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1. THEORY QUESTIONS [50 Marks – one mark/minute]

Note, to obtain maximum points for each problem clearly motivate solutions and equations used.

Question 1 [40 marks]:

The contractor of the construction of a structure is unfortunately forced to weld a socket to the flange of the I-beam as shown in the figure below. Use the nominal stress approach and EN 1993-1-9 and determine the fatigue life (for 95% probability of survival) of the beam under a bending moment that varies between -10 and 35 kNm. The material is 350W steel with yield strength 350 MPa and ultimate tensile strength 470 MPa. The thickness of the attachment is 20 mm and is joined by means of an 8 mm double fillet weld. The steel is used inside the plant and is properly surface protected. Assume a design philosophy of damage tolerant with low consequence of failure.

The following is an extraction from EN 1993-1-9 for more information:

Web thickness = 10 mm

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1.1. Solution

Detail category

From information supplied in class, the detail category for the joint is 80.

Partial factor for fatigue

No specific information was given and Investmech used 𝛾𝑀𝑓 = 1.0.

Bending stress in the flange

The second moment of area was calculated as:

𝐼𝑥𝑥 =1

12𝑡𝑤ℎ𝑤

3 + 2 [𝑡𝑓𝑊𝑓 (ℎ𝑓

2)

2

]

Where: ℎ𝑓 = 𝐻 − 𝑡𝑓 and ℎ𝑤 = 𝐻 − 2𝑡𝑓

Calculations

The fatigue life is 75 182 cycles.

H 180 mm Moment-max 35000 Nm

tw 12 mm Moment-min -10000 Nm

tf 12 mm Delta-Moment 45000 Nm

Wf 100 mm Delta-S 239 MPa

hf 168 mm Sigma_C 80 MPa

hw 156 mm N_C 2.00E+06 cycles

Ixx 16 958 736 mm4 N_D 5.00E+06 cycles

Z 188 430 mm3 N_L 1.00E+08

m1 3.0

m2 5.0

Delta-Sigma_D 58.9

Delta-Sigma_L 32.4

N_R 75 182 cycles

2 Marks

1 Mark

1 Mark

4 Marks

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Question 2 [5 Marks]

You have complete joint penetration butt weld that is cyclically loaded transverse to the weld at a stress

range equal to 1.5Δ𝜎𝐶, where Δ𝜎𝐶 is the characteristic strength of the applicable Sr-N curve. Assume no modifying effects and calculate the fatigue life of the joint for 5% probability of crack initiation

according to EN 1993-1-9. Use partial factor for fatigue 𝛾𝑀𝑓 = 1. The slopes of the Sr-N curve is 3 and

5 respectively.

Answer: Start: 20:48, End: 20:51, Duration: 3 min. Allow: 6 min.

For stress ranges less than 1.5 the yield strength and the constant amplitude fatigue limit, the equation for the S-N curve according to EN 1993-1-9 is:

𝑁𝑅 = (Δ𝜎𝐶

Δ𝜎𝑅)

3

𝑁𝐶

𝑁𝐶 = 2 × 106 Δ𝜎𝑅 = 1.5Δ𝜎𝐶 𝑁𝑅 = 593 000 𝑐𝑦𝑐𝑙𝑒𝑠

Question 3 [5 Marks]

A joint detail is subjected to a cyclic load with stress range 0.57Δ𝜎𝐶, where Δ𝜎𝐶 is the characteristic strength of the applicable detail category dependent Sr-N curve. What is the fatigue life in number of

cycles for a 5% probability of crack initiation according EN 1993-1-9? Use partial factor for fatigue 𝛾𝑀𝑓 =

1. The slopes of the Sr-N curve is 3 and 5 respectively.

Answer. Start 20:56, End: 20:59, Duration 3 minutes, Allow 6 min.

The constant amplitude fatigue limit is:

Δ𝜎𝐷 = Δ𝜎𝐶 (𝑁𝐶

𝑁𝐷)

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𝑁𝐶 = 2 × 106 𝑁𝐷 = 5 × 106 Δ𝜎𝐷 = 0.74Δ𝜎𝐶

The applied stress range of 0.57Δ𝜎𝐶 is lower than the constant amplitude fatigue limit. Therefore, the joint will have infinite life.

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2. FRACTURE MECHANICS [20 Marks – one mark/minute]

Explain, by making use of sketches,

the total life if a component, and,

the principles of fracture control.

Your discussion should at least include the following:

1. Fatigue & crack initiation and Miner’s damage rule.

2. Fracture & plastic collapse.

3. Crack propagation & Paris rule.

4. Inspection intervals.

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3. ADVANCED WELD FATIGUE DECISION-MAKING [45 Marks – one mark/minute]

Note, to obtain maximum points for each problem clearly motivate solutions and equations used.

3.1. Problem Statement

A propane gas cylinder has a nominal diameter of 250 mm and a wall thickness of 3 mm. The semi-

elliptical heads are welded on using a complete joint penetration groove weld (See figure below). This

is the only weld type in the pressure vessel and there are no longitudinal welds.

The maximum design pressure is 4.0 MPa.

Determine the number of times that the cylinder can be filled for a 95% probability of survival (5%

probability of crack initiation) according to EN 1993-1-9.

Assume that suitable surface protection is applied. Failure of this gas cylinder will have a high

consequence. Further, for extra safety, provision shall be made for safe life. That is, the cylinder shall

not be assumed damage tolerant. Assume that the longitudinal and circumferential pressure are given

by:

𝜎𝑙𝑜𝑛𝑔 =𝑃𝑟

2𝑡

𝜎𝑐𝑖𝑟𝑐 =𝑃𝑟

𝑡

On the next page, typical detail categories are shown for welds subject to parallel and transverse stress (EN 1993-1-9 relevant tables).

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3.2. Solution

Stress range

The stress that will cause crack initiation at the weld toe in this joint is the longitudinal stress. However, the weld is also loaded in parallel by the circumferential stress. Therefore, both cases need to be calculated and compared.

Δ𝜎𝑙𝑜𝑛𝑔 =𝑃𝑟

2𝑡− 0

Δ𝜎𝑐𝑖𝑟𝑛 =𝑃𝑟

𝑡− 0

Detail category

Longitudinal stress:

4 Marks

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Circumferential stress:

The detail categories for this problem from supplied slides are 80 for longitudinal stress and 100 for circumferential stress. However, circumferential stress is twice the longitudinal stress. Therefore, the weld will fail for the circumferential stress loading the detail category 100 weld.

Partial factor for fatigue

The partial factor for fatigue is 𝛾𝑀𝑓 = 1.35

Calculations

Based on the calculations below, the cylinder can tolerate 312 695 cycles before there is a 75% confidence level of a 95% probability of crack initiation. The crack is expected to initiate transverse to the weld because it is dominated by the parallel circumferential stress.

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D 250 mm

t 3 mm

P 4.00E+06 Pa

Delta-S 167 MPa

Det.Cat. 100

gamm_Mf 1.35

Sigma_C 74.1 MPa

N_C 2.00E+06 cycles

N_D 5.00E+06 cycles

N_L 1.00E+08

m1 3.0

m2 5.0

Delta-Sigma_D 54.6

Delta-Sigma_L 30.0

N_R 175 583 cycles

3 Marks 1 Mark for correct procedure, but wrong answer.

1 Mark

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4. PRESSURE EQUIPMENT [45 Marks – one mark/minute]

4.1. Problem Statement

A pressure vessel is required that will contain air at design pressure 1 000 kPa and temperature 150 °C. The cylindrical pressure vessel must have a diameter of 2 m and length 3 m. Due to the available plate sizes, the pressure vessel shell will have circumferential and longitudinal welded joints. The cylindrical shell will have two 2:1 semi-elliptical heads. The corrosion allowance that must be designed for is 2 mm. The plates that must be used in the construction of the pressure vessel is as summarised in Table 1. Complete joint penetration butt welds (welded from both sides) are used in all joints. Sufficient sections shall be selected for radiographic examination (according to UW-52) to comply with the spot degree of radiographic examination.

Table 1: Properties of two steels

Spec No. Thickness P-No. Group No. 𝒇𝒖,𝒎𝒊𝒏

[MPa]

𝒇𝒚,𝒎𝒊𝒏

[MPa]

𝑺

[MPa]

150°C 315°C

SA-455 ≤10 mm 1 2 525 266 150 135

SA-299 >25.4 mm 1 2 525 280 150 143

Source: ASME Section II, Part D, Subpart 1 (1999:20).

Questions:

1. What minimum thickness would you recommend to use for the pressure vessel shell? [15 Marks]

2. What minimum thickness would you recommend for the 2:1 ration semi-elliptical heads? [10 Marks]

3. Make sketch of the weld design that you would recommend at the circumferential joint between the shell and heads. [5 Marks]

4. What hydrostatic test pressure will you prescribe for this pressure vessel according to OHS Act No. 85? [5 Marks]

5. What is the hazard category for this pressure vessel according to SANS 347? [10 Marks]

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4.2. Solution

Question 1 Start: 17:57, End 18:04, Duration 7 min. Allow 15 min.

The joint efficiency in this case is 𝐸 = 0.85.

The longitudinal welds are subject to the highest stress.

The wall thickness for circumferential stress is:

𝑡𝑚𝑖𝑛 =𝑃𝑅

𝑆𝐸 − 0.6𝑃

In this case:

𝑃 = 1 × 106 𝑆 = 150 × 106 𝐸 = 0.85

From which:

>> P=1e6;

>> S=150e6;

>> E=0.85;

>> R=1;

>> t=P*R/(S*E-0.6*P)

and

𝑡𝑚𝑖𝑛 = 0.00788 m = 7.9 mm

The corrosion allowance is 2 mm, and therefore, the minimum plate thickness to use for the shell is 10 mm.

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Question 2. Start: 18:05, End 18:10, Duration 00:05, Allow 10 min.

The thickness of the 2:1 semi-elliptical dome is:

𝑡𝑝 =𝑃𝐷

2𝑆𝐸 − 0.2𝑃

𝑃 = 1 × 106 𝑆 = 150 × 106 𝐸 = 0.85 𝑡𝑝 = 7.8 𝑚𝑚

Make allowance for corrosion and use plate with minimum thickness: 𝑡 = 7.8 + 2 = 9.8 𝑚𝑚.

Question 3. Allow 5 min

Allow the following options:

Question 4. Allow 5 min.

The hydrostatic test pressure according to OHS Act No. 85 is 1.25 × design pressure. In this case, it will be 1 250 kPa.

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Question 5, Start 18:18. End: 18:22, Duration 4 min, Allow 10 min.

The pressure vessel will be used for a non-dangerous gas, air. The volume of the pressure vessel is:

𝑉 =𝜋

4𝐷2𝐿

𝐷 = 2 𝑚 𝐿 = 3 𝑚 𝑡ℎ𝑒𝑛 𝑉 = 9.42 𝑚3 = 9 425 ℓ

>> V=pi/4*D^2*3

V =

9.4248

Based on the figure shown below, this pressure vessel is in Hazard Category IV.