4.12 The mass of a wet sample of soil and its container is 0.33 kg. The dry mass of the soil and its container is 0.29 kg. The mass of the container is 0.06 kg and its volume is 0.15 x 10-3 m3. Determine the following.(a) The bulk, dry, and saturated unit weights of the soil.(b) The void ratio and the degree of saturation.(c) How much air void is in the soil?(d) The weight of water required to saturate 1m3 of this soil. I only need the last question answered part D Step 2Solu tiona)i) buck unit werght is given bymoss of Soid and canteainer)- eCantoiner massCa ntainer Volume0. 33 - 0.06O.15 XI63- 1800 kg /m3bulk unit weight is 1800 kg/m3ii) dny unit we'ght is given- byYd=I+ w0:33-0.290. 29 - 006: 0.1731800rd=I+o 173Stop 3 Step 31534.526 Kglm3- PuDry unit weight is 1534.526 kg/m3So turated unit werght is given by,(G te) YwYs =(Ite)I+ e= GrwGYwe:-1Yd2.7X1O001534.526e = 0-759Ye= (2.7+0.75)1000I+0.759Ys = 1966. 458 kg/m3Saturated unit werght is 1966.458 Kglm3b)void ratio and the degree Of satunatron

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Answered: (d) The weight of water required to…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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4.12 The mass of a wet sample of soil and its container is 0.33 kg. The dry mass of the soil and its container is 0.29 kg. The mass of the container is 0.06 kg and its volume is 0.15 x 10^-3 m³. Determine the following.

(a) The bulk, dry, and saturated unit weights of the soil.

(b) The void ratio and the degree of saturation.

(d) The weight of water required to saturate 1m³ of this soil.

I only need the last question answered part D

Step 2
Solu tion
a)
i) buck unit werght is given by
moss of Soid and canteainer)
- e
Cantoiner mass
Ca ntainer Volume
0. 33 - 0.06
O.15 XI63
- 1800 kg /m3
bulk unit weight is 1800 kg/m3
ii) dny unit we'ght is given- by
Yd=
I+ w
0:33-0.29
0. 29 - 006
: 0.173
1800
rd=
I+o 173
Stop 3

Step 3
1534.526 Kglm3
- Pu
Dry unit weight is 1534.526 kg/m3
So turated unit werght is given by,
(G te) Yw
Ys =
(Ite)
I+ e= Grw
GYw
e:
-1
Yd
2.7X1O00
1534.526
e = 0-759
Ye= (2.7+0.75)1000
I+0.759
Ys = 1966. 458 kg/m3
Saturated unit werght is 1966.458 Kglm3
b)
void ratio and the degree Of satunatron

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