Question

Q-1. You are testing a circular beam of two metals, 1045 steel and 2014-T6 aluminum, in a rotating-cantilever beam fatigue test. If you need to choose one of these metals so that you can essentially design a part made of that metal, and subject the part to a repetitive stress-amplitude without ever failing due to fatigue, which one would you choose. (See the attached figure) Q-2. What is the fatigue strength for the 1045 steel alloy, if you expect to survive 100,000 cycles of repetitive stress? Answer: Q. What is the stress below which you could essentially claim that a part made of 4340 steel alloy would not fail due to fatigue? Answer: Q-4. What is the stress below which you could essentially claim that a part made of 2014-T6 Al alloy would not fail due to fatigue? Answer: Q-s. Plastic deformation of a metal at a low (-room) temperature is (a) A mechanism for strengthening a metal (b) A manufacturing method for producing metal parts (c) Both a manufacturing method and a strengthening mechanism

Answer 1

5. C Plastic deformation is both a strengthening and a manufacturing method.For example drawing and cold rolling.

6. C A larger value of strain hardening exponent means good stretch formability i.e,more the capacity of the material to delay the neckingi.e, more strength

7, 8,9 in image

10.B In cold working process, generally strength increases with decrease in ductility.Option C is wrong though it seems right because at higher temperatures any materials probability of failing because of creep is high.

11.C Bascially cold working is a simultaneous process of deformation and strengthening of a metal.But one of its disadvantages is the ductility and strain hardening of the work metal limit the amount of forming that can be done to the part.

7 1 4 Cold Work - + - ++ X100 ti 2.3-0.7 X 100 2.3 = 69.56% 8. Foy Calculating total cold Work, in steps; Reference in Each step We must take initial thickness as 4-2 4-2 x 100 = 507 4 4-1 X 100 757 4 - 1 4 470.4 4-0.4 X100 4 907 The total CW is actually 90%.

9. Ą - A f X 100 Ao | - df 1- 0.005 x 100 Z do N 3 X100 = 75 57 4