Understanding Shear Force in Simply Supported Beams

A simply supported beam rests on supports at both ends. When a load is applied to the beam, internal forces are generated, including shear force and bending moment. The shear force at any section of the beam is the algebraic sum of the vertical forces acting to the left or right of that section.

Calculating Reactions

First, we need to find the reactions at the supports. Let the beam be denoted as AB, with support A on the left and B on the right. The total length is 5 m. A point load P is applied at a distance of 2 m from the right end (B). This means the load is at a distance of \(5 - 2 = 3\) m from the left end (A). Let \(R_A\) be the reaction at A and \(R_B\) be the reaction at B.

For equilibrium, the sum of vertical forces must be zero, and the sum of moments about any point must be zero.

• Sum of vertical forces: \(R_A + R_B - P = 0 \implies R_A + R_B = P\)
• Taking moments about point A: \((R_B \times 5) - (P \times 3) = 0\)

From the moment equation:

\(5 R_B = 3P\)

\(R_B = \frac{3P}{5}\)

Substituting \(R_B\) into the vertical force equation:

\(R_A + \frac{3P}{5} = P\)

\(R_A = P - \frac{3P}{5}\)

\(R_A = \frac{5P - 3P}{5} = \frac{2P}{5}\)

So, the reactions are \(R_A = \frac{2P}{5}\) and \(R_B = \frac{3P}{5}\).

Determining Shear Force Along the Beam

The shear force changes value only at the locations of point loads and reactions. We can calculate the shear force in the sections of the beam:

• Consider a section between A and the point load (from x=0 to x=3 m from A): The shear force is equal to the reaction at A.
• SF\(_{left of load}\) = \(+R_A = +\frac{2P}{5}\)
• Consider a section between the point load and B (from x=3 m to x=5 m from A): The shear force is the reaction at A minus the point load P.
• SF\(_{right of load}\) = \(+R_A - P = +\frac{2P}{5} - P = \frac{2P - 5P}{5} = -\frac{3P}{5}\)

Finding Maximum Shear Force

The shear force values in the beam are \(+\frac{2P}{5}\) and \(-\frac{3P}{5}\). The maximum shear force is the largest absolute value of shear force along the beam.

• Absolute value of SF\(_{left of load}\) = \(\left|+\frac{2P}{5}\right| = \frac{2P}{5}\)
• Absolute value of SF\(_{right of load}\) = \(\left|-\frac{3P}{5}\right| = \frac{3P}{5}\)

Comparing the two values, \(\frac{3P}{5}\) is greater than \(\frac{2P}{5}\).

\(\frac{3}{5} > \frac{2}{5}\)

Therefore, the maximum shear force in the beam is \(\frac{3P}{5}\).