Class Test Problems
Question 1:
Following data were
obtained from the stability analysis of a concrete gravity dam:
(i)
Total
overturning moment about toe = 1 × 105 t-m
(ii)
Total
resisting moment about toe = 2 × 105 t-m
(iii)
Total
vertical force above base = 5000 t
(iv)
Base
width of the dam = 50 m
(v)
Slope
of the d/s face = 0.8(H): 1(V)
Calculate the maximum and minimum vertical stress to
which the foundation will be subjected to. What is the maximum principal stress
at toe? Assume there is no tail water.
Solution:
Here,
MOTM = 1 × 105
t-m
MRM = 2 × 105
t-m
B = 50 m
V = 5000 t
We know,
𝑿 = ∑ 𝑴/ ∑ 𝑽
= [2 × 105 - 1 × 105] / 5000 = 20 m
𝒆 = B/𝟐 – 𝑿
= 50/2 – 20 =5 m
Vertical
stresses at the toe
Pv =
∑V / B * [1 ± 6 e / B]
= 5000/50 * [ 1± 6*5/50]
Maximum vertical stress
Pv max = 5000/50 * [ 1+ 6*5/50] = 160 t/m2
Minimum vertical stress
Pv min = 5000/50 * [ 1- 6*5/50] = 40 t/m2
The maximum principal stress at the toe
σat
Toe = Pv. Sec2α – (P’ – Pe’)
tan2α
σat
Toe = Pv. Sec2α [ As No tail water and
no earthquake effect]
= 160*1.64 [tanα = 0.8, sec2α
= 1 + tan2α]
= 262.4 t/m2
Question 2:
A
hydraulic structure with length of horizontal floor in alluvial soil 15 m and 3
m deep vertical sheet pile is attached at its downstream end and the head of
water is 4.0 m (see Figure below). Find the thickness of the floor (using
Khosla’s theory). Is the structure safe against the exit gradient? (F = 8, Gf =
2.45).
Solution:
Floor thickness:
tE = hE / (Gf - 1)
∅
is the ratio of the residual seepage head (h) to the total seepage head (HL),
thus
∅
= ℎ/HL
h = 0.265 x 4 = 1.06
tE = 1.06 / [2.45 – 1] = 0.73 m
Exit Gradient
Safety factor F
F = 1 / Gf = 1/0.243 = 4.11 < 8, The
structure is unsafe.
Question 3:
For the earth dam of homogeneous section with a horizontal filter as shown in Fig. below, if the coefficient of permeability of the soil material used in the dam is 5 *10-4 cm/sec., find the seepage flow per unit length of the dam.
Solution:
At C,
FC’ = 146-30-72+21.6 = 65.6 m
CC’ = 18 m
√(x2+y2) = x + s
√ (65.62 + 182) =
65.6 + S
S = 2.42 m
Now,
y = √(2xs+s2) = √ (2*2.42*x + 2.422)
= √ (4.84*x + 5.86)
Now, the coordinates of the parabola
are tabulated below
|
x |
y |
|
0 |
2.42 |
|
5 |
5.48 |
|
10 |
7.36 |
|
15 |
8.86 |
|
20 |
10.11 |
|
25 |
11.26 |
|
30 |
12.30 |
|
35 |
13.24 |
|
40 |
14.10 |
|
45 |
14.96 |
|
50 |
15.70 |
|
55 |
16.50 |
|
60 |
17.20 |
|
65.6 |
18 |
The base parabola is drawn and correction at the entry point for the curve BI is made by free hand sketching, in such a way that the phreatic line becomes perpendicular to u/s face GB of the dam. The exit point should also be corrected such that the phreatic line meets perpendicular to base line FOD. The final phreatic line BIO is thus drawn as shown below.
