Contents preface (VII) introduction 1—37

Example 15.4 A fill is placed at an initial saturation of 80% (S_o) and initial porosity of 37.5% (n_o). The average unit weight of the compacted fill is 20,000 N/m³. The dam is raised to a height of 15 m in the first stage after which there is a gap during which one-third of pore pressure will have dissipated. In the next stage, the dam is raised to a height of 35 m. Plot u v/s σ. The variation of ∆V/V_o v/s σ′ is as follows:

∆V/Vo(%)	1		2			3	4	4.5	4.75
σ′ (N/cm^2)	5.0		12.5	23	40	54.5	66.0
Solution: During the first stage:	F	∆V I
						G		J ^po
	∆u = –				H	Vo K
		F	∆V I
		G		J	+ no (1 − So + So H)
		H	Vo K

with po = atmosheric pressure = 10.3 N/cm^2 and H = 0.02		F	∆V I
									F	∆V I
								10.3 G		J		10.3 G			J
								H	Vo K		H	Vo K
	= –										= –					.
		∆V								F	∆V I
					+ 0.375 (1	− 0.80 + 0.80 × 0.02)		0.0798 + G			J
				Vo
													H	Vo K
Using this equation and the curve	∆V	v/s σ′, the following table can be prepared.
V
						o
	∆V/Vo(%)	∆u						σ′		σ
	1	1.48				5.0		6.48
	2	3.45				12.5		15.95
	3	6.21				23.0		29.21
	4	10.35				40.0		50.35
	4.5	13.32				54.5		67.82
	4.75	15.15				66.0		81.15

520 IRRIGATION AND WATER RESOURCES ENGINEERING
At the end of the first stage of construction (1 on Fig. 15.23)

• 
  = γ H = 20000 × 15 N/m² = 300000 N/m²
  
• 
  = 30 N/cm²
  

For

σ = 30 N/cm^2

∆V

= 3.06%

from graph

V

o

∴

σ′ = 23.3 N/cm^2

from graph

and

u = 6.7 N/cm^2

from graph

After one-third of pore pressure dissipation ( = 2.23 N/cm^2)