Beams must be strong enough and stiff enough. Choose the support and load case, then enter the load, span, Young's modulus and moment of inertia to get the maximum deflection and bending moment.
Choose the beam and load case.
Enter the load, span, E and I.
Read the deflection and bending moment.
SS, point: δ = WL³ ÷ 48EI
SS, uniform: δ = 5wL⁴ ÷ 384EI
Cantilever, point: δ = WL³ ÷ 3EI
Every loaded beam sags a little. This calculator finds how much, and the largest bending moment, for the four standard textbook cases: a simply supported beam with a point load at mid-span, a simply supported beam with a uniform load, a cantilever with a point load at its free end, and a cantilever with a uniform load. You give the load in kN or kN/m, the span, Young's modulus and the section's moment of inertia.
Engineers check two things for a beam: that it is strong enough not to fail, which depends on bending moment, and stiff enough not to sag visibly or crack finishes, which depends on deflection. The calculator reports deflection as a fraction of span too, so you can compare it with common limits of span/250 to span/360 used in design practice and Indian standards.
1. Identify the case: support type (simply supported or cantilever) and load type (point load W in kN or uniform load w in kN/m).
2. Write the span L in metres, E in GPa and I in cm⁴, then convert to SI: EI = E × 10⁹ × I × 10⁻⁸ in N·m².
3. Pick the deflection formula: WL³ ÷ 48EI for simply supported with central point load, 5wL⁴ ÷ 384EI for simply supported with uniform load, WL³ ÷ 3EI for a cantilever with end load, and wL⁴ ÷ 8EI for a cantilever with uniform load.
4. Pick the bending moment: WL ÷ 4, wL² ÷ 8, WL and wL² ÷ 2 respectively.
5. Convert loads to newtons before calculating, then express deflection in mm and moment in kN·m.
6. Compare span ÷ deflection with the allowable limit for your application.
Euler–Bernoulli beam theory says the curvature of a beam at any point equals the bending moment there divided by EI. Integrating this twice along the span, with the support conditions fixing the constants, gives the deflected shape. For a simply supported beam with a central load the moment rises linearly to WL/4 at mid-span, and the double integration yields a maximum sag of WL³/48EI. The other cases follow the same way. The theory assumes small deflections, a straight prismatic beam and linear elastic material.
Deflection grows with the cube of span for point loads and the fourth power for uniform loads. Doubling the span of a uniformly loaded beam, with the load per metre unchanged, makes it sag sixteen times as much. Bending moment grows with the square of span. That is why long spans need much deeper sections, and why adding an intermediate support is so effective. Supports matter too: a cantilever with an end load deflects 16 times as much as a simply supported beam of the same span with a central load.
Stiffness depends on the product EI. E is set by the material, about 200 GPa for steel and roughly 25 to 30 GPa for concrete. I depends on the shape of the cross-section and grows with the cube of depth, I = bd³/12 for a rectangle. Doubling a beam's depth makes it eight times stiffer, while doubling its width only doubles stiffness. This is why joists are placed on edge and why I-sections put most of their material far from the centre.
A fabricator in Coimbatore is checking a 6 m simply supported steel I-beam (E = 200 GPa, I = 8,600 cm⁴) that will carry a uniform load of 12 kN/m for a mezzanine floor.
Flexural rigidity: EI = 200 × 10⁹ × 8600 × 10⁻⁸ = 1.7200 × 10^7 N·m²
Maximum deflection: δ = 5wL⁴ ÷ 384EI = 11.773 mm
Maximum bending moment: M = wL² ÷ 8 = 54 kN·m
Span ÷ deflection: 6000 mm ÷ 11.773 mm = L/510
Answer: Maximum deflection 11.773 mm; Maximum bending moment 54 kN·m
Entering a total uniform load in kN when the formula needs load per metre in kN/m.
Using I in mm⁴ or m⁴ instead of cm⁴, which changes the answer by factors of 10⁴ or more.
Choosing the simply supported formula for a beam that is actually fixed or continuous, which overestimates deflection.
Forgetting the beam's own weight, which should be added as a uniform load.
Checking only deflection and not bending strength, or the other way round.
Preliminary sizing of steel and timber beams for mezzanines, sheds and lofts.
Engineering coursework in strength of materials and structural analysis.
Checking shelves, cantilever balconies and signboard arms for excessive sag.
Comparing sections from steel tables to meet a span/250 or span/360 limit.
Quick checks before a detailed design by a structural engineer.
Where do I find I?
Steel section tables list Ixx in cm⁴; for a rectangle, I = bd³ ÷ 12.
Is self-weight included?
No. Add it as a uniform load if it matters.