Engineering, 17.09.2019 22:00 keigleyhannah30

1.the moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kn/m^3, degree of saturation (1) = 50%; moist unit weight (2) = 17.71 kn/m^3, degree of saturation (2) = 75%. determine a) e (void ratio) and b) gs (specific gravity)

Answers: 3

Show answers

Answers

Answer from: haylieshaven1

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

$\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}$

where, $\gamma_m$ = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

$\gamma_w$ = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation, $\gamma_m = 16.62 kN/m3$

puttng all value to get Gs value;

$16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}$

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

$17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}$

Gs - 1.055*e = 1.805 .........(2)

solving both equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

Answer from: kholman

The void ratio and specific gravity of this question are;

Void ratio; e = 0.7999

Specific Gravity; G_s = 2.65

We are given;

Moist unit weight 1; γ₁ = 16.62 kN/m³

degree of saturation 1; S₁ = 50% = 0.5

Moist unit weight 2; γ₂ = 17.71 kN/m³

degree of saturation 1; S₂ = 75%

The formula for moist unit weight is;

γ = (G_s + Se)γ_w/(1 + e)

where;

G_s is specific gravity

e is void ratio

S is degree of saturation

γ_w is unit weight of water = 9.81 kN/m³

Thus, for γ₁ = 16.62 kN/m³ and S₁ = 0.5, we have;

16.62 = 9.81(G_s + 0.5e)/(1 + e)

⇒ 16.62(1 + e) = 9.81G_s + 4.905e

⇒ 16.62 + 16.62e = 9.81G_s + 4.905e

⇒ 11.715e + 16.62 = 9.81G_s

divide through by 9.81 to get;

1.1942e + 1.6942 = G_s ---eq 1

for γ₂ = 17.71 kN/m³ and S₂ = 0.75, we have;

17.71 = 9.81(G_s + 0.75e)/(1 + e)

⇒ 17.71(1 + e) = 9.81G_s + 7.3575e

⇒ 17.71 + 17.71e = 9.81G_s + 7.3575e

⇒ 10.3525e + 17.71 = 9.81G_s

divide through by 9.81 to get;

1.0553e + 1.8053 = G_s ---eq 2

Putting 1.0553e + 1.8053 for G_s in eq 1;

1.1942e + 1.6942 = 1.0553e + 1.8053

1.1942e - 1.0553e = 1.8053 - 1.6942

0.1389e = 0.1111

e = 0.1111/0.1389

e = 0.7999

Thus;

G_s = 1.0553(0.7999) + 1.8053

G_s = 2.65

Another question on Engineering

Engineering, 04.07.2019 18:10

Aloaded platform of total mass 500 kg is supported by a dashpot and by a set of springs of effective stiffness 72 kn/m. it is observed that when the platform is depressed through a distance x = 12.5 cm below its equilibrium position and then released without any initial velocity; it reaches its equilibrium position in the shortest possible time without overshoot. find the position and velocity of the loaded platform 0.10 sec. after its release. if a further load of 400 kg is added to the platform, find, i) the frequency of damped vibrations, and i) the amplitude of vibration after 2 complete oscillations, given that the initial amplitude is 15 cm.

Answers: 1

Answer

Engineering, 04.07.2019 18:10

The higher the astm grain size number, the finer the gran is. a)-true b)-false

Answers: 2

Answer

Engineering, 04.07.2019 18:10

Manometers are good examples of measuring instruments, nowadays they are not as common as before. a)-capacitive probe gauges b)-gravitational gauges deformation ) gauges d)-digital gauges

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Which of the following refers to refers to how well the control system responds to sudden changes in the system. a)-transient regulation b)- distributed regulation c)-constant regulation d)-steady-state regulation

Answers: 1

Answer

You know the right answer?

1.the moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16....

Questions