Ask question

Ask question

All categories

3 years ago

14

g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima

te strength before and after the cold-work operation. Determine the percent increase in each strength. (b) Determine the ratios of ultimate strength to yield strength before and after the cold-work operation. What does the result indicate about the change of ductility of the part

1 answer:

nikdorinn [45]3 years ago

6 0

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

You might be interested in

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at

BartSMP [9]

Answer:

a) C= 1/120

b) P(X>=5) = 0.333

Explanation:

The attached file contains the explanation for the answers

7 0

3 years ago

Read 2 more answers

An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust

notka56 [123]

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

5 0

3 years ago

Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho

marshall27 [118]

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

8 0

3 years ago

A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o

Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0

3 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

4 0

3 years ago