Question

Repeat problem 8.2 using Schmertmann’s method, compute the settlement of this footing att = 50 years. (divide the soil into 6 layers, 3 ft thickness for each layer)

8.2) A 250-k column load is to be supported on a 9 ft square footing embedded 2 ft below the ground surface. The underlying soil is a silty sand with an average N60 of 32 and a unit weight of 129 lb/ft3. The groundwater table is at a depth of 35 ft. Estimate the undrained settlement of the footing using the generalized elastic method with Christian and Carrier’s (1978) influence factors

Ans) Given, Column load = 250 kips Size of footing = 9 ft Depth of footing = 2 ft Average N60 = 32 Unit weight of soil = 129 lb/ft3 We know, s = \mu1\muo q B/E where,\mu1 and \muo are influence factor for thickness and depth of footing respectively q = net allowable pressure B = width of footing E = elastic modulus of soil According to Christian and Carrier (1978) , influence factors \mu1 and \muo can be calculated by plots formed by them.For square footing with L/B = 1 and H/B = (35-2)/9 = 3.67 , \mu1 = 0.53 and \muo = 0.94 Also, q = Load / Area = 250000 / (9 x 9) = 3086.42 lb/ft2 E = Bo(OCR)0.5 + B1N60 where, B1 and Bo are correction factors For silty sand , Bo = 2500 and B1 = 600 = 2500(1)0.5 + 600(32) = 21700 kN/m2 or 453213.92 lb/ft2 Putting values, s = 0.53(0.94)(3086.42) (9) / 453213.92 = 0.0305 ft or 0.36 in

Answer #1

Ans) We know, according to Schmertmann method, settlement (s) ,

s = C1 C2 q (
I_{z}
z / E_{s})

where, C1 = correction factor for foundation depth

C2 = correction factor for time

I = strain influence factor

z = depth of each layer

Es = Modulus of elasticity

Also, C1 = 1 - 0.5 (P_{f} / q)

where, P_{f} = effective stress at depth of foundation =
z = 129 pcf x 2 ft = 258 psf

q = Stress at base of footing = 250 k / (9 ft x 9 ft) = 3.086 kip/ft2 or 3086 lb/ft2

=> C1 = 1 - 0.5 ( 258 / 3086) = 0.958

Also, C2 = 1 - 0.2 Log ( t / 0.1) ,

where, t = time in years

=> C2 = 1 - 0.2 Log (50 / 0.1) = 0.46

Now , plot Influence factor vs depth curve for square footing as suggested by Schmertmann ,

Let us divide soil layer into 6 layer with each layer 3 ft thick. So, z = 3 ft

=> z / B = 3 ft / 9 ft = 0.333

According to above curve, for z/B = 0.333, strain influence factor (I) = 0.40

Also, modulus of elasticity (E) for silty sand = 4 N60 = 4(32) = 128 tsf or 256000 psf

Since, each layer has same values of I and E, we can multiply I z /E with 6 to determine settlement of whole layer

Putting values in formula,

=> s = 0.958 x 0.46 x 3086 x 6 x (0.40 x 3 / 256000)

**=> s = 0.038 ft or 0.45 in**

**Hence, settlement of footing after 50 years is 0.45
in**

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