A 10 m x 6 m mat foundation is placed at 2.0 m depth in sand where the average value of N60 is 23. Determine the allowable net pressure that would limit the settlement to 75 mm, using the following equations. Ngo (B + 0.32Inca (kN/m²) =0.08Fa(for B > 1.22 m)B25whereF = depth factor = 1 + 0.33(D;/ B)B = foundation width, in metersS = settlement, in mmwhereNg = standard penetration resistanceB = width (m)F = 1 + 0.33(D,/B) < 1.33S, = settlement, (mm)When the width B is large, the preceding equation can be approximated asIna (kN/m²)NgoFa0.0825D;1 + 0.33BSe(mm)0.0825[S,(mm) |< 16.63N025

Given data: Length (L) = 10 m Width (B) = 6 m Foundation Depth (Df) = 2 m N60 = 23 Settlement (Se) =…

Answered: A 10 m x 6 m mat foundation is placed…

A 10 m x 6 m mat foundation is placed at 2.0 m depth in sand where the average value of N60 is 23. Determine the allowable net pressure that would limit the settlement to 75 mm,

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

2. What FS reading is required to establish the correct excavation depth? EL= 120.532 EL = 290.667 EL= 221.981 1.591 2.022 ? 118.025 BOTTOM EXC. EL= 287.959 BOTTOM ERC. EL= 219,003 BOTTOM EXC.
Question Attached
B- Describe standard penetration test and its application in foundation engineering Values of N P = 35' %3D |的 8 Ng b. 40 Ny 35 30 25 20 15 Na 10 2. 3. 4. Values of N, and N., 20 8. 9. %24 Angle of shearing resistance, o (degrees)
CECE 3223 Construction Measurement-IV Problem Set: 1 a. Estimate the quantity of earthwork for a portion of road from the following data: Chainage 1 3 4 5 6 8 9 Reduced 7.50 7.70 7.50 7.25 6.85 6.95 6.70 6.45 6.30 5.95 Level The formation level at chainage 0 is 8.00 m and having falling gradient of 1 in 100.The top width is 12.00 m and side slopes 12 horizontal to 1 vertical assuming the transverse direction is in level. Calculate the quantity of earth work by using Trapezoidal & Prismoidal formula. Take 1 chain length = 20.00 m. Reduced Reduced Bd Sd? Bd+Sd? Chainage Distance level of level of (m) Ground(m) Formation(m) (m²) (m²) (m²) 7.50 1 20 7.70 40 7.50 3 60 7.25 4 80 6.85 100 6.95 6. 120 6.70 7 140 6.45 8. 160 6.30 9 180 5.95 10 2.
3500 kN 3 m x 3 m 0.6 m Ya = 17.5 kN/m³ Sand 0.6 m GWT Sand Ysat = 18.3 kN/m3 1.2 m Y sat = 17.1 kN/m³ C, = 0.06 Clay 1.2 m e, = 1.4 Cc = 0.38 %3D
Q3: Shallow footing shown in Figure below calculate the: 文 1 m 7-100KN/m2 Es KNIM2 ImX 2m lo00 Ms =0-3 m I, 12 b00 - 2 0.611 0.o38 2 0.641 0.031 fo00 IF=0-685 3. 11 Rock
Subject: Geotechnical engineering Give me right solution according to the question. Please give me all solutions.. urgent... Don't give me wrong solution
10-) Determine the factor of safety for the figure below and state that if the design is acceptable or not. Qal e = 0.35m 0.5m Yd = 15 kN/m³ WG.W.T 1.5 m 3m 009. + = 30° C = 30 kN/m² Y sar = 21 kN/m³ rectangular footing (2.5m x 3m) 410 kN/m² 9max =
Chapter 5: Settlement of Buildings EXE-3) Determine the total immediate settlement of the rectangular footing shown in figure below after 2 months. 1200 kN G.S. 1.0m 3m x 4m 8000.kN/m², ..Y = 20.kN/m 0.5B =1.5m 0.6 2.0m Clay 0.533 Iz(avg.) Sand 3.0m E = 20000.kN/m? 4.5m 0.133 Rock 2B = 6m 2B – 0.6 Iz redo problem (5.3) but with sand instead of clay as shown in the figure below (Ans:Si(total)=5.75mm
Determine the flexural stresses using the transformed area method. Show complete solution
Subject: Geotechnical engineering Please give me right solution according to the question.. urgent
Q 11.2