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This section includes 10 Mcqs, each offering curated multiple-choice questions to sharpen your Mechanical Metallurgy knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A uniform cylinder of length L is compressed to half of its original length. Calculate true strain and engineering strain for the cylinder. |
| A. | Engineering strain = 1, True strain = 0.69 |
| B. | Engineering strain = 0.5, True strain = 0.69 |
| C. | Engineering strain = -0.5, True strain = -0.69 |
| D. | Engineering strain = -1, True strain = -0.69 |
| Answer» D. Engineering strain = -1, True strain = -0.69 | |
| 2. |
A tensile specimen of 6 mm diameter and gauge length 25 mm reached a maximum load of 45 kN and fractured at 35 kN. The maximum diameter at fracture is 5mm. Determine the true stress at the fracture point. |
| A. | 445 MPa |
| B. | 371.13 MPa |
| C. | 398 MPa |
| D. | 518.87 MPa |
| Answer» E. | |
| 3. |
A tensile specimen of 6 mm diameter and gauge length 25 mm reached a maximum load of 45 kN and fractured at 35 kN. The maximum diameter at fracture is 5mm. Determine the engineering stress at maximum load (Ultimate tensile strength). |
| A. | 796 MPa |
| B. | 398 MPa |
| C. | 512 MPa |
| D. | 52 MPa |
| Answer» C. 512 MPa | |
| 4. |
What is the relationship between engineering stress and true stress?Given that; s = engineering stress, = true stress, e = engineering strain, = true strain. |
| A. | = s(e+1) |
| B. | = ln(s) |
| C. | = ln[s(e+1)] |
| D. | = se+1 |
| Answer» B. = ln(s) | |
| 5. |
For perfectly plastic material, Poisson s ratio is equal to ________ |
| A. | 1 |
| B. | 0.5 |
| C. | 0.33 |
| D. | 0 |
| Answer» C. 0.33 | |
| 6. |
The advantage of using true strain is that the total true strain is equal to the sum of the incremental strains. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 7. |
A uniform cylinder of length L is elongated to twice of its original length. Calculate true strain and engineering strain for the cylinder in percentage? |
| A. | Engineering strain=100%, True strain=69.31% |
| B. | Engineering strain=69.31%, True strain=100% |
| C. | Engineering strain=50%, True strain=50% |
| D. | Engineering strain=50%, True strain=100% |
| Answer» B. Engineering strain=69.31%, True strain=100% | |
| 8. |
The relationship between the true strain and engineering strain is given as _____Where e=conventional strain, = Engineering strain |
| A. | =e |
| B. | =ln(e) |
| C. | =ln(e+1) |
| D. | =ln(1/(1+e)) |
| Answer» D. =ln(1/(1+e)) | |
| 9. |
The initial length of sample is Lo and instantaneous length of the sample is L. The true strain of the material will be equal to _________ |
| A. | L/L<sub>o</sub> |
| B. | L<sub>o</sub>/L |
| C. | ln{(L-L<sub>o</sub>)/L} |
| D. | ln{L/L<sub>o</sub>} |
| Answer» E. | |
| 10. |
The engineering stress-strain curve does not give an accurate indication of the deformation characteristic of the material because it s calculation is based on the original dimension of the specimen. These dimensions continuously change during the test. |
| A. | True |
| B. | False |
| Answer» B. False | |