Answered: Problem (5.3): (Total immediate…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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resolve problem: Redo problem (5.3) but with sand instead of clay as shown in the figure below.
(Ans.: S_i(Total) = 5.75 mm)

"please resolve the problem 5.3 but sand instead of clay"

Transcribed Image Text:Problem (5.3): (Total immediate settlement) 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 9 = 8000KN/m²y = 20.kN/m³ 0.6 2.0m Clay 0.5B -1.5m 0.633 Izavg) 3.0m Sand E = 20000KN/m? 4.5m 0.133 Rock 2B - 6m 2B – 0.6 Iz Solution: Since the soil profile is made up of two different soils, then the total immediate settlement will be: Si(Total) = Sa(clay) + S;(sand) Immediate Settlement of clay by Bjerrum's method: q.B Si(average )flexible =HoH1 -(5.7) From Fig.(5.4): for Dę/B = 1/3 = 0.33 and L/B = 4/3 = 1.33; Ho=0.93 for H/B = 2/3 = 0.66 and L/B = 1.33; H1=0.38 (1200/3x4)(3)(1000), S(average)flexible = (0.93)(0.38)- Immediate Settlement of sand by Schmertmann's method: 6.6 mm (2x8x1000) For square foundation: C;C2 I,Az S; = 2.5 (5.6a) C = 1-0.50 20.5 Ap At foundation level: 1200 P = D5.y = 1(20) = 20 kN/m², AP = P/A - P% = On sand surface: 20 = 80 kN/m Зx4 (80)(3)(4) P% = D5.y = 3(20) = 60 kN/m², = 32 kN/m² (2:1 method) ΔΡ (3 + 2)(4 + 2) 60 C = 1-0.5 = 0.06 <0.5 32 .. Use C = 0.5 2/12 C2 = 1+0.2log 10 - =1+ 0.2 log10 0.1 =1.04 0.1 0.533 + 0.133 Iz Az _ (0.333)(3) = 4.9.x.10-5 Iz(avg.) = 0.333, 2 E 20000 Si(sand) = (0.5)(1.04)(32)(4.9.x.10-³) =0.815 mm * Si(Total) = 6.6 + 0.815 = 7.415 mm

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