|

Pole Embedment Calculator (Brinch Hansen): Required Setting Depth for Directly Embedded Poles in Layered Soil

Table of Contents

Engineering Disclaimer: This content is for educational/informational purposes only. Always verify calculations and designs with a licensed engineer before applying them to actual construction or projects.
01

Pole Embedment Calculator

Pick Load example in the toolbar to see a complete case, then change one value at a time. Press the ? beside any input for its meaning and typical values. Click into a soil row, then a soil type in the guide panel, to fill φ, c and γ. Switch to Check an embedment when you already have a setting depth. Report / Save PDF gives you the inputs, both drawings and every calculation step; Clear empties the form for your own data.

Pole Embedment Calculator Brinch Hansen · ENA C(b)1 App. B

Find how deep to set a directly embedded pole so the soil can resist the overturning load, from your load, pole size and borehole layers. Click any ? for typical values.

Mode

m

1 · Pole & load

kN
×
m
m
m
m
m

2 · Soil profile

Soil typeBottom mφ °c kPaγ kN/m³γsat kN/m³
1
2
3
4
5
6

The last layer continues to any depth. Click a row, then a soil type in the guide, to fill it.

m

3 · Design factors

×
Solving rigid-pole equilibrium…
Reminder: This content is for educational/informational purposes only. Always verify calculations and designs with a licensed engineer before applying them to actual construction or projects. The calculator gives the ultimate lateral capacity of a pole that rotates as a rigid body. It does not check pole deflection or rotation at working load, the pole section, the concrete plug or pole–concrete bond, base and side friction, or long-term creep.
02

What to Enter: Load, Pole, Backfill and Soil

The form follows the order you’d work through a pole foundation: the load from your pole-loading calculation, the pole and hole, then the soil from the borehole log. Both drawings update as you type, so a wrong layer depth or a hole smaller than the pole shows up in red before you calculate.

Which load do I enter?

  • Horizontal load P: the unfactored resultant of wind on the pole and conductors, line-angle and dead-end tension, as one force.
  • Load factor (OCF): multiplies P to the ultimate load Pu the soil must resist at failure. The guide lists the NESC 2017 Table 253-1 factors (2.50 Grade B wind, 2.20 Grade C at crossings, 1.75 Grade C elsewhere); check them against the edition and standard you design to.
  • Load position: either a fixed height above groundline, or an offset below the pole top. With the offset, the load height drops as the pole is set deeper, which is how the pole really behaves.
  • Pole length: only needed for the offset option and for the comparison with the 10% rule and the RUS formula.

How do the hole and backfill change the calculation?

Enter the augered or excavated hole diameter for every backfill; it’s drawn on both figures and checked against the pole diameter. What it does to the calculation depends on the backfill:

  • Concrete, soil-cement or flowable fill: the backfill bonds with the pole into one shaft, so the effective diameter D is the hole diameter and the soil fails around the hole.
  • Compacted excavated soil or crushed rock: D is the pole diameter, and the layer values you enter must describe the soil as compacted around the pole.
Compaction matters more than the formula. In full-scale tests on the same steel pole in similar holes, very loose sand backfill gave an ultimate moment of 137 kN·m, loose sand 253 kN·m and dense native soil 674 kN·m. If your crews don’t compact in thin lifts, enter loose values (φ ≈ 29°, c = 0) for the backfilled depth.

What soil data do I need?

  • Layers: up to six, top down, each with its bottom depth below groundline. The last layer continues to any depth.
  • φ and c: for sand, the drained friction angle with c = 0; for clay under short-term load, the undrained strength cu with φ = 0. The quick-fill buttons use mid-range values from ENA C(b)1 Tables B3 to B5.
  • γ and γsat: bulk unit weight above the water table and saturated weight below it. Leave γsat blank to use γ.
  • Soil type: sets the fill pattern in the drawing and the name in the report. The analysis uses only φ, c and γ.
  • Groundwater: the highest seasonal level. Below it the calculator uses buoyant weight, which cuts the resistance of granular soil.
Spend your site-investigation effort on the top 2 m. In the worked example below, all of F₁, the resistance in front of the pole, comes from the top 1.78 m, and the 1.5 m clay crust alone supplies 161 kN of its 205 kN. A borehole that starts logging at 1.5 m misses the soil that matters most.
03

The Brinch Hansen Method Behind the Numbers

Brinch Hansen (1961) gave the ultimate lateral earth pressure on a short rigid pile in soil with any combination of friction and cohesion. ENA C(b)1-2006 Appendix B applies it to directly embedded poles: the embedded length rotates about a point zr, soil in front of the pole resists above that point and soil behind it resists below.

What is the ultimate soil reaction at each depth?

Ultimate lateral pressure and reaction per metre of pole (Brinch Hansen 1961)
\( p_z = q_z\,K_{q,z} + c\,K_{c,z} \qquad w_z = p_z\,D \qquad q_z = \sum \gamma_i\,h_i \)

Variables

  • q_z: vertical effective overburden at depth z (kPa), using γ′ = γsat − 9.81 below the water table
  • c: cohesion of the layer at depth z (kPa); D: effective diameter (m)
  • K_q,z, K_c,z: earth-pressure coefficients for the layer’s φ at relative depth z/D
Variation of the coefficients with depth (ENA C(b)1 App. B)
\( K_{q,z} = \dfrac{K_q^0 + K_q^\infty\,\alpha_q\,\dfrac{z}{D}}{1 + \alpha_q\,\dfrac{z}{D}} \qquad K_{c,z} = \dfrac{K_c^0 + K_c^\infty\,\alpha_c\,\dfrac{z}{D}}{1 + \alpha_c\,\dfrac{z}{D}} \)
Coefficients at the surface
\( K_q^0 = e^{(\pi/2+\varphi)\tan\varphi}\cos\varphi\tan\left(45^\circ+\dfrac{\varphi}{2}\right) – e^{-(\pi/2-\varphi)\tan\varphi}\cos\varphi\tan\left(45^\circ-\dfrac{\varphi}{2}\right) \)

\( K_c^0 = \left[e^{(\pi/2+\varphi)\tan\varphi}\cos\varphi\tan\left(45^\circ+\dfrac{\varphi}{2}\right) – 1\right]\cot\varphi \)
Coefficients at great depth and the depth factors
\( K_c^\infty = N_c\,d_c^\infty \qquad K_q^\infty = K_c^\infty K_0 \tan\varphi \qquad N_c = \left[e^{\pi\tan\varphi}\tan^2\left(45^\circ+\dfrac{\varphi}{2}\right) – 1\right]\cot\varphi \)

\( d_c^\infty = 1.58 + 4.09\tan^4\varphi \qquad K_0 = 1 – \sin\varphi \)

\( \alpha_q = \dfrac{K_q^0}{K_q^\infty – K_q^0}\,\dfrac{K_0\sin\varphi}{\sin(45^\circ+\varphi/2)} \qquad \alpha_c = \dfrac{K_c^0}{K_c^\infty – K_c^0}\,2\sin\left(45^\circ+\dfrac{\varphi}{2}\right) \)

These reproduce ENA C(b)1 Tables B1 and B2 exactly, including clay with φ = 0, which the calculator evaluates as the limit at a very small φ.

How is the rotation point and the capacity found?

Rigid-pole equilibrium (ENA C(b)1 App. B3)
\( F_1 = \int_0^{z_r} w_z\,dz \qquad F_2 = \int_{z_r}^{L} w_z\,dz \qquad F_1\,(e + z_1) = F_2\,(e + z_2) \qquad H_u = F_1 – F_2 \)

Variables

  • e: height of the load above groundline; L: embedment depth
  • z1, z2: depths of the resultants F₁ and F₂
  • The moment equation (about the load point) fixes zr; the horizontal balance then gives the ultimate load Hu
Design check, solved for the smallest L
\( \phi_s\,k\,H_u(L) \ge P_u = \text{OCF}\times P \)

The calculator integrates w in 2 mm steps, splitting at every layer boundary, and bisects on L. The required depth is rounded up to the next 0.1 m, and the sizing table lists every 0.1 m step around it. k is the optional ENA calibration: 1.01 for undrained (clay) parameters, 0.60 for drained parameters, or 1.0 for the raw Hansen value.

Where is the largest bending moment in the pole?

Zero-shear depth and maximum moment under Pu
\( \int_0^{z_0} w_z\,dz = P_u \qquad M_{max} = P_u\,(e + z_0) – \int_0^{z_0} w_z\,(z_0 – z)\,dz \)

The maximum moment sits a little below grade, not at the groundline. Check the pole section at z0.

04

Worked Example: 45 ft Steel Pole in Clay over Sand

This is the first entry in Load example. Load it and you’ll get the same numbers.

Example inputs
Pole13.72 m (45 ft) steel pole, 0.30 m at groundline
LoadP = 8 kN applied 0.61 m below the top, load factor 2.5 → Pu = 20 kN
Hole0.60 m augered hole, concrete backfill → D = 0.60 m
Soil0–1.5 m firm clay, cu = 37 kPa, γ = 17 / γsat = 18 kN/m³; below 1.5 m medium dense sand, φ = 33°, γ = 18.5 / γsat = 20 kN/m³
Groundwater1.5 m below grade
Factorsφs = 1.0, no ENA calibration

What does the calculator return?

Results — required embedment
QuantityValue
Load height above ground at that depth, e10.46 m
Rotation point, zr1.78 m (zr/L = 0.67)
F₁ (in front, above zr)205.0 kN at z₁ = 1.01 m
F₂ (behind, below zr)185.0 kN at z₂ = 2.25 m
Ultimate capacity, Hu = F₁ − F₂20.0 kN = Pu
Groundline moment, Pu·e209.2 kN·m
Maximum moment in the pole212.1 kN·m at z₀ = 0.28 m below grade
Layer share of the soil reactionClay 41% (all in F₁); sand 59% (governs)
Required embedment2.65 m (8.70 ft) → set at 2.70 m
Hu is the small difference of two large forces. F₁ and F₂ are each about ten times the 20 kN load, and the capacity is what’s left after one cancels most of the other. That’s why a small change in depth moves the capacity a lot: from 2.60 m to 2.70 m, Hu rises from 19.0 to 21.0 kN.

How does the sizing table pick the setting depth?

Embedment in 0.1 m steps (Pu = 20.0 kN)
L (m)e (m)zr (m)Hu (kN)Margin
2.5010.611.6617.1−14.3%fails
2.6010.511.7419.0−5.0%fails
2.7010.411.8221.0+4.9%pick
2.8010.311.9023.1+15.5%ok
2.9010.211.9825.4+26.8%ok

“Check this depth in the form” sends 2.70 m to Check mode: utilisation 95.3%, safety factor 2.62 on the unfactored 8 kN load.

Would the 10% + 0.61 m rule have been enough?

Same pole, setting by rule of thumb
L = 0.10 × 13.72 + 0.61 = 1.98 m → utilisation 213.5%
= the pole needs twice the capacity the rule gives it
The rule ignores the load and the soil. It works for light loads in good ground, which is where it came from. Here, with a soft-ish clay crust, a water table at 1.5 m and a Grade B load factor, it falls 0.7 m short. The RUS formula lands between the two: 2.04 m for good soil, 2.44 m for average and 2.91 m for poor.
05

What Moves the Embedment Depth Most?

Each row changes one input from the worked example (required depth 2.65 m) and keeps everything else the same.

How sensitive is the depth to backfill, soil, water and load?

45 ft pole, clay over sand — base case 2.65 m
ChangeRequired Lvs base
Compacted soil or crushed rock backfill (D = 0.30 m pole)3.19 m+0.54 m
Concrete in a 0.75 m / 0.90 m hole2.49 m / 2.37 m−0.16 / −0.28 m
Top clay soft, cu = 18 kPa3.06 m+0.41 m
Top clay stiff, cu = 75 kPa2.26 m−0.39 m
Sand loose (φ = 29°) / dense (φ = 38°)2.79 m / 2.50 m+0.14 / −0.15 m
No groundwater / groundwater at the surface2.59 m / 2.85 m−0.06 / +0.20 m
Load factor 1.75 (Grade C) / 4.02.31 m / 3.16 m−0.34 / +0.51 m
ENA drained calibration (× 0.60)3.21 m+0.56 m
ENA undrained calibration (× 1.01)2.64 m−0.01 m
Backfill and the top layer come first. Losing the concrete costs more depth (+0.54 m) than doubling the hole saves (−0.28 m), and the strength of the top 1.5 m of clay moves the answer by ±0.4 m. The water table and the sand density matter less here because most of the resistance is in front of the pole above the rotation point.
Choose the ENA calibration honestly. For drained (sand) parameters ENA reduces Hansen’s capacity to 60%, which adds more than half a metre in this case. Leaving it at “None” reproduces the raw Hansen value and many spreadsheets, but it isn’t conservative in granular soil.
06

Why Results Differ from Rules of Thumb and Spreadsheets

When your number disagrees with a standard setting depth, an old spreadsheet or another program, check these first.

Which assumptions usually explain the difference?

  • Effective diameter: pole diameter or hole diameter. Concrete backfill is the only case where using the hole is justified.
  • Overburden: some spreadsheets multiply the current layer’s γ by the full depth and ignore the water table. This calculator adds each layer at its own weight and uses buoyant weight below the water.
  • Adequacy check: an “assume a depth, compare moments” check is not the same as Hansen’s equilibrium. In one such workbook, 10.6 ft passed its own check, but solving F₁(e + z₁) = F₂(e + z₂) gives 11.09 ft (111% utilisation at 10.6 ft).
  • c–φ soils: Hansen adds the cohesion and friction terms in full. A large c with a large φ in the same layer can overstate capacity; the calculator warns when you enter one.
  • Load factor and calibration: a legacy OCF of 4.0 against a modern 2.5 load factor, or ENA’s 0.60 drained factor, change the depth more than any refinement in the method.
Validation. The coefficients reproduce ENA C(b)1 Tables B1 and B2. For a full-scale field test on a directly embedded steel pole in dense native-soil backfill (φ = 42.7°, 0.76 m pole, 2.59 m embedment, load 22.86 m above ground), the calculator predicts an ultimate moment of 468 kN·m. The paper’s own Hansen calculation gave 453 kN·m and the pole failed at 674 kN·m, so Hansen was conservative by about a third there. The same paper found Hansen’s mean prediction at 0.84 × measured over all its tests, with a lot of scatter. The calculator engine also agrees with an independent Python implementation to the third decimal.
07

Frequently Asked Questions

Why doesn’t a wider concrete hole save more depth?

A wider shaft pushes against more soil, but Hansen’s coefficients depend on z/D, so they fall as D grows at the same depth. The two effects partly cancel: in the worked example, going from a 0.60 m to a 0.90 m hole (50% wider) saves only 0.28 m, about 11%. Concrete’s bigger benefit is that it removes the uncertainty of compaction: in the field tests, flowable fill reached 1,056 kN·m at a shorter embedment than any compacted backfill.

Can I model a ring of compacted backfill around the pole?

Not as a separate ring. With soil or crushed-rock backfill the calculator uses the pole diameter and the layer values you enter. When the hole is much wider than the pole, part of the failure surface runs through the backfill, so the calculator warns you to describe the top layer as placed, not as undisturbed soil.

Does it check how far the pole will lean?

No. Hansen gives the ultimate capacity, the load at which the soil fails. Rotation at working load needs a stiffness-based method; ENA treats it separately, and deflection-based foundation programs cover it. If pole tilt matters for your structures (dead-ends, angle poles, clearance-critical spans), check it separately.

Why does groundwater increase the depth?

Below the water table the soil is buoyed up, so its effective weight drops from about 20 to about 10 kN/m³. Granular resistance is proportional to that overburden, so it roughly halves below the water. Clay under undrained load is barely affected, because its resistance comes from cu. Use the wet-season level; in low-lying sites that is often near the surface.

Why Brinch Hansen and not Broms?

Broms’ method handles either pure sand or pure clay. Brinch Hansen handles soils with both c and φ and lets the resistance vary layer by layer, which matches a real borehole log. In the full-scale field tests above, Broms overpredicted capacity on average (1.22 × measured), while Hansen underpredicted (0.84 × measured). Both had a coefficient of variation of about 0.4, so neither method is precise.

08

Summary of Equations

Ultimate soil reaction
\( p_z = q_z K_{q,z} + c\,K_{c,z} \qquad w_z = p_z D \qquad q_z = \sum \gamma_i h_i \)
Coefficients with depth
\( K_{q,z} = \dfrac{K_q^0 + K_q^\infty \alpha_q z/D}{1 + \alpha_q z/D} \qquad K_{c,z} = \dfrac{K_c^0 + K_c^\infty \alpha_c z/D}{1 + \alpha_c z/D} \)
Rigid-pole equilibrium and design check
\( F_1 (e + z_1) = F_2 (e + z_2) \qquad H_u = F_1 – F_2 \qquad \phi_s\,k\,H_u \ge \text{OCF}\times P \)
Maximum moment in the embedded pole
\( M_{max} = P_u (e + z_0) – \int_0^{z_0} w_z (z_0 – z)\,dz \qquad \int_0^{z_0} w_z\,dz = P_u \)
Comparison rules
\( L = 0.10\,\ell + 0.61\ \text{m} \qquad H = \dfrac{S_e L^{3.75}}{\ell – 2 – 0.662 L}\ \ (\text{RUS; H in lb, lengths in ft}) \)
Summary: A directly embedded pole resists overturning by rotating about a point below grade, with the soil in front of it above that point and behind it below. Brinch Hansen’s coefficients give the soil’s ultimate resistance layer by layer, and the depth that balances it against the factored load is usually deeper than the 10% + 0.61 m rule in weak or wet ground. Get the top 2 m of soil, the backfill and the load factor right; they move the answer far more than refinements in the method.

References: J. Brinch Hansen, “The ultimate resistance of rigid piles against transversal forces,” Danish Geotechnical Institute Bulletin No. 12, 1961. Energy Networks Association, ENA C(b)1-2006, Guidelines for Design and Maintenance of Overhead Distribution and Transmission Lines, Appendix B. A. Haldar, Y. V. S. N. Prasad and T. R. Chari, “Full-scale field tests on directly embedded steel pole foundations,” Canadian Geotechnical Journal, vol. 37, 2000. USDA Rural Utilities Service, Bulletin 1724E-200, Design Manual for High Voltage Transmission Lines. IEEE C2-2017, National Electrical Safety Code. CIGRE, The Design of Transmission Line Support Foundations: An Overview.

Similar Posts

Leave a Reply

Your email address will not be published. Required fields are marked *