Simply Supported Beam with Center Point Load
Calculate support reactions, maximum bending moment and midspan deflection for a simply supported beam carrying a point load at its center. Enter span, load, Young's modulus and moment of inertia.
Beam Reactions, Bending Moment and Deflection
Center-Loaded Simply Supported Beam Formula
Where:
- \(R\) — support reaction at each end (N)
- \(P\) — point load applied at midspan (N)
- \(L\) — span between the two supports (m)
- \(M_{max}\) — maximum bending moment, occurring under the load (N·m)
- \(\delta_{max}\) — maximum deflection at midspan (m)
- \(E\) — Young's modulus of the beam material (Pa)
- \(I\) — second moment of area of the cross-section (m⁴)
Because the load acts exactly halfway between two identical supports, symmetry splits it equally between them. The bending moment and deflection both peak at midspan, where the load is applied.
Center-Loaded Simply Supported Beam – Calculation Example
Example: a steel beam with a span of 4 m carries a 10 kN point load at midspan. The beam has Young's modulus \(E = 200\) GPa and a second moment of area \(I = 1000\) cm⁴ = \(1 \times 10^{-5}\) m⁴.
\[ R = \frac{P}{2} = \frac{10\,000\ \text{N}}{2} = 5\ \text{kN} \]
\[ M_{max} = \frac{P L}{4} = \frac{10\,000 \times 4}{4} = 10\ \text{kN·m} \]
\[ \delta_{max} = \frac{P L^3}{48 E I} = \frac{10\,000 \times 4^3}{48 \times 200 \times 10^9 \times 1 \times 10^{-5}} \approx 6.67 \times 10^{-3}\ \text{m} \]
Result: each support carries 5 kN, the maximum bending moment is 10 kN·m at midspan, and the beam deflects approximately 6.67 mm (about 0.26 in) at the point of loading.
How to use the calculator
- Enter the Span (L) — the distance between the two supports — and choose the unit (m or ft).
- Enter the Point load (P) applied at midspan and select its unit (kN, N or lbf).
- Enter the material's Young's modulus (E). Steel is about 200 GPa, aluminum about 69 GPa, wood about 11 GPa.
- Enter the Second moment of area (I) of the beam cross-section about the bending axis, in cm⁴, mm⁴ or in⁴.
- Read the results: the support reactions, the maximum bending moment and the maximum deflection at midspan.
What is a center-loaded simply supported beam?
A simply supported beam rests on two supports — typically a pin at one end and a roller at the other. Neither support resists rotation, so the beam is free to bend and its ends can rotate. This is one of the most common idealizations in structural and mechanical engineering: it closely models floor joists on two walls, shafts on two bearings, and many machine and frame elements.
When a single point load \(P\) is applied exactly at midspan, the problem becomes symmetric. Each support carries exactly half the load, so \(R = P/2\). The bending moment grows linearly from zero at each support to a peak of \(M_{max} = PL/4\) directly under the load — this is the point of maximum stress in the beam, where failure or yielding would begin.
The deflection follows a cubic relationship with the span: \(\delta_{max} = PL^3/(48EI)\). Doubling the span multiplies the deflection by eight, which is why long spans are so much harder to keep stiff. The product \(EI\), called the flexural rigidity, measures the beam's resistance to bending: a stiffer material (higher \(E\)) or a deeper cross-section (higher \(I\)) both reduce deflection. If you need to work with material stiffness separately, the Young's modulus calculator covers related material property calculations.
When interpreting the result, compare the deflection against your application's allowable limit. Many building codes limit live-load deflection to \(L/360\) or \(L/240\) of the span; for the 4 m example above, \(L/360 \approx 11.1\) mm, so a 6.67 mm deflection would pass.
Assumptions and limitations
- The load must act exactly at midspan. For a load at an arbitrary position, the formulas change — reactions become \(R_1 = Pb/L\), \(R_2 = Pa/L\).
- The beam is assumed prismatic (constant cross-section), homogeneous, and linearly elastic — deflections are small compared to the span.
- Self-weight of the beam is ignored. If the beam is heavy relative to the applied load, add it as a distributed load in a separate analysis.
- Supports are ideal: they carry vertical load only and provide no rotational restraint or fixity.
- Only vertical loading is considered; horizontal loads, torsion and dynamic effects are outside the scope of these formulas.
For dynamic situations — a mass supported by a beam acting as a spring, for example — the static result here is only the starting point. See the spring mass calculator and the spring resonant frequency calculator for vibration analysis.
Frequently Asked Questions
How do I calculate the midspan deflection by hand?
Multiply the load by the span cubed, then divide by 48 times the flexural rigidity: \(\delta = P L^3 / (48 E I)\). First convert everything to consistent base units — newtons, meters, pascals and m⁴. For example, with \(P = 10\,000\) N, \(L = 4\) m, \(E = 200\times10^9\) Pa and \(I = 1\times10^{-5}\) m⁴, you get \(10\,000 \times 64 / (48 \times 2\times10^6) \approx 6.67\) mm. Always check that \(I\) is in m⁴ before plugging it in.
What is an acceptable amount of beam deflection?
It depends on the application, but common building-code limits are \(L/360\) for live loads on floors (where plaster or brittle finishes could crack) and \(L/240\) where appearance is less critical. For a 5 m span, \(L/360 \approx 13.9\) mm. Machinery shafts and precision structures often demand much tighter limits. A deflection within the code limit does not automatically mean the beam is safe — you must also check bending stress and shear against the material's allowable values.
How do I find the second moment of area (I) for my beam?
For a rectangular cross-section of width \(b\) and height \(h\) bending about the horizontal axis, \(I = bh^3/12\). For a solid circular section, \(I = \pi d^4/64\). For standard steel, aluminum or timber profiles, look up the tabulated value. Use the axis the beam actually bends about — for a beam loaded vertically, that is the horizontal centroidal axis. Mixing up the strong and weak axis is one of the most common errors and can throw the deflection off by an order of magnitude.
Why is my calculated deflection wildly too large or too small?
Almost always a unit mismatch. The formula requires \(I\) in m⁴, and the conversion factors are brutal: \(1\ \text{cm}^4 = 10^{-8}\ \text{m}^4\) and \(1\ \text{mm}^4 = 10^{-12}\ \text{m}^4\). Entering 1000 mm⁴ as if it were cm⁴ makes deflection 10,000 times too large. Similarly, \(E\) must be in pascals — 200 GPa is \(2\times10^{11}\) Pa, not \(200\times10^9\) MPa. This calculator handles the conversions automatically when you select the correct input units.
Can I use this calculator for a load that is not at the center, or for two point loads?
No — these formulas are only valid for a single load at exact midspan. For an off-center load at distance \(a\) from one support, the reactions become \(R_1 = Pb/L\) and \(R_2 = Pa/L\), and the deflection formula changes. However, you can use superposition: for two symmetrically placed equal loads, calculate each load's effect separately and add the deflections. For a single midspan load plus the beam's own weight, add the distributed-load deflection \(5wL^4/(384EI)\) to the point-load result.
How do I calculate the bending stress from the maximum moment?
Use the flexure formula \(\sigma = M c / I\), where \(c\) is the distance from the neutral axis to the outermost fiber — half the height for a symmetric section. For the example beam with \(M = 10\) kN·m and \(I = 1\times10^{-5}\) m⁴, a 200 mm deep section gives \(\sigma = 10\,000 \times 0.1 / 10^{-5} = 100\) MPa, well below structural steel's yield strength of about 250 MPa. Always verify stress separately — a stiff beam can still be overstressed.
Does the beam's own weight affect the results?
Yes, but often only slightly. Self-weight acts as a uniformly distributed load \(w\) (N/m), which adds reactions of \(wL/2\) at each end, a midspan moment of \(wL^2/8\), and a midspan deflection of \(5wL^4/(384EI)\). For a 4 m steel beam weighing 30 kg/m, that adds only about 0.59 kN·m of moment — small next to a 10 kN point load, but significant for heavy concrete or long-span beams. Evaluate it separately and add it to the point-load results.