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

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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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"

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
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
1200 kN
G.S.
1.0m
Зт x 4m
2.0m
9c = 8000 .kN / m²,..y = 20.kN / m`
sand
3.0m
clay
E= 20000...kN/m?
Rock
Transcribed Image Text:1200 kN G.S. 1.0m Зт x 4m 2.0m 9c = 8000 .kN / m²,..y = 20.kN / m` sand 3.0m clay E= 20000...kN/m? Rock
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