Draw The Shear And Bending Moment Diagrams For The Beam

To draw the shear andbending moment diagrams for the beam, you must first understand the internal forces that develop when external loads are applied. This article walks you through a systematic procedure, explains the underlying physics, and provides a worked example that you can adapt to any simply supported or cantilever beam. By following the steps and using the highlighted tips, you will be able to produce accurate diagrams that clearly illustrate how shear force and bending moment vary along the length of the beam.

## Understanding the Basics

What is Shear Force?

Shear force at a cross‑section of a beam is the algebraic sum of all vertical forces acting on either side of that section. It represents the internal resistance that prevents the beam from sliding apart.

What is Bending Moment?

Bending moment is the algebraic sum of moments about the same cross‑section caused by external forces. It causes the beam to bend and is a critical parameter in design because it determines the stress distribution within the material.

Why Draw Both Diagrams?

Drawing the shear force diagram (SFD) and bending moment diagram (BMD) together provides a visual representation of how internal forces change along the beam. These diagrams are essential for:

• Selecting appropriate beam dimensions and materials
• Identifying critical sections where failure is most likely
• Communicating design intentions to engineers and stakeholders

## Step‑by‑Step Procedure### 1. Identify the Loading Conditions* Point loads – Concentrated forces applied at specific locations.

• Uniformly distributed loads (UDL) – Loads spread evenly over a length, expressed as w (N/m).
• Triangular or trapezoidal loads – Varying intensity that may require integration.

2. Compute Support Reactions

Use equilibrium equations:

• ΣFy = 0 → Sum of vertical forces = 0
• ΣMx = 0 → Sum of moments about any point = 0

For a simply supported beam with a central point load P, each support carries P/2. For a cantilever with an end load P, the fixed support must provide a reaction of P and a moment of P·L.

3. Choose a Cutting Plane

Select a section x at a distance from the left support. The location of x determines which portion of the beam you will analyze.

4. Determine Shear Force (V) at the Cut

• Sum all vertical forces to the left (or right) of the cut.
• Include reactions, point loads, and the resultants of distributed loads that act on that side.

5. Determine Bending Moment (M) at the Cut

• Sum all moments about the cut caused by forces on one side. * Use the formula M = Σ(Mi) where each moment is the product of a force and its perpendicular distance from the cut.

6. Plot the Diagrams

• Shear Force Diagram – Plot V versus x. The curve is piecewise linear for point loads and parabolic for UDLs.
• Bending Moment Diagram – Plot M versus x. The curve is quadratic for point loads and cubic for UDLs.

7. Verify ResultsCheck that the SFD returns to zero at the far end of a simply supported beam, and that the BMD is zero at free ends of cantilevers. The area under the SFD should equal the change in BMD between two points.

## Detailed Example: Simply Supported Beam with a Central Point Load

Consider a beam of length L = 10 m simply supported at both ends, carrying a point load P = 8 kN at the mid‑span.

Compute Reactions

[ R_A = R_B = \frac{P}{2} = \frac{8}{2} = 4 \text{kN} ]

Shear Force Calculation

The SFD is a horizontal line at +4 kN from the left support to the load, then drops to –4 kN until the right support.

Bending Moment CalculationFor 0 ≤ x ≤ 5 m:

[ M(x) = R_A \cdot x = 4x]

For 5 m ≤ x ≤ 10 m:

[M(x) = R_A \cdot x - P \cdot (x-5) = 4x - 8(x-5) = 40 - 4x ]

The BMD is a triangular shape, peaking at the centre:

• At x = 5 m: (M_{max} = 4 \times 5 = 20 \text{kN·m})

Sketching the Diagrams

1. Draw a horizontal axis representing x from 0 to 10 m.
2. Plot the SFD: start at +4 kN, remain constant until x = 5 m, then jump to –4 kN and stay constant.
3. Plot the BMD: start at 0, rise linearly to 20 kN·m at x = 5 m, then fall linearly back to 0 at x = 10 m.

## Common Scenarios and Variations

Uniformly Distributed Load (UDL)

For a beam with a UDL w (N/m) over the entire span L:

• Reactions: (R_A = R_B = \frac{wL}{2})
• Shear force varies linearly: (V(x) = R_A - wx)
• Bending moment is parabolic: (M(x) = R_A x - \frac{w x^2}{2})

Triangular Load

When the load intensity varies linearly from zero at one end to w_max at the other, the resultant force acts at a distance of L/3 from the zero‑intensity end. The shear and moment equations become cubic and quadratic, respectively.

Cantilever Beam

A cantilever fixed at the left end and free at the right end:

• Shear force is constant and equal to the applied load (or resultant of distributed loads).
• Bending moment varies linearly from zero at the free end to a maximum at the fixed support: (

8. Cantilever‑type loading

When the structural element is fixed at one end and free at the other, the internal distribution of forces differs markedly from simply‑supported cases.

Point load at the free tip – Let a concentrated force P act downward at the far end of a cantilever of length L.

• Shear force is constant along the entire span and equals ‑P (the negative sign indicates that the internal shear opposes the external load).
• Bending moment varies linearly from zero at the free surface to a maximum of ‑P·L at the fixed support. The moment diagram is a straight line descending from the origin to the fixed root.

Uniformly distributed load over the whole length – If a constant load intensity w (N/m) acts downward from the fixed section to the free tip:

• Shear force decreases linearly from ‑wL at the fixed end to zero at the free end.
• Bending moment follows a quadratic variation, attaining a peak of ‑wL²/2 at the support. The diagram is a parabola opening downward.

Triangular or trapezoidal distributed loading – When the load intensity changes linearly along the cantilever, the shear diagram becomes a second‑order curve and the moment diagram a third‑order curve. The resultant force acts at a distance of 2L/3 from the fixed support, and the maximum moment can be obtained by integrating the shear expression once more.

Mixed loading – A common real‑world scenario combines a point load at some intermediate position with a UDL over the remaining portion. In such cases the shear diagram consists of two linear segments separated by a jump equal to the applied point load, while the moment diagram is assembled by adding the contributions from each load component.

9. Practical workflow for any beam configuration

1. Identify support conditions – pin, roller, fixed, or cantilever. 2. Select a coordinate origin – typically at the leftmost support or at the fixed root for cantilevers.
2. Determine reaction forces – apply equilibrium equations, remembering that a fixed support can furnish both vertical and moment reactions.
3. Write shear expressions – break the span into segments where the loading type changes; integrate the distributed load to obtain shear in each segment.
4. Integrate shear to obtain moment – each integration introduces a constant that is fixed by boundary conditions (zero moment at free ends, known moment at fixed ends).
5. Plot the diagrams – use the derived piecewise functions to draw step‑wise or continuous curves, marking key values such as maximum shear, peak moment, and locations of sign changes.
6. Validate – confirm that the shear diagram returns to zero at the free end of a cantilever or at the far support of a simply supported beam, and that the moment diagram satisfies the appropriate boundary values.

10. Concluding remarks

Constructing shear and bending‑moment diagrams is a systematic exercise that blends static equilibrium with calculus. By breaking the structure into regions of constant loading, writing the appropriate expressions, and applying the correct boundary conditions, engineers can predict how internal forces will redistribute under any prescribed set of external loads. Mastery of these diagrams not only safeguards against excessive deformation or failure but also provides a clear visual language for communicating design decisions to colleagues, clients, and inspectors. Ultimately, the ability to translate a physical loading scenario into accurate SFD and BMD representations is a cornerstone of structural analysis and a vital skill for anyone working in civil, mechanical, or mechanical‑systems engineering.