In-class Exercise Problem: Select load factors for the Strength I and Service I Limit States for the problem illustrated below. Loading Diagram for Student Exercise For this exercise, complete the following table with the appropriate load factors Load Effect Limit State and Performance Limit Strength I Service I Sliding Bearing Eccentricity Settlement Vertical Live Load Surcharge, LS V Horizontal Live Load Surcharge, LS H Horizontal Earth Load, EH Vertical Earth Load, EV Concrete Dead Load, DC 1
In-class Exercise Problem: Select load factors for the Strength I and Service I Limit States for the problem illustrated below. Loading Diagram for Student Exercise For this exercise, complete the following table with the appropriate load factors Load Effect Limit State and Performance Limit Strength I Service I Sliding Bearing Eccentricity Settlement Vertical Live Load Surcharge, LS V 1.75 1.75 1.75 1.00_ Horizontal Live Load Surcharge, LS H 1.75 1.75 1.75 1.00_ Horizontal Earth Load, EH _1.50 1.50 1.50 1.00_ Vertical Earth Load, EV _1.00 1.35 1.00 1.00_ Concrete Dead Load, DC _0.90 1.25 0.90 1.00_ 2
Student In-Class Exercise Shallow Foundation The soil profile, with SPT N-value and N1 60, is provided in the schematic diagram below: a) Compute the settlement of a strip footing under the vertical load of 10 k/ft using the Hough chart of bearing capacity Index c. Use 2:1 distribution of stress method to calculate change in vertical stress, σ z. b) Use f f =30 and γ total =120 psf for the silty sand gravel. Assuming the depth of embedment D f = 5 ft, compute the nominal bearing capacity. Note: Bearing capacity factors N c, N q, and N γ, for f f =30, are 30.1, 18.4 and 22.4, respectively. (AASHTO Table 10.6.3.1.2a-1) 1
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b) The nominal bearing capacity equation is: q n = cn c + γd f N q C wq + 0.5γB f N γ C wγ For the given information, variables for this equation are defined as follows: c = 0 psf Cohesion intercept D f =5 ft Depth of embedment of the footing γ=120 pcf Density of the soil above the groundwater table B f =5 ft Width of the footing Cw γ = 1.0 Groundwater table is greater than 1.5 times footing width below bearing level, therefore correction factor is 1.0. If groundwater table is less than 1.5 times footing width below bearing level, correction factor = 0.5 Solving the nominal bearing capacity equation for this footing gives a bearing capacity of: q n = 0 + (120 pcf)(5 ft)(18.4)(1) + 0.5( l20 pcf)(5 ft)(22.4)(1.0) = 17760 psf a) Settlement of strip footing and effective stress diagram, 3
10 q = ksf 5 q = 2000 psf 2 layer method @ 5 ft σ' vo =600 psf B 5 σ v = q, σ v = 2000 = 1000 psf B + Z (5 + 5) 1 σ o + σ v H = H log c C σ o 1 600 + 1000 H = 10 log = 0.077 ft 55 600 @ 15 ft σ' o =1488 psf 4
5 σ v = 2000 = 500 psf (5 + 15) 1 1488 + 500 H = 10 log = 0.024 ft 52 1488 The total settlement = 0.077 ft + 0.024 ft = 0.101 ft = 1.2 in 5
In- Class Exercise Given: Total factored load Q = 160 kips/pile The soil profile with pertinent soil properties are shown in Fig. (1) Required : a) Based on ODOT BDM Article 202.2.3.2.6, select a suitable pile size (diameter) for friction pile b) Determine the estimated length for the selected friction pile size using R ndr =R s +R P Where: R s : Unfactored Resistance provided along the side of pile, LRFD 10.7.3.8.6 R P : Unfactored Resistance provided at the pile tip, LRFD 10.7.3.8.6 The DRIVEN program is used to generate the skin friction and end bearing values for 12 inch and 16 inch pipe pile in the given soil profile. The DRIVEN analysis results are given in Tables (1 & 2) and plotted in Figures (2 & 3). c) Assume downdrag force in the clay layer is fully developed, (i.e., DD = S dd = 31 kips for 12 inch pipe pile), estimate pile length to account for downdrag force. d) Check the adequacy of 12 inch pipe pile regarding the downdrag force. If it is not adequate, check the adequacy of using 16 inch pie pile and estimate the required pile length. Assume the downdrag force (DD) = 41 kips. 1
Q 20 Clay S u = 500 psf γ = 120 lb/ft 3 Sand f= 35 o γ = 125 lb/ft 3 Pipe Pile Figure (1) 2
Table 1: Driving Summary of Capacity for 12 inch Pipe pile Table 2: Driving Summary of Capacity for 16 inch Pipe pile 3
Figure 2: Bearing Capacity Graph, 12 inch Pipe Pile Figure 3: Bearing Capacity Graph, 16 inch Pipe Pile 4
Solution: (a) The nominal driving resistance required to support the total factored load (Q) is Rndr = Q φ DYN Where R ndr = Ultimate Bearing Value (Kips) Q = Total factored load for the highest loaded pile at each substructure unit (Kips) φ = Resistance factor for driven piles DYN Q R = 16 0 φ = 230 Kips ndr = DYN 0.7 Referring to ODOT BDM Article 202.2.3.2b Pipe Pile Diameter Maximum R ndr 12 inch 250 kips 14 inch 300 kips 16 inch 390 kips Use 12 inch pipe pile Max. R ndr =250 Kips >230 Kips OK. 5
(b) The estimated length for the friction pile using Tale (1) and Fig. (2) is L= 42 feet (c) The total nominal driving resistance, R ndr, for 12 inch pile required is R = R + ndr Sdd R n Where R Sdd = skin friction which must be overcome during driving through downdrag zone (kips) R n γ iqi = ( ) φ dyn + γ DD p φ dyn From Table (1) and Fig. (2), R Sdd = 31 kips [ ] 160 1.4(31) + + 31 = 322 R = kips > 250 kips.n.g. ndr 0.7 0.7 6
(d) Use 16 inch pipe pile Max R ndr = 390 kips The total nominal driving resistance, R ndr, required is R = R + ndr Sdd R n Where R Sdd = skin friction which must be overcome during driving through downdrag zone (kips) R n γ iqi = ( ) φ dyn + γ DD p φ dyn From Table (2) and Fig. (3), R Sdd = 41 kips [ ] 160 1.4(41) + + 41 = 352 R = kips < 390 kips OK. ndr 0.7 0.7 The estimated length for the friction pile using Tale (2) and Fig. (3) is L = 37 feet 7