75W = 3W $

$ R_A = 3W - 1.75W = 1.25W $

Determining Shear Force Values

The shear force $V(x)$ is calculated as follows:

• At the left support (x=0): Shear Force $V_A$ = $R_A = 1.25W$.
• In the first half span ($0 < x < L/2$): $V(x) = R_A - w_1 \times x = 1.25W - \frac{2W}{L} \times x$.
• At the midpoint (x=L/2): $V(L/2) = 1.25W - \frac{2W}{L} \times \frac{L}{2} = 1.25W - W = 0.25W$.
• In the second half span ($L/2 < x < L$): $V(x) = R_A - W - w_2 \times (x - L/2) = 1.25W - W - \frac{4W}{L} \times (x - L/2) = 0.25W - \frac{4W}{L} \times (x - L/2)$.
• At the right support (x=L): $V(L) = 0.25W - \frac{4W}{L} \times (L - L/2) = 0.25W - \frac{4W}{L} \times \frac{L}{2} = 0.25W - 2W = -1.75W$. This equals $-R_B$, as expected.

Maximum Shear Force Identification

The shear forces at critical points are:

• $V_A = 1.25W$
• $V_{L/2} = 0.25W$
• $V_B = -1.75W$

The maximum absolute shear force is the largest magnitude among these values:

$ \text{Max Shear Force} = \max(|1.25W|, |0.25W|, |-1.75W|) = 1.75W $