Question 1. Consider a structural material with a yield strength of 80 ksi¹ and an ultimate tensile strength of 100 ksi. A very wide plate of this material 2 inches thick contains an edge crack 1.0 inch long. The part is loaded axially with a normal stress σ. Call the value of this stress far away from the crack the "nominal stress."

a) Determine the nominal stress σ at failure if the material has a toughness of K_k = 60 ksi√in?

b) If the nominal stress is 20 ksi, determine the critical crack length acr at which fracture would occur.

Fatigue, Goodman Relation, Miner's Rule

Quesiton 2. Consider a part made of Aluminum 7075-T6 alloy. What number of cycles to failure (using the mean value from the S-N-P curves, P = 0.50) is expected for the following cases. Use the Goodman relation.

• S_m = 0, S_a = 35 ksi: Compare 10%, 50% and 90% probability of failure

• S_m = 20 ksi, S_a = 35 ksi: Compare 10%, 50% and 90% probability of failure

• S_m = -20 ksi, S_a = 35 ksi: Compare 10%, 50% and 90% probability of failure

- Comment on the comparisons.

- Note that N is on a logarithmic scale. Be careful about how you read numbers off the chart!

- Be careful to handle the negative mean stress case appropriately. See class notes.

- Include a marked copy of the S-N diagram with your solution.

Question 3. Consider a member made of 2024-T4 aluminum (yield strength 325 MPa, ultimate strength 470 MPa) if the fully-reversed S-N diagram is represented as σ_a = 839N_f^-0.102

At a location of interest the material is repeatedly subjected to the uniaxial stress history shown in the figure below. Estimate the number of repetitions of this loading history necessary to cause fatigue failure using the Palmgren-Miner rule.

• Think carefully how the cycles in the "repetition" should be counted. Required: as part of your solution, make a table showing the mean stress, stress amplitude, number of cycles for each cycle type, and number of cycles to failure Nf for each type of cycle contained in the operational cycle.

• Analyze your results and comment. Which portions of the loading history most contribute to the fatigue failure of the part?

Question 4. Suppose we are given the following data about the fracture toughness and fatigue crack growth behavior of a certain type of aluminum:

• Uniaxial yield strength σ_Y = 60 ksi

• Ultimate tensile strength σ_u = 70 ksi

• Plane strain fracture toughness K_k = 45 Ksi√in

• Paris Law valid for zero-to-max loading: da/dN = 0.70 x 10^-9(ΔK)³⁶ , where K is in units of ksi√in and a is in units of inch.

• Recall the following formula for stress intensity factor for double edge cracks: k₁ = 1.12σ_o√Πa, valid for a ratio of a/b < 0.6. For larger a/b ratios, use formula from Dowling available in the class notes.

Consider a sample of this material containing double edge cracks as shown in the sketch. given: b = 3 inches and a thickness of 0.5 inches

a) Determine the maximum quasistatic load P that the specimen can support when a = 1 inch. Consider both fracture and ductile collapse.

b) Now, if the part is loaded with a cyclic load as shown below, estimate the number of cycles of loading it would take for a crack to grow from a = 1.00 inches to 1.01 inches, assuming da/dN remains constant.

c) For a crack loaded as shown below starting at length of 1.0 inches and ending at the critical crack length (i.e. the crack length at which failure would occur up reaching P=50kip), plot the crack length versus total number of cycles of loading for the same loading condition as used in part b (this graph will show how long the crack will be for any total number of cycles counted during a fatigue test to failure). Note that da/dN will not remain constant as the crack grows.