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Temka [501]
3 years ago
10

Please help i will give brainilest

Engineering
2 answers:
Eduardwww [97]3 years ago
8 0

Answer:

I hot u

Explanation:

it makes things function better and more fluently than a manual version of the certain object

Troyanec [42]3 years ago
3 0

Answer:

l m a o

Explanationl

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A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str
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Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is \frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}

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3 0
3 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603                         225

0.203                         368

0.162                         442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

Bending Stress = \frac{1}{(Mirror Radius)^{n}}

At mirror radius 1 = 0.603 mm   Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm  Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm   Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}

\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}

0.6114 = (0.3366)^{n_1}

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}

\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}

0.8326 = (0.7980)^{n_2}

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}

\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}

0.5090 = (0.2687)^{n_3}

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

n = \frac{n_1 + n_2 + n_3}{3}

n = \frac{0.452 +0.821 + 0.514}{3}

n = 0.596

Hence to get bending stress x at mirror radius 0.796

\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}

\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}

\frac{Stress x}{225} = 0.8475

stress x = 191 MPa

3 0
4 years ago
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