Understanding the Shear Force Diagram for a Simply Supported Beam with UDL

Let's analyze the shear force diagram for a simply supported beam carrying a uniformly distributed load (UDL) of intensity \(w\) per unit length over its entire span.

A simply supported beam is supported at both ends, typically with one pin support and one roller support, allowing it to carry vertical loads.

A uniformly distributed load is a load that is spread evenly over a section of the beam, meaning the load intensity \(w\) is constant along that section.

Calculating Reactions

For a simply supported beam of length \(L\) subjected to a UDL \(w\) over its full length, the total load is \(wL\). Due to symmetry, the vertical reactions at both supports are equal.

Let \(R_A\) and \(R_B\) be the reactions at supports A and B respectively.

Total downward force = \(wL\)

Total upward force = \(R_A + R_B\)

For vertical equilibrium, \(R_A + R_B = wL\). Since the loading is symmetric, \(R_A = R_B\).

Therefore, \(R_A = R_B = \frac{wL}{2}\).

Shear Force Calculation Along the Beam

The shear force at any section of the beam is the algebraic sum of the vertical forces to the left or right of that section.

Let's consider a section at a distance \(x\) from the left support (A).

The forces to the left of this section are:

• The upward reaction at A: \(R_A = \frac{wL}{2}\)
• The downward distributed load over the length \(x\): \(w \times x\)

Considering upward forces as positive and downward forces as negative, the shear force \(V(x)\) at distance \(x\) from A is:

\(V(x) = R_A - wx = \frac{wL}{2} - wx\)

Plotting the Shear Force Diagram

The equation \(V(x) = \frac{wL}{2} - wx\) shows that the shear force varies linearly with \(x\). This means the shear force diagram will be a straight line.

• At the left support (x=0): \(V(0) = \frac{wL}{2} - w(0) = \frac{wL}{2}\). The shear force starts with a positive value equal to the reaction.
• At the right support (x=L): \(V(L) = \frac{wL}{2} - w(L) = \frac{wL}{2} - wL = -\frac{wL}{2}\). The shear force ends with a negative value equal to the reaction at B (but acting upwards, hence the sign change when looking from the left).

The shear force changes linearly from \(+\frac{wL}{2}\) at one end to \(-\frac{wL}{2}\) at the other end.

The point where the shear force is zero can be found by setting \(V(x) = 0\):

\(\frac{wL}{2} - wx = 0\)

\(wx = \frac{wL}{2}\)

\(x = \frac{L}{2}\)

This means the shear force is zero at the mid-span of the beam.

Shape of the Shear Force Diagram

The shear force diagram is a graph of \(V(x)\) versus \(x\). Since \(V(x)\) is a linear function of \(x\), the graph is a straight line with a negative slope (\(-w\)).

The diagram starts at \(x=0\) with a value of \(+\frac{wL}{2}\) and ends at \(x=L\) with a value of \(-\frac{wL}{2}\), passing through zero at \(x=\frac{L}{2}\).

This linear variation results in the shear force diagram being composed of two parts:

• From \(x=0\) to \(x=\frac{L}{2}\), the shear force decreases from \(+\frac{wL}{2}\) to 0. This forms a triangle above the x-axis.
• From \(x=\frac{L}{2}\) to \(x=L\), the shear force decreases from 0 to \(-\frac{wL}{2}\). This forms a triangle below the x-axis.

When plotted against the horizontal beam axis (x-axis) and the vertical shear force axis (y-axis), these two parts form triangles. Since the axes are perpendicular, these triangles are right-angled triangles.

Both triangles have a base of length \(\frac{L}{2}\) and a height of \(\frac{wL}{2}\).

Therefore, the shear force diagram for a simply supported beam carrying a uniformly distributed load consists of two right-angled triangles.

Revision Table: Simply Supported Beam UDL Properties

Additional Information: Related Beam Loading Cases

Understanding shear force diagrams is crucial in structural analysis. The shape of the diagram depends heavily on the type of loading and supports. Here are some other common cases:

• Simply Supported Beam with a Point Load at Mid-span: The shear force diagram consists of two rectangles of equal height but opposite signs, with a sudden drop at the point load.
• Cantilever Beam with a Point Load at the Free End: The shear force is constant along the beam (equal to the load) and forms a rectangle.
• Cantilever Beam with UDL: The shear force starts at zero at the free end and increases linearly to \(wL\) at the fixed end, forming a single right-angled triangle.

Comparing these cases helps illustrate how different loading conditions result in distinct shear force diagram shapes, from rectangles (constant shear) to triangles (linearly varying shear) to parabolas (for bending moment with UDL, but not shear force in this case).