Simply Supported Beam with Uniform Load
Calculates the support reactions, maximum bending moment and midspan deflection of a simply supported beam carrying a uniformly distributed load. Enter the span, load, Young's modulus and second moment of area.
Beam Reactions, Bending Moment and Deflection
Uniformly Loaded Simply Supported Beam Formula
Where:
- \(w\) — distributed load per unit length (N/m)
- \(L\) — span, the distance between the two supports (m)
- \(E\) — Young's modulus of the beam material (Pa)
- \(I\) — second moment of area of the cross-section about the bending axis (m⁴)
- \(R\) — reaction at each support (N), \(M_{max}\) — maximum bending moment (N·m), \(\delta_{max}\) — maximum deflection (m)
These are the classic closed-form results for a simply supported beam with a uniformly distributed load over the entire span. The total load \(wL\) splits equally between the two supports, the bending moment peaks at midspan, and the maximum deflection also occurs at midspan because the loading and the supports are symmetric.
Uniformly Loaded Simply Supported Beam – Calculation Example
Example: a steel beam with a span of \(L = 4\ \text{m}\) carries a uniform load of \(w = 10\ \text{kN/m}\). The material has \(E = 200\ \text{GPa}\) and the cross-section has \(I = 1000\ \text{cm}^4 = 1.0 \times 10^{-5}\ \text{m}^4\).
\[ R = \frac{wL}{2} = \frac{10 \times 4}{2} = 20\ \text{kN} \]
\[ M_{max} = \frac{wL^2}{8} = \frac{10 \times 4^2}{8} = 20\ \text{kN}\cdot\text{m} \]
\[ \delta_{max} = \frac{5wL^4}{384EI} = \frac{5 \times 10{,}000 \times 4^4}{384 \times (200 \times 10^9) \times (1.0 \times 10^{-5})} \approx 0.0167\ \text{m} \]
Result: each support reaction is 20 kN, the maximum bending moment is 20 kN·m at midspan, and the maximum deflection is about 16.7 mm (≈ 0.656 in) at midspan.
How to Use the Calculator
- Enter the span between the two supports in the Span (L) field and choose metres or feet from the unit selector.
- Enter the load per unit length in the Distributed load (w) field and select kN/m, N/m or lbf/ft. This is the total weight carried by the beam divided by its length.
- Enter Young's modulus (E) for the beam material — steel ≈ 200 GPa, aluminum ≈ 69 GPa, wood ≈ 11 GPa — and pick GPa, MPa or psi.
- Enter Second moment of area (I) for the cross-section (check a profile table for standard beams) and select cm⁴, mm⁴ or in⁴.
- Read the results: the reaction at each support, the maximum bending moment at midspan, and the maximum deflection at midspan. Outputs are shown in the unit system implied by your span unit, with the other system in parentheses.
What Is a Uniformly Loaded Simply Supported Beam?
A simply supported beam rests on two supports — usually a pin at one end and a roller at the other — so it can rotate freely at both ends but cannot move vertically. It is the most common idealization in building design: floor joists, roof purlins and many structural members behave essentially this way. A uniformly distributed load (UDL) is a load spread evenly along the whole length, such as the self-weight of the beam, a floor slab it carries, or snow on a roof.
Because the loading is symmetric, each support carries exactly half of the total load: \(R = wL/2\). The bending moment varies parabolically along the span, starting at zero at each support and reaching \(M_{max} = wL^2/8\) at midspan. This midspan moment is what governs the bending stress in the beam — combined with the section modulus it tells you whether the member is strong enough. For comparison, the same total load concentrated at midspan (see the center-loaded simply supported beam) produces twice the moment, so distributing a load is structurally far more efficient.
The deflection \(\delta_{max} = 5wL^4/(384EI)\) also occurs at midspan and follows the elastic curve of the beam. Note the strong influence of the span: deflection grows with the fourth power of \(L\), so doubling the span increases deflection sixteen-fold. Stiffness depends on the product \(EI\): a stiffer material (higher \(E\)) or a deeper cross-section (much higher \(I\), since \(I\) scales roughly with the cube of depth) both reduce deflection. If you are interested in how Young's modulus also controls dynamic behaviour, see the Young's modulus spring resonant frequency calculator.
Assumptions and Practical Tips
- Small deflections, linear elasticity: the formulas assume the material stays below its yield point and deflections are small relative to the span, so superposition and Euler–Bernoulli beam theory apply.
- Ideal supports: the beam is treated as pin-supported at one end and roller-supported at the other, with no fixity. Real connections that restrain rotation reduce both moment and deflection, so this is a conservative idealization for deflection.
- Load covers the whole span: partial-length loads, point loads, or overhangs need different (or superposed) formulas. A central point load alone is handled by the center point load calculator.
- Self-weight counts: include the beam's own weight in \(w\) unless the load is specified as an imposed (superimposed) load only.
- Check deflection limits: many codes limit live-load deflection to \(L/360\) (e.g. floors) or \(L/240\). For the worked example, \(L/360 = 4000/360 \approx 11\ \text{mm}\), so a 16.7 mm deflection would exceed a typical floor limit and a stiffer section would be needed.
- Units must be consistent: the calculator converts everything internally, but if you use the formulas by hand, keep \(w\), \(L\), \(E\) and \(I\) in a single coherent system (N, m, Pa, m⁴) before converting the result.
Frequently Asked Questions
How do I calculate the deflection by hand without the calculator?
Convert everything to consistent units first: load in N/m, span in m, E in Pa, I in m⁴. Then apply \( \delta = 5wL^4/(384EI) \). The most common by-hand error is forgetting to convert \(I\) from cm⁴ or mm⁴ to m⁴ (1 cm⁴ = 10⁻⁸ m⁴, 1 mm⁴ = 10⁻¹² m⁴) and E from GPa to Pa (multiply by 10⁹). Doing the arithmetic in one coherent unit system gives the deflection directly in metres.
What is an acceptable deflection for a beam?
Most building codes limit live-load deflection to \( L/360 \) for floors and plastered ceilings, and \( L/240 \) for roofs or members carrying less brittle finishes. For a 6 m floor joist, \( L/360 \) means about 16.7 mm. These are general guidelines — actual limits depend on your local code, the finish materials and whether the load is short-term or long-term, so always verify against the applicable standard for your project.
How do I find the second moment of area (I) for my beam?
For standard rolled steel sections, look up \(I\) in the manufacturer's profile table and enter it directly. For a solid rectangular section, use \( I = bh^3/12 \), where \(b\) is the width and \(h\) the depth in the bending direction. For a circular section, \( I = \pi d^4/64 \). Note that \(I\) grows with the cube of depth, so a slightly deeper beam is dramatically stiffer than a wider one.
Does the beam's own weight count in the distributed load?
Yes. Unless your load is explicitly specified as a superimposed (imposed) load, add the self-weight: multiply the beam's mass per metre by 9.81 m/s² (or its tabulated weight per unit length) and include it in \(w\). For heavy steel or concrete members the self-weight can be a significant fraction of the total load, and omitting it underestimates both the moment and the deflection.
Can I use this calculator with US imperial units?
Yes — select feet for the span and enter the load in lbf/ft. Young's modulus in psi (steel ≈ 29,000,000 psi) and \(I\) in in⁴ work with these units. Alternatively, you can mix selectors freely: the calculator converts everything internally. If you compute by hand in imperial units, the same formulas hold as long as force, length and their derivatives stay consistent (lbf, ft or inches throughout).
How do I get the bending stress from the maximum moment?
The maximum bending stress at midspan is \( \sigma = M_{max} \cdot c / I \), where \(c\) is the distance from the neutral axis to the extreme fibre (half the depth for a symmetric section). Dividing \(I\) by \(c\) gives the section modulus \(S\), so \( \sigma = M/S \). Compare the result with the material's allowable or yield stress, applying the appropriate safety factor — this calculator gives the moment, not the stress check.
What happens if my load only covers part of the span or includes point loads?
These closed-form formulas apply only to a uniform load over the full span between two simple supports. For partial-span loads, point loads or overhangs you need different formulas or the superposition method (adding results of individual load cases). If the total load acts as a single point load at midspan, the moment doubles to \( wL^2/4 \) and deflection rises to \( 5wL^4/(384EI) \times 8/5 \) — use the dedicated center point load calculator for that case.