Beam Analysis: Shear and Moment Diagrams
Key Takeaways
- Shear V and bending moment M vary along a beam and are found by sectioning and applying ΣF = 0 and ΣM = 0.
- The fundamental relations are dV/dx = −w(x) and dM/dx = V(x): load drives shear, shear drives moment.
- A concentrated load makes the shear diagram jump by the load; a concentrated moment makes the moment diagram jump by that moment.
- Maximum bending moment occurs where shear crosses zero, or at a fixed support where a reaction moment exists.
- Simply supported, central load P: M_max = PL/4. Simply supported, UDL w: M_max = wL²/8. Cantilever, end load P: M_max = PL.
- The area under the shear diagram between two points equals the change in moment between them — a fast graphical check.
Why Shear and Moment Diagrams Matter
Before you can size a beam you must know the internal shear force V and bending moment M at every cross-section, because bending stress depends on M and transverse shear stress depends on V. A shear and moment diagram plots these along the beam's length so the critical section — where stress is largest — jumps out visually.
Sign Convention
| Quantity | Positive | Negative |
|---|---|---|
| Shear V | Rotates the element clockwise | Counterclockwise |
| Moment M | Concave-up bending ("smile," sagging) | Concave-down ("frown," hogging) |
Pick a convention and hold it for the whole problem. A positive moment puts the bottom fiber in tension.
The Calculus Relationships
The load w(x), shear V(x), and moment M(x) are linked by two derivatives:
dV/dx = −w(x) and dM/dx = V(x)
Integrating gives two graphical rules that solve most FE problems without calculus:
- Change in shear between two points = −(area under the load curve)
- Change in moment between two points = (area under the shear diagram)
These also fix the shape of each diagram:
| Loading region | Shear diagram | Moment diagram |
|---|---|---|
| No load | Constant (horizontal) | Linear (sloped) |
| Uniform load w | Linear (sloped) | Parabolic |
| Triangular load | Parabolic | Cubic |
| Concentrated force P | Vertical jump of P | Slope changes (kink) |
| Concentrated moment M₀ | No change | Vertical jump of M₀ |
Finding the Maximum Moment
Because dM/dx = V, the moment is stationary (a max or min) exactly where V = 0. So the procedure is: (1) find reactions, (2) build the shear diagram, (3) locate where V crosses zero, (4) compute M there by summing the shear area up to that point. Also check fixed supports, where a reaction moment can be the largest M.
Standard Beam Formulas (memorize the top three)
Simply supported — central point load P
| Quantity | Formula |
|---|---|
| Reactions | R_A = R_B = P/2 |
| Max shear | V = P/2 (at supports) |
| Max moment | M_max = PL/4 (at midspan) |
| Max deflection | δ_max = PL³/(48EI) |
Simply supported — uniform distributed load w
| Quantity | Formula |
|---|---|
| Reactions | R_A = R_B = wL/2 |
| Max shear | V = wL/2 (at supports) |
| Max moment | M_max = wL²/8 (at midspan) |
| Max deflection | δ_max = 5wL⁴/(384EI) |
Cantilever — end point load P
| Quantity | Formula |
|---|---|
| Reaction | R = P; wall moment = PL |
| Max shear | V = P (constant) |
| Max moment | M_max = PL (at the wall) |
| Max deflection | δ_max = PL³/(3EI) (free end) |
Cantilever — uniform distributed load w
| Quantity | Formula |
|---|---|
| Reaction | R = wL; wall moment = wL²/2 |
| Max moment | M_max = wL²/2 (at the wall) |
| Max deflection | δ_max = wL⁴/(8EI) (free end) |
Worked Example — Simply Supported Beam with UDL
A 8 m simply supported beam carries w = 5 kN/m over its full length. Find R_A, the location of M_max, and M_max.
- By symmetry, total load = wL = 5(8) = 40 kN, so R_A = R_B = 20 kN.
- Shear starts at +20 kN at the left support and decreases at −w = −5 kN/m, so it reaches zero at x = 20/5 = 4 m (midspan), as symmetry predicts.
- M_max = area of the shear triangle from 0 to 4 m = ½(4)(20) = 40 kN·m.
- Cross-check with the formula: M_max = wL²/8 = 5(8²)/8 = 5(64)/8 = 40 kN·m. ✓
The moment diagram is a parabola peaking at midspan and zero at both pinned supports.
Worked Example — Overhanging / Point-Load Check
For a simply supported beam with a single central load P = 12 kN over L = 6 m: reactions R_A = R_B = 6 kN. Shear is +6 kN from the left support to midspan, then jumps down by 12 kN to −6 kN at the load, and stays −6 kN to the right support. The moment rises linearly to M_max = PL/4 = 12(6)/4 = 18 kN·m at midspan, then falls linearly to zero — a triangular moment diagram. Notice the shear jump of 12 kN at the load equals the load, confirming the concentrated-force rule.
Common Traps
- Forgetting reactions first. Every diagram starts from correct support reactions; an error there propagates everywhere.
- Wrong M_max location. For an off-center point load or combined loading, M_max is wherever V = 0, not automatically at midspan.
- Distributed-load resultant. A UDL of intensity w over length a acts as a single force wa at the centroid (a/2 from either end) when computing reactions.
- Sign of cantilever moment. At a fixed wall the reaction moment is the maximum; treat it as negative (hogging) under downward load.
Quick reference
- dV/dx = −w; dM/dx = V; ΔM = area under V-diagram.
- M_max where V = 0 (or at a fixed support).
- SS central load: M = PL/4. SS UDL: M = wL²/8. Cantilever end load: M = PL; cantilever UDL: M = wL²/2.
A simply supported beam of length 8 m carries a uniform distributed load of 5 kN/m. What is the maximum bending moment?
Where does the maximum bending moment occur along a beam?
A 6 m simply supported beam carries a single central point load of 12 kN. What is the maximum bending moment?
A 3 m cantilever carries a uniform load w = 4 kN/m over its entire length. What is the maximum bending moment, and where?