The direction of fracture surface respectively to tensile loading direction in the fracture test.

Question

Want to see the full answer?

Answer 1

Question

Chapter 11, Problem 12CQ

To determine

The direction of fracture surface respectively to tensile loading direction in the fracture test.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

2. a) As an engineer, choose a desired fracture mode in metals with one (1) reason. b) Sketch the fracture behaviour in metals. c) i. The modulus of elasticity of a metal alloy A is 260 GPa. Compute the specific surface energy if propagation of an internal crack of length 0.40 mm is observed when a stress of 63 MPa is applied to the alloy. ii. The elastic deformation energy of alloy A is 3.0 J/m². Based on the answer obtained in c)(i), show whether alloy A is an elastic deformation or a plastic deformation. iii. Interpret if alloy A can be classified as a brittle or a ductile material. d) i. A metal with an internal crack is loaded with a tensile stress of 15 MPa. If the crack length and the radius of curvature are 2.6 x 10-2 mm and 1.1 x 10-4 mm, compute its maximum stress. ii. The modulus of elasticity of the metal is 90 GPa and the specific surface energy is 2.6 J/m2. Based on its critical stress, show that the crack will not grow when a tensile stress of 15 MPa is loaded on it.

2. Please estimate the number of cycles to failure of a steel specimen under tensile fatigue loading with the following parameters. The R ratio is 3, mean stress 200 MPa, yield strength 450 MPa, ultimate tensile strength 560 MPa, Young’s modulus 200 GPa, KIC = 140 MPa . Assume the initial crack length is 0.1 mm.

A brass specimen of the circular cross-section is fractured at 151 kN force and the final length of the specimen at fracture is 49 mm. The fracture strength of the specimen is found to be 74 kN/mm2. The percentage of elongation of the specimen is 42 %. Determine the following (i) Diameter of the specimen ii) Initial length of the specimen iii) Stress under an elastic load of 16 kN iv) Young's Modulus if the elongation is 1.6 mm at 16 kN (v) Final diameter if the percentage of reduction in area is 20 % solve: Initial Cross-sectional Area (in mm2) = The Diameter of the Specimen (in mm) = Initial Length of the Specimen (in mm) =