STUDENT EXERCISE #2

STUDENT EXERCISE #2STUDENT EXERCISE #2

Use the α-Method described in Section 9.7.1.2a and the Nordlund Method

described in Section 9.7.1.1c to calculate the ultimate pile capacity and

the allowable design load for a 12.75 inch O.D. closed end pipe pile

driven into the soil profile described below. The trial pile length for the

calculation is 63 feet below the bottom of pile cap excavation which

extends 3 feet below grade. The pipe pile has a pile-soil surface area of

3.38 ft2/ft and a pile toe area of 0.89 ft2. Use Figure 9.18 to calculate the

shaft resistance in the clay layer. The pile volume is 0.89 ft3/ft. The

effective overburden at 56 feet, the midpoint of the pile shaft in the sand

layer is 3.73 ksf, and the effective overburden pressure at the pile toe is

4.31 ksf. Remember, the soil strengths provided are unconfined

compression test results (cu = qu / 2).

46 ft

20 ft

Silty Clay

= 127 lbs / ft3

qu = 5.46 ksf

Set-up Factor = 1.75

Dense, Silty F-M Sand

= 120 lbs / ft3

= 35˚

Set-up Factor = 1.0

3 ft

Soil ProfileSoil Profile

STEP 1 Delineate the soil profile and determine the pile adhesion from Figure 9.18.

Layer 1:qu = 5.46 ksf so cu =

D/b =

Therefore ca from Figure 9.18 =

Calculate the Shaft Resistance in the Clay Calculate the Shaft Resistance in the Clay

Layer Using Layer Using αα-Method-Method

2.73 ksf2.73 ksf

43 ft / 12.75 in = 40.543 ft / 12.75 in = 40.5

1.47 ksf1.47 ksf

Concrete, Timber, Corrugated Steel Piles

Smooth Steel Pilesb = Pile Diameter

D = distance from ground surface to bottom of clay layer or pile toe, whichever is less

9-45 Figure 9.18

cu = 2.73 ksf

ca = 1.47 ksf

STEP 2 Compute the unit shaft resistance, fs, for each

soil layer.

STEP 3 Compute the shaft resistance in the clay layer.

Layer 1: Rs1 = ( fs1 )( As )( D1) =

Calculate the Shaft Resistance in the Clay Calculate the Shaft Resistance in the Clay

Layer Using Layer Using αα-Method-Method

RRs1s1 = (1.47 ksf)(3.38 ft = (1.47 ksf)(3.38 ft22/ft)(43 ft) /ft)(43 ft)

= 213.6 kips= 213.6 kips

ffss = c = caa = 1.47 ksf = 1.47 ksf

Calculate the Shaft Resistance in the Sand Calculate the Shaft Resistance in the Sand

Layer Using the Nordlund MethodLayer Using the Nordlund Method

STEP 1 The po diagram, soil layer determination, and the soil

friction angle, , for each soil layer were presented in the problem introduction.

STEP 2 Determine .

a. Compute volume of soil displaced per unit length of pile, V.

V = 0.89 ft3/ft (per problem description)

b. Determine / from Figure 9.10.

V = 0.89 ft3/ft / = or =

/ = 0.62

V = 0.89

a – closed-end pipe and non-tapered Monotube pilesb – timber pilesc – pre-cast concrete pilesd – Raymond Step-Taper piles

e – Raymond uniform pilesf – H-pilesg – tapered portion of Monotube piles

Relationship Between Soil Displacement, V, and Relationship Between Soil Displacement, V, and //



STEP 1 The po diagram, soil layer determination, and the soil

friction angle, , for each soil layer were presented in the

problem introduction.

STEP 2 Determine .

a. Compute volume of soil displaced per unit length of pile, V.

V = 0.89 ft3/ft (per problem description)

b. Determine / from Figure 9.10.

V = 0.89 ft3/ft / = or = =0.620.62 0.620.62 0.62 (350.62 (35˚̊) = 21.7) = 21.7˚̊

STEP 3 Determine K* for each soil layer based on displaced volume, V,

and pile taper angle, .

Layer 2: For = 35˚, V = 0.89 ft3/ft and = 0˚

From Figure 9.13: K = 1.15 for V = 0.10 ft3/ft K = 1.75 for V = 1.00 ft3/ft

Using log linear interpolation K = 1.72 for V = 0.89 ft3/ft

0.620.62

STEP 4 Determine correction factor, CF, to be applied to K when ≠ .

(Figure 9.15.)

Layer 2: = 35˚ and / = CF =



Figure 9.15

Correction Factor for KCorrection Factor for K when when

= 35˚

CF = 0.78

STEP 3 Determine K* for each soil layer based on displaced volume, V,

and pile taper angle, .

Layer 2: For = 35˚, V = 0.89 ft3/ft and = 0˚

From Figure 9.13: K = 1.15 for V = 0.10 ft3/ft K = 1.75 for V = 1.00 ft3/ft

Using log linear interpolation K = 1.72 for V = 0.89 ft3/ft

0.620.62

STEP 4 Determine correction factor, CF, to be applied to K when ≠ .

Layer 2: = 35˚ and / = CF = 0.780.78



STEP 5 Compute effective overburden pressure at midpoint of each soil layer, pd.

From problem description, pd for layer 2 is 3.73 ksf.

STEP 6 Compute the shaft resistance for each soil layer.

Rs2=K CF pd sin Cd D

=

= 125.1 kips125.1 kips



(1.72) (0.78) (3.73 ksf) (sin 21.7˚) (3.38 ft(1.72) (0.78) (3.73 ksf) (sin 21.7˚) (3.38 ft22/ft) (20 ft)/ft) (20 ft)

Rs = Rs1 + Rs2

Rs =

Rs =

Compute the Ultimate Shaft Compute the Ultimate Shaft Resistance, RResistance, Rss

213.6 kips + 125.1 kips213.6 kips + 125.1 kips

338.7 kips338.7 kips

STEP 7 Determine αt coefficient and bearing capacity factor

N'q from angle of 35˚ at pile toe and Figures

9.16(a) and 9.16(b)

At pile toe depth D/b =

From Figure 9.16(a) αt =

From Figure 9.16(b) N'q =

Compute the Ultimate Toe Compute the Ultimate Toe Resistance, RResistance, Rtt

66 ft / 12.75 in. = 6266 ft / 12.75 in. = 62

(degrees)

t

Figure 9.16a

ααtt Coefficient versus Coefficient versus

= 35˚

0.67

Figure 9.16b

65

STEP 7 Determine αt coefficient and bearing capacity factor

N'q from angle of 35˚ at pile toe and Figures

9.16(a) and 9.16(b)

At pile toe depth D/b = 6262

From Figure 9.16(a) αt = 0.670.67

From Figure 9.16(b) N'q = 6565

STEP 8 Compute effective overburden pressure at pile toe.

pt =


4.31 ksf. However, maximum of 3.0 ksf governs.4.31 ksf. However, maximum of 3.0 ksf governs.

STEP 9 Compute the ultimate toe resistance, Rt.

a. Rt =αt N'q At pt

b. Rt =qL At (qL determined from Figure 9.17)

c. Use lesser value of Rt from Step 9a and 9b.

Therefore, Rt =


= (0.67)(65)(0.89 ft= (0.67)(65)(0.89 ft22)(3.0 ksf) = 116.3 kips)(3.0 ksf) = 116.3 kips

Limiting Unit Toe Resistance Limiting Unit Toe Resistance

Figure 9.17

105

STEP 9 Compute the ultimate toe resistance, Rt.

a. Rt =αt N'q At pt

b. Rt =qL At (qL determined from Figure 9.17)

c. Use lesser value of Rt from Step 9a and 9b.

Therefore, Rt =


= (0.67)(65)(0.89 ft= (0.67)(65)(0.89 ft22)(3.0 ksf) = 116.3 kips)(3.0 ksf) = 116.3 kips

= (105 ksf)(0.89 ft= (105 ksf)(0.89 ft22) = 93.5 kips) = 93.5 kips

93.5 kips93.5 kips

Compute the Ultimate Pile Capacity, Compute the Ultimate Pile Capacity, QQuu

STEP 10

Qu = Rs + Rt = 338.7 + 93.5 kips = 432.2 kips= 338.7 + 93.5 kips = 432.2 kips

STUDENT EXERCISE #2

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