Which method and strength parameter are used for the bearing capacity of a mat on cohesive soil (phi = 0)?

Skempton's Nc with undrained cohesion cu

Terzaghi's Nq with friction angle phi

Meyerhof's SPT formula with N value

For a wide mat on sand, which criterion usually governs the design?

Allowable settlement, since qu is very large

Shear failure, since qu is small

How does the width B of the foundation affect ultimate bearing capacity in the two soils?

No effect in clay, increases it in sand

Increases it in clay, none in sand

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Mat bearing capacity: cohesive vs cohesionless

Short answer

Cohesive soil (φ=0): Use Skempton: q_nu = c_u N_c, N_c for square = 6(1+0.2D_f/B) ≤ 9, rectangle adjusted; safe q_ns = q_nu/F. Cohesionless: Use Terzaghi with φ; q_u = q N_q + 0.5γ B N_γ, but settlement governs, use SPT-based Meyerhof formula. Key difference: cohesive q_u independent of width, shear controls; cohesionless q_u increases with width, settlement controls.

Describe the procedure of determining the bearing capacity of cohesive and cohesionless soil in case of mat foundation. What is the basic difference between them?

Cohesive soil (clay, $\phi = 0$ )

Find undrained cohesion $c_u$ (from $q_u/2$ of an unconfined test or a vane shear test).
Use Skempton's bearing capacity factor: for a square mat $N_c = 6\left(1+0.2\dfrac{D_f}{B}\right)\le 9$ , and for a rectangle $N_c = \left(0.84+0.16\dfrac{B}{L}\right)N_{c(sq)}$ .
Net ultimate capacity $q_{nu} = c_uN_c$ ; net safe capacity $q_{ns} = q_{nu}/F$ ( $F \ge 3$ ).
Check that the net pressure $\left(\dfrac{Q}{A} - \gamma D_f\right) \le q_{ns}$ and then check consolidation settlement.

Cohesionless soil (sand, gravel)

Find $\phi$ from SPT $N$ (corrected for overburden) or direct tests.
Use Terzaghi's equation with shape factors, $q_u = qN_q + 0.5\gamma BN_\gamma$ (use $\gamma'$ and correction factors if the water table is near).
Obtain the safe pressure: $q_s = q_u/F$ with $F = 3$ .
As $q_u$ is very large for a wide mat, the allowable settlement governs. Use the SPT-based formula $q_{na} = 19.16\,N\,F_d\,\dfrac{S_a}{25.4}$ (Meyerhof, for mats) with $F_d = 1 + 0.33D_f/B \le 1.33$ .

Basic difference

Point	Cohesive soil	Cohesionless soil
Strength parameter	$c_u$ ( $\phi=0$ )	$\phi$ (from $N$ )
Effect of width $B$	None on $q_u$	$q_u$ increases with $B$
Controlling criterion	Shear or consolidation settlement	Settlement (usually)
Method	Skempton	Terzaghi/Meyerhof, SPT

✓ Answer checked by Claude Sonnet 5.5

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Mat bearing capacity: cohesive vs cohesionless

Cohesive soil (clay, ϕ=0\phi = 0ϕ=0)

Cohesionless soil (sand, gravel)

Basic difference

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