Shear Force & Bending Moment Diagrams
Step-by-Step Construction — Sign Convention, Relationships, Beam Types & Solved Problems
Last Updated: March 2026
📌 Key Takeaways
- Shear Force (V): The algebraic sum of all transverse forces on one side of a section.
- Bending Moment (M): The algebraic sum of moments of all forces on one side of a section about that section.
- Key relationship: dM/dx = V — the slope of the BMD at any point equals the shear force at that point.
- Maximum bending moment occurs where V = 0 (shear force changes sign).
- Point load → jump in SFD, kink in BMD. UDL → linear SFD, parabolic BMD.
- Always find support reactions first before drawing diagrams.
1. What Are SFD and BMD?
When a beam carries external loads, internal forces develop at every cross-section to maintain equilibrium. At any section, cutting the beam into two parts reveals two types of internal forces:
- Shear force (V): The internal force acting perpendicular to the beam’s axis. It resists the tendency of one part of the beam to slide vertically relative to the other.
- Bending moment (M): The internal moment (couple) that resists the tendency of the beam to bend. It is what causes the beam to curve and develop bending stresses.
The Shear Force Diagram (SFD) plots V along the beam length. The Bending Moment Diagram (BMD) plots M along the beam length. Together, they reveal where the beam is most heavily loaded — and therefore where it is most likely to fail.
Drawing SFD and BMD accurately is one of the most essential skills in SOM. It is tested in virtually every GATE ME paper and is the prerequisite for bending stress calculations.
2. Sign Convention
| Quantity | Positive (+) | Negative (−) |
|---|---|---|
| Shear Force | Left side: upward force. Right side: downward force. (Clockwise rotation tendency) | Left side: downward force. Right side: upward force. (Anticlockwise rotation tendency) |
| Bending Moment | Sagging — beam bends concave up (smile shape). Top in compression, bottom in tension. | Hogging — beam bends concave down (frown shape). Top in tension, bottom in compression. |
Consistency is critical. Choose one sign convention and stick with it throughout the problem. The convention above is used in most Indian textbooks and GATE papers.
3. Relationships: Load ↔ Shear ↔ Moment
These differential relationships connect the load diagram, SFD, and BMD:
Fundamental Relationships
dV/dx = −w(x) — Rate of change of shear = negative of distributed load intensity
dM/dx = V(x) — Rate of change of moment = shear force
d²M/dx² = −w(x) — Second derivative of moment = negative load
Practical consequences:
| Loading | SFD Shape | BMD Shape |
|---|---|---|
| No load (w = 0) | Horizontal (constant V) | Straight line (linear M) |
| UDL (w = constant) | Straight line (linear V) | Parabola (quadratic M) |
| Triangular load (w varies linearly) | Parabola | Cubic curve |
| Point load P | Sudden jump of magnitude P | Kink (change in slope) of magnitude P |
| Point moment M₀ | No change in SFD | Sudden jump of magnitude M₀ |
Critical rule: The bending moment is maximum (or minimum) where the shear force is zero or changes sign. This is the most important point for bending stress design.
4. Step-by-Step Procedure
- Draw the free body diagram of the entire beam showing all loads, supports, and reactions.
- Find support reactions using equilibrium equations: ΣFy = 0 and ΣM = 0.
- Start from the left end and move rightward. At each point, calculate the running sum of vertical forces (= V) and the running sum of moments (= M).
- At point loads: V jumps by the load value. Upward forces add; downward forces subtract.
- Under UDL: V changes linearly (slope = −w). M changes parabolically.
- At point moments: M jumps by the moment value. V is unaffected.
- Mark key values: V and M at supports, load points, and where V = 0 (location of Mmax).
- Check: V should return to zero at the right end (if all reactions are correct). M should be zero at simple supports and free ends.
5. Standard Cases — Quick Reference
| Beam & Loading | Max Shear (Vmax) | Max Moment (Mmax) | Location of Mmax |
|---|---|---|---|
| Simply supported, central point load P | P/2 | PL/4 | At midspan |
| Simply supported, UDL w over full span | wL/2 | wL²/8 | At midspan |
| Cantilever, point load P at free end | P | PL | At fixed end |
| Cantilever, UDL w over full span | wL | wL²/2 | At fixed end |
| Simply supported, point load P at distance a from left | Pb/L or Pa/L | Pab/L | Under the load |
Exam tip: Memorise these five standard cases. Most exam problems are either one of these or a combination (superposition).
6. Worked Examples
Example 1: Simply Supported Beam with Central Point Load
Problem: A simply supported beam of span 6 m carries a point load of 12 kN at midspan. Draw SFD and BMD. Find maximum shear force and maximum bending moment.
Solution
Reactions: By symmetry, RA = RB = 12/2 = 6 kN (upward)
SFD: Starting from left: V = +6 kN (constant from A to midspan). At midspan, V jumps by −12 kN → V = 6 − 12 = −6 kN (constant from midspan to B).
BMD: At A: M = 0. At midspan: M = 6 × 3 = 18 kN·m. At B: M = 0. Linear from A to midspan, linear from midspan to B.
Vmax = 6 kN, Mmax = 18 kN·m = PL/4 = 12 × 6/4 = 18 ✓
Example 2: Simply Supported Beam with UDL
Problem: A beam of span 8 m carries a UDL of 5 kN/m over the entire span. Find Mmax.
Solution
RA = RB = wL/2 = 5 × 8/2 = 20 kN
Mmax = wL²/8 = 5 × 8²/8 = 5 × 64/8 = 40 kN·m at midspan
SFD: Linear from +20 kN at A to −20 kN at B, passing through zero at midspan.
BMD: Parabolic, maximum at midspan (where V = 0).
Example 3: Cantilever with Point Load at Free End
Problem: A cantilever beam of length 4 m has a point load of 10 kN at the free end. Find the SFD, BMD, and maximum bending moment.
Solution
SFD: V = −10 kN (constant along entire length — the only transverse force is the 10 kN at the free end).
BMD: At free end: M = 0. At fixed end: M = −10 × 4 = −40 kN·m (hogging). Linear variation.
Mmax = 40 kN·m (hogging, at the fixed support)
7. Common Mistakes Students Make
- Wrong support reactions: If reactions are incorrect, every subsequent value in the SFD and BMD will be wrong. Always verify: ΣFy = 0 AND ΣM = 0 (check moments about a different point as verification).
- Inconsistent sign convention: Switching between sign conventions mid-problem is the fastest way to get wrong diagrams. Define your convention at the start and stick to it.
- Forgetting that Mmax occurs where V = 0: Many students just compute M at the loads and supports. But under UDL or combined loading, the maximum moment can occur between load points — specifically where the shear force is zero or changes sign.
- Drawing BMD as linear under UDL: Under UDL, the SFD is linear and the BMD is parabolic (quadratic). Drawing a straight-line BMD under UDL is incorrect.
- Not checking end conditions: For a simply supported beam: M = 0 at both supports. For a cantilever: V = 0 and M = 0 at the free end. If your diagram violates these, there is an error.
8. Frequently Asked Questions
What is a shear force diagram?
A shear force diagram (SFD) shows how the internal shear force varies along the length of a beam. It is constructed by calculating the algebraic sum of all transverse forces to one side of each section. The SFD reveals the maximum shear force and its location, which is needed for shear stress calculations and for finding where the bending moment is maximum.
What is a bending moment diagram?
A bending moment diagram (BMD) shows how the internal bending moment varies along the beam. It is constructed by summing moments of all forces on one side of each section. The maximum bending moment determines the maximum bending stress (via σ = My/I), making the BMD the most critical diagram for beam design.
What is the relationship between load, shear force, and bending moment?
The relationships are: dV/dx = −w (shear changes at a rate equal to the negative of load intensity) and dM/dx = V (bending moment changes at a rate equal to the shear force). This means: no load → constant shear, linear moment. UDL → linear shear, parabolic moment. Maximum moment occurs where V = 0.