All categories

3 years ago

A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).

Physics

1 answer:

Tems11 [23]3 years ago

8 0

Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

$\sigma_{max}$ = Maximum stress

$\sigma_{min}$ = Minimum stress

$\sigma_m$ = Mean stress = 50 MPa

$\sigma_a$ = Stress amplitude = 225 MPa

Mean stress is given by

$\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}$

Stress amplitude is given by

$\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa$

$\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa$

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

$R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636$

The stress ratio is -0.63636

Stress range is given by

$\sigma_{max}-\sigma_{min}=450\ MPa$

Magnitude of the stress range is 450 MPa

You might be interested in

When doing numerical calculations involving temperature, you need to pay particular attention to the temperature scale you are u

Mrac [35]

1) $293 ^{\circ}C$

2) $859^{\circ}C$

Explanation:

The average kinetic energy of the molecules of an ideal gas is directly related to the Kelvin temperature of the gas, by the formula

$KE=\frac{3}{2}kT$

where

KE is the kinetic energy

k is the Boltzmann constant

T is the Kelvin temperature

We can say therefore that the average kinetic energy of the particles is directly proportional to the absolute temperature of the gas; so, we can write:

$KE\propto T$

And therefore

$\frac{KE_1}{KE_2}=\frac{T_1}{T_2}$ (1)

In this problem, we have:

$KE_1 = K_{10}$ is the initial kinetic energy of the molecules when the temperature of the gas is

$T_1=10^{\circ}+273=283 K$

Here we want to find the temperature $T_2$ at which the average kinetic energy of the particles is

$KE_2=2K_{10}$

So, twice the initial value. Substituting into eq.(1) and solving for T2, we find:

$T_2=\frac{T_1 KE_2}{KE_1}=\frac{(283)(2K_{10})}{K_{10}}=566 K$

Converting into Celsius degrees,

$T_2=566-273=293 ^{\circ}C$

The root-mean-square (rms) speed of the molecules in a gas is given by the equation

$v=\sqrt{\frac{3kT}{m}}$

where

k is the Boltzmann constant

T is the Kelvin temperature of the gas

m is the mass of each molecule

Therefore, from the equation we can say that the rms speed is proportional to the square root of the temperature:

$v\propto \sqrt{T}$

So we can write:

$\frac{v_1}{v_2}=\frac{\sqrt{T_1}}{\sqrt{T_2}}$ (2)

where in this problem:

$v_1 = v_{rms}$ is the rms speed of the molecules when the temperature is

$T_1=10^{\circ}C+273=283 K$

$v_2=2v_{rms}$ is the final rms speed of the molecules

Solving eq.(2), we find the temperature at which the rms speed is twice the initial value:

$T_2=T_1 (\frac{v_2}{v_1})^2=(283)(\frac{2v_{rms}}{v_{rms}})^2=1132 K$

Converting into Celsius degrees,

$T_2=1132-273=859^{\circ}C$

8 0

3 years ago

Monochromatic light with a wavelength of 6.4 E -7 meter passes through two narrow slits, producing an interference pattern on a

seropon [69]

Answer:

The distance between the slits is given by 1.3 × $10^{-4}$ m

Given:

$\lambda = 6.4 \times 10^{-7} m$

D = 4 m

y = $2 \times 10^{-2} m$

m = 1

To find:

distance between slits, d = ?

Formula used:

y = $\frac{m \times \lambda \times D}{d}$

y = distance of first bright band from central maxima

D = distance between screen and source

d = distance between slits

$\lambda$ = wavelength

Solution:

distance of first bright band from central maxima is given by,

y = $\frac{m \times \lambda \times D}{d}$

y = distance of first bright band from central maxima

D = distance between screen and source

d = distance between slits

$\lambda$ = wavelength

Thus,

d = $\frac{m \times \lambda \times D}{y}$

d = $\frac{1 \times 6.4 \times 10^{-7} \times 4 }{2 \times 10^{-2} }$

d = 1.28 × $10^{-4}$

d = 1.3 × $10^{-4}$ m

The distance between the slits is given by 1.3 × $10^{-4}$ m

8 0

3 years ago

1. How do populations of invasive species continue to grow, and disrupt ecosystems?

olganol [36]

2 -An invasive species is a species that is not native to a specific location (an introduced species), and that has a tendency to spread to a degree believed to cause damage to the environment, human economy or human health.[2]

The term as most often used applies to introduced species that adversely affect the habitats and bioregions they invade economically, environmentally, or ecologically. Such species may be either plants or animals and may disrupt by dominating a region, wilderness areas, particular habitats, or wildland–urban interface land from loss of natural controls (such as predators or herbivores). This includes plant species labeled as exotic pest plants and invasive exotics growing in native plant communities.[3][4][5][6] The European Union defines "Invasive Alien Species" as those that are, firstly, outside their natural distribution area, and secondly, threaten biological diversity.[7][8] The term is also used by land managers, botanists, researchers, horticulturalists, conservationists, and the public for noxious weeds.[9]

6 0

3 years ago

Read 2 more answers

Given 368172 hours, convert to days.

dezoksy [38]

Answer:

15340.5

Explanation:

368172/24 hours

8 0

3 years ago

An airplane is flying horizontally with a constant momentum during a time interval ?t. (a) Is there a net impulse acting on the

goldenfox [79]

Answer:

Explanation:

Given

Airplane is flying with horizontally with a constant momentum during time interval $\Delta t$

Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant

(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction

7 0

3 years ago