Beam Bending and Deflection
Reactions, shear and moment diagrams, bending stress, deflection formulas for the common load cases, and how much deflection is acceptable.
Beams, supports and loads
A beam is a member that is loaded sideways, across its length, so that it bends. Shelves, shafts, machine frames, axles, bridges and floor joists are all beams. The way a beam is supported changes the answer enormously:
| Support | Symbol | Reactions | Allows |
|---|---|---|---|
| Pin | Triangle | A force in two directions | Rotation |
| Roller | Circle on a line | One force, perpendicular to the surface | Rotation and sliding |
| Fixed (built-in) | Hatched wall | Two forces and a moment | Nothing |
A beam on a pin at one end and a roller at the other is simply supported; a beam fixed at one end and free at the other is a cantilever; fixed at both ends it is fixed–fixed (built-in). Loads are point loads P (N), distributed loads w (N per unit length, with a total of w L for a uniform load), or moments.
Reactions, shear and moment diagrams
The first job is to find the reactions at the supports from equilibrium: the sum of vertical forces is zero and the sum of moments about any point is zero. Then the shear force V(x) (the net vertical force on one side of a cut) and the bending moment M(x) (the net moment on one side) are drawn along the beam. The two are related: dM/dx = V and dV/dx = −w, so the moment is largest where the shear passes through zero.
Worked example: reactions of a loaded beam
- A 2.0 m simply supported beam carries a 3.0 kN point load 0.5 m from the left support A and a uniform load of 2.0 kN/m over its whole length (total 4.0 kN).
- Moments about A: RB × 2.0 = 3.0 × 0.5 + 4.0 × 1.0 gives RB = 2.75 kN. Vertical equilibrium: RA = 7.0 − 2.75 = 4.25 kN.
- The moment under the point load is M = RA × 0.5 − w × 0.5² ÷ 2 = 1.88 kN·m. The maximum moment occurs where V = 0, which for this loading is a little to the right of the point load (check by plotting V along the beam).
Bending stress
σ = M y ÷ I σmax = M c ÷ I = M ÷ S S = I ÷ c
M is the bending moment at the section, y the distance from the neutral axis, I the second moment of area of the section about the neutral axis (mm⁴) and S the section modulus (mm³). The stress is zero at the neutral axis and largest at the outer fiber, at distance c. To keep the stress low, make S large: material far from the neutral axis is what counts, which is why tubes and I-beams are efficient (see Section Properties and Structural Profiles for the formulas). Maximum moments for the common cases:
| Case | Maximum moment Mmax | Where |
|---|---|---|
| Simply supported, central load P | P L ÷ 4 | Mid-span |
| Simply supported, uniform load w | w L² ÷ 8 | Mid-span |
| Simply supported, load P at distance a from A | P a (L − a) ÷ L | Under the load |
| Cantilever, end load P | P L | At the fixed end |
| Cantilever, uniform load w | w L² ÷ 2 | At the fixed end |
| Fixed–fixed, central load P | P L ÷ 8 | At the ends and the middle |
| Fixed–fixed, uniform load w | w L² ÷ 12 | At the ends |
Shear stress in a beam
The shear force V also produces a shear stress, τ = V Q ÷ (I b), where Q is the first moment of the area beyond the point and b the width there. It is largest at the neutral axis and zero at the outer faces. For a rectangular section the maximum is 1.5 V ÷ A (1.5 times the average); for a solid round section 4/3 V ÷ A; for an I-beam most of the shear is carried by the web, about V ÷ (web area). Shear is usually small in long beams but governs short, heavily loaded ones (a bracket, a deep girder near the support).
Example: a 1 000 N shear on a 40 × 60 mm rectangle gives τmax = 1.5 × 1 000 ÷ 2 400 = 0.63 MPa, negligible against the yield strength of steel.
Deflection: the common load cases
A beam must not only be strong but stiff: a shelf that bends visibly, a machine frame that sags, a shaft that deflects at its gears (see Shafts, Keys and Couplings) has failed even though it has not broken. The curve of the beam follows E I y″ = M. The standard results (L is the span, E the modulus, I the second moment of area):
| Case | Maximum deflection δ | Slope at the support or tip θ |
|---|---|---|
| Simply supported, central load P | P L³ ÷ (48 E I) | P L² ÷ (16 E I) at the ends |
| Simply supported, uniform load w | 5 w L⁴ ÷ (384 E I) | w L³ ÷ (24 E I) at the ends |
| Cantilever, end load P | P L³ ÷ (3 E I) | P L² ÷ (2 E I) at the tip |
| Cantilever, uniform load w | w L⁴ ÷ (8 E I) | w L³ ÷ (6 E I) at the tip |
| Fixed–fixed, central load P | P L³ ÷ (192 E I) | zero at the ends |
| Fixed–fixed, uniform load w | w L⁴ ÷ (384 E I) | zero at the ends |
Notice that deflection depends on L³ or L⁴ and on E I, which is why stiffening a beam by making it deeper (I rises with the cube of the depth for a rectangle) is so effective, and why a long, thin member sags so much. Use these as the building blocks of any loading with superposition (below).
Worked examples
Worked example: a steel tube as a simply supported beam
- A 40 × 40 × 3 mm square steel tube (E = 200 GPa) spans 1 200 mm, simply supported, carrying 1.0 kN at mid-span.
- Section properties: inside dimension 34 mm. I = (40⁴ − 34⁴) ÷ 12 = 101 972 mm⁴ and S = I ÷ 20 = 5 099 mm³.
- Stress: M = P L ÷ 4 = 1 000 × 1 200 ÷ 4 = 300 000 N·mm; σ = M ÷ S = 300 000 ÷ 5 099 = 58.8 MPa: far below the yield of mild steel (about 250 MPa).
- Deflection: δ = P L³ ÷ (48 E I) = 1 000 × 1 200³ ÷ (48 × 200 000 × 101 972) = 1.77 mm, i.e. L ÷ 680. Against the common limit L ÷ 360 = 3.33 mm it passes. Stiffness, not strength, would govern if the load were larger.
- If the ends were fixed (welded rigidly into the supports) the deflection would be one quarter: 0.44 mm, and the end moment P L ÷ 8 = 150 000 N·mm would be carried at the ends (a stress of 29.4 MPa).
Worked example: an aluminum bar as a cantilever
- A 20 × 10 mm aluminum flat bar (E = 69 GPa) projects 250 mm from a wall and carries 50 N at the tip.
- Flat (the 10 mm side vertical, I = b h³ ÷ 12 = 20 × 10³ ÷ 12 = 1 667 mm⁴): δ = P L³ ÷ (3 E I) = 50 × 250³ ÷ (3 × 69 000 × 1 667) = 2.26 mm, with σ = M c ÷ I = 37.5 MPa.
- On edge (the 20 mm side vertical, I = 6 667 mm⁴): δ = 0.57 mm, four times stiffer, with σ = 18.8 MPa. Turning the bar on its edge multiplies the stiffness by (h2 ÷ h1)² = (20 ÷ 10)² = 4 for free, a first lesson in structural design (but an edge-loaded thin bar can twist or buckle sideways: see Columns and Buckling).
How much deflection is acceptable?
| Application | Typical limit | Why |
|---|---|---|
| Roof members (no plaster) | L ÷ 180 to L ÷ 240 | Appearance, ponding |
| Floors, shelves, general beams | L ÷ 360 | Cracking of finishes, comfort |
| Machine members, frames | L ÷ 1 000 to L ÷ 3 000 | Accuracy of the machine |
| Crane and gantry girders | L ÷ 600 to L ÷ 1 000 | Smooth running of the trolley |
| Shafts carrying gears | 0.025 mm at the gear, slope 0.001 rad at the bearings | Tooth contact and bearing life |
| Precision tables and guideways | a few micrometers | Measurement and machining accuracy |
| Springs, flexures | by design: large deflections | The deflection is the function |
Fixed ends, continuous beams and superposition
Statically determinate beams (simply supported, cantilever) have reactions that equilibrium alone fixes. Adding supports (fixed ends, a middle support) makes the beam statically indeterminate: there are more reactions than equations, and the extra ones come from compatibility of the deflection. The benefit is lower moments and deflections: a fixed–fixed beam under a central load has the maximum moment P L ÷ 8 instead of P L ÷ 4, and a quarter of the deflection. Standard tables (Roark, AISC, Machinery’s Handbook) give the results for common cases.
Superposition: for a linear elastic beam the deflection and the moment due to several loads are the sum of those due to each load acting alone. This allows any combination of the standard cases above: a tube with a central point load plus its own weight (uniform) deflects by δ = P L³ ÷ (48 E I) + 5 w L⁴ ÷ (384 E I). Remember that fixed supports are never perfectly rigid: a welded end in a thin plate is something between pinned and fixed, so design for both extremes.
Beam design checklist
- Loads: the point and distributed loads, the self-weight, impact and the worst position of a moving load.
- Supports: pinned, fixed or in between; reactions by equilibrium.
- Diagrams: V and M along the beam; find Mmax and where V = 0.
- Section: choose the shape and size so that σ = M ÷ S ≤ Sy ÷ n (see Section Properties and Structural Profiles).
- Shear and local effects: web shear, bearing under loads and at supports, and the buckling of thin walls (web crippling).
- Deflection: compute and compare with the limit; it often governs.
- Lateral stability: a deep, narrow beam can twist and buckle sideways: brace the compression flange.
- Connections and stress concentrations: holes, notches and welds reduce the strength (see Fatigue and Stress Concentration).
- Dynamic and repeated loads: check the fatigue life and vibration (the natural frequency of a beam rises with √(E I ÷ (m L³))).
| Mistake | Result | Fix |
|---|---|---|
| Wrong axis for I | The beam is two to ten times weaker than expected | Take I about the axis of bending (the neutral axis) |
| Mixing units (m and mm) | An answer off by a factor of 1 000 or more | Work in N and mm throughout |
| Forgetting the self-weight | Undersized long members | Add w = ρ A g |
| Checking strength but not stiffness | A beam that bends too much | Always compute the deflection |
| Ignoring the support flexibility | Deflections and moments wrong | Analyze the pinned and fixed extremes |
Related
- Structures and Strength of Materials: the overview of this guide.
- Stress, Strain and Safety Factors: the stress concepts used here.
- Section Properties and Structural Profiles: I, S and the standard profiles.
- Columns and Buckling and Torsion, Shear and Combined Loads: other loading cases.
- Worked Design Checks: a worked bracket check.
FAQ
What is the second moment of area?
A property of the cross-section (I, in mm⁴) that measures how well its shape resists bending: it sums the area times the square of its distance from the neutral axis. A deep section has a large I even with little material.
Why does a thin deep beam need lateral bracing?
The compression flange of a deep beam behaves like a column and can buckle sideways, twisting the beam, long before the material yields. Bracing the compression side, or using a wide or closed section, prevents it.
Can I use these formulas for a 3D-printed or plastic part?
Yes, with the right modulus: plastics are much less stiff (2 to 4 GPa) and creep, and printed parts are anisotropic (weaker between layers). Use data for the printed orientation and keep the stress well below the yield.
How do I find the natural frequency of a beam?
For a simply supported beam with a central mass m the frequency is f = (1 ÷ 2π) √(k ÷ m), with the stiffness k = 48 E I ÷ L³ from the deflection formula; the more flexible the beam, the lower the frequency.
References
- Beer, F. P., Johnston, E. R. and DeWolf, J. T., Mechanics of Materials, McGraw-Hill: beams, shear and moment diagrams, deflection.
- Young, W. C. and Budynas, R. G., Roark’s Formulas for Stress and Strain, McGraw-Hill: beam formulas and tables.
- AISC, Steel Construction Manual; IBC Table 1604.3 (deflection limits); Machinery’s Handbook (beam tables).
Last reviewed: October 2026. Formulas assume linear elastic behavior and small deflections; check the limits of the theory and use the applicable code for structures.