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 = 8000KN/m²,.y = 20KN/m³ 2.0m Clay 0.5B -1.5m 0.6 0.533 Izgavg) 3.0m Sand E = 20000KN/m² 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(Tonl) = Suclay) + S%cand) Immediate Settlement of clay by Bjerrum's method: qB Si(average )flexzable = HotI En -(5.7) From Fig.(5.4): for Dę/B = 1/3 = 0.33 and L/B = 4/3 = 1.33; µo=0.93 for H/B = 2/3 = 0.66 and L/B = 1.33; H1=0.38 Saverage)flexible =(0.93)(0.38) d200/3x4)(3)(1000) (2x8x1000) - 6.6 mm Immediate Settlement of sand by Schmertmann's method: For square foundation: S; = Gc, - ApE 2.5 -(5.62) q = 1-0.520.5 Ap At foundation level: P% = D5y = 1(20) = 20 kN/m², AP = P/A - P% =- On sand surface: 1200 - 20 = 80 kN/m². Зx4 (80)(3)(4) (3+ 2)(4+2) P% = D5y = 3(20) = 60 kN/m², = 32 kN/m² (2:1 method) AP C = 1-0.5= 0.06 < 0.5 32 .: Use C1 = 0.5 C2 = 1+ 0.21og 10 -=1+0.2 log10 0.1 2/12 =1.04 0.1 0.533 +0.133 = 0.333 , E I̟ Az _ (0.333)(3) - 4.9.x.10-5 2 20000 Si(sand) = (0.5)(1.04)(32)(4.9.x.10-³)=0.815 mm . SiTotal) = 6.6+ 0.815 = 7.415 mm
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 = 8000KN/m²,.y = 20KN/m³ 2.0m Clay 0.5B -1.5m 0.6 0.533 Izgavg) 3.0m Sand E = 20000KN/m² 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(Tonl) = Suclay) + S%cand) Immediate Settlement of clay by Bjerrum's method: qB Si(average )flexzable = HotI En -(5.7) From Fig.(5.4): for Dę/B = 1/3 = 0.33 and L/B = 4/3 = 1.33; µo=0.93 for H/B = 2/3 = 0.66 and L/B = 1.33; H1=0.38 Saverage)flexible =(0.93)(0.38) d200/3x4)(3)(1000) (2x8x1000) - 6.6 mm Immediate Settlement of sand by Schmertmann's method: For square foundation: S; = Gc, - ApE 2.5 -(5.62) q = 1-0.520.5 Ap At foundation level: P% = D5y = 1(20) = 20 kN/m², AP = P/A - P% =- On sand surface: 1200 - 20 = 80 kN/m². Зx4 (80)(3)(4) (3+ 2)(4+2) P% = D5y = 3(20) = 60 kN/m², = 32 kN/m² (2:1 method) AP C = 1-0.5= 0.06 < 0.5 32 .: Use C1 = 0.5 C2 = 1+ 0.21og 10 -=1+0.2 log10 0.1 2/12 =1.04 0.1 0.533 +0.133 = 0.333 , E I̟ Az _ (0.333)(3) - 4.9.x.10-5 2 20000 Si(sand) = (0.5)(1.04)(32)(4.9.x.10-³)=0.815 mm . SiTotal) = 6.6+ 0.815 = 7.415 mm
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.: Si(Total) = 5.75 mm)
"please resolve the problem 5.3 but sand instead of clay"
"solve all problem"
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