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4 years ago

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The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single cry

stal of Fe pulled in tension.

1 answer:

jasenka [17]4 years ago

7 0

Answer:

Answer is mentioned below.

Explanation:

Answer: Calculation of maximum yield strength for a single crystal of Fe pulled in tension is shown in the image attached .

value of max. Yield strength= 54MPa

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / $0.02^{0.22$

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

$l_i$ = $l_0e^{ET$

given that $l_0$ = 480 mm

we substitute

$l_i$ = $480mm$ × $e^{0.04481$

$l_i$ = 501.998 mm

Now we find the elongation;

Elongation = $l_i - l_0$

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0

3 years ago

A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of

OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0

3 years ago

Free brainlist because im new and i just want to but you have t friend me first

Amiraneli [1.4K]

Okay sure.

I’ll 1)chords
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4) the answer is C
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Pretty sure those are the answers

4 0

3 years ago

Help I need to know if it’s true or false

e-lub [12.9K]

Answer: False

explanation: for a bloodborne pathogen to spread you would have to have an open wound as well as the blood would have to get in it.

3 0

3 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

8 0

3 years ago