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evablogger [386]
4 years ago
7

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Engineering
1 answer:
Kamila [148]4 years ago
7 0

Answer:

(b)False

Explanation:

I_{xy} defined as

      I_{xy} =\int \left (x\cdot y\right )dA

Where x is the distance from centroidal x-axis

           y is the distance from centroidal y-axis

          dA is the elemental area.

The product of x and y can be positive or negative ,so the value of  I_{xy} can be positive as well as negative .

So from the above expressions we can say that the product of I_{x},I_y is different from I_{xy} .

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Answer:

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Explanation:

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3 years ago
Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di
blsea [12.9K]

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )

And \tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

4 0
3 years ago
Sarah needs to create an architectural drawing for a museum building with an inclined surface. Which presentation view will be t
prohojiy [21]

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Explanation: got it right

7 0
3 years ago
Explain when it is appropriate to use Tier II?
garik1379 [7]

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3 0
3 years ago
The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large
SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

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Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

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C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0
3 years ago
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