Example Calculation — Default Settings

Important: This example uses the default soil profile and pile parameters in dpmPILE. It is intended for demonstration and training only. For design, users must input site-specific soil parameters and verify all assumptions.

1. Problem Statement

Compute the axial factored resistance of a bored cast-in-place concrete pile using the default soil profile and parameters in dpmPILE.

Pile and analysis settings
Pile type: Friction + End Bearing
Pile length: \( L = 15.0\,\text{m} \)
Pile diameter: \( D = 1.0\,\text{m} \)
Ignore shaft friction (top / bottom): 0.0 / 0.0 m
Groundwater: considered, at ground surface (top of Layer 1)
Unit weight of water: \( \gamma_w = 9.81\,\text{kN/m}^3 \)
Limiting effective overburden depth: \( z_{\text{lim}} = 30D = 30\,\text{m} \)
Pile tip depth: \( z_{\text{tip}} = 15\,\text{m} < z_{\text{lim}} \Rightarrow \) capping is not engaged.

2. Soil Profile

The default soil profile in this example consists of three layers:

Layer 1: Clay       0 – 3 m   (thickness = 3 m)
Layer 2: Sand       3 – 9 m   (thickness = 6 m)
Layer 3: Dense sand 9 – 20 m  (pile penetrates to 15 m)

Layer 1 — Clay
Thickness: 3 m (0–3 m)
Total unit weight: \( \gamma = 18\,\text{kN/m}^3 \)
Undrained shear strength: \( S_u = 45\,\text{kPa} \)
Type: cohesive
SPT \( N_1(60) = 8 \) (not used directly in the \( \alpha \)-method here)
Layer 2 — Sand
Thickness: 6 m (3–9 m)
Total unit weight: \( \gamma = 19\,\text{kN/m}^3 \)
Friction angle: \( \varphi = 32^\circ \)
Type: cohesionless
SPT \( N_1(60) = 18 \)
Layer 3 — Dense Sand
Thickness: 11 m (9–20 m), pile penetrates first 6 m (9–15 m)
Total unit weight: \( \gamma = 20\,\text{kN/m}^3 \)
Friction angle: \( \varphi = 36^\circ \)
Type: cohesionless
SPT \( N_1(60) = 30 \)

3. Pile Geometry

Cross-sectional area and perimeter of a circular pile:

Area:

\[ A_p = \frac{\pi D^2}{4} = \frac{\pi \times 1.0^2}{4} \approx 0.785\,\text{m}^2 \]

Perimeter:

\[ P = \pi D = \pi \times 1.0 \approx 3.142\,\text{m} \]

4. Effective Unit Weights

The water table is at the ground surface, so all depths are considered submerged. Effective unit weights:

Clay:       γ' = 18.0 − 9.81 =  8.19 kN/m³
Sand:       γ' = 19.0 − 9.81 =  9.19 kN/m³
Dense sand: γ' = 20.0 − 9.81 = 10.19 kN/m³

5. Effective Overburden Stresses at Layer Mid-Depth

The effective vertical stress is computed by:

\[ \sigma'_v(z) = \sum_i \gamma'_i t_i \]

5.1 Layer 1 — Clay (mid-depth \( z = 1.5\,\text{m} \))

Only submerged clay contributes from 0 to 1.5 m:

\[ \sigma'_v(1.5) = \gamma'_\text{clay} \times 1.5 \approx 8.19 \times 1.5 \approx 12.3\,\text{kPa} \]

5.2 Layer 2 — Sand (mid-depth \( z = 6.0\,\text{m} \))

From 0–3 m clay and 3–6 m sand:

\[ \begin{aligned} \sigma'_v(6) &= \gamma'_\text{clay} \times 3 + \gamma'_\text{sand} \times 3 \\ &\approx 8.19 \times 3 + 9.19 \times 3 \\ &\approx 52.1\,\text{kPa} \end{aligned} \]

5.3 Layer 3 — Dense Sand (mid-depth \( z = 12.0\,\text{m} \))

From 0–3 m clay, 3–9 m sand, and 9–12 m dense sand:

\[ \begin{aligned} \sigma'_v(12) &= \gamma'_\text{clay} \times 3 + \gamma'_\text{sand} \times 6 + \gamma'_\text{dense} \times 3 \\ &\approx 8.19\times 3 + 9.19\times 6 + 10.19\times 3 \\ &\approx 110.3\,\text{kPa} \end{aligned} \]

6. Shaft Resistance

The pile passes through:

3 m of clay (0–3 m)
6 m of sand (3–9 m)
6 m of dense sand (9–15 m, part of the 9–20 m layer)

6.1 Layer 1 — Clay (\( \alpha \)-method)

For undrained clay, dpmPILE uses the \( \alpha \)-method consistent with AASHTO:

\[ \alpha = f\!\left( \frac{S_u}{p_a} \right), \quad p_a \approx 101\,\text{kPa} \]

With \( S_u = 45\,\text{kPa} \):

\[ \frac{S_u}{p_a} \approx \frac{45}{101} \approx 0.45 \le 1.5 \Rightarrow \alpha = 0.55 \]

Unit shaft resistance:

\[ q_{s,1} = \alpha S_u \approx 0.55 \times 45 \approx 24.75\,\text{kPa} \]

Shaft resistance in Layer 1:

\[ F_{s1} = q_{s,1} \; P \; t = 24.75 \times 3.142 \times 3 \approx 233\,\text{kN} \]

6.2 Layer 2 — Sand (\( \beta \)-method)

dpmPILE implements the AASHTO \( \beta \)-method for sand, using:

Atmospheric pressure: \( p_a = 101\,\text{kPa} \)
SPT: \( N_{60} = 18 \)
Friction angle: \( \varphi = 32^\circ \)
Exponent \( m = 0.8 \) (when the "Use \( m = 0.8 \)" option is checked)

Reference pressure:

\[ \sigma_p = 0.47 \; N_{60}^m \; p_a \approx 0.47 \times 18^{0.8} \times 101 \approx 480\,\text{kPa} \]

At mid-depth of Layer 2, \( \sigma'_v \approx 52.1\,\text{kPa} \). The \( \beta \) factor is calculated internally as:

\[ \beta = (1 - \sin\varphi)\,\left(\frac{\sigma_p}{\sigma'_v}\right)^{\sin\varphi} \tan\varphi \]

For \( \varphi = 32^\circ \), this yields approximately:

\[ \beta \approx 0.95 \]

Unit shaft resistance:

\[ q_{s,2} = \beta \sigma'_v \approx 0.95 \times 52.1 \approx 49.6\,\text{kPa} \]

Shaft resistance in Layer 2 (thickness 6 m):

\[ F_{s2} = q_{s,2} \; P \; t \approx 49.6 \times 3.142 \times 6 \approx 935\,\text{kN} \]

6.3 Layer 3 — Dense Sand (\( \beta \)-method, partial layer)

Pile penetrates 6 m into Layer 3 (from 9 to 15 m). Parameters:

\( N_{60} = 30 \)
\( \varphi = 36^\circ \)
\( m = 0.8 \)

Reference pressure:

\[ \sigma_p = 0.47 \; N_{60}^m \; p_a \approx 0.47 \times 30^{0.8} \times 101 \approx 720\,\text{kPa} \]

At mid-depth (\( z = 12\,\text{m} \)), \( \sigma'_v \approx 110.3\,\text{kPa} \). The \( \beta \) value is:

\[ \beta \approx 0.90 \]

Unit shaft resistance:

\[ q_{s,3} = \beta \sigma'_v \approx 0.90 \times 110.3 \approx 99.6\,\text{kPa} \]

Shaft resistance in the 6 m penetration of Layer 3:

\[ F_{s3} = q_{s,3} \; P \; t \approx 99.6 \times 3.142 \times 6 \approx 1878\,\text{kN} \]

6.4 Total Shaft Resistance (Unfactored)

Summing all layers:

\[ \begin{aligned} F_s &= F_{s1} + F_{s2} + F_{s3} \\ &\approx 233 + 935 + 1878 \\ &\approx 3046\,\text{kN} \end{aligned} \]

7. End Bearing Resistance

The pile tip is in the dense sand at \( z = 15\,\text{m} \).

7.1 SPT-based Tip Pressure (Reese & O'Neill)

The program uses the global tip \( N_{60} \) from User Options. For this example:

\[ N_{60,\text{tip}} = 50 \]

Reese & O'Neill (1988) sand tip formula implemented in dpmPILE:

\[ q_p = 0.057 \; N_{60} \times 1000 \]

So:

\[ q_p = 0.057 \times 50 \times 1000 = 2850\,\text{kPa} \]

(Program caps at \( 3000\,\text{kPa} \) if necessary; here no capping occurs.)

7.2 Tip Resistance

Tip resistance at the pile base:

\[ Q_b = q_p A_p = 2850 \times 0.785 \approx 2238\,\text{kN} \]

8. Nominal Axial Resistance

The nominal axial resistance is the sum of total shaft and tip:

\[ \begin{aligned} R_n &= F_s + Q_b \\ &\approx 3046 + 2238 \\ &\approx 5285\,\text{kN} \end{aligned} \]

9. Factored Resistance (LRFD)

dpmPILE uses separate resistance factors for clay shaft, sand shaft, and tip (consistent with AASHTO LRFD):

\( \phi_{\text{clay}} = 0.45 \)
\( \phi_{\text{sand}} = 0.55 \)
\( \phi_{\text{tip}} = 0.50 \)

Clay shaft (Layer 1):

\[ \phi_{\text{clay}} F_{s,\text{clay}} \approx 0.45 \times 233 \approx 105\,\text{kN} \]

Sand shaft (Layers 2 + 3):

\[ F_{s,\text{sand}} = F_{s2} + F_{s3} \approx 935 + 1878 = 2813\,\text{kN} \]

\[ \phi_{\text{sand}} F_{s,\text{sand}} \approx 0.55 \times 2813 \approx 1547\,\text{kN} \]

Total factored shaft:

\[ F_{s,\text{factored}} \approx 105 + 1547 = 1652\,\text{kN} \]

Factored tip resistance:

\[ Q_{b,\text{factored}} = \phi_{\text{tip}} Q_b \approx 0.5 \times 2238 \approx 1119\,\text{kN} \]

Total factored resistance:

\[ \begin{aligned} R_r &= F_{s,\text{factored}} + Q_{b,\text{factored}} \\ &\approx 1652 + 1119 \\ &\approx 2771\,\text{kN} \end{aligned} \]

10. Design Check

Default applied load in the User Options:

\[ Q_{\text{applied}} = 2000\,\text{kN} \]

Factored resistance:

\[ R_r \approx 2771\,\text{kN} \]

Since

\[ R_r = 2771\,\text{kN} > Q_{\text{applied}} = 2000\,\text{kN}, \]

the pile satisfies the LRFD axial capacity requirement for the default settings. In dpmPILE, this appears as an OK design check for the default example.

Note: Numerical values here are rounded for readability. The internal program calculations may show slightly different values due to more precise floating point operations.