Lest check Bearing Capacity for a rectangular foundation first Assume 2m x 3m

# Lest check bearing capacity for a rectangular

This preview shows page 186 - 195 out of 206 pages.

Lest check Bearing Capacity for a rectangular foundation first Assume 2m x 3m rectangular footing B=2m L=3m We are assuming the depth of footing D f =2m inorder to pass soft Organic top soil qu = cNc ( 1 + 0.3 B L ) + γDfNq + 1 2 γBNγ ( 1 0.2 B L ) Eq.134 C: Cohesion from laboratory testing (0,93kg/cm3) : 91,22 kN/m2 ? : unit weight of soil at foundation level from laboratory test : 20.11 kN/m N c , N q and N ɣ are the bearing capacity factors. They are functions of the angle of friction 𝝓. Figure 215 Hawever Before that we need to get Anfle of Shear Resistance 186 Figure 216 We can get the blow N number from SPT log which is 14 Figure 217 Figure 218 N c (38) , N q (25) and N ɣ (25) qu = cNc ( 1 + 0.3 B L ) + γDfNq + 1 2 γBNγ ( 1 0.2 B L ) qu = 91.22 38 ( 1 + 0.3 2 3 ) + 20.11 2 25 + 1 2 20.11 25 ( 1 0.2 3 2 ) qu = 5382.13 kN m 2 qnu = 5382.13 γDf ( 20.11 2 ) = 5341.91 kN m 2 qna = 5341.91 3 = 1780 kN m 2 q = Fd + Wf A γDf Eq. 135 q = Structural Loading G = 4190.15 Tn = 41750.72 kN (From Static Calculations) Q = 948.86 Tn = 9454.45 kN Fd = 1.4G + 1.6Q = 58451 + 15127 = 73578 kN A = Area under the foundation(2*3) 6 m 2 187 ? D f = 21.11*2 = 42.22 kN/m 2 W f = (B*L*D* ? con ) = 2*3*2*23.6 = 319.2 kN (unit weight of concrete = 23.6 kN/m 3 ) q = (73578 + 319.2)/6 - 42.22 = 12273 kN/m 2 qna ( 1780 kN/m 2 ) < q (12273 kN/m 2 ) NOT ACCEPTABLE !!! 11.11 TRY RAFT FOUNDATION We used modifide equation of Meyerhof (1963) qn ( u ) = cu c Fcs Fcd ¿ { cu Nc | 1 + B L ( Nq Nc ) | | 1 + 0.4 ( Df B ) | } Eq. 136 qn ( a ) = { cu Nc | 1 B L ( Nq Nc ) | | 1 + 0.4 ( Df B ) | } Fs Eq. 137 Where q n(U) = net ultimate bearing capacity q n(a) = net allowable bearing capacity c u = undrained cohesion N c ,N q = bearing capacity factors with respect to cohesion and surcharge respectively 188 F cs = shape factor with to cohesion F cd = shape factor with respect to depth B,L = breadth and length of foundation respectively D f = depth of foundation FS = factor of safety From Structural Engineers Suggestions we use; Df : 1,5m B : 14m L : 28m Cu : 91,22 kN/m2 Nc (38) , Nq(25) and Nɣ (25) qnu = 91.22 38 [ 1 + 14 28 ( 25 38 ) ] [ 1 + 0.4 ( 1.5 14 ) ] = 4804 kN m 2 qna = 4804 3 = 1600 kN m 2 q = Fd + Wf A γDf q = Structural Loading G = 4190.15 Tn = 41750.72 kN (From Static Calculations) Q = 948.86 Tn = 9454.45 kN Fd = 1.4G + 1.6Q = 58451 + 15127 = 73578 kN A = Area under the foundation(14.71*28) 411 m 2 ? D f = 21.11*2 = 42.22 kN/m 2 189 W f = (B*L*D* ? con ) = 14*28*1.5*23.6 = 14580 kN (unit weight of concrete = 23.6 kN/m 3 ) q = (73578 + 14580)/411 - 42.22 = 172 kN/m 2 qna ( 1600 kN/m 2 ) > q (172 kN/m 2 ) NOT ACCEPTABLE !!! 11.12 SETTLEMENT 11.12.1 Elastic Settlement (Si) Jambu. Bjerrum and Kjaernslı formulation for unitial settlement Se = qo ( α B ' ) ( 1 μs 2 ) Es IsIf Eq. 138 q o = qn from raft foundation bearing capacity calculation = 172 kN/m 2 α = 4 for the centre of the foundation B’ = B/2 for the centre of foundation = 14/2 = 7m µ s = 0,3 poisons ratio E s : Elastic modulus Butler and Bjerrum relationship for consolidated clays : Es/Cu = 400 E s = 400 x 91,22 = 36488 kN/m 2 For example ; 190 If = 0.97 Df/B = 1.5/14 = 0.11 L/B = 28/14 = 2 Figure 219 m ' = L B = 28 4 = 2 n ' = H B 2 = 2 14 2 = 0.28 F 1 = 0.022 F2 = 0.060 Is = F 1 + 1 2 μs 1 μs F 2 Eq. 139 Is = 0.022 + 1 2 0.3 1 0.3 0.060 = 0.056 Se = 172 4 14 2 1 0.3 2 36488 0.056 0.97 = 0.006 m = 6 mm 11.12.2 Consolidation Settlement (Sc) Sc = mv Δσ ' H Eq. 140 191 Mv = 1 1 + e 0 ( e 0 e 1 ) σ 1 σ 0 Eq. 141 Mv = 1 1 + e 0 ( e 0 e 1 ) σ 1 σ 0 = 1 1 + 0.79 0.79 0.59 294 29 = 4.2 10 4 m 2 kN = 0.42 m 2 MN Şekil 220 H= 2m settlement will occur in that layer from log Figure 221 192 Figure 222 m = 2 m mz = 14/2 = 7 m = 3.5 nz = 7/2 = 3.5 n = 1.75 I r = 0.23 ?? ' 0 = 4*0.23*172 = 158.24 kN/m 2 S c = 0.42*158.24*2 = 132 mm (Eq.140) Total Settlement ST = S e + S c = 132+6 = 138 mm 193 CHAPTER 12 PILE FOUNDATIONS 12.1 OUTLINE of LECTURES Pile Types Selection Capacity of Single Piles Capacity of Pile Groups Settlement Considerations 12.2 LOAD/SETTLEMENT RESPONSE  #### You've reached the end of your free preview.

Want to read all 206 pages?

• Winter '19
• • • 