Steel Grating Wheel Load Check: Transverse Distribution Width, Impact Factor and Load Limits

ASPER - Engineering calculation notes for metal products - 2026-10-21 - Steel gratingConcentrated loadYB/T 4001.1-2019Impact factor
Every number in this note is reproducible from the following basis. Steel grade Q235 with E = 206 000 N/mm2 and bending design strength f = 215 MPa (GB 50017-2017, t not exceeding 16 mm). Reference panel G405/30/100: bearing bar 40 x 5 mm at 30 mm centres, twisted square cross bar 6 x 6 mm at 100 mm centres. Reference span L = 1000 mm. Deflection limit per YB/T 4001.1-2019 clause 8.7.7, min(L/200, 4 mm) = 4.0 mm. Bearing bar I_b = 26 667 mm4 and W_b = 1 333 mm3; cross bar I_c = 108 mm4. The transverse distribution model is an elastically coupled beam approximation that assumes the cross bar is fixed against rotation at both ends by the bearing bars; it therefore yields a theoretical upper bound on the distribution width, and design values must still be checked against the code figure.

1. Where the problem starts: 4 kN/m2 does not mean wheel-tight

Designing a maintenance platform for a 4 kN/m2 uniform load, as clause 8.7.3 of YB/T 4001.1-2019 prescribes, is standard practice. That magnitude is adequate for foot traffic and for tools set down loose. It is a completely different quantity from a wheel.

Compare three cases on the same G405/30/100 panel at a 1000 mm span.

CaseLoad per bearing barMoment M1Stress sigmaFraction of f
Uniform 4 kN/m20.12 kN/m x 1 m = 0.12 kN15 000 N mm11.25 MPa5.2 percent
Trolley wheel 1.2 kN (static)1.2 / 12.29 = 0.0977 kN23 942 N mm17.96 MPa8.4 percent
Forklift wheel 15 kN (static)15 / 12.29 = 1.221 kN275 488 N mm206.72 MPa96.1 percent

The uniform case uses 5.2 percent of the design strength. A single forklift wheel reaches 96.1 percent before the truck even moves. The gap is a factor of 18.4, and impact has not been added yet.

There is a second trap inside the code itself. Clause 8.7.2 of YB/T 4001.1-2019 requires that "at the most unfavourable position, a 200 mm x 200 mm area shall be capable of carrying a concentrated load of 1.5 kN". 1.5 kN is roughly one adult standing on one foot. It is the minimum requirement for a walking platform, not a criterion for a wheel traffic route. Applying it to pallet trucks or forklifts is the most common mismatch in the industry.

2. Stiffness baseline: how stiff is one bar, and how stiff is the cross bar

Split the panel into two stiffnesses.

Bearing bar, simply supported with a midspan point load: k_b = 48 E I_b / L^3 = 48 x 206 000 x 26 667 / 1000^3 = 263.7 N/mm
Cross bar coupling, segment length equal to the bar pitch s, both ends fixed: k_c = 12 E I_c / s^3 = 12 x 206 000 x 108 / 30^3 = 9 888 N/mm

k_c is 37.5 times k_b. That single ratio drives every counter-intuitive result that follows: the cross bar is far stiffer than the bearing bar itself, so transverse spreading is governed by the cross bar, not by the bar.

3. How many bars does a wheel really engage: the coupled beam model

Index the bearing bars by i with deflection delta_i and apply the wheel load P only at i = 0. Each bar carries its own bending stiffness k_b plus the coupling stiffness k_c of the cross bars on either side:

k_b delta_i + k_c (2 delta_i - delta_(i-1) - delta_(i+1)) = P_i

Take the decaying form delta_i = delta_0 r^|i|. Substituting for i not equal to 0 gives the characteristic equation:

r + 1/r = 2 + beta, where beta = k_b / k_c
n_eff = (1 + r) / (1 - r) (number of bars effectively engaged)
k_sys = k_b + 2 k_c (1 - r) = k_b n_eff

For the reference case, beta = 263.7 / 9 888 = 0.026667. Solving r + 1/r = 2.026667 gives r = 0.84949, hence:

n_eff = 1.84949 / 0.15051 = 12.29 bars
Effective distribution width b_e = 12.29 x 30 = 368.6 mm
System stiffness k_sys = 263.7 x 12.29 = 3 240 N/mm

Against the naive estimate: a contact width of 15 to 77 mm divided by a 30 mm pitch suggests only 1.5 to 3.6 bars are touched. The number actually mobilised is 12.3, which is 3.4 to 8 times the naive figure. The cross bars spread the load across nearly 370 mm.

How span changes the distribution width

Span L (mm)k_b (N/mm)betarn_effb_e (mm)k_sys (N/mm)
6001 220.70.123460.704985.78173.47 055
800515.00.052080.796348.82264.64 542
1000263.70.026670.8494912.29368.63 240
1200152.60.015430.8832516.13483.92 461
150078.10.007900.9149722.52675.71 760

Longer spans mobilise more bars, which produces a result that runs against intuition.

Going from a 600 mm to a 1500 mm span (2.5 times) would be predicted by a single-beam model to raise deflection by 15.6 times. The real increase is 4.01 times. Since deflection scales as L^3 / n_eff and n_eff grows from 5.78 to 22.52 over the same range, the effective exponent falls from L^3 to L^1.51 (ln 4.01 / ln 2.5). Span extrapolations built on a single isolated bearing bar badly overstate the penalty of long spans.

4. Contact geometry of three typical wheels

Contact area follows from the tyre contact pressure: A = P / p. Take the patch as a rectangle with the long side a along the span direction and the short side b_0 across it, at an aspect ratio of 2.5, which is typical for a cylindrical wheel on a flat surface, so b_0 = sqrt(A / 2.5).

WheelP per wheel (kN)Contact pressure p (MPa)A = P/p (mm2)b_0 (mm)a (mm)Bars directly under the patch
Hand trolley, polyurethane1.22.060015.4938.71.52
Pallet truck, nylon3.04.075017.3243.31.58
Forklift, solid rubber15.01.015 00077.46193.63.58

The nylon wheel is the hardest at p = 4.0 MPa and has the smallest patch, making it the sharpest of the three loads. The forklift's solid rubber is the softest with the largest patch at 15 000 mm2, yet because the total force is 12.5 times greater it remains the governing case in absolute terms.

US practice carries one hard warning here: Note 4 of the Vulcraft vehicular load tables states that wheeled vehicles with urethane tires should never be used on open grid bar grating. The trolley case in this note is exactly that condition, a polyurethane wheel at 2.0 MPa over only 600 mm2, which after impact becomes a local peak on a single bar. Vulcraft gives the prohibition; this note gives the reproducible reason: the concentrated pressure of a small hard patch runs 53 times above the 0.0375 MPa implied by the YB/T 4001.1-2019 concentrated load criterion, and no current impact factor covers the light wheel plus step combination (section 6).

5. Moment and stress: why M = PL/4 is not enough

When the contact length a along the bar cannot be neglected, the load must be treated as partially distributed. Equation 12 of the NAAMM MBG 534 Metal Bar Grating Engineering Design Manual gives:

M = P (2L - a) / 8

The expression is self-consistent at both limits: as a approaches 0 it reduces to the point load result M = PL/4, and at a = L it reduces to the full-span uniform result M = PL/8. For the forklift with a = 193.6 mm the difference from PL/4 is 9.7 percent, which is not negligible.

Dividing by n_eff gives the moment per bar, M1 = M / n_eff, and the stress sigma = M1 / W_b.

WheelM = P(2L-a)/8 (N mm)M1 (N mm)sigma (MPa)sigma / 215Static deflection P/k_sys (mm)
Trolley 1.2 kN294 19123 94217.968.4 percent0.370
Pallet truck 3.0 kN733 76259 71344.7820.8 percent0.926
Forklift 15 kN3 386 908275 488206.7296.1 percent4.629

The forklift's static deflection of 4.63 mm already exceeds the 4 mm limit of YB/T 4001.1-2019, and that is with the truck standing still.

6. Impact amplification: a 2 mm step makes light wheels worse than heavy ones

No grating surface is perfectly flat. Steps between adjacent panels, offsets between panel and support, and the valleys of serrated bar all create a height difference h. A wheel dropping onto the lower surface is a free-fall impact, for which the energy method gives:

phi = 1 + sqrt(1 + 2h / delta_st), with delta_st = P / k_sys

phi is the dynamic amplification factor; delta_max = phi delta_st and the equivalent static load is P_eff = phi P. Taking h = 2 mm:

Wheeldelta_st (mm)phi at h = 2 mmP_eff (kN)sigma with impact (MPa)delta_max (mm)Verdict (215 MPa, 4 mm)
Trolley 1.2 kN0.3704.4355.3279.61.643Passes
Pallet truck 3.0 kN0.9263.3079.92148.13.061Passes, 23 percent reserve
Forklift 15 kN4.6292.36535.48488.910.950Stress 2.27x, deflection 2.74x over
Counter-intuitive result: the lighter the wheel, the larger the impact factor. The trolley reaches phi = 4.435 while the forklift only reaches phi = 2.365. The reason is that a light wheel has a tiny static deflection, 0.370 mm, so a 2 mm step is 5.4 times that value and a very large share of the drop energy is converted. The heavy wheel already deflects 4.63 mm, so the same 2 mm is only 43 percent of it. The widely used rule of thumb "apply an impact factor of 1.5 to 2.0 for forklifts" fails worst exactly where it is applied most casually, on light wheels.

Sensitivity to step height

Step h (mm)Trolley phiPallet truck phiForklift phi
1.03.5302.7782.197
2.04.4353.3072.365
3.05.1473.7352.515

Reducing the step from 3 mm to 1 mm drops the trolley impact factor from 5.15 to 3.53, a 31.4 percent reduction, and the equivalent static load from 6.18 kN to 4.24 kN. Controlling installation flatness is cheaper and more effective than upsizing the panel, a point that reappears in the efficiency comparison below.

Reconciling with current impact factor practice

Current practice puts the impact factor between 1.3 and 2.0. AASHTO and NAAMM MBG 534 take 30 percent for vehicular loads, which represents the vibration component on a paved level surface; YB/T 4001.1-2007 Annex B takes i = 0.4 for transverse trench covers; industry rules of thumb take 1.4 to 2.0 for forklift routes. The energy method here models a different mechanism, the step drop. At h = 0 it gives phi = 1.0, no additional impact, which does not conflict with the AASHTO vibration component and should be considered alongside it. At h = 2 mm the light wheel already rises by 344 percent, beyond every current criterion.

Every existing impact factor is calibrated for heavy loads on level surfaces; the combination of a light wheel and a step is covered by none of them. A wheel route that passes an AASHTO 30 percent or an industry 1.5 to 2.0 check can carry several times of unaccounted impact once installation steps go out of control, which makes the step height a measurable acceptance item.

7. Reconciling with YB/T 4001.1-2019: where the 200 mm comes from

The standard gives the concentrated load design value as:

F = 2 n f b d^2 / (3L)

where n is the number of bearing bars assumed to carry the load, b is bar thickness and d is bar depth. This follows directly from sigma = M / (n W_b) with M = FL/4 and W_b = b d^2/6, so it shares its origin with the model used here. Taking n = 200 / 30 = 6.667 from the 200 mm distribution area in clause 8.7.2:

Strength: F_sigma = 2 x 6.667 x 215 x 5 x 40^2 / (3 x 1000) = 7 645 N = 7.64 kN
Deflection: I_g = 6.667 x 26 667 = 177 778 mm4
delta = F L^3 / (48 E I_g) = F / 1 757.9, so at delta = 4 mm, F_delta = 7 032 N = 7.03 kN

Deflection governs, giving an allowable concentrated load of 7.03 kN under the code method. The same panel through the model in this note, with n_eff = 12.29:

F_sigma = 2 x 12.29 x 215 x 5 x 40^2 / (3 x 1000) = 14.09 kN
F_delta = 4 mm x 3 240 N/mm = 12.96 kN, so deflection governs at 12.96 kN

The ratio is 12.96 / 7.03 = 1.843. The code is 84 percent more conservative than the model.

That 1.843 is not a coincidence

Now consider the wheel at a panel edge, where only one side of bars is available to spread into:

n_eff (edge) = 1 / (1 - r) = 1 / 0.15051 = 6.64 bars, so b_e = 6.64 x 30 = 199.3 mm
n_eff (middle) / n_eff (edge) = 12.29 / 6.64 = 1.849

199.3 mm matches the code's 200 mm, and 1.849 matches the conservatism factor of 1.843, both within 0.3 percent.

This shows that the 200 mm distribution area in YB/T 4001.1-2019 corresponds to the worst case of a wheel at the panel edge, not the middle of the panel. The code provides a conservative fixed value that is independent of span and of cross bar size. The model here shows that mid-panel spreading actually reaches 369 mm at a 1000 mm span and 676 mm at 1500 mm, so the code's conservatism widens further to 3.4 times at long spans.

One operational rule follows directly: locate wheel routes over the middle of panels, or place the panel joint at a support beam. Running along an edge is 85 percent worse than running down the middle.

8. Comparison with NAAMM MBG 534: the US method spreads even wider

Example 7 of NAAMM MBG 534, for AASHTO H-20 vehicular loading, uses:

ParameterNAAMM valueMetric
Wheel load P_w20 800 lb92.5 kN
Distribution length a, parallel to bars20 in508 mm
Distribution width c, perpendicular to bars22.75 in578 mm
Bearing bar spacing A_w1.375 in34.9 mm
Bars engaged N = c / A_w16.545-

The US method adopts a 578 mm distribution width and 16.5 bars, larger than the 369 mm and 12.3 bars from this model at a 1000 mm span. Two things follow. First, the order of magnitude of the distribution width derived here is sound and not an overestimate. Second, the real difference between the Chinese and American approaches lies in whether the distribution width is treated as a function of span: NAAMM derives a geometric width from tyre contact and surfacing spread, YB/T fixes 200 mm, and the model here gives an analytical width that varies with span and cross bar stiffness.

One detail worth noting is that NAAMM computes M = (P_w/a)(L^2/8), spreading the wheel load along a and then treating it as a uniform load. That applies when a is close to or greater than the span, as the manual states explicitly ("when the tire contact length exceeds the clear span"). Where a is less than L, the correct expression is M = P(2L - a)/8, which is what section 5 uses.

9. Three ways to strengthen: cross bars beat bearing bars by 38 times

The three routes to a higher wheel load capacity differ enormously in cost. The baseline is G405/30/100 at L = 1000 mm.

Measuren_effk_sys (N/mm)F_sigma (kN)F_delta (kN)Strength gainDeflection gainAdded steelEfficiency
Baseline G405/30/10012.293 24014.0912.96----
Bar depth 40 to 508.824 54215.8018.17+12.2 percent-28.7 percent+13.08 kg/m20.93 percent/kg
Cross bar 6x6 to 8x821.805 74724.9922.99+77.4 percent-43.6 percent+2.20 kg/m235.20 percent/kg
Bar pitch 30 to 2022.525 93925.8323.75+83.3 percent-45.4 percent+26.17 kg/m23.18 percent/kg
Going from a 40x5 to a 50x5 bearing bar grows the section modulus W by 56.3 percent, from 1 333 to 2 083 mm3, yet the allowable concentrated load rises only 12.2 percent. A stiffer bar attracts more load, so the number of bars sharing it falls from 12.29 to 8.82 and the gain is diluted by transverse distribution. A single isolated bar model cannot reveal this, and it is the root cause of the widespread but wrong intuition that stepping up one size always solves the problem.

Changing the cross bar from 6x6 to 8x8 costs only 2.20 kg/m2 and buys a 77.4 percent gain, making it 37.8 times more efficient per kilogram than deepening the bearing bar. The side effect is small too: the overall open area falls from 78.3 to 76.7 percent, a loss of 1.6 percentage points that barely affects ventilation, drainage or light transmission.

Closing up the bar pitch from 30 to 20 mm gives the largest absolute gain at +83.3 percent but consumes 26.17 kg/m2, so its efficiency is only 3.18 percent/kg. It suits cases where a smaller opening is needed anyway, for example to stop small components falling through or to meet high-heel and accessibility requirements. For capacity alone, change the cross bar first.

10. The changeover span: L* = sqrt(24 E h / f)

Which criterion governs switches with span. Setting sigma/f equal to delta/delta_lim with the constant limit delta_lim = 4 mm (the region L greater than 800 mm), substituting sigma = P(2L - a)/(8 n_eff W_b) and delta = P L^3/(48 E I_b n_eff), cancelling P and n_eff since they are common to both criteria, and taking a much less than 2L:

f L^3 = 24 E h L, so L* = sqrt(24 E h / f)

The result scales only with the square root of bar depth and is independent of bar thickness, span, bar pitch and cross bar size.

Bar depth h (mm)Wheel load L* (mm)Uniform load L* = sqrt(19.2 E h / f) (mm)
32 (G325)857.8767.3
40 (G405)959.1857.8
50 (G505)1 072.3959.1

Below L* strength governs; above it deflection governs. For G405 the common 1000 mm span sits just to the right of its own changeover at 959 mm. Under uniform load the changeover is 858 mm, so a 1000 mm span is already deflection governed, while under a wheel load it has only just crossed its own switch point. At a 1000 mm span a wheel load design must therefore be checked against stress and deflection together, never against one alone.

Keeping the forklift contact length a = 193.6 mm without simplification gives f L^3 = 12 E h (2L - a), whose numerical solution is L* = 906 mm. The simplified expression is conservative by about 5.9 percent.

11. Serrated anti-slip and wheel traffic are in conflict

Clause 8.5.1 of YB/T 4001.1-2019 recommends serrated grating for working platforms with a slope not exceeding 10 degrees, or where liquid or oil accumulates; clause 8.5.2 requires more substantial anti-slip measures above 10 degrees. Serrated bar must have at least 5 teeth per 100 mm with a tooth depth of at least 1.5 mm.

Converting the 10 degree threshold into a friction demand, with walking safety requiring tan theta not exceeding mu / K and K = 2:

Surface conditionFriction coefficient muAllowable tan thetaAllowable slope
Dry (steel against rubber sole)0.650.32518.0 degrees
Wet0.350.1759.9 degrees
Oily0.150.0754.3 degrees

The wet condition works back to 9.9 degrees, almost exactly the 10 degrees in the code, which shows the code threshold was set for the wet condition. By implication an oily surface is already out of range at 4.3 degrees, so serration is a necessary but not sufficient measure there and must be paired with a cleaning regime.

The conflict is that serrated grating is hostile to trolley wheels. For a 1.2 kN wheel on a 10 degree slope with a rolling resistance coefficient of 0.03, the pushing force is 1.2 sin 10 degrees + 1.2 x 0.03 = 0.244 kN, which is 20.4 percent of the wheel load, and serration pushes rolling resistance higher still. Hence:

Selection rule: walkways where pedestrian safety dominates use serrated bar with a slope not exceeding 5 degrees. Routes carrying trolleys or pallet trucks use plain bar and rely on slope control and drainage for slip resistance, with local anti-slip strips outside the wheel tracks if needed. Where both demands occur on one platform, zoning the specification by wheel track is both cheaper and safer than applying one upgraded size across the whole area.

12. Cross-check against external data: two concentrated load criteria 55 times apart

Converting the Chinese and American concentrated load criteria into the same dimension, contact pressure, exposes a fact that is routinely overlooked.

CriterionLoadAreaContact pressureWhat it represents
YB/T 4001.1-2019 clause 8.7.21.5 kN200 x 200 = 40 000 mm20.0375 MPaA whole sole, large soft contact
OSHA walkway concentrated load300 lbf = 1.335 kN1 in2 = 645 mm22.07 MPaA heel, small hard contact
Trolley polyurethane wheel (this note)1.2 kN600 mm22.0 MPaWheel
Pallet truck nylon wheel (this note)3.0 kN750 mm24.0 MPaWheel
Forklift solid rubber wheel (this note)15.0 kN15 000 mm21.0 MPaWheel
The OSHA criterion carries 55 times the contact pressure of the YB/T criterion. The YB/T value of 1.5 kN over 200 x 200 mm is essentially one sole spread across 40 000 mm2, a pressure of only 0.0375 MPa. Real wheel contact pressure runs from 1.0 to 4.0 MPa, which is 27 to 107 times higher. Checking a wheel load against the YB/T concentrated load criterion silently assumes an area magnification of several tens of times, and that is the quantified root of the belief that a panel passing the uniform load check will carry traffic.

The same comparison applied to impact factors. Industry practice commonly takes 1.5 to 2.0 for forklifts. The energy method here gives 2.365 for a forklift at h = 2 mm, 3.307 for a pallet truck and 4.435 for a trolley. The rule of thumb is slightly optimistic for the forklift, 2.0 against 2.365, and badly inadequate for light wheels, 2.0 against 4.435. This confirms section 6: impact factor rules of thumb are calibrated on heavy loads and systematically underestimate light ones.

Chinese suppliers advise that vehicular areas "should use G405 or higher, with bearing bar depth not less than 40 mm and thickness not less than 5 mm, and cross bar spacing not too wide". The direction is right, but the emphasis falls on the bearing bar. Section 9 shows the cross bar should be changed first: raising the allowable concentrated load by 77 percent costs 2.20 kg/m2 through the cross bar, while deepening the bearing bar costs 13.08 kg/m2 and delivers only 12 percent.

Three actions for site

ActionQuantified targetBasis
Control step between adjacent panelsh not exceeding 1 mm (typically 2 to 3 mm)Step from 3 to 1 mm takes the trolley impact factor from 5.15 to 3.53, cutting equivalent static load by 31 percent
Route wheels over mid-panelAt least 200 mm from a panel edgeEdge n_eff is only 6.64, so stress is 1.85 times the mid-panel value
Enlarge the cross bar first on wheel routes6x6 to 8x8Costs 2.20 kg/m2, raises allowable concentrated load 77.4 percent, reduces open area by only 1.6 percentage points

13. Quick selection table

TrafficP per wheel (kN)P_eff after impact, h = 2 mm (kN)G405/30/100 at 1000 mmG405/30/100 at 800 mmRecommendation
Single pedestrian1.5 (code basis)-PassesPasses4 kN/m2 uniform is sufficient
Hand trolley1.25.32Passes (79.6 MPa, 1.64 mm)PassesKeep steps within 2 mm
Pallet truck3.09.92Passes, 23 percent reservePassesRun mid-panel; steps within 1 mm
3 t forklift15.035.48Fails (2.3 to 2.7 times over)FailsUse heavy duty type or add a support beam under the wheel track

Upsizing the grating is not the right way to force a forklift route through. To bring a 15 kN wheel load, 35.48 kN after impact, inside a 4 mm deflection limit requires k_sys of at least 35 480 / 4 = 8 870 N/mm, which is 2.74 times the baseline 3 240 N/mm. An 8x8 cross bar only reaches 5 747 N/mm, still short. The correct answer is to add a support beam beneath the wheel track and bring the span below 600 mm, while remembering from section 7 that short spans weaken transverse distribution, since n_eff drops from 12.29 to 5.78. The shortened span must therefore be recalculated rather than assumed safe.

14. Where this model applies and where it does not

The coupled beam model rests on four simplifications. Confirm that the case satisfies them before use.

AssumptionDirection of errorWhat to do when it is violated
Cross bar rotationally fixed by the barsOverestimates k_c and hence n_effPressure-locked rather than welded grating has lower joint rotational stiffness; measured n_eff will fall below the calculated value, so check against the code's 200 mm
Bars simply supported, load at midspanConservative, midspan is worstContinuous spans may be reduced, but YB/T 4001.1-2019 assumes simple support, so stay consistent
Bar lateral torsional movement ignoredOverestimates capacityAt 100 mm cross bar spacing lateral restraint is adequate; at 200 mm spacing check lateral stability separately
Contact patch aspect ratio 2.5Affects a and therefore M by about 10 percentRecheck with 2.0 for wide solid rubber and 3.0 for narrow hard wheels

The model gives a theoretical upper bound on the distribution width. Design should treat the 200 mm of YB/T 4001.1-2019 as the lower bound and the value calculated here as a reference upper bound. The gap between them, from 1.84 to 3.38 times, is precisely the reserve the code keeps for construction uncertainty.