3. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Figure (5-1) Sand boiling
Figure (5-2) Upheaval failure
Problem 5-1
For the excavation shown in the figure below, when the depth of excavation
reached 8.0 m the excavation bed rose and was flooded with mixture of sand
and water.
a) Find the depth of ground table before the excavation started.
b) Calculate the depth at which G.W.T must be lowered to provide a factor of
safety against ground heave = 1.5.
4. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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c) Calculate the factor of safety against heave if the depth of excavation = 6
m.
Solution
a) Up lift pressure = Hw x γw =10 Hw kN/m²
Soil pressure = h xγsat = (14-8) x 19 = 114 kN/m²
Up lift pressure = Soil pressure
10 Hw =114 → Hw =11.4m.
b) Factor of safety against heave = total stress/up lift pressure
Total stress= Soil pressure=114 kN/m².
1.5= 114/ 10 Hw → Hw = 7.6 m.
The drawdown = 11.4 -7.6=3.8m.
c) Factor of safety against heave = total stress/up lift pressure
Total stress=(14-6)x19=152 kN/m²
Factor of safety = 152/114=1.33.
5-3 Data required to design dewatering system
1-The site dimensions. الموقع أبعاد
2- The purpose of dewatering. التخفيض من الغرض
3- Site investigation report. الموقع استكشاف تقريرالتربة وطبقات المياه عمق به موضح
4- The results of permeability tests in laboratory and field. النفاذية اختبارات نتائج
( جدول ويوضح1-5. التربة أنواع من نوع لكل المقابل النفاذية معامل قيمة )
8. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Figure (5-5) Ditches and open sump system
Advantages of Open Sump and Ditches
1. Widely used method.
2. Most economical method for installation and maintenance.
3. Can be applied for most soil and rock conditions.
Disadvantages of Open Sump and Ditches
Ground water flows towards the excavation with high head or a steep slope
and hence there is a risk of collapse of sides.
5-4-2Well point system األبرية اآلبار طريقة
( الصغيرة األقطار ذات المواسير من مجموعة عن عبارة8-5صفوف هيئة على تنفذ التى )سم(شكل
5-6)أسفل حفر وألعماق محددة بأبعاد موقع أو ممتد مواسير خط هناك كان حالة فى يستخدم وهو
13. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Where:
λ= معاملعلي الحصول يمكن( الشكل من ه11-5)
Figure (5-11) Factor λ versus ratio w/D
Problem 5-2
Design the well point system shown in the figure below. Knowing that: total
line discharge =60 m³/h, K=10-4
m/sec.
14. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Solution
Assume the down draw at well point (H-ho) = 6.0m→ho=3.0 m
L=1500(H-ho)√k = 1500x6√10-4
=90 m
hD = 3(1.48/90(6)+1) = 3.3 m.
K= 10-4
m/sec =10-4
(60)*(60) m/h= 0.36 m/h
60= (0.73+0.27 (6/9)(0.36 *X/180) (9²-3²).
X= 457.9 m. Assume distance between wells =2.0m.
No. of well point = X/ 2 =457.9/2=228.9
Take 230 wells.
5-4-3Deep wells العميقة اآلبار
When water has to be extracted from depths greater than 8 m and it is not
feasible to lower the type of pump and suction piping used in shallow wells to
gain a few extra meters of depth the deep wells are such and submersible
pumps installed within them.
16. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Design of deep wells system
Drawdown of the water table at a point produces a cone of depression and the
radius of influence (R) is a function of the drawdown (h) and the permeability
(k) of the soil as shown in figure (5-14). More permeable the soil means
greater the radius of influence.
Figure (5-14) Cone of depression resulting from drawdown
The proposed equation to calculate R (Sichardt ,1928)
R = C (H-hw)√ k R, H, hw in meters and k in m/s
Where
C = Factor equal to 3000 for radial flow to pumped wells and between 1500
and 2000 for line flow to trenches or to a line of well points.
17. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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تا بعدها يتالشى التى الدائرة قطر نصف هو للبئر التأثير دائرة قطر نصفبتخفيض يقوم البئروال ثير
.الجوفية المياه منسوب
For unconfined aquifers and fully penetration well:
Figure (5-17) Drawdown in free aquifers for the fully penetration well
For Multi wells:
18. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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For confined aquifers and fully penetration:
For Multi wells:
Figure (5-18) Drawdown curve for fully penetration in confined aquifers
20. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Problem 5-2
Design the dewatering system for the excavation shown in figure below to
lower the ground water table to 2.0 m below the bottom of excavation.
Assume the quantity of well discharge = 60 m³/h.
Solution
1-Equivalent radius of excavation
re=√(a b/π)
re= √(300x100/3.14)= 97.7 m
2-Height of water level in well
hw=6+46-25-2=25 m
3-Influence range
R = 3000(H-hw)√ k = 3000(46-25)x√(4.7x10-5
)=431.9m ≈432m
21. الرءوف عبد السيد /مؤمن د.م–هندسةب األزهرقنا
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Q=3.14x4.7x10-5
(462
-252
)/ ln(432/97.7) =0.148 m³/sec
Quantity of well discharge = 60 m³/h =60/(60x60)=0.0167 m³/sec
No. of wells= 0.148/0.0167 =8.86 wells
Take 9 wells
من (فى الموقع فى حرجة نقطة أكثر عند قيمته بحساب المطلوب التخفيض من التحقق يتمتالموقنع صنف
).
Lay out of deep wells
No. of wells ri = √(x²+y²) ln(Ri/ri) n* ln(Ri/ri)
1 150 1.057 1.057
2,5,6,9 123.11 1.255 5.02
3,4,7,8 62.5 1.93 7.73
∑ 13.807