Question

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a. A 0.5 meter diameter, 10 meters long , concrete pile is driven into a sandy soil stratum. Groundwater is located 3 m below the surface. Compute the ultimate bearing capacity of the pile. Sand 17.3 kN/m3 Pile length - 10 m Pile diameter 0.5 m Friction angle (0) -30° (110 lbs/ft) (32.8 ft) (1.6 ft) 7 m b. Determine required length of same concrete pile with an ultimate bearing capacity of 250 kips

Answer 1

Both sections are done.

Necessary equations are specified in the solution.

Solution is uploaded below.
factors ). aq (qz ou xNq) [Neglecting all other minor a= bearing stressay in kN/m2 Tuz effective vertical pressure at tipi -C{1x6, + Pisot xha) - rsatz &bulk +rw = 17.3 K1/3+ 9.81 kN/m² > 27.11. KNIm3 ov ={(17.3 x 3) + (27, 4x 7)] KN/m² • 67 Knim? ) for $'=300 & for driven piles 2 Nq=21 q- 241.67*212 5075.07 kN/m2 1..? UlX7 Ultimate bearing Capacity a e q x (Aread Qu = 5075.07 Tx0.5) = 996.488 KN CSScanned with CamScanner
A)
YO - 110 lbs/et3. & sat = Obrow el10+ 62.4 = 172.4 lb/ft3 Q2 250 kips'e 250x103 lb 250x103 12 = 124340.8 en 1 lb/et? 61.6) 2 = 124340.8 Le - 9.8452 Sauti ove from the moo)][a.sisz sist su CD -9.8452) f330++72.40€ - 517.2}={172340$ - 619.34057 Na z 21 .q=ov XNq 614.3405 124340. 8 172.4 De 18 21 . टिस्टार पर e Dea 37.91 ft CScanned with corner
A)996.488 KN is the ultimate bearing capacity of the pile.

B)Df=37.91 ft is the required length of same pile to have a ultimate bearing capacity of 250 kips.