The Lugeon Test: Methodology, Interpretation, and Applications in Geotechnical Engineering

Abstract

This worked example designs a three-hinged parabolic arch for a single symmetric span. We select geometry and loads, compute support reactions and horizontal thrust, show that with the parabolic profile the arch carries the chosen uniform load entirely in compression (no bending), size a rectangular concrete arch rib, and check basic stresses and bearing. All calculations are shown step-by-step.

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1. Problem statement and assumptions

Design a three-hinged parabolic arch with:

• Span (horizontal projection) $L = 30.0\ \text{m}$.

• Rise at crown $f = 6.0\ \text{m}$ (rise/span = 6/30 = 1/5).

• External uniform vertical load on the span (superimposed dead + live) $w_{sup} = 20.0\ \text{kN/m}$ (per horizontal metre).

• Arch rib cross-section chosen as rectangular, width (out of plane) $b = 0.60\ \text{m}$, depth (thickness in vertical plane) $t = 1.00\ \text{m}$ so area $A = b \times t = 0.60\times1.00 = 0.60\ \text{m}^2$.

• Material: reinforced concrete, characteristic compressive strength $f'_c = 25\ \text{MPa}$ (we will use a conservative allowable compressive stress for service checks).

• Self-weight of the arch rib included as uniform load: concrete density $\gamma_c = 25.0\ \text{kN/m}^3$. Self-weight linear load $w_{self} = \gamma_c \times A$ (per horizontal metre).

• Total uniform load $w = w_{sup} + w_{self}$.

• Three-hinged arch (hinges at both supports and crown) — statically determinate.

• Parabolic arch geometry follows the equation of a parabola so that with a uniform horizontal projection load the arch can act funicularly (no bending) if the horizontal thrust is the correct value.

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2. Step 1 — compute self-weight (linear load)

Area $A = 0.60\ \text{m}^2$ (given).

Concrete unit weight $\gamma_c = 25.0\ \text{kN/m}^3$.

Self-weight per horizontal metre:

$$w_{self} = \gamma_c \times A.$$

Compute digit-by-digit:

• $A = 0.60$ (given).

• $w_{self} = 25.0 \times 0.60$.
  $25.0\times0.60 = 25.0\times6/10 = (25.0\times6)/10 = 150.0/10 = 15.0$.

So $w_{self} = 15.0\ \text{kN/m}$.

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3. Step 2 — total uniform load on horizontal span

Superimposed load $w_{sup} = 20.0\ \text{kN/m}$.

Total uniform load:

$$w = w_{sup} + w_{self} = 20.0 + 15.0 = 35.0\ \text{kN/m}.$$

So $w = 35.0\ \text{kN/m}$.

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4. Step 3 — vertical support reactions

For a symmetric uniformly loaded span on a three-hinged arch, vertical reactions at supports are simply half the total vertical load:

Total vertical load on span:

$$W = w \times L = 35.0\ \text{kN/m} \times 30.0\ \text{m}.$$

Compute:

• $35.0\times30.0 = 35.0\times3\times10 = 105.0\times10 = 1050.0$.

So $W = 1050.0\ \text{kN}$.

Each support vertical reaction:

$$R_v = \frac{W}{2} = \frac{1050.0}{2} = 525.0\ \text{kN}.$$

So vertical reactions are $525.0\ \text{kN}$ at each support.

Check: $525.0 + 525.0 = 1050.0\ \text{kN}$ — matches total.

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5. Step 4 — horizontal thrust for parabolic arch under uniform horizontal projection load

For a parabolic arch loaded with a uniform vertical load distributed over the horizontal span, the horizontal thrust $H$ required so that the bending moment everywhere is zero (i.e., the arch is the funicular) is:

$$H = \frac{w L^2}{8 f}.$$

Plug values:

• $w = 35.0\ \text{kN/m}$.

• $L = 30.0\ \text{m}$ → $L^2 = 30.0^2 = 900.0\ \text{m}^2$.

• $f = 6.0\ \text{m}$.

Compute numerator:

$$w L^2 = 35.0 \times 900.0 = 31500.0\ \text{kN}\cdot\text{m}.$$

(Compute: $35\times900 = 35\times9\times100 = 315\times100 = 31500$.)

Compute denominator:

$$8f = 8\times6.0 = 48.0.$$

Now $H = 31500.0 / 48.0$.

Divide stepwise:

• $48 \times 600 = 28800$.

• Remainder: $31500 - 28800 = 2700$.

• $48 \times 50 = 2400$. Remainder now $2700 - 2400 = 300$.

• $48 \times 6 = 288$. Remainder $300 - 288 = 12$.

• So far quotient = $600 + 50 + 6 = 656$ with remainder $12$.

• Remainder $12 / 48 = 0.25$.

Hence $H = 656.25\ \text{kN}$.

So the horizontal thrust is $\boxed{H = 656.25\ \text{kN}}.$

This value is the thrust that makes the parabolic arch carry the uniform horizontal-projection load in pure compression (zero bending moment everywhere).

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6. Step 5 — check bending moment (verification that M = 0)

For a parabolic three-hinged arch with the computed $H = wL^2/(8f)$, the bending moment at any section is zero (the arch becomes a funicular line for that loading). We can show this briefly at a representative section $x$ measured from the left support.

Parabola equation (vertical coordinate $y$, crown at $x = L/2$):

$$y(x) = \frac{4f}{L^2}\, x(L - x).$$

Compute coefficient:

$$\frac{4f}{L^2} = \frac{4\times6.0}{900.0} = \frac{24.0}{900.0} = 0.0266666667\ (\text{m}^{-1}).$$

Internal bending moment at section $x$ (left portion) is:

$$M(x) = H\,y(x) - R_v\,x + \frac{w x^2}{2}.$$

With $H = \dfrac{wL^2}{8f}$ and the parabolic $y(x)$, substitution simplifies to $M(x)=0$ for all $x$. (We confirm numerically at a few points if desired; e.g., at $x=5$, $x=7.5$, $x=15$ all give $M=0$.)

Thus the arch is axially loaded in compression only for this uniform load and parabolic shape.

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7. Step 6 — axial force and compressive stress checks

7.1 Axial force at crown (midspan)

At the crown (arch crown slope = 0), the axial force equals the horizontal thrust $H$ (since vertical component of axial force is zero there for symmetric load). So:

$$N_{crown} = H = 656.25\ \text{kN}.$$

Axial compressive stress at crown (average):

$$\sigma_{crown} = \frac{N_{crown}}{A} = \frac{656.25\ \text{kN}}{0.60\ \text{m}^2}.$$

Compute:

• $656.25 / 0.60 = 656.25 \div 0.6$.

• Dividing by 0.6 is same as multiplying by $10/6 = 1.666666667$.

• $656.25 \times 1.666666667 = 1093.75$ (step: $656.25\times1 = 656.25$; $656.25\times0.666666667 = 437.5$; sum = 1093.75).

So

$$\sigma_{crown} = 1093.75\ \text{kN/m}^2 = 1093.75\ \text{kPa} = 1.09375\ \text{MPa}.$$

7.2 Compare with allowable concrete compressive stress (service check)

Take a conservative service allowable compressive stress for concrete: for checking service stress we might use an allowable $\sigma_{allow} \approx 0.4\,f'_c$. With $f'_c=25\ \text{MPa}$:

$$\sigma_{allow} = 0.4\times25 = 10.0\ \text{MPa} = 10,000\ \text{kPa}.$$

Our computed compressive stress $1.09375\ \text{MPa}$ is much less than $10.0\ \text{MPa}$. Factor:

$$\text{Utilization} = \frac{1.09375}{10.0} = 0.109375 \approx 10.9\%.$$

So the section is more than adequate in direct compression capacity.

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8. Step 7 — axial force at springing and bearing pressure check

At the springing (support), the axial force magnitude equals $N_{support} = \sqrt{H^2 + V_{support}^2}$ where $V_{support}$ is the vertical reaction at each support (for symmetric loading). Calculate:

• $H = 656.25\ \text{kN}$.

• $V_{support} = R_v = 525.0\ \text{kN}$.

Compute $N_{support}$:

Square terms:

• $H^2 = 656.25^2$. Compute stepwise:

  $656.25 \times 656.25$.

  It's convenient to note 656.25 = 656 + 0.25. However easier: $656.25 = 656.25$. Use multiplication:

  $656.25^2 = (656 + 0.25)^2 = 656^2 + 2\times656\times0.25 + 0.25^2$.

  Compute:

  • 656^2 = 656\times656 = (600+56)^2 = 600^2 + 2\times600\times56 + 56^2 = 360000 + 67200 + 3136 = 430, (let's do accurately) 360000 + 67200 = 427200; \(427200 + 3136 = 430336. So $656^2 = 430336$.

  • $2\times656\times0.25 = 2\times656\times1/4 = 656\times0.5 = 328.0$.

  • $0.25^2 = 0.0625$.

  Sum: $430336 + 328.0 + 0.0625 = 430664.0625$.

Thus $H^2 = 430664.0625\ \text{(kN)}^2$.

• $V_{support}^2 = 525.0^2 = 525\times525$.

  Compute $500\times550?$ Better: $525^2 = (500+25)^2 = 500^2 + 2\times500\times25 + 25^2 = 250000 + 25000 + 625 = 275625$.

So sum:

$$N_{support}^2 = 430664.0625 + 275625 = 706289.0625.$$

Now $N_{support} = \sqrt{706289.0625}$.

Compute square root — approximate:

• $840^2 = 705600$. Remainder $706289.0625 - 705600 = 689.0625$.

• $841^2 = 707281$ (since $841^2 = 840^2 + 2*840 +1 = 705600 +1680 +1 = 707281$) which is larger than our number. So root is between 840 and 841.

Linear interpolation:

• Difference between squares: $707281 - 705600 = 1681$.

• Our offset above 705600 is $689.0625$.

• Fraction = $689.0625/1681 \approx 0.4099.$

Thus approximate root = $840 + 0.4099 \approx 840.4099\ \text{kN}$.

We can give $N_{support} \approx 840.41\ \text{kN}$.

So axial force at support ~ $840.41\ \text{kN}$.

Bearing pressure check

If bearing area under base of arch footing is (example) $A_{bearing} = b \times l_{base}$. Suppose we provide a footing that gives bearing area of $1.5\ \text{m} \times 1.0\ \text{m} = 1.5\ \text{m}^2$ under each support (this is a chosen design variable — arch rib cannot directly transfer all vertical load to the soil without a footing).

Total vertical reaction to be supported $= V_{support} = 525.0\ \text{kN}$. Bearing pressure $q = \dfrac{525.0}{1.5} = 350.0\ \text{kN/m}^2 = 350.0\ \text{kPa}.$

Typical allowable soil bearing for good soil might be 200–300 kPa; for improved soil or shallow footings maybe higher. If allowable $q_{allow} = 250\ \text{kPa}$, then $q = 350\ \text{kPa}$ would be too large and a larger footing would be needed. For example, required bearing area to limit $q$ to 250 kPa:

$$A_{req} = \frac{525.0}{250.0} = 2.10\ \text{m}^2.$$

So a footing base, e.g., $2.10\ \text{m}^2$ (say $1.05\ \text{m}$ by $2.0\ \text{m}$ or $1.5\ \text{m} \times 1.4\ \text{m}$) would be provided.

Conclusion: Provide footing sized to keep bearing pressure within allowable soil values.

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9. Step 8 — slenderness and stability notes

Although direct compressive stresses are small (≈1.09 MPa), arch members that are slender in the lateral direction or have large unsupported lengths can be susceptible to buckling or local instability. For this rectangular rib of depth $t=1.00\ \text{m}$ and out-of-plane width $b=0.60\ \text{m}$, lateral buckling of the rib around the out-of-plane axis is unlikely at these stress levels, but stability checks (local buckling, lateral torsional buckling for steel arches, and second-order effects for very shallow arches) should be done for final design. When the arch is acting purely axially, buckling checks for the rib element and global stability of the arch-footing system are required in detailed design.

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10. Step 9 — reinforcement and concrete detailing (summary guidance)

Because compressive stresses are low, the concrete section has ample compression capacity. However, practical reinforced-concrete arch design must include:

• Longitudinal reinforcement to resist tension that may occur under transient or asymmetric loads (wind, eccentric live loads) and to provide ductility. For purely compressive service cases, minimum longitudinal reinforcement is still required by code (typical minimum area ~0.2–0.5% of gross area depending on code). For our section $A=0.60\ \text{m}^2$, 0.5% gives steel area $=0.005\times0.60 = 0.003\ \text{m}^2 = 3000\ \text{mm}^2$ (e.g., 6 bars of 25 mm dia gives area $=6\times490.9 \approx 2945\ \text{mm}^2$, close to 3000). Final reinforcement must follow code limits, cover, spacing, and durability requirements.

• Transverse (shear/ties) reinforcement to prevent splitting and provide confinement at supports and crown.

• Adequate bearing plates or footings to distribute support reactions to soil.

• Construction provision: centering/formwork for concrete until arch gains strength, or use precast segments with temporary supports.

Detailed reinforcement design is code-driven and depends on serviceability, ductility requirements, and seismic detailing rules — follow the relevant concrete code (e.g., Eurocode, ACI, IS codes).

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11. Summary of results (worked example)

• Geometry: $L=30.0\ \text{m},\ f=6.0\ \text{m}.$

• Cross-section: rectangular $b=0.60\ \text{m},\ t=1.00\ \text{m},\ A=0.60\ \text{m}^2.$

• Self-weight: $w_{self} = 15.0\ \text{kN/m}.$

• Superimposed load: $w_{sup} = 20.0\ \text{kN/m}.$

• Total uniform load: $w = 35.0\ \text{kN/m}.$

• Total vertical load $W = 1050.0\ \text{kN}.$

• Vertical reactions: $R_v = 525.0\ \text{kN}$ each.

• Horizontal thrust required (funicular): $H = 656.25\ \text{kN}.$

• Arch carries load in pure compression (bending moment $M(x)=0$).

• Axial force at crown $N_{crown} = 656.25\ \text{kN}$.

• Average compressive stress at crown $\sigma_{crown} = 1.09375\ \text{MPa}$ — well within typical allowable service stress.

• Axial resultant at support $N_{support} \approx 840.41\ \text{kN}$. Footing area chosen must limit bearing pressure (e.g., $A_{req} \ge 2.10\ \text{m}^2$ for 250 kPa allowable).

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12. Practical design remarks

1. Why a three-hinged parabolic arch? — The three-hinged configuration is statically determinate (easy analysis), less sensitive to differential settlement (hinges accommodate rotation), and the parabolic geometry is the funicular shape for uniformly distributed horizontal-projection loads so internal bending vanishes when the correct $H$ is present.

2. Load variations and safety: Real structures are subject to nonuniform loads, asymmetric loads, wind, seismic forces, and construction-stage conditions. These introduce bending and additional demands. The design above is an idealized base case — perform additional load combination checks, second-order analysis, and limit state checks per the applicable code.

3. Footings and abutment design: The arch thrust and vertical reactions must be transferred to earth through properly designed abutments and footings. Ensure soil bearing, sliding, overturning, and durability checks.

4. Construction sequence: Reinforced concrete arches require centering until concrete attains strength. Precast or steel arches use different erection sequences; temporary bracing might be needed.

5. Detailing for durability and maintenance: Provide adequate cover, drainage, waterproofing of fill above the arch (if any), and access for inspection.

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Conclusion

This worked example demonstrates the classical design workflow for a three-hinged parabolic arch: select geometry and section, compute loads (including self-weight), determine reactions and horizontal thrust, verify internal force state (here pure compression), and perform basic section and bearing checks. The numbers show a conservative concrete section performs well for the chosen loads, but final design must include detailed reinforcement, stability/buckling checks, footing design, and code-based load combinations.

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FAQ

Q1: Why choose a parabolic arch instead of another shape?
A parabolic arch is the funicular shape for a uniform vertical load over the horizontal span. If you choose the parabolic profile and the horizontal thrust $H = wL^2/(8f)$, bending moments are zero and the arch works purely in compression — this is the most material-efficient shape for that loading.

Q2: Why use three hinges rather than two or fixed?
A three-hinged arch is statically determinate (simpler analysis), tolerant to differential settlements (hinges allow rotation), and easier to evaluate for moving/asymmetric loads. Two-hinged and fixed arches can be more efficient but require indeterminate analysis and are more sensitive to support movements.

Q3: I computed a small compressive stress — am I safe from buckling?
Low compressive stress is good, but buckling depends on geometry and slenderness as well as axial load. Always check local and global buckling (member slenderness, lateral restraint). Provide sufficient cross-section and bracing or limit unsupported spans of the rib.

Q4: What if actual loads are nonuniform (e.g., point loads, traffic)?
Nonuniform loads produce bending. The parabolic/funicular ideal eliminates bending only for the specific uniform load case. Real designs must analyze asymmetric and moving loads, and design reinforcement and sections to resist bending and combined axial+moment stresses.

Q5: How should footings be sized?
Size footings to keep bearing pressure ≤ allowable soil bearing capacity (determine from geotechnical investigation), and check for sliding and overturning given the horizontal thrust. In the example, to limit bearing pressure to 250 kPa we needed ≈2.10 m² area per support (for the given reactions).