A linear frequency-modulated signal makes a good test for aliasing, because the frequency moves over a range. This signal is

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4

krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

$\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}$

<u>Where:</u>

<em>ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)</em>

<em>σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi) </em>

<em>2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in) </em>

Hence, the maximum stress (σ) is:

$\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa$

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0

4 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

3 years ago

Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.

Liono4ka [1.6K]

Answer: Eye injury

Explanation: small material such as dust, dirt, and metal shards can harm your eyes with potential blindness or infection.

7 0

2 years ago

The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, de

garri49 [273]

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

<h3>The static equilibrium is given as:</h3>

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = $\frac{\pi d^{2} }{4}$

d = 1.5

$\frac{\pi *1.5^{2} }{4}$

= 1.767 inches²

<h3>Solve for inertia</h3>

$\frac{\pi *0.75^4}{4}$

= 0.2485inches⁴

Solve for the tensile force from here

$\frac{F}{A} +\frac{Mc}{I}$

30x10³ = $\frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\$

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

4 0

2 years ago

can anyone give me tips on adding HP to my 2014 dodge charger SE? it uses the 3.6L220 CI 24 valve v6 vvt (variable valve timing)

marissa [1.9K]

<h2>ANSWER</h2><h2></h2>

I had a couple of answers for this, but when I checked nothing

was right, so im not sure.

5 0

3 years ago