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Reference worksheet · one static point load

Documented Point-Load Shaft-Deflection Worksheet

Calculate the small-deflection Euler–Bernoulli response of a constant-EI member for one transverse point load: at any position between ideal simple supports, or at the free end of an ideal cantilever.

General beam equationsNo generated allowanceStrict input validation
Scope, not acceptance. This is a general engineering beam worksheet, not an ISO, API, ANSI/ASME or DIN acceptance calculation. It does not generate a permissible shaft deflection, bearing/seal clearance margin, critical speed, runout, fatigue life or safe operating verdict.

Documented inputs

Use a decimal point or decimal comma; thousands separators and scientific notation are intentionally rejected.

Reference results

Maximum downward deflection
Location of maximum from left/fixed end
Deflection at the load point
Largest support/end rotation magnitude
Maximum bending-moment magnitude
Maximum elastic surface bending stress
Second moment used
Vertical reactions

Equations and definitions

Let L be span, a the distance from the left support to P, b = L − a, E Young’s modulus, I the centroidal area second moment about the bending axis, and x the distance from the left support. Magnitudes are reported for one transverse load.

Simply supported member; point load at a

Rₐ = Pb/L; Rᵦ = Pa/L; Mmax = Pab/L
δ(a) = Pa²b²/(3EIL)
0 ≤ x ≤ a: δ(x) = Pbx(L² − b² − x²)/(6LEI)
a ≤ x ≤ L: δ(x) = Pa(L − x)[L² − a² − (L − x)²]/(6LEI)

The maximum is found from dδ/dx = 0 in the valid side of the piecewise field. If a ≤ L/2, x最大 = L − √[(L² − a²)/3]; if a ≥ L/2, x最大 = √[(L² − b²)/3]. Therefore δ(a) is generally 不是 the maximum when a ≠ L/2.

|θₐ| = Pab(L + b)/(6LEI)
|θᵦ| = Pab(L + a)/(6LEI)
|θ|max = max(|θₐ|, |θᵦ|)

Fixed cantilever; point load at free end

δ(x) = Px²(3L − x)/(6EI)
δmax = δ(L) = PL³/(3EI); |θ(L)| = PL²/(2EI); Mmax = PL

Section quantities

Solid circular section: I = πd⁴/64; σmax = Mmax(d/2)/I = 32Mmax/(πd³)

In direct-I mode the worksheet does not calculate surface stress because I alone does not define the extreme-fibre distance or section modulus. The displayed stress is an ideal elastic bending stress only; it omits stress concentrations, keyways, shoulders, residual stress, axial/torsional/shear stress and combined-stress criteria.

Dimensional check

P has dimension force, E force/area and I length⁴, so PL³/(EI) has dimension length. M has force·length and Mc/I has dimension force/area. These equations are not formulas issued by ISO.

Model boundary

  • No distributed load or shaft self-weight is included. For a verified full-span uniform-load/self-weight worksheet, use the related uniform-load deflection calculator.
  • Stepped/hollow shafts, overhangs, disks, multiple loads, bearing stiffness, shear deformation, contact, large displacement and rotor-dynamic effects require an appropriate beam, finite-element or rotordynamic model.
  • The former universal L/3000, L/5000 and L/10000 labels are not used: an allowable value must come from the controlled design basis for the actual bearings, seals, gears, couplings, process and operating state.
  • No critical speed is inferred from an arbitrary static load. A gravity-deflection relation is valid only under its own mass/load and modal assumptions; this worksheet does not establish them.
  • For a solid circle, changing d while P, E and L stay fixed makes deflection scale as 1/d⁴. That statement does not apply unchanged to self-weight because the load then also changes with area.

Source traceability

Claim分类Evidence
Piecewise deflection and true maximum for a point load between simple supports; end-loaded cantilever field and maximumGeneral Euler–Bernoulli beam solutions; not ISO formulasMIT, Engineering Mechanics for Structures, §10.1, pp. 268–270
Published center-load benchmark: P = 10 kip, L = 60 in, E = 30,000 ksi, I = 416.7 in⁴ gives 0.0036 in by PL³/(48EI)Published numerical exampleNASA NTRS document 19700016059, report page 75 (PDF page 81), equation (53)
1 in = 0.0254 m; 1 lbf = 4.4482216152605 N under standard gravityExact SI conversionNIST SP 811, Appendix B.8 and footnotes

Accessed: 15 July 2026. No standard-based permissible-deflection table was verified for this generic model. Any claimed universal allowance remains NEEDS_LICENSED_SOURCE until the controlled equipment/project criterion is supplied.

Published benchmark entry

Select simple supports and direct I. Enter L = 60 in, a = 30 in, I = 416.7 in⁴, E = 30 Mpsi and P = 10000 lbf. The worksheet should return approximately 0.003599712 in, which rounds to the report’s 0.0036 in.

Questions

Why is the offset-load maximum not at the load?

The load point is where bending moment is largest, but maximum deflection occurs where slope is zero. Except at midspan, those locations differ; the worksheet evaluates the correct side of the piecewise elastic curve.

Can this determine a shaft critical speed?

No. This is a static one-load beam calculation. Rotor critical speed depends on the distributed mass, disks, bearings/support coefficients, gyroscopic effects and mode shape. Applying a gravity-deflection shortcut to an unrelated load case is not justified.

Does the calculated stress prove the shaft is safe?

No. It is only the nominal elastic surface stress for a uniform solid circle in the stated bending plane. Strength and fatigue checks need material allowables, combined loads, stress concentrations, mean/alternating components, surface/size/reliability effects and the controlled design method.

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