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Viktor [21]
3 years ago
7

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

tained in the field were as follows:
0-6 in. 4 blows; 6-12 in.: 6 blows; 12-18 in.: 6 blows. The tests were conducted using a USA-style donut hammer in a 6-in diameter boring using a standard sampler with the liner installed. The vertical effective stress at the test depth was 1500 lb/ft^2. Determine N1.60.
Engineering
1 answer:
scoray [572]3 years ago
8 0

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

                                             $ \sim 72 \ kN/m^2 $

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 = $ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

                      = 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                        = 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                         = 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

       = 13.14

       = 13

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avanturin [10]

Answer:

sa china po

Explanation:

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6 0
2 years ago
A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-v
adelina 88 [10]

Answer:

The power produced by the turbine is 23309.1856 kW

Explanation:

h₁ = 3755.39

s₁ = 7.0955

s₂ = sf + x₂sfg  =

Interpolating fot the pressure at 3.25 bar gives;

570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175

2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345

h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg

Power output of the turbine formula =

Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1}  \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]

Which gives;

560 - \dot{W } = 8\left [ \left (2599.2418-3755.39  \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]

= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856

- \dot{W } = -22749.1856 - 560 = -23309.1856 kJ

\dot{W } = 23309.1856 kJ

Power produced by the turbine = Work done per second = 23309.1856 kW.

5 0
3 years ago
What steps would you take to design an improved toothpaste container?
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Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

7 0
3 years ago
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str
Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

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$J=\frac{\pi}{32}\times (46)^4$

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So the shear stress at point  A is :

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Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

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The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

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