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3 years ago

Question # 3

Engineering

2 answers:

oksian1 [2.3K]3 years ago

7 0

False. Because large corporation hires the general contractor

tino4ka555 [31]3 years ago

4 0

Answer:

I think it's False.

Apologies if I am wrong.

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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

A person’s ability to use understand and relate to technology is known as :

ratelena [41]

Answer:

A: Technology literacy

I took a computer class in the 8th grade.

Explanation:

hope this helps. . . <3

good luck! uωu

3 0

3 years ago

True or False Most coolants you put into your car are toxic.

swat32

Answer: True.

Explanation: Coolant is flammable and toxic.

8 0

3 years ago

Read 2 more answers

If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det

Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

solution:

to find the shear stress

T_allow=T_fail/F.S

=120 Mpa/2.5

=48 Mpa

we know that,

V=P

Area for shear head:

A(head)=π×d×t

=π×0.04×0.075

=0.003×πm^2

Area for plate:

A(plate)=π×d×t

=π×0.08×0.03

=0.0024×πm^2

now we have to find shear stress for both head and plate

For head:

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

For plate:

T_allow=V/A(plate)

48 Mpa=P/0.0024×π ..(V=P)

P =48 Mpa×0.0024×π

=361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

P=361.91 KN

note:

find the attached pic

7 0

4 years ago

A rotating cup viscometer has an inner cylinder diameter of 2.00 in., and the gap between cups is 0.2 in. The inner cylinder len

Lesechka [4]

Answer:

The dynamic viscosity and kinematic viscosity are $1.3374\times 10^{-6}$ lb-s/in2 and $1.4012\times 10^{-3}$ in2/s.

Explanation:

Step1

Given:

Inner diameter is 2.00 in.

Gap between cups is 0.2 in.

Length of the cylinder is 2.5 in.

Rotation of cylinder is 10 rev/min.

Torque is 0.00011 in-lbf.

Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.

Step2

Calculation:

Tangential force is calculated as follows:

T= Fr

$0.00011 = F\times(\frac{2}{2})$

F = 0.00011 lb.

Step3

Tangential velocity is calculated as follows:

$V=\omega r$

$V=(\frac{2\pi N}{60})r$

$V=(\frac{2\pi \times10}{60})\times1$

V=1.0472 in/s.

Step4

Apply Newton’s law of viscosity for dynamic viscosity as follows:

$F=\mu A\frac{V}{y}$

$F=\mu (\pi dl)\frac{V}{y}$

$0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}$

$\mu =1.3374\times 10^{-6}$ lb-s/in².

Step5

Kinematic viscosity is calculated as follows:

$\upsilon=\frac{\mu}{\rho}$

$\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}$

$\upsilon=1.4012\times 10^{-3}$ in2/s.

Thus, the dynamic viscosity and kinematic viscosity are $1.3374\times 10^{-6}$ lb-s/in2 and $1.4012\times 10^{-3}$ in2/s.

4 0

4 years ago