Calculate the tensile modulus of elasticity for a laminated composite consisting of 62 percent by volume of unidirectional carbon fibers and an epoxy matrix under isostress conditions. The tensile modulus of elasticity of the carbon fibers is 340 GPa and that of the epoxy is 4.50  103 MPa.

Answer:

silos to store grains

Explanation:

I got it from PLATO... hope this helps...

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Answer:

Ceramics are generally made by taking mixtures of clay, earthen elements, powders, and water and shaping them into desired forms. Once the ceramic has been shaped, it is fired in a high temperature oven known as a kiln. Often, ceramics are covered in decorative, waterproof, paint-like substances known as glazes.

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Answer:

made essentially from a nonmetallic mineral (such as clay) by firing at a high temperature.

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Answer:

for 13

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Answer:

You cant update it if there isin´t any update available lol

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4

or

22

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The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

[tex]\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant[/tex]

Differentiating the above equation:

[tex]\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\[/tex]

[tex]\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\[/tex]

        [tex]= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}[/tex]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is [tex]$\frac{3}{8}$[/tex]  in.

A) Determine the maximum shear stress developed in the shaft.

[tex]$\tau_{max}$[/tex] = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

[tex]$ \phi $[/tex] = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              [tex]$ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$[/tex]

                              [tex]$= 56.67 \pi \text{ rad/s}$[/tex]

Power [tex]$= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$[/tex]

          = 1424500 ft. lb/s

Torque, [tex]$T = \frac{P}{\omega}$[/tex]

                 [tex]$=\frac{1424500}{56.67 \pi}$[/tex]

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

[tex]$\tau_{max} = \frac{T_c}{J}$[/tex]

        [tex]$=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$[/tex]

      = 2.93 ksi

B). Angle of twist :

     [tex]$\phi = \frac{TL}{JG}$[/tex]

         [tex]$=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$[/tex]

         = 0.08002 rad

         = 4.58°