A screw is a simple machine that can be described as an inclined plane wrapped around a cylinder.

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

Primary mental skills needed for riding a motorcycle

Irina-Kira [14]

Answer:

You have to know the basic mechanics to help you correctly use the motorcycle, and you also have to have the right mindset - knowing you can ride it, and having determination. All of these factors are the primary mental skills you need for riding a motorcycle.

5 0

2 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

3 years ago

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

sergejj [24]

Answer:

T₂ =93.77 °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure ,P₂= 367+1 = 368 kPa

Lets take temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

${T_2}=T_1\times \dfrac{P_2}{P_1}$

Now by putting the values in the above equation we get

${T_2}=300\times \dfrac{368}{301}\ K$

The temperature in °C

T₂ = 366.77 - 273 °C

T₂ =93.77 °C

8 0

3 years ago

Some of our modern kitchen cookware is made of ceramic materials. (a) List at least three important characteristics required of

Alisiya [41]

Answer:

A)

It should be Non- toxic

It should possess high Thermal conductivity

It should have the Required Thermal diffusivity

B)

stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
porcelain: mostly used for mugs and it is non-toxic
Pyrex : posses good thermal conductivity used in oven

C) All the materials are suitable because they serve different purposes when making modern kitchen cookware

Explanation:

A) characteristics required of a ceramic material to be used as a kitchen cookware

It should be Non- toxic
It should possess high Thermal conductivity
It should have the Required Thermal diffusivity

B) comparison of three ceramic materials as to their relative properties

stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
porcelain: mostly used for mugs and it is non-toxic
Pyrex : posses good thermal conductivity used in ovens

C) material most suitable for the cookware.

All the materials are suitable because they serve different purposes when making modern kitchen cookware

8 0

3 years ago