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3 years ago

What determines the depth of the foundation excavation

Engineering

2 answers:

WARRIOR [948]3 years ago

7 0

Answer:

The required depth of any foundation can depend on several factors: Soil bearing capacity. This determines how much load (weight or force) the existing soil can withstand. Soil type.

Explanation:

Klio2033 [76]3 years ago

6 0

They determine based on ground movement studies in that area

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Used ______ must be hot drained for 12 hours or crushed before disposal.

WITCHER [35]

Answer:

Explanation:

The answer is towels because towels after a little bit of just sitting around have a chemical reaction that cam cause them to spontaneously combust

5 0

3 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

3 years ago

A metal specimen with an original diameter of 0.50 in. and a gauge length of 2.75 in. is tested in tension until a fracture occu

tatiyna

Answer:

Percent Elongation = 52.72%

Percent Reduction in Area = 64%

Explanation:

First we find percent elongation:

Percent Elongation = {Final Gage Length - Initial Gauge Length/Initial Guage Length} x 100%

Percent Elongation = {(4.20 in - 2.75 in)/2.75 in} x 100%

<u>Percent Elongation = 52.72%</u>

Now, for the percent reduction in area:

Percent Reduction in Area = {Final Cross Sectional Area - Initial Cross Sectional Area|/Initial Cross Sectional Area Length} x 100%

Percent Reduction in Area = {π(0.3 in)² - π(0.5 in)²/π(0.5 in)²} x 100%

<u>Percent Reduction in Area = - 64%</u>

here, negative sign shows a decrease in area.

5 0

3 years ago

11. What are restrictions when building or completing a challenge?

katrin [286]

Explanation:

The minimum exterior open spaces around buildings that are 55 metres or more, should be 16 metres. On sides where no habitable rooms face, a minimum space of 9 metres shall be left for heights above 27 metres.

3 0

2 years ago

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressur

Veseljchak [2.6K]

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

7 0

3 years ago