REASSESSMENT COURSEWORK

Reassessment Coursework

Part A

Question 1

Stainless steel wire of 0.500 mm diameter has been produced by wire drawing. To achieve this size, the material of initial yield strength 630 MPa, was cold worked by 20%, what was the drawing force acting on this initial material?

Solution

Pressure = Force/Area

A = pr2

Radius = D/2

= (0.500 x 10-3)/2

Radius = 25 mm

Area = 1.97 x 10-7 m

20% of 630 MPa = 126 MPa

Calculating Drawing Force = Pressure/ Area

=126 x 106/1.97 x 10-7

Drawing Force = 6.34 Newton

Question 2

On annealing a cold metal, the following data were obtained:

Annealing Temperature

(C)

Residual Stresses (MPa)

Tensile Strength

(MPa)

Grain Size (mm)

250

145

360

0.076

275

145

360

0.076

300

35

360

0.076

325

0

360

0.076

350

0

235

0.025

375

0

200

0.025

400

0

185

0.089

425

0

170

0.183

Task 1

Present the data graphically onto one graph where the left size vertical axis can be used for the stress values and a right hand axis used to plot the grain size data.

Figure 1: Grain Size and Temperature

Question 2 Why is the metal of initial high strength? Explain briefly the underlying reason of this strength being developed during the previous processing of the metal.

Question 3

Estimate the recovery temperature.

The recovery temperature is 250K, because at this temperature, it started to rise up. While calculating the recovery temperature latent heat of fusion and latent heat of vaporization must be considered, only then appropriate values are obtained, we have done the same, and these values can be verified by looking at the above graph.

Estimate the recrystallization temperature.

The recrystallization temperature is 400K, as at this temperature, the strength and stress would be of the value, such that the crystals will begin to form.

Estimate the grain growth temperature.

The grain growth temperature is estimated to be 350k

Question 4 Recommend a suitable temperature for obtaining a high strength, but high electrical conductivity wire.

1900F.

Part B

Here Stress is being calclulated by using the formula

Stress = Force /Area

Area is used as 0.04 m, as per the given dimensions

Stress (Mpa) at indicated temperatures (0C)

Strain (%)

-20

+20

40

60

80

100

140

0

0

0

0

0

0

0

0

0.1

3200

3300

2600

2200

1300

800

500

0.2

6400

6400

5200

4300

2400

1500

900

Stress (Mpa) at indicated temperatures (0C)

Strain (%)

-57

-50

-40

-20

0

30

70

0

0

0

0

0

0

0

0

0.1

4925

4875

4850

1925

975

625

550

0.2

9750

9750

9200

3700

1875

1275

1125

Next, we calculate the percentage modulus depending on the above calculations. Young's Modulus is calculated by using the following formula:

Young's Modulus = (Stress / Strain) * 100

Temperature (0C)

Dry PA66

Wet PA66

-20

32000

-

20

32000

-

40

26000

60

21500

-

80

12000

-

100

7500

-

140

4500

-

-57

-

48750

-50

-

48750

-40

-

46000

-20

-

18500

0

-

9375

30

-

6375

70

-

5625

Figure 2 shows the graph between Temperature and 0.2% modulus for dry PA66.

Figure 2: Temperature and 0.2% Modulus for dry PA66

Figure 3 shows the graph between Temperature and 0.2% modulus for wet PA66.

Figure 3: Temperature and 0.2% Modulus for Wet PA66

The weld is more prone to the crack as it contains internal flaws. Also, the weld creates external stress which encourages the crack to initiate, as the welding process introduces the residual stresses in the welded region. There are four types of cracks that may occur in the welded steel structures.

The steel structure, which is subject to the fluctuating loads mainly, fails due to fatigue (Shine U.P., 2008, pp.1). After being subjected to the fluctuated loads for a continuous period of time, the welded steel structure often breaks into two pieces. The fatigue occurs when the cyclic stress is less than the material's tensile strength (Craig B.D., 2010, pp.13).

The creep occurs at the micro scale ...