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Assoli18 [71]
3 years ago
15

The tensile strength of brittle materials may be determined using a variation of Equation 8.1. Compute the critical crack tip ra

dius for an Al2O3 specimen that experiences tensile fracture at an applied stress of 275 MPa (40,000 psi). Assume a critical surface crack length of 2 × 10−3 mm and a theoretical fracture strength of E/10, where E is the modulus of elasticity.
Engineering
1 answer:
meriva3 years ago
6 0

Answer:

0.39 nM

Explanation:

The critical crack tip radius from the equation

attached below

for fracture to occur the бm will have to reach the fracture strength of material

i.e.  бm =  E/10

from equation 2 attached below

St = ( 400 * a * бo^2 ) / E^2 ------ 3

Given <em>: a = 2*10^-3 mm , бo = 275 MPa, E = 393 * 10^3 MPa ( gotten from table)</em>

Input given values into equation 3

St = 3.9 * 10^-7 ≈ 0.39 nM

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Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

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Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

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Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

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Part C:

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New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

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