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3 years ago

Trashekf wfpnof alaalalalallaalalalalalalllalal

Engineering

2 answers:

STatiana [176]3 years ago

8 0

Answer:

Explanatiowfpnofn:

Alex73 [517]3 years ago

3 0

Answer:

I HAVE NO IDEA WHAT YOU JUST SAID BUT THANKS FOR THE POINTS!!!!

Explanation:

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A strain gauge with a 4 mm gauge length gives a displacement reading of 1.5 um. Calculate the stress at the location of the stra

Art [367]

Answer:

1) 75Mpa

2) 1.125 MPa

Explanation:

given data:

gauge length = 4 mm

displacement $= 1.5\mu m = 1.5\times 10^{-3} m$

a) structural steel

Young modulus for steel is $200 GPa = 200\times 10^3 MPa$

we know that

$E =\frac{stress}{strain}$

$Stress = 200\times 10^3 \times \frac{1.5\times 10^{-3}}{4}$

= 75Mpa

b) PMMA

Young's modulus $= 3GPa = 3\times10^3 MPa$

$stress = 3\times \frac{1.5\times10^{-3}}{4}$

stress = 1.125 MPa

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3 years ago

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 deliv

Nataliya [291]

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 $\frac{3}{8}$ in.

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

$\tau_{max}$ = ?

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

$\phi$ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

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

$= 56.67 \pi \text{ rad/s}$

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

= 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

$=\frac{1424500}{56.67 \pi}$

= 8001.27 lb.ft

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

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

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

= 2.93 ksi

B). Angle of twist :

$\phi = \frac{TL}{JG}$

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

= 0.08002 rad

= 4.58°

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3 years ago

A data bus can be visualized as a multilane highway

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Answer:

B. with each component having an individual address

Explanation:

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Answer:

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Answer:

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You could be trapped in your car for days

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