Physics : asked on mooncake9090
 22.12.2020

Mechanism of crack propagation in brittle and ductile

. 4

Faq

Engineering
Step-by-step answer
P Answered by PhD

Theoretical fracture strength is; σ_m = 13,685 MPa

None of the options are correct

Explanation:

We are given;

Nominal applied stress;σ_o = 1060 MPa

Length of surface crack; a = 0.25 mm

Tip radius of curvature; ρ_t = 0.006 mm

Now, the formula for the theoretical fracture strength which is the maximum stress at the crack tip is given as;

σ_m = 2σ_o[a/ρ_t]^(½)

Plugging in the relevant values, we have;

σ_m = 2(1060)[0.25/0.006]^(½)

σ_m = 13,685 MPa

Engineering
Step-by-step answer
P Answered by Specialist

the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Explanation:

Given the data in the question;

Length of surface crack α = 0.25 mm

tip radius ρ_t = 1.2 × 10⁻³ mm

applied stress σ₀ = 1200 MPa

the theoretical fracture strength of a brittle material = ?

To determine the the theoretical fracture strength or maximum stress at crack tip, we use the following formula;

σ_m = 2σ₀( α / ρ_t)^{\frac{1}{2}

where α is the Length of surface crack,

ρ_t is the tip radius,

and σ₀ is the applied stress.

so we substitute

σ_m = (2 × 1200 MPa)( 0.25 mm / ( 1.2 × 10⁻³ mm ) )^{\frac{1}{2}

σ_m = 2400 MPa × ( 208.3333 )^{\frac{1}{2}

σ_m = 2400 MPa × 14.43375

σ_m = 34641 MPa

σ_m = ( 34641 × 145 )psi

σ_m = 5.02 × 10⁶  psi

Therefore, the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Engineering
Step-by-step answer
P Answered by Specialist

critical stress = 17.899 MPa

Explanation:

given data

specific surface energy = 0.33 J/m²

flexural strength = 88.1 MPa

elastic modulus values E = 61 GPa

surface crack of length = 0.04 mm

to find out

critical stress (in MPa)

solution

we get here critical stress that is express as

critical stress = \sqrt{\frac{2E\gamma}{\pi a}}      .....................1

here E is elastic modulus values and a is surface crack of length and γ is specific surface energy

put here value we get

critical stress = \sqrt{\frac{2*(61*10^9)*0.33}{\pi *0.04*10^{-3}}}

critical stress = 1.7899 × 10^{7} N/m²

critical stress = 17.899 MPa

Engineering
Step-by-step answer
P Answered by PhD

Theoretical fracture strength is; σ_m = 13,685 MPa

None of the options are correct

Explanation:

We are given;

Nominal applied stress;σ_o = 1060 MPa

Length of surface crack; a = 0.25 mm

Tip radius of curvature; ρ_t = 0.006 mm

Now, the formula for the theoretical fracture strength which is the maximum stress at the crack tip is given as;

σ_m = 2σ_o[a/ρ_t]^(½)

Plugging in the relevant values, we have;

σ_m = 2(1060)[0.25/0.006]^(½)

σ_m = 13,685 MPa

Engineering
Step-by-step answer
P Answered by Master

theoretical fracture strength is 9916.58 MPa

Explanation:

given data

surface crack of length L = 0.26 mm

radius of curvature r = 0.004 mm

stress So = 1230 MPa

solution

we get here theoretical fracture strength S that is express as

S  =    S_{0} \times \sqrt{\frac{L}{r} }    .................1

here So is stress and L is length and r is radius

put here value and we get

S  =    1230 \times \sqrt{\frac{0.26}{0.004} }  

solve it we get

S  =  9916.58 MPa

so theoretical fracture strength is 9916.58 MPa

Engineering
Step-by-step answer
P Answered by PhD

The theoretical fracture strength of the brittle material is 11864.5 MPa

Explanation:

Fracture strength is the ability of a material to withstand fracture. It is also known as the breaking stress, it is the stress at which the material fails as a result of fracture. It usually determined from the stress-strain curve after performing a tensile test.

Given that:

Length (L) = 0.15 mm = 0.15 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1370 MPa = 1370 × 10⁶ Pa

theoretical fracture strength (s) = ?

The theoretical fracture strength is given as:

s=s_{0} .\sqrt{\frac{L}{r} }

Substituting values:

s=1370*10^6 .\sqrt{\frac{0.15*10^{-3}}{0.002*10^{-3}} }\\s=1370*10^6 *8.66=11864.5*10^6\\s=11864.5*10^6

s = 11864.5 MPa

The theoretical fracture strength of the brittle material is 11864.5 MPa

Engineering
Step-by-step answer
P Answered by Master

critical stress = 17.899 MPa

Explanation:

given data

specific surface energy = 0.33 J/m²

flexural strength = 88.1 MPa

elastic modulus values E = 61 GPa

surface crack of length = 0.04 mm

to find out

critical stress (in MPa)

solution

we get here critical stress that is express as

critical stress = \sqrt{\frac{2E\gamma}{\pi a}}      .....................1

here E is elastic modulus values and a is surface crack of length and γ is specific surface energy

put here value we get

critical stress = \sqrt{\frac{2*(61*10^9)*0.33}{\pi *0.04*10^{-3}}}

critical stress = 1.7899 × 10^{7} N/m²

critical stress = 17.899 MPa

Physics
Step-by-step answer
P Answered by Master

34.64Gpa

Explanation:

We can calculate the theoretical fracture strenght by,

\sigma_m = 2\sigma_0 (\frac{\alpha}{\rho})^(1/2)

Here we have,

\sigma_0 = Applied stress

\rho =Tip radius

\alpha = lenght of surface crack or half of the lenght of internal crack

Replacing,

\sigma_m = 2*1200*(\frac{0.25}{1.2*10^{-3}})^(1/2)

\sigma_m = 34641Mpa = 34.64Gpa

Therefore the maximum stress at the crack tip is 34.64Gpa

Physics
Step-by-step answer
P Answered by Master

The theoretical fracture strength of a brittle material is 26465.29 MPa.

Explanation:

Given that,

Length = 0.54 mm

Radius curvature R=3.4\times10^{-3}\ mm

Stress = 1050 MPa

We need to calculate the theoretical fracture strength of a brittle material

Using formula of strength

\sigma_{m}=2\sigma_{0}(\dfrac{l}{\rho_{t}})^{\frac{1}{2}}

Where, l = length of the crack

\rho_{t}= tip radius of curvature

Put the value into the formula

\sigma_{m}=2\times1050\times(\dfrac{0.54}{3.4\times10^{-3}})^{\frac{1}{2}}

\sigma_{m}=26465.29\ MPa

Hence, The theoretical fracture strength of a brittle material is 26465.29 MPa.

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