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4 years ago

Line of Action of the Axial Forces for a Uniform Stress Distribution

Engineering

1 answer:

nikklg [1K]4 years ago

4 0

Answer:

Line of action of axial force for a uniform stress distribution should pass through the centroid of the cross sectional area.

Explanation:

If the line of action of the force is along the centroidal axis of the cross sectional area there is no eccentricity in the line of application of force hence no moment is generated in the cross sectional area hence we get a uniform stress distribution as theorized by hookes law.

In case of eccentric force there is an additional moment in addition of the force. This induced moment induces bending in the section thus giving a non uniform stress distribution in the section.

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Tanya Pierce, President and owner of Florida Now Real Estate is seeking your assistance in designing a database for her business

const2013 [10]

Answer:

the answer is attributes for each entity

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4 years ago

A mixture of air and methane is formed in the inlet manifold of a natural gas-fueled internal combustion engine. The mole fracti

german

Answer:

The mass flow rate of the mixture in the manifold is 6.654 kg/min

Explanation;

In this question, we are asked to calculate mass flow rate of the mixture in the manifold

Please check attachment for complete solution and step by step explanation.

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3 years ago

The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in

amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be $Q\frac{gallons}{minute}$

Now replacing the gallon and the minute by the above values we get

$Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}$

Thus $Q'=0.631Q\frac{liters}{second}$

Now since we know that 1 liter = $0.0353ft^{3}$

Using this in above relation we get

$Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q$

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0

4 years ago

A controlled process is described by the closed-loop transfer function G(s).

MissTica

Answer:

The answer is "Option B".

Explanation:

Given equation:

$G(s) =\frac{K(s + 1)}{2s^2 + (K-1)s + (K-1)}\\\\$

$\to 2s^2 + (K-1)s + (K-1)=0$

Calculating by the Routh's Hurwitz table:

$\to s^2 \ \ \ \ \ 2 \ \ \ \ \ \ K-1 \\\\\to s^2 \ \ \ \ \ K-1 \ \ \ \ \ \ \\\\\to s^0 \ \ ( \frac{(K-1)(K-1)(-2) (0)}{K-1} \\\\ \ \ \ \ = (K-1) )$

Form the above table:

$\to K-1 > 0 \\\\ \to K > 1$

In the above, the value of k is greater than 1.

3 0

3 years ago

T he area of a circle is pr 2. Define r as 5, then find the area of a circle,using MATLAB®.(b) The surface area of a sphere is 4

aksik [14]

Answer:

Area of Circle = 78.5398

Surface Area of Sphere = 1.2566 x 10^3 = 1256.6 ft

Volume of Sphere = 33.5103 ft

Explanation:

Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.

r=5; % define r as 5

a=pi*r^2;% calculate the area of the circle

AreaOfCircle=a

r=10; % define r and 10 ft

sa=4*pi*r^2; %Calculate the surface area of the sphere

SphereSurfaceArea=sa

r=2;% define r as 2 ft

vs=(4/3)*pi*r^3;% Calculate the volume of the sphere

VolumeShere=vs

3 0

3 years ago

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