Question 4 of 9: Why a plate load test in saturated clay is size-independent

Question 4: Why a plate load test in saturated clay is size-independent (7 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A saturated clay tested undrained, so $\phi_u = 0$ and $c = c_u$. A rigid bearing plate of width B_P and a proposed foundation of width B_F, both founded at the same depth D_f in the same clay, so both carry the same surcharge $q = \gamma D_f$. Terzaghi's equation is supplied as $q_{ult} = cN_c + qN_q + 0.5\gamma B N_{\gamma}$.

Find. Show that $(q_{ult})_P = (q_{ult})_F$ for any pair of widths B_P and B_F, and state the assumptions the proof needs.

Approach. Evaluate Terzaghi's three bearing-capacity factors at phi = 0, show that the only width-dependent term is multiplied by N_γ = 0, and conclude that the surviving expression contains no B.

1. Write the bearing-capacity factors at phi_u = 0. Terzaghi's factors are $N_q = \dfrac{e^{2\left(3\pi/4-\phi/2\right)\tan\phi}}{2\cos^{2}\!\left(45+\phi/2\right)}$, $N_c = (N_q-1)\cot\phi$ and $N_{\gamma} = \tfrac{1}{2}\left(\dfrac{K_{p\gamma}}{\cos^{2}\phi}-1\right)\tan\phi$. Substituting phi = 0 gives $$N_q = \frac{e^{0}}{2\cos^{2}45^{\circ}} = \frac{1}{2(0.5)} = 1, \qquad N_c = 5.7 \ \text{(limiting value)}, \qquad N_{\gamma} = 0 .$$ The vanishing of N_γ is not an approximation: every term in it carries a factor tan phi, which is identically zero.
2. Substitute into Terzaghi's equation. With $c = c_u$, $N_q = 1$ and $N_{\gamma} = 0$, $$q_{ult} = c_u N_c + q N_q + 0.5\,\gamma B \,(0) = \boxed{\,q_{ult} = 5.7\,c_u + \gamma D_f\,}$$ The width B has disappeared from the expression altogether — it survived only in the self-weight term, and that term is now zero.
3. Apply the result to the plate and to the foundation. Because both are founded at the same depth in the same clay, both see the same $q = \gamma D_f$ and the same c_u, so $$(q_{ult})_P = 5.7\,c_u + \gamma D_f = (q_{ult})_F \quad \text{for all } B_P,\ B_F .$$ As a numerical illustration, take $c_u = 60$ kPa, $\gamma = 18$ kN/m$^3$ and $D_f = 1.5$ m, so $q = 27.0$ kPa. Then a 0.30 m plate, a 0.60 m plate, a 1.0 m footing and a 4.0 m footing all return $q_{ult} = 5.7(60) + 27.0 = 369.0$ kPa.
4. Check the square-footing form as well. Terzaghi's square-footing version, $q_{ult} = 1.3cN_c + qN_q + 0.4\gamma B N_{\gamma}$, behaves identically: the 0.4 gamma B term is again multiplied by zero and $$q_{ult} = 1.3(5.7)c_u + \gamma D_f = 7.41\,c_u + \gamma D_f,$$ which for the same illustrative data is $7.41(60) + 27.0 = 471.6$ kPa for a 0.30 m square plate and for a 3.0 m square footing alike. The shape constant matters; the size does not.

The physical reading is that in a phi = 0 material the shear strength on the failure surface is the same everywhere, whatever the confining stress. Enlarging the footing enlarges the failure surface and the volume of soil mobilised in exact proportion, so the average pressure at failure is unchanged. In a frictional soil the strength grows with depth, so the deeper, larger mechanism under a wide footing is stronger and capacity does increase with B.