CHEMN-1. A large thick plate is fabricated from a

Solution

1

Identify the relevant data: Based on the last digit of your student ID, select the corresponding values from the table. For example, if the last digit is 4, use

K_{Ic} = 50

MPa√m,

\sigma_y = 950

MPa,

x = 0.99

mm, and

y = 1.75

mm

2

Calculate the critical crack length for fast fracture: Use the formula for critical crack length

a_c

in plane-strain conditions:

a_c = \left( \frac{K_{Ic}}{\sigma_y} \right)^2 \frac{1}{\pi}

Substitute the values:

a_c = \left( \frac{50 \, \text{MPa} \sqrt{\text{m}}}{950 \, \text{MPa}} \right)^2 \frac{1}{\pi} \approx 0.00088 \, \text{m} \approx 0.88 \, \text{mm}

3

Compare crack lengths: Compare the calculated critical crack length

a_c

with the given crack lengths

x

and

y

. If

x

or

y

is greater than

a_c

, fast fracture will occur before general yielding. In this case,

x = 0.99

mm and

y = 1.75

mm are both greater than

a_c = 0.88

mm

4

Conclusion: Since both

x

and

y

are greater than

a_c

, the plate will fail by fast fracture before general yielding occurs

Answer

The plate will fail by fast fracture before general yielding occurs.

Key Concept

Critical crack length for fast fracture

Explanation

The critical crack length

a_c

is calculated using the plane-strain fracture toughness

K_{Ic}

and yield strength

\sigma_y

. If the actual crack lengths

x

or

y

exceed

a_c

, fast fracture will occur before general yielding.

Solution

1

Identify the parameters: Based on the last two digits of your student ID, select the corresponding values from the table. For example, if the last two digits are between 00-19, use

L = 0.95 \, \text{m}

,

D = 10 \, \text{mm}

,

\eta = 10.1 \, \text{GPa s}

, and

E = 96 \, \text{MPa}

2

Calculate the load at 0 seconds: The initial elongation is

20 \, \text{mm}

. Using Hooke's Law,

F = k \Delta L

, where

k = \frac{E A}{L}

and

A = \pi \left(\frac{D}{2}\right)^2

. Therefore,

F = \frac{E \pi \left(\frac{D}{2}\right)^2 \Delta L}{L}

3

Calculate the load at 30 seconds (20 mm elongation): The load remains the same as at 0 seconds because the elongation is maintained at

20 \, \text{mm}

4

Calculate the load at 30 seconds (30 mm elongation): The elongation is increased to

30 \, \text{mm}

. Using the same formula,

F = \frac{E \pi \left(\frac{D}{2}\right)^2 \Delta L}{L}

, substitute

\Delta L = 30 \, \text{mm}

5

Calculate the load at 60 seconds: The elongation is maintained at

30 \, \text{mm}

, so the load remains the same as at 30 seconds with

30 \, \text{mm}

elongation

6

Draw a schematic diagram: Plot the load as a function of time, showing the changes at 0, 30, and 60 seconds

Answer

The load at 0 seconds is calculated using the initial elongation of

20 \, \text{mm}

. The load at 30 seconds remains the same for

20 \, \text{mm}

elongation and changes when the elongation increases to

30 \, \text{mm}

. The load at 60 seconds remains the same as at 30 seconds with

30 \, \text{mm}

elongation.

Key Concept

Stress relaxation in polymers

Explanation

Stress relaxation is the gradual decrease in stress under a constant strain, modeled using a spring and dashpot system. The load changes with elongation and time, following Hooke's Law and the properties of the material.

Sure, let's break down the problem and solve it step by step.

Solution

Application a: A component with a fixed length required to have minimum weight with a prescribed stiffness when loaded in compression.

1

Define the terms: -

L

: Length of the component -

W

: Weight of the component -

E

: Young's modulus (stiffness) -

A

: Cross-sectional area -

\rho

: Density of the material -

P

: Load applied in compression -

\delta

: Deformation under load

2

Derive the material performance index: - Stiffness in compression is given by

P = \frac{EA}{L} \delta

- For minimum weight,

W = \rho A L

- Combining these, we get

P = \frac{E}{\rho} \frac{W}{L^2} \delta

- The material performance index for minimum weight with prescribed stiffness is

\frac{E}{\rho}

Application b: A component with a fixed length required to have minimum weight with a prescribed yield load

1

Define the terms: -

L

: Length of the component -

W

: Weight of the component -

\sigma_y

: Yield strength -

A

: Cross-sectional area -

\rho

: Density of the material -

P

: Load applied

2

Derive the material performance index: - Yield load is given by

P = \sigma_y A

- For minimum weight,

W = \rho A L

- Combining these, we get

P = \sigma_y \frac{W}{\rho L}

- The material performance index for minimum weight with prescribed yield load is

\frac{\sigma_y}{\rho}

Application c: A beam with a fixed length and a fixed width to height ratio required to have low weight with a prescribed stiffness when subject to three point bending

1

Define the terms: -

L

: Length of the beam -

W

: Weight of the beam -

E

: Young's modulus (stiffness) -

I

: Moment of inertia -

\rho

: Density of the material -

F

: Load applied -

\delta

: Deformation under load

2

Derive the material performance index: - Stiffness in bending is given by

F = \frac{48EI}{L^3} \delta

- For minimum weight,

W = \rho A L

- Moment of inertia

I = \frac{b h^3}{12}

, where

b

is width and

h

is height - Combining these, we get

F = \frac{48E}{\rho} \frac{I}{L^4} \delta

- The material performance index for minimum weight with prescribed stiffness in bending is

\frac{E}{\rho}

Application d: A thin-wall cylindrical pressure vessel designed for "leak before break"

1

Define the terms: -

t

: Wall thickness -

r

: Radius of the cylinder -

\sigma_t

: Tensile strength -

\rho

: Density of the material -

P

: Internal pressure

2

Derive the material performance index: - For "leak before break", the vessel must withstand internal pressure without catastrophic failure - The stress in the wall is given by

\sigma_t = \frac{P r}{t}

- For minimum weight,

W = \rho 2 \pi r t L

- Combining these, we get

\sigma_t = \frac{P r}{\rho 2 \pi r L} W

- The material performance index for "leak before break" is

\frac{\sigma_t}{\rho}

Answer

The material performance indices for the given applications are: - Application a:

\frac{E}{\rho}

- Application b:

\frac{\sigma_y}{\rho}

- Application c:

\frac{E}{\rho}

- Application d:

\frac{\sigma_t}{\rho}

Key Concept

Material performance indices help in selecting materials that optimize specific properties like stiffness, yield strength, and tensile strength relative to density.

Explanation

By deriving the material performance indices, we can identify the best materials for each application based on their mechanical properties and density, ensuring optimal performance and minimum weight.

Sure, let's address the questions one by one. Question 1(a) Describe, with the aid of sketches, the blow moulding process used to manufacture glass bottles.

Solution

1

Identify the process: Blow moulding is a manufacturing process used to create hollow glass bottles by inflating a heated glass preform inside a mould

2

Sketch the process: - Step 1: A glass preform is heated until it becomes pliable. - Step 2: The preform is placed inside a mould. - Step 3: Air is blown into the preform, causing it to expand and take the shape of the mould. - Step 4: The mould is cooled, and the formed glass bottle is removed

Answer

Blow moulding involves heating a glass preform, placing it in a mould, inflating it with air to take the shape of the mould, and then cooling it to form a glass bottle.

Key Concept

Blow moulding process

Explanation

Blow moulding is used to create hollow glass bottles by inflating a heated preform inside a mould.

Question 1(b) Describe the process of thermal tempering of glass and explain how it leads to a distribution of residual stress in the workpiece that increases its resistance to impact.

Solution

1

Identify the process: Thermal tempering involves heating glass to a high temperature and then rapidly cooling it

2

Explain the steps: - Step 1: Heat the glass to a temperature above its annealing point. - Step 2: Rapidly cool the surface of the glass using air jets. - Step 3: The surface cools and solidifies faster than the interior, creating compressive stress on the surface and tensile stress in the interior

3

Impact on resistance: The compressive stress on the surface makes the glass more resistant to impact because cracks are less likely to propagate through the compressive layer

Answer

Thermal tempering involves heating glass and rapidly cooling it to create compressive stress on the surface and tensile stress in the interior, increasing its resistance to impact.

Key Concept

Thermal tempering of glass

Explanation

Thermal tempering creates a distribution of residual stress that enhances the glass's resistance to impact by making it harder for cracks to propagate.

Question 1(c) A large plate is fabricated from a ceramic material that has a plane strain fracture toughness of $3.84 \mathrm{MPa} \sqrt{\mathrm{m}}$. If the surface cracks are shorter than $105 \mu \mathrm{m}$ and the internal cracks are shorter than $170 \mu \mathrm{m}$, determine the maximum tensile stress the plate can withstand without breaking. Assume a value of 1.0 for the geometric factor $Y$.

Solution

1

Identify the given values: - Plane strain fracture toughness,

K_{IC} = 3.84 \mathrm{MPa} \sqrt{\mathrm{m}}

- Surface crack length,

a = 105 \mu \mathrm{m} = 105 \times 10^{-6} \mathrm{m}

- Internal crack length,

a = 170 \mu \mathrm{m} = 170 \times 10^{-6} \mathrm{m}

- Geometric factor,

Y = 1.0

2

Use the fracture toughness equation:

K_{IC} = Y \sigma \sqrt{\pi a}

\sigma = \frac{K_{IC}}{Y \sqrt{\pi a}}

3

Calculate the maximum tensile stress for surface cracks:

\sigma_{surface} = \frac{3.84 \mathrm{MPa} \sqrt{\mathrm{m}}}{1.0 \sqrt{\pi \times 105 \times 10^{-6} \mathrm{m}}}

\sigma_{surface} \approx 66.7 \mathrm{MPa}

4

Calculate the maximum tensile stress for internal cracks:

\sigma_{internal} = \frac{3.84 \mathrm{MPa} \sqrt{\mathrm{m}}}{1.0 \sqrt{\pi \times 170 \times 10^{-6} \mathrm{m}}}

\sigma_{internal} \approx 49.8 \mathrm{MPa}

5

Determine the limiting factor: The plate can withstand the lower of the two stresses, which is

49.8 \mathrm{MPa}

Answer

The maximum tensile stress the plate can withstand without breaking is

49.8 \mathrm{MPa}

.

Key Concept

Fracture toughness and tensile stress

Explanation

The maximum tensile stress is determined by the fracture toughness and the length of the cracks, with the lower stress value being the limiting factor.

Question 1(d) A three-point bending test was performed on an aluminium oxide specimen of $80 \mathrm{~mm}$ long, having a rectangular cross section with the width of $8 \mathrm{~mm}$ and height of $6 \mathrm{~mm}$. The distance between support points was $50 \mathrm{~mm}$. The reported flexural strength of aluminium oxide is $370 \mathrm{MPa}$. Would you expect the specimen to fracture when a load of $2000 \mathrm{~N}$ is applied? Justify your prediction.

Solution

1

Identify the given values: - Length of specimen,

L = 80 \mathrm{~mm}

- Width of specimen,

b = 8 \mathrm{~mm}

- Height of specimen,

d = 6 \mathrm{~mm}

- Distance between support points,

L_s = 50 \mathrm{~mm}

- Flexural strength,

\sigma_{fs} = 370 \mathrm{MPa}

- Applied load,

F = 2000 \mathrm{~N}

2

Use the flexural strength formula:

\sigma_{fs} = \frac{3 F L_s}{2 b d^2}

\sigma_{fs} = \frac{3 \times 2000 \mathrm{~N} \times 50 \mathrm{~mm}}{2 \times 8 \mathrm{~mm} \times (6 \mathrm{~mm})^2}

\sigma_{fs} = \frac{300000 \mathrm{~N \cdot mm}}{576 \mathrm{~mm^3}}

\sigma_{fs} \approx 520.8 \mathrm{MPa}

3

Compare with reported flexural strength: The calculated flexural strength (

520.8 \mathrm{MPa}

) is greater than the reported flexural strength (

370 \mathrm{MPa}

)

4

Conclusion: The specimen is expected to fracture under the applied load of

2000 \mathrm{~N}

Answer

The specimen is expected to fracture when a load of

2000 \mathrm{~N}

is applied.

Key Concept

Flexural strength in three-point bending test

Explanation

The calculated flexural strength exceeds the reported value, indicating that the specimen will fracture under the given load.

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