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STUDENT EXERCISE #2 Use the α-Method described in Section a and the Nordlund Method described in Section c to calculate the ultimate pile capacity and the allowable design load for a 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).

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**Soil Profile 3 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

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**Calculate the Shaft Resistance in the Clay Layer Using α-Method**

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 = 2.73 ksf 43 ft / in = 40.5 1.47 ksf

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ca = 1.47 ksf cu = 2.73 ksf Concrete, Timber, Corrugated Steel Piles Smooth Steel Piles b = Pile Diameter D = distance from ground surface to bottom of clay layer or pile toe, whichever is less 9-45 Figure 9.18

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**Calculate the Shaft Resistance in the Clay Layer Using α-Method**

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) = fs = ca = 1.47 ksf Rs1 = (1.47 ksf)(3.38 ft2/ft)(43 ft) = kips

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**Calculate the Shaft Resistance in the Sand Layer Using the Nordlund Method**

STEP 1 The po diagram, soil layer determination, and the soil friction angle, N, 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 */N from Figure 9.10. V = 0.89 ft3/ft 6 */N = or * = N

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

/ = 0.62 a – closed-end pipe and non-tapered Monotube piles b – timber piles c – pre-cast concrete piles d – Raymond Step-Taper piles e – Raymond uniform piles f – H-piles g – tapered portion of Monotube piles

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**Calculate the Shaft Resistance in the Sand Layer 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 */N from Figure 9.10. V = 0.89 ft3/ft 6 */N = or * = N = 0.62 0.62 0.62 (35˚) = 21.7˚

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**Calculate the Shaft Resistance in the Sand Layer Using the Nordlund Method**

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 = for V = 0.10 ft3/ft K = for V = 1.00 ft3/ft Using log linear interpolation K = for V = 0.89 ft3/ft STEP 4 Determine correction factor, CF, to be applied to K when ≠ . (Figure 9.15.) Layer 2: = 35˚ and / = CF = 0.62

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**Correction Factor for K when **

CF = 0.78 = 35˚ Figure 9.15

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**Calculate the Shaft Resistance in the Sand Layer Using the Nordlund Method**

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 = for V = 0.10 ft3/ft K = for V = 1.00 ft3/ft Using log linear interpolation K = for V = 0.89 ft3/ft STEP 4 Determine correction factor, CF, to be applied to K when ≠ . Layer 2: = 35˚ and / = CF = 0.62 0.78

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**Calculate the Shaft Resistance in the Sand Layer Using the Nordlund Method**

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 = = kips (1.72) (0.78) (3.73 ksf) (sin 21.7˚) (3.38 ft2/ft) (20 ft)

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**Compute the Ultimate Shaft Resistance, Rs**

Rs = Rs1 + Rs2 Rs = 213.6 kips kips 338.7 kips

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**Compute the Ultimate Toe Resistance, Rt**

STEP 7 Determine αt coefficient and bearing capacity factor N'q from angle of 35˚ at pile toe and Figures (a) and 9.16(b) At pile toe depth D/b = From Figure 9.16(a) αt = From Figure 9.16(b) N'q = 66 ft / in. = 62

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**αt Coefficient versus **

0.67 t = 35˚ (degrees) Figure 9.16a

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Figure 9.16 Chart for Estimating αt Coefficient and Bearing Capacity Factor N'q (Chart modified from Bowles, 1977) 65 Figure 9.16b

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**Compute the Ultimate Toe Resistance, Rt**

STEP 7 Determine αt coefficient and bearing capacity factor N'q from angle of 35˚ at pile toe and Figures (a) and 9.16(b) At pile toe depth D/b = 62 From Figure 9.16(a) αt = 0.67 From Figure 9.16(b) N'q = 65 STEP 8 Compute effective overburden pressure at pile toe. pt = 4.31 ksf. However, maximum of 3.0 ksf governs.

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**Compute the Ultimate Toe Resistance, Rt**

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 ft2)(3.0 ksf) = kips

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**Limiting Unit Toe Resistance**

105 Figure 9.17

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**Compute the Ultimate Toe Resistance, Rt**

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 ft2)(3.0 ksf) = kips = (105 ksf)(0.89 ft2) = 93.5 kips 93.5 kips

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**Compute the Ultimate Pile Capacity, Qu**

STEP 10 Qu = Rs + Rt = kips = kips

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