Express the internal shear and moment in terms of x and then draw the shear and moment diagrams. You must use the 'BRUTE FORCE' method of sections, drawing your section FBDs for ea... Express the internal shear and moment in terms of x and then draw the shear and moment diagrams. You must use the 'BRUTE FORCE' method of sections, drawing your section FBDs for each region, and coming up with an exact formula for V(x) and M(x) in each region. Read more

Steps to Solve

1. Determine Reaction Forces

First, find the support reactions at points A and B. For a uniformly distributed load $w_0$ over the beam of length $L$, the total load is $W = w_0 \cdot L$.

Using static equilibrium:

• The sum of vertical forces: $R_A + R_B - W = 0$
• The sum of moments about point A: $\sum M_A = 0 \Rightarrow R_B \cdot L - W \cdot \frac{L}{2} = 0$

From the moment equation, we can solve for $R_B$: $$ R_B = \frac{W \cdot \frac{L}{2}}{L} = \frac{w_0 L}{2} $$

Now substituting $R_B$ back to find $R_A$: $$ R_A = W - R_B = w_0 L - \frac{w_0 L}{2} = \frac{w_0 L}{2} $$

2. Divide the Beam into Sections

Next, divide the beam into two sections to analyze separately:

• Section 1: from A to a point $x$ where $0 < x < L/2$
• Section 2: from $x$ to B where $L/2 < x < L$

3. Shear Force for Section 1 (A to x)

For Section 1:

• The internal shear force $V(x)$ can be determined by taking moments about point A.

Using equilibrium: $$ \sum F_y = 0: R_A - w_0 x - V(x) = 0 $$

Which gives: $$ V(x) = R_A - w_0 x = \frac{w_0 L}{2} - w_0 x $$

4. Bending Moment for Section 1 (A to x)

The internal moment $M(x)$ can be found from: $$ M(x) = R_A \cdot x - \frac{w_0 x^2}{2} $$

Substituting $R_A$: $$ M(x) = \frac{w_0 L}{2} \cdot x - \frac{w_0 x^2}{2} $$

5. Shear Force for Section 2 (x to B)

For Section 2: Assume the distance from A to B is $L$. The internal shear force $V(x)$ is given by: $$ \sum F_y = 0: V(x) + R_B - w_0 \cdot \left(L - x\right) = 0 $$

Substituting for $R_B$: $$ V(x) = -\frac{w_0 L}{2} + w_0 \cdot (L - x) $$

6. Bending Moment for Section 2 (x to B)

For the bending moment: $$ M(x) = M_{B} - V(x) \cdot (L - x) $$

Since the moment at point B is 0 (simply supported), we have: $$ M(x) = 0 - \left(-\frac{w_0 L}{2} + w_0 (L - x)\right)(L - x) $$

Calculating this will give the moment in terms of $x$.