Aluminium alloy AA7475 is applied as a high strength aluminium sheet for aerospace. In preparation for its use a number of tests are carried out on sample sheets including analysis of fracture toughness. A central through-thickness crack is introduced in the sheets. The suppliers of the material quote a fracture toughness of 42 MPa m1/2. Dimensions of the sheet are demonstrated in Figure Q3, where Width W = 50 mm and thickness t = 1 mm. A stress of 447 MPa is applied. Calculate the critical crack half length ac. You may assume a value of Y = 1. Answers should be given in metres (without units) to 2 significant figures Stress intensity K = Υσ πα t W 2a Į F Figure Q3 - a plate of width W containing a central through-thickness crack of length 2a

Consider a metal rod with a critical resolved shear stress of 40 MPa for a specific crystallographic plane (hkl). The angle between the normal to the plane and the applied force is 30 degrees, and the angle between the slip direction and the force is 15 degrees. Two force levels of F1 = 75 N and F2 = 40 N is applied. The cross-sectional area of the rod is 1 mm^(2). Will the slip initiate at the two force levels? Select one: a. F1: Yes, F2: No b. F1: No, F2: No OC. F1: Yes, F2: Yes Od. F1: No, F2: Yes

Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…

A Tensile Tests of structural steel indicate that it fails whe the axial tensile strain is 1.80x10 μ. 3 Let maximum change in length that 0.25 [in] diamoren in tension car with stand before failure Length Find Structural Steel rad it It is 24 [in] in ni sa 9102