Presented by: Civil Engineering Academy

Bearing CapacityPresented by: Civil Engineering Academy

Different types of foundations◦ Shallow – spread footings, drilled footings, mats.

◦ Deep – piles, drilled shafts, and piers

Shallow foundations – depth is shallower than width.

Deep foundations – depth is larger than width.

The ultimate bearing capacity or pressure is the pressure that would result in shear failure. To avoid this, applied pressures are limited to allowable bearing pressures (factor safety is applied).

Eq 36.1(b) qult = 0.5𝛾BN𝛾S+cNcS+(pq+ 𝛾Df)Nq

density

width

bearing factor (Table 36.2)

shape factor (Table 36.5)cohesion

bearing factor (Table 36.5)

shape factor (Table 36.4)

surface surcharge (if any)

depth of footing

bearing factor (T. 36.5)

See CERM Table 36.2

Clays (internal friction=0, N𝛾=0, Nq=1)◦ Eq. 36.6(b): qult = cNc+𝛾Df

◦ Eq. 36.5: Su=c=Suc/2

Sands (c=0)◦ Eq. 36.9(b): qult = 0.5𝛾BN𝛾S+(pq+ 𝛾Df)Nq

P

D

B

Wc

Stress at bottom of footing: qLoad= 𝑃+𝑊𝑠𝑜𝑖𝑙+𝑊𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒

𝐵𝑥𝐿

qallowable= 𝑞𝑢𝑙𝑡 −𝛾𝐷

𝐹𝑆

qnet= (𝑞𝑢𝑙𝑡 −𝛾𝐷) Eq. 36.3(b)

Eq. 36.4

Case1: ◦ If the water table is located where Df ≥ D1 ≥ 0

Then q = D1𝛾+D2(𝛾sat-𝛾w)

Case 2: ◦ If the water table is located where B ≥ d ≥ 0

Then q = Df𝛾 and the value of 𝛾 in the N𝛾 term is replaced with 𝛾 = 𝛾’ + d/B(𝛾-𝛾’)

Case 3: ◦ If the water table is located where d ≥ B, there is no

effect.

If a foundation has a moment and a vertical load then a eccentric load is created.

Eccentricity, 𝜖 = M/P

qmax = 𝑃

𝐵𝐿(1 +

6𝑒

𝐵) ; qmin =

𝑃

𝐵𝐿(1 −

6𝑒

𝐵)

If eccentricity is X-direction: B’=B-2𝜖 ; L’=L

If eccentricity is Y-direction: B’=B ; L’=L-2𝜖Use new B’ and L’ in equations (except 36.20).

Eq. 36.20

q = 𝑃

𝐵′𝐿′

Assuming an eccentricity of 3 inches in the direction of B for the

footing shown below, use the Terzaghi bearing capacity factors in

determining the applied load (P). Assume a factor of safety of 2. Use

qnet=qult- 𝛾Df when solving for P.

Solution:

Eq 36.1(b) qult = 0.5𝛾BN𝛾(S)+cNc(S)+𝛾DfNq

B’ = B-2𝜖 = 3 ft – 2(3/12) ft = 2.5 ft L’ = 3 ft

Area of footing = B’x L’ for eccentric loaded.

For 𝜱 = 30°; N𝛾 = 19.7, Nc = 37.2,Nq = 22.5 (Table 36.2), S = 0.85 (Table 36.5)

qult = 0.5(120)(2.5)(19.7)(0.85)+0+(120)(2.5)(22.5) = 9261.75 lb/ft²

qnet = qult - 𝛾Df = 9261.75 – (120)(2.5) = 8961.75 lb/ft²

qallow = qnet/FS = 8961.75/2 = 4480.88 lb/ft²

q = P/A’

4480.88=P/(3x2.5); P= 33606 or 33.6 kips

L

B

Shape factor Table 36.5 for square

More bearing capacity problems!

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