Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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- The velocity gradient is 1000 m/s. The viscosity u= is 1.2*10^-4* N-s/(m^2). The shear stress is (a) 0.12 N/m² (b) 12N/ (m^2) (c) 1.2x 107 N/m² (d) 12 * 10^-5 * N/(m^2)arrow_forward(Use The Figure (10.20) to find the solution of this question) An embankment load on a silty clay soil layer as shown below. Determine the stress increase under the embankment at points A and B that are loaded at a depth of 6 m below the ground surface. 5m 1H:2V 1H:2V 1H:1V 8 m y-18 kN/m y-18 kN/m 6 m 4. y Figure 10.19 Embankment loading 050 TTTT 3.0 20 16 045 14 12 LO 040 09 07 035 06 030 05 04 - 025 03 020 02 015 a.10 Figure 10.20 Osterberg's chart for determination of vertical stress 0.00 due to embank- 0.01 0.1 10 100 ment loading 327arrow_forward7. 4) Determine the increase in vertical stress at a depth (2) m belowpoint A due to surface loads shown in Figure (2). m or n L.O 10.0 9.0 8.0 7.0 6.0 5.0 4.0 E 3.0 2.5 2.0 G.S.L xxxx 1.5 z 1.5 -0.006- -0.007 0.006 0.009 0.15 19 -0.011 0.010 Section b-b 2 160 kPa 0.2 0.012 0.014 m or n 2.5 3 G014 0.016 0.020 & 0.25 0.3 0,024 120 kPa 4 2.000 0.4 5 -0.04 6 0.06 S b + 7 8 9 10 0.13 ats 0.34 0.5 0.6 0.7 0.8 0.9 1.0 m or n (a) 2.5 m 160 kPa 120 kPa ais. 1.5 . 1.5 2 2.5 3 2.5 T 3 1.5 m 2 m 0.5 m 4 4 5 Fig.(2) 6 7 8 9 . 1 10 11.0 0.9 0.8 0.7 0.6 40.5 10.4 0.3 0.25 0.2 m of n E 0.15 0.10 .9 10arrow_forward
- Subject: Soil Mechanicsarrow_forwardH.Q 5 Figure below shows an embankment load on a silty clay soil layer. Determine the stress increase at points A, B, and C that are located at a depth of 5 m below the ground surface. + 6 m I Center line IV:2H IV:2H 10m y=17 kN/marrow_forwardFor linear stress variation over the base of the dam of the figure, find where the resultant crosses the base and compute the maximum and minimum pressure intensity at the base. Neglect hydrostatic uplift. Ans.Pmax=555 kPa; Pmin=129 kPa Note: unit weight of concrete is 2.5 times unit weight of water. 6m 19m 1 Y'=2.5Y 1 30 40 ¹3 mi 4 m -11 marrow_forward
- An embankment load on a silty clay soil layer as shown below. Determine the stress increase under the embankment at points A and B that are loaded at a depth of 6 m below the ground surface, please use figure (10.20). 5 m 1Н:2V, 1H:2V 1H:1V 8 m y=18 kN/m y=18 kN/m" 6 m : B Figure 10.19 Embankment loading 0.50- 3.0 20 16 0.45- 14 1.2 LO 0.40 - 0.9 0.8 0.7 0.35 - 0.6 030- 05 04 * 025 - 03 0.20 - 0.2 0.15 - 0.10 - 0.1 0.05- Figure 10.20 Osterberg's chart for determination of vertical stress 0.00 TI TT - due to embank- 01 0.1 10 ment loading 100 327arrow_forwardConsider the soil profile depicted in the figure below. The water table is located 3 m below the ground surface. Calculate the vertical effective stress at Point A in kPa (acceptable tolerance = 2%).arrow_forwardEXAMPLE 10.15 An embankment is shown in Figure 10.29a. Determine the stress increase under the embankment at points A₁ and A₂. 14 m- 5m 11.5 m- 5 m č A₂ At point A₁ -14 m- → 5m → ← 5m 5 m A₂ 11.5 m At point A₂ H=7m ► 5 m 8 A₁ 5m+¦+ H=7m >· 5 m X A₁ 14 m 16.5 m 5 m 14 m EXAMPLE 10.15 An embankment is shown in Figure 10.29a. Determine the stress increase under the embankment at points A, and A₂. 14 m 16.5 m y= 17.5 kN/m³ 5m9 = (2.5 m) x (17.5 - A₂ H A0₂ (1) kN/m³) = 43.75 kN/m2 Aσ₂ (1) y= 17.5 kN/m² B A₁ 1 90 = 122.5 kN/m2 + + T 5m 9=(7 m) x (17.5 kN/m³)= 122.5 kN/m² % = 122.5 kN/m² A₂ 2.5 m →→ ● A0₂ (2) A₁ 14 m- A₂ (2) 9 (4.5 m) x (17.5 kN/m³)= 78.75 kN/m² Aar: (33) A₂ 14 m 9 m ➜ 14 m 1arrow_forward
- Given The uniformly loaded area shown below is built on the ground surface and carries a load of 160 kPa. D1-8 m D2 = 12 m D3 = 4 m D4-5 m D5 = 3 m D1 D3 D2 D5 A D4 Required Determine the vertical stress increment at a depth of 10 m below Point A. Provide answer in kN/m², to the nearest 100th.arrow_forwardI need the answer as soon as possiblearrow_forwardQ-1: For the following soil system determine the effective stress and pore-water pressure at the point “A" when i) water table is at ground surface ii) water table is 1m below the ground surface (consider ya = 16 kN/m³) 1 m Ysat = 18 kN/m 5 m 5 m Ysat= 18 kN/m Yw = 10 kN/m A Y = 10 kN/m³ (i) (ii)arrow_forward
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