W²-5w+14

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

3 years ago

The status of which of these determines the sequence in which output devices, such as solenoid values and motor contactors, are

Mkey [24]

A. Physical I/O sensors

Safety switches, operator inputs, travel limit switches etc

5 0

3 years ago

A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th

ddd [48]

emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

=1000*3.14*0.5*10-4

=.0785

E=0.0785(.2/10*10-3)

=1.57 V

Thus, the emf generated is 1.57 V

4 0

3 years ago

214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle,

Zolol [24]

Answer:

(C) ln [Bi]

Explanation:

Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.

3 0

3 years ago

Read 2 more answers

A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum

adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress $\sigma$ = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder there exist both the circumferential stress and the longitudinal stress.

Circumferential stress $\sigma = \dfrac{pd}{2t}$

Making thickness t the subject; we have

$t = \dfrac{pd}{2* \sigma}$

$t = \dfrac{560000*3}{2*150000000}$

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

$\sigma = \dfrac{pd}{4t}$

$t= \dfrac{pd}{4*\sigma }$

$t = \dfrac{560000*3}{4*150000000}$

t = 0.0028 mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm

8 0

3 years ago