Simply Supported Beam Triangular Load Analysis

This solution explains how to find the maximum bending moment for a simply supported beam subjected to a specific type of load distribution. The beam has a total length denoted by L. It is supported at both ends.

The load applied is a triangular load distribution. This means the load intensity varies linearly across the beam. Specifically, the load intensity is zero at both supports (at $x=0$ and $x=L$) and reaches its maximum value of w per unit length at the center of the beam (at $x=L/2$).

Calculating Support Reactions

Due to the symmetric nature of the beam and the applied load, the vertical reactions at the supports will be equal. Let the reactions at the left support (A) and the right support (B) be $R_A$ and $R_B$, respectively.

The total load on the beam is the area under the load distribution curve. The load distribution forms two triangles, each with a base of $L/2$ and a maximum height of $w$.

The area of one triangle is calculated as:

Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{L}{2} \times w = \frac{wL}{4}$

The total load on the beam is the sum of the areas of these two triangles:

Total Load = $2 \times \frac{wL}{4} = \frac{wL}{2}$

For static equilibrium, the sum of vertical forces must be zero. Therefore, $R_A + R_B = \text{Total Load}$.

Since the loading and supports are symmetric, the reactions are equal: $R_A = R_B$.

Substituting $R_B = R_A$ into the equilibrium equation:

$R_A + R_A = \frac{wL}{2}$

$2R_A = \frac{wL}{2}$

Solving for $R_A$ gives:

$R_A = \frac{wL}{4}$

Thus, the reactions at both supports are $R_A = R_B = \frac{wL}{4}$.

Deriving the Bending Moment Equation

To find the bending moment $M(x)$ at any section located at a distance $x$ from the left support (A), we consider the forces acting on the segment of the beam from the left support to the section at $x$.

The load intensity function $p(x)$ for the triangular load from $x=0$ to $x=L/2$ is given by the equation of a line passing through the points $(0, 0)$ and $(L/2, w)$. The slope is $m = \frac{w - 0}{L/2 - 0} = \frac{w}{L/2} = \frac{2w}{L}$.

The load intensity function is:

$p(x) = \frac{2w}{L}x$ (for $0 \le x \le L/2$)

The bending moment $M(x)$ at section $x$ is the algebraic sum of the moments caused by the support reaction and the distributed load acting on the segment from $0$ to $x$.

$M(x) = (\text{Moment due to } R_A) - (\text{Moment due to distributed load})$

$M(x) = R_A \times x - \int_{0}^{x} p(\xi)(x - \xi) d\xi$

Substitute the known values of $R_A$ and $p(\xi)$:

$M(x) = \left(\frac{wL}{4}\right)x - \int_{0}^{x} \left(\frac{2w\xi}{L}\right)(x - \xi) d\xi$

Expand the integral term:

$M(x) = \frac{wLx}{4} - \frac{2w}{L} \int_{0}^{x} (x\xi - \xi^2) d\xi$

Evaluate the integral:

$M(x) = \frac{wLx}{4} - \frac{2w}{L} \left[ \frac{x\xi^2}{2} - \frac{\xi^3}{3} \right]_{0}^{x}$

$M(x) = \frac{wLx}{4} - \frac{2w}{L} \left( \frac{x(x^2)}{2} - \frac{x^3}{3} \right)$

$M(x) = \frac{wLx}{4} - \frac{2w}{L} \left( \frac{x^3}{2} - \frac{x^3}{3} \right)$

Combine terms inside the parenthesis:

$M(x) = \frac{wLx}{4} - \frac{2w}{L} \left( \frac{3x^3 - 2x^3}{6} \right)$

$M(x) = \frac{wLx}{4} - \frac{2w}{L} \left( \frac{x^3}{6} \right)$

Simplify the expression:

$M(x) = \frac{wLx}{4} - \frac{wx^3}{3L}$

Finding the Location of Maximum Moment

The maximum bending moment occurs at the point where the shear force $V(x)$ is zero. The shear force $V(x)$ is found by differentiating the moment equation $M(x)$ with respect to $x$, or by calculating it as the difference between the upward reaction and the downward load up to section $x$.

$V(x) = R_A - \int_{0}^{x} p(\xi) d\xi$

$V(x) = \frac{wL}{4} - \int_{0}^{x} \frac{2w\xi}{L} d\xi$

$V(x) = \frac{wL}{4} - \frac{2w}{L} \left[ \frac{\xi^2}{2} \right]_{0}^{x}$

$V(x) = \frac{wL}{4} - \frac{2w}{L} \left( \frac{x^2}{2} \right)$

$V(x) = \frac{wL}{4} - \frac{wx^2}{L}$

Set the shear force $V(x)$ to zero to find the location of maximum bending moment:

$\frac{wL}{4} - \frac{wx^2}{L} = 0$

Rearrange the equation:

$\frac{wL}{4} = \frac{wx^2}{L}$

Cancel out $w$ from both sides:

$\frac{L}{4} = \frac{x^2}{L}$

Solve for $x^2$:

$x^2 = \frac{L^2}{4}$

Take the square root to find $x$:

$x = \sqrt{\frac{L^2}{4}} = \frac{L}{2}$

This calculation confirms that the maximum bending moment occurs at the center of the beam ($x = L/2$).

Calculating the Maximum Bending Moment Value

Substitute the value $x = L/2$ back into the bending moment equation $M(x)$:

$M_{max} = M\left(\frac{L}{2}\right) = \frac{wL(L/2)}{4} - \frac{w(L/2)^3}{3L}$

Simplify the terms:

$M_{max} = \frac{wL^2}{8} - \frac{w(L^3/8)}{3L}$

$M_{max} = \frac{wL^2}{8} - \frac{wL^3}{24L}$

$M_{max} = \frac{wL^2}{8} - \frac{wL^2}{24}$

To subtract these fractions, find a common denominator, which is 24:

$M_{max} = \frac{3 \times wL^2}{3 \times 8} - \frac{wL^2}{24}$

$M_{max} = \frac{3wL^2}{24} - \frac{wL^2}{24}$

$M_{max} = \frac{3wL^2 - wL^2}{24}$

$M_{max} = \frac{2wL^2}{24}$

Simplify the final expression:

$M_{max} = \frac{wL^2}{12}$

Thus, the maximum bending moment for a simply supported beam with a triangular load peaking at $w$ at the center is $\frac{wL^2}{12}$.