EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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Chapter 3, Problem 3.4Q
To determine
The significance of a slip system.
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Students have asked these similar questions
(a) Define a slip system.
(b) Do all metals have the same
slip system? Why or why not?
1. There are three slip systems on a FCC octahedral plane, i.e. one of the {111} planes. Assume a2 MPa tensile stress is applied along the [100] direction of a gold single crystal, whose criticalresolved shear stress is 0.91 MPa at room temperature. Demonstrate quantitatively thatmeasurable slip will not occur on any of the three slip systems containing the (111) plane as aresult of this applied tensile stress. (Hint: Schmid’s law)
2. Identify the Burgers vector (using vector notation) of a screw dislocation that can cross-slipbetween (111) and (111) planes of an FCC crystal (Hint: Use Weiss zone law).
3. Why are dislocations not considered thermodynamic equilibrium defects like vacancies?Explain by describing the relative interplay between entropy and enthalpy affecting the Gibbsfree energy of the dislocations-containing material system.
Metal y (mJ/m)n
Slip Character
(a) Discuss why Planar slip is associated with a small stacking fault energy, as shown in the
table below. (b) Sketch the shape of a single-crystal shear stress-strain curve for Al and
Stainless Steel, (1label) stage I, II and III portions of deformation, which one end stage II and
begin stage III sooner?.
S.Steel
<10
-0.45
Planar
Cu
-90
-0.3
Planar/ wayy
Al
-250
-0.15
wayy
Chapter 3 Solutions
EBK MANUFACTURING PROCESSES FOR ENGINEE
Ch. 3 - Prob. 3.1QCh. 3 - Prob. 3.2QCh. 3 - Prob. 3.3QCh. 3 - Prob. 3.4QCh. 3 - Prob. 3.5QCh. 3 - Prob. 3.6QCh. 3 - Prob. 3.7QCh. 3 - Prob. 3.8QCh. 3 - Prob. 3.9QCh. 3 - Prob. 3.10Q
Ch. 3 - Prob. 3.11QCh. 3 - Prob. 3.12QCh. 3 - Prob. 3.13QCh. 3 - Prob. 3.14QCh. 3 - Prob. 3.15QCh. 3 - Prob. 3.16QCh. 3 - Prob. 3.17QCh. 3 - Prob. 3.18QCh. 3 - Prob. 3.19QCh. 3 - Prob. 3.20QCh. 3 - Prob. 3.21QCh. 3 - Prob. 3.22QCh. 3 - Prob. 3.23QCh. 3 - Prob. 3.24QCh. 3 - Prob. 3.25QCh. 3 - Prob. 3.26QCh. 3 - Prob. 3.27QCh. 3 - Prob. 3.28QCh. 3 - Prob. 3.29QCh. 3 - Prob. 3.30QCh. 3 - Prob. 3.31QCh. 3 - Prob. 3.32QCh. 3 - Prob. 3.33QCh. 3 - Prob. 3.34QCh. 3 - Prob. 3.35QCh. 3 - Prob. 3.36QCh. 3 - Prob. 3.37QCh. 3 - Prob. 3.38QCh. 3 - Prob. 3.39QCh. 3 - Prob. 3.40QCh. 3 - Prob. 3.41QCh. 3 - Prob. 3.42QCh. 3 - Prob. 3.43PCh. 3 - Prob. 3.44PCh. 3 - Prob. 3.45PCh. 3 - Prob. 3.46PCh. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54P
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- You are given 1) the normal to the slip plane and the normal to the slip direction angle with the tensile axis, 2) the critical resolved shear stress, and 3) applied stress. How would you determine whether the given applied stress will cause the single crystal to yield?arrow_forwardWhat does the slip being between 0,1-1 mean?arrow_forwardWhat is a consequence of strain hardening?arrow_forward
- Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, respectively, with the tensile axis. If the critical resolved shear stress is 27.1 MPa, what applied stress (in MPa) will be necessary to cause the single crystal to yield?arrow_forwardQ3 contd. (d) The yield strength values of pure aluminium (Al) and pure copper (Cu) are 25 MPa and 20 MPa, respectively; whereas the yield strength values of cold rolled Al-Mn-Mg alloy and cast 60-40 Brass (60% Cu, 40% Zn) are 200 MPa and 105 MPa, respectively. With aid of schematics, explain the main mechanisms account for the increases in the strengths. (e) A cylindrical tie rod with a diameter of 18.4 mm is subjected to cyclic loading. The stress range is +/- 200 kN. Figure Q3.3 shows the S-N curve of the material of which the rod is made, how many cycles will this rod survive? Stress amplitude O₂ (MPa) 1500 1400 1300 1200 1100 1000 900 800 700 600 500 400 10² 10³ 104 4340 low-alloy steel Stress ratio = -1 Fig. Q3.3 105 106 Number of cycles to failure, Nf 107 108arrow_forwardCold working of metals will give better final properties than hot working. Select one: True Falsearrow_forward
- From the work of D. C. Jillson, Trans. AIME 188, 1129 (1950), the following data were taken relating to the deformation of zinc single crystals. The crystals have a normal cross-sectional area of 122 x 10-6 m2. ϕ = angle between loading axis and normal to slip plane λ = angle between loading axis and slip direction F = force acting on crystal when yielding begins a. Identify the slip system for this material. b. Calculate the resolved shear τRSS and normal σn stresses acting onthe slip plane when yielding begins. c. From your calculations, does τRSS or σn control yielding? d. Plot the Schmid factor versus the normal stress P/ A0 acting on the rod. At what Schmid factor value are these experimentally-measured yield loads at a minimum? Does this make sense?arrow_forwardWhat are slipbands and slip lines? What causes the formation of slipbands on a metal surface? What are the principal slip planes and slip directions for FCC, BCC and HCP metals?arrow_forward4. Assume the critical resolved shear stress of a single crystal metal is 5.5 MPa on the slip system. Please (a) calculate the corresponding tensile stress applied along the direction that can cause such slip, and (b) what is the corresponding Schimd factor?arrow_forward
- Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1° and 47.9°, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa (3000 psi), will an applied stress of 45 MPa (6500 psi) cause the single crystal to yield? If not, what stress will be necessary?arrow_forwardA single crystal of aluminum is loaded under a stress state such that ơ11 = 250 kPa, 022 = 50 kPa , o33 = 10 kPa and T23 = T 31 = T 12 = 0 Assume that slip occurred on the (111) plane in the [1 -1 0] direction and only on this slip system. What is the shear stress on this slip system?arrow_forwardWhat is the effect of solid solution strengthening on Tensile strength (TS) and Ductility (%EL,%AR)?arrow_forward
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