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Kaylis [27]
3 years ago
13

A freshly annealed glass containing flaws of maximum length of 0.1 microns breaks under a tensile stress of 120 MPa. If a sample

of this glass is now subjected to stress of 30 MPa, failure is found to occur after 10 days. Assuming that the specific surface energy (or fracture toughness) does not change, the average rate at which the crack has grown during the period of the test is
Engineering
1 answer:
almond37 [142]3 years ago
6 0

Answer:

0.16 micron per day

Explanation:

Given:

The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m

Initial tensile stress, σ₁ = 120 MPa

Final stress = 30 MPa

now from Griffith's equation, we have

\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}

where,

Gc and E are the material constants

now,

for the initial stage

120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}  ........{1}

and for the final case

30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}   ............{2}

on dividing 1 by 2, we get

\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}

or

a₂ = 4² × 0.1 × 10⁻⁶ m

or

a₂ = 1.6 micron

Now,

the change from 0.1 micron to 1.6 micron took place in 10 days

therefore, the rate at which the crack is growing = \frac{1.6-0.1}{10}

or

average rate of change of crack = 0.16 micron per day

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Answer:

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Explanation:

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G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

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With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

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St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

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but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

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multiplying through by 300

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Explanation:

Solution

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Studentka2010 [4]

Answer:

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A = D/2

A = 2/2

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Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
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