Engineering Environmental Sustainability- Take Home FINAL EXAM - Assignment Example

13-14, pages 507- 520=13 Both Closed-end steel pipe piles and drilled shafts are under consideration in a project. As shown in Fig. 13-53, the piles will be installed through a 10-m- thick, natural clay layer with ( Su/σ’v)NC= 0.25 into a sand layer with energy corrected, normalized below counts N60 of 25, 29, 32, 40, 42 and 45 for the 6m into unit layer. Sand has unit weight equal to 20 kN/ m3, and the clay has unit weight equal to 17kN/m3. The critical- state friction angle of the sand is 320 and K0= 0.45. The water table is at the surface.

In the past the soil profile had been subjected to uniform surcharge of 50kPa applied on the surface of soil deposit, which was later removed causing both the sand and clay layers in the current states to be over consolidated. For a 15-m- long, 500-mm drilled shaft and a geometrically identical closed-end steel pipe pile, calculate (a) the shaft capacity due to the clay layer ( divide the clay into ten sub layers of equal thickness in your calculations), (b) The shaft capacity due to the sand layer, (c) the total shaft capacity, (d) the ultimate base capacity, (e)the ultimate load capacity of the pile, (f) the allowable load based on a suitable factor of safety (without consideration of the strength of the pile cross-section), and (g) the allowable load if the compressive strength of the concrete is 15 MPa.

Figure 13-53 Pile and soil profile for problem 13-14SOLUTION:(a) The shaft capacity due to the clay layerClosed end pipe pile Let us first divide the clay layer into 10 sub layers. The current vertical effective stress at each layer and past maximum vertical effective stress for each sub layer can be calculated from given data.For example, at z= 5.5m, the current vertical effective stress σ’v and past maximum vertical effective stress σ’vp are given byσ’v = (17- 9.81) ˣ 5.5 = 39.

5kPaσ’vp = 50 + 39.5kPaTherefore, OCR= σ’vp/ σ’v = 89.5/ 39.5 =2.3. In addition, from (6.54), undrained shear strength at a depth of 5.5m below the ground surface isSu= σ’v ( Su/ σ’v )NC OCR 0.8=39.5ˣ 2.30.8 = 19.2 kPaTherefore, Su/ σ’v for this layer is given by Su/ σ’v = 19.2/ 39.5= 0.49 ≤ 1:Now that α = ( Su/ σ’v )-0.5 NC = (0.25)-0.5= 0.71Finally, the limited unit shaft resistance is calculated by qsL ǀat 5.5m for the pipe pile = αSu = 0.71ˣ 19.2kPa = 13.

6kPaSi1milarly, we can finish calculations for other sub- layers and prepare the following table.S- Table 13-9LayerZtout(m)Zbot (m)MidDepth(m)Δhi(m) σv’(kPa)σ vp’(kPa) OCRSu(kPa)αSu/ σ’vαqsLi(kPa)qsLiΔhi (kN/m)1010.513.653.614.97.82.1704.13.23.22121.5110.860.85.610.70.990.505.45.43232.5118.068.03.813.10.730.597.77.74343.5125.275.2315.20.490.659.99.95454.5132.482.42.516.90.520.6911.711.76565.5139.589.52.319.20.490.7113.613.67676.5146.796.72.121.10.450.7515.815.88787.5153.9103.91.922.50.420.7717.317.39898.5161.1111.11.824.40.400.7919.319.3109109.5168.3118.31.726.10.380.8121.121.1 Σ qsLi Δhi =125.

0 kN/mTherefore, the shaft capacity of closed- ended pipe pile due to clay later is Qsl= Σ qsLiAsi = ∏BΣ qsLi Δhi= ∏ 0.5mˣ125kNm =196.3 ˣ 196 kN answer to (a) Drilled shaftThe value α is calculated from (13.50). For example, the value α for the 3rd layer is calculated by α =0.4 [1-0.12ln(su/ pa)]= 0.4 [1-0.12ln(13.1/1oo)]= 0.50Now, we can simply prepare the following table: S- Table 13-10LayerZtot(m)Zbot (m)midDepth(m)Δhi(m) σv’(kPa)σ vp’(kPa) OCRSu(kPa)αSu/ σ’vαqsLi(kPa)qsLiΔhi (kN/m)1010.513.653.614.97.80.524.14.14.12121.5110.860.85.610.70.510.505.55.53232.5118.068.03.813.10.500.596.66.64343.5125.275.2315.20.490.497.47.45454.5132.482.42.516.90.520.498.38.36565.5139.589.52.319.20.490.489.29.27676.5146.796.72.121.10.450.479.99.98787.5153.9103.91.922.50.420.4710.610.69898.5161.1111.11.824.40.400.4711.511.5109109.5168.3118.31.726.10.380.4612.012.

o Σ qsLi Δhi =85.1 kN/mNote that the same equation was used regardless of OCR values. The overall result is on the conservation side. Therefore, the shaft capacity of the drilled shaft due to the clay layer is qsl= Σ qsLiAsi = ∏BΣ qsLi Δhi= ∏ 0.5mˣ85.1Nm =133.7 ͌ 134 kNanswer(b) Shaft capacity due to the sand layerClosed- end pipe pileLet us first calculate the relative density of each soil layer. Using (7.6), for example, the relative density at z= 12.5m is calculated asσ’v= (17-9.81) ˣ 10+ (20-9.81) ˣ2.5 = 97.

4 kPaσ’v= 97.4 + 50= 147.4 kPaOCR= 147.4/ 97.4= 1.5 C= (Ko/ KoNC)=√OCR=√1.51= 1.22DR = 100√ [N60/ A+BC (σ’v/ PA)= 100√ 32/ 36.5 + 27 ˣ1.22 97.4/ 100 = 68.3%]σ ‘h=Koσ’v= 0.47ˣ 97.4kPa= 43.8kPaThe unit pile base capacity at z= 12.5m is calculated using (13.25):qbl= 1.64pA exp[0.1041Фc+(0.0264 – 0.0002Фc) DR]( σ ‘h/ PA)0.841- 0.0047DR=164exp[0.1041 ˣ32+(0.0264- 0.0002 ˣ32) ˣ168.3](43.8/100)0.841-0.0047 68.3= 11705.3kPa.Now, taking the interface friction angle between pile and soil as 0.

85Фc for the steel pile unit shaft capacity is calculated from (13.35):V qsL at 12.5m for the pipe pile = 0.02tanδ[ 1.02- 0.0051DR]qbL = 0.02ˣ tan (0.85ˣ 320)[ 1.02- 0.0051ˣ 68.3] 11705.3= 80.8kPaSimilarly, we can finish calculation for other sub- layers and prepare the following table. S- Table 13-11 Σ qsLi Δhi =412.9kN/mTherefore, the shaft capacity of closed-end steel pipe pile due to sand layer is.Qsl= Σ qsLiAsi = ∏BΣ qsLi Δhi= ∏ 0.5m 412.9kN/ m 640kN Answer to (b) The coefficient of lateral earth pressure K is calculated using (13.32).For example, at z= 125K= 0.

7Ko Experise [{o.0114- 0.0022ln [{0.0114- 0.0022ln(σ’v/ PA)} DR]= 0.7K0exp[{o.o114- 0.0022ln( 97.4]/ 100)} 68.3= 0.69Taking δ= Фc, the limit unit shaft capacity is calculated byqsL at 12.5m for drilled shaft =K σ’v tan δ= 0.67ˣ 97.4ˣ tan 32 = 42.kPaSimilarly we can finish calculations on other sub systems and following table:S-Table 13-12 Σ qsLi Δhi =213.4kN/mTherefore the fraction of the shaft capacity of the drilled shaft due to the sand layer is QsL = Σ qsLiAsi= ∏B Σ qsLi Δhi= ∏ˣ 0.5 ˣ 213.4= 335kNanswer to (b)(c)Total shaft capacityThe total shaft capacity is the sum of the shaft capacity due to the sand and clay layers.

Closed-end pipe pileQsL= QsL, CLAY + QsL, SAND= 196.3 + 648.6kN= 844.9 ͌845kNanswer to (c)(d) Based capacityClosed-end pipe pileWe will calculate the base capacity using soil variables. Let us first calculate relative density of soil layer below pile base. Using (7.6), the relative density at z=15m is calculated as σ’v= (17-9.81) ˣ 10+ (20-9.81) ˣ5 = 122.9kPaσ’vp= 122.9 + 50= 172.9kPaOCR= 172.9/ 122.9= 1.41 C= (Ko/ KoNC)=√OCR=√1.41= 1.19DR = 100√ [N60/ A+BC (σ’v/ PA)= [100√ 36.

5 + 27 ˣ1.19ˣ 122.9/ 100] =77%σ’h= K0 σ’v0.45ˣ 122.9kPa =55.3kPaThe limit unit pile base capacity at z= 15m is calculated using (13.25):QbL=164PAexp[0.1041Фc + (0.0264- 0.0002 Фc)DR]( σ’h/PA)0.841-0.0047DR =164expp0.1041 ˣ 32+ (0.0264- 0.0002 ˣ 32)77](55.3/100) 0.841-0.0047ˣ77=16110.9kPaThe ultimate unit base resistance of the full displacement pile is obtained byQb, 10%=[1.02- 0.0051DR]qbL=[1.02- 0.005 ˣ 77]16110.910106.4kPaTherefore the ultimate base resistance is calculated asQb, ult=qb, 10%Ab=10106.

4kPa ˣ ∏/ 4 ˣ (0.5m)2 ͌ 1984kNanswer to (d)(e) Qult = QsL + Qult =845kN + 1984kN= 2829kNanswer to (e)Drilled shaftQult = QsL + Q ult=469kN +438kN= 907kNanswer to (e)(f) Allowable loadClosed-ended pipe pileWith a FS= 2.5, the allowable load is found asQall= Qult/ FS= 2829/2.5 ͌ 1132kNanswer to (f)Drilled shaftQall= Qult/ FS= 907/2.5 ͌ 363kNanswer to (f)(g) Allowable load for the drilled shaft if the comprehensive strength of the concrete is 15MPaWe also must check the integrity of the cross section of the drilled shaft.

with a FS= 3.0, the design comprehensive strength is obtained from (13.13):Fcd = f’c/ (FS)LA-2= 15000kPa/3= 5000kPaTherefore, the allowable shaft load to ensure integrity across section is:Fall = fcdA= 5000kPaˣ ∏/ 4 ˣ (0.5m)2 ͌ 982kNThe allowable axial load for the drilled shaft from geotechnical consideration is less than the allowable structural load, so, integrity of the cross section is not a concern. Therefore, the final allowable load of drilled shaft is obtained as Qall = min (geotechnical capacity, structural capacity)= min(363kN, 982kN)= 363kNanswer to (g) 3-15, pages 520- 524=5SolutionCalculation of shaft resistance We will consider the first two layers as clay layers and sand given in table 13-9 and 13-5.

We can now calculate the fundamental soil properties (undrained shear strength for clay and relative density for the sand layers.) Let us first calculate undrained shear strength of the clay layers. Following (7.22),Su= qc- = σv /Nk The undrained shear strength at the mid depth of the clay layers are calculated as shown in the following table.S- Table 13-13LayerAverage qc (kPa)Zmid-depth (m)σv (kPa)σv (kPa)Su at the mid-depth (kPa)Su/ σv13001.525.525.522.90.90250046959.235.90.61Now, relative densities for sand layers can be calculated using Eq. (7.20)DR = ln(qc/PA)- 0.4947-0.1041Фc- o.

841ln (σv/PA)] / [0.0264- 0.0002 Фc – 0.0047ln(σ’h/PA)]S- Table 13-14LayerAverage qc (kPa)Zmid-depth (m)σv (kPa)σh(kPa)DR (%)36500785.838.650.8410.5116.410.552.426.25800012.5133.36048.2Now, from the Table 13-9 and Table 13-5, the unit shaft resistance can be calculated as follows:For ClayQsL = αsu,Where as α=√(Su/ σv)NC/(Su/ σv ) for Su/ σv ≤=√1For sand qsL =0.02 tan [1.02- 0.051DR]qbLWhere δ = 0.95 Фc for concrete pile, and q­bL = qcThe values of different unit shaft resistance for different layers are presented 8n the table below:S- Table 13-14LayerAverage qc (kPa)ΔHi(m)= α of clay playerDR (%)For sand layersqsLi (kPa)qsLiΔHi(kN/m)130030.58-13.339.9250020.70-25.150.2365004-50.855.9223.

6440003-26.238.5115.5580001-48.269.969.9 ∑5i=1qsLi ΔHi=499.1kN/m QsL=4B∑5i=1qsLi ΔHi=499.1= 598.9 ͌ 599 kNCalculating the base resistanceAt the depth of 13m, σv=137.9, σh=62.1, and qc = 8000kPa Eq. (7.19), DR=47.3%.Table 13-5Qb, 10%= [1.02- 0.0051DR]qc%= [1.02- 0.0051 ˣ 47.3]8000=6230.2kPaQ ult = QsL + Qb, ult = 598.9 + 560.7 = 1159.6kNThe allowable vertical capacityQall = Q ult/ FS =1159.6/2.5 = 463.8kN ͌ 464kN17-8, pages 703 - 705 =3 (FS)OMS = ∑Ni=1 [cli + (Wicosαi) tanФ ]/ ∑Ni=1 Wisinαi =893.86/538.02= 1.66 ͌ 1.7S- Table 17-2SliceY1 (top)Y2 (top)Y1 (bottom)Y2 (bottom)X1X2Wi(kN/m)αi(rad)sin αi10.0003.3330.0000.3750.000 5903.333 0.11298.5900.1120.11223.3336.6670.3751.5633.3336.667268.7870.3420.33636.66710.0001.5633.8206.66710.000376.0960.5950.561410.00010.0003.8206.09510.00012.071208.8600.8320.739510.00010.0006.09510.00012.07114.14280.8731.0830.883SliceWisinαi (kN/m) CosαiN (Wicosαi)Lic Li+NtanФ111.0230.99497.9723.354128.300290.2200.942253.1933.539228.9993210.8790.828311.4134.025275.0984154.4490.673140.5993.076149.386571.4470.46937.8914.420112.081∑538.0893.

917-9, pages 705- 706=2(FS)OMS = [∑Ncbi i=1 + (WitanФ ]/ mα]/ [∑Ncbii=1 Wsinαi Mα= cosαi ( 1+ tanαi tanФ/ FS)(FS)BSM was determined to be 1.74 by the iteration.SliceWi(kN) αi ( rad)SinαiWisinαi bic bi+WitanФTry FS= 1.6Try FS= 1.7Try FS= 41.7mαc bi+WitanФ/ mαimαc bi+WitanФ/ mαimαc bi+WitanФ/ mαi198.5900.1120.11211.o233.333128.2661.0371.035123.9491.034124.0622268.7870.3240.33690.2203.334234.6731.073218.6591.065220.2411.063220.8293376.0960.5950.561210.8793.333310.6711.047288.131.034291.7191.029293.0624208.8600.8320.739154.4492.071171.9300.962178.7270.945181.940.939183.151580.8731.0830.88371.4472.07191.9550.814113.0280.793115.920.786117.020∑538.02922.19933.77938.13FSFS1.71FS1.74FS1.74