Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 6.13, Problem 91SEP
To determine
The reason for the higher value of resolved shear stress of BCC than FCC Metals.
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Chapter 6 Solutions
Foundations of Materials Science and Engineering
Ch. 6.13 - (a) How are metal alloys made by the casting...Ch. 6.13 - Why are cast metal sheet ingots hot-rolled first...Ch. 6.13 - What type of heat treatment is given to the rolled...Ch. 6.13 - Describe and illustrate the following types of...Ch. 6.13 - Describe the forging process. What is the...Ch. 6.13 - What is the difference between open-die and...Ch. 6.13 - Describe the wire-drawing process. Why is it...Ch. 6.13 - Distinguish between elastic and plastic...Ch. 6.13 - Define (a) engineering stress and strain and (b)...Ch. 6.13 - Define (a) modulus of elasticity, (b) yield...
Ch. 6.13 - (a) Define the hardness of a metal. (b) How is the...Ch. 6.13 - What types of indenters are used in (a) the...Ch. 6.13 - What are slipbands and slip lines? What causes the...Ch. 6.13 - Describe the slip mechanism that enables a metal...Ch. 6.13 - (a) Why does slip in metals usually take place on...Ch. 6.13 - Prob. 16KCPCh. 6.13 - What other types of slip planes are important...Ch. 6.13 - Define the critical resolved shear stress for a...Ch. 6.13 - Describe the deformation twinning process that...Ch. 6.13 - What is the difference between the slip and...Ch. 6.13 - Prob. 21KCPCh. 6.13 - Prob. 22KCPCh. 6.13 - What experimental evidence shows that grain...Ch. 6.13 - (a) Describe the grain shape changes that occur...Ch. 6.13 - How is the ductility of a metal normally affected...Ch. 6.13 - (a) What is solid-solution strengthening? Describe...Ch. 6.13 - What are the three main metallurgical stages that...Ch. 6.13 - Describe the microstructure of a heavily...Ch. 6.13 - Describe what occurs microscopically when a...Ch. 6.13 - When a cold-worked metal is heated into the...Ch. 6.13 - Describe what occurs microscopically when a...Ch. 6.13 - When a cold-worked metal is heated into the...Ch. 6.13 - Prob. 33KCPCh. 6.13 - Prob. 34KCPCh. 6.13 - Prob. 35KCPCh. 6.13 - Prob. 36KCPCh. 6.13 - Prob. 37KCPCh. 6.13 - Why are nanocrystalline materials stronger? Answer...Ch. 6.13 - A 70% Cu30% Zn brass sheet is 0.0955 cm thick and...Ch. 6.13 - A sheet of aluminum alloy is cold-rolled 30% to a...Ch. 6.13 - Calculate the percent cold reduction when an...Ch. 6.13 - Prob. 42AAPCh. 6.13 - What is the relationship between engineering...Ch. 6.13 - A tensile specimen of cartridge brass sheet has a...Ch. 6.13 - A 0.505-in.-diameter rod of an aluminum alloy is...Ch. 6.13 - In Figure 6.23, estimate the toughness of SAE 1340...Ch. 6.13 - The following engineering stress-strain data were...Ch. 6.13 - Prob. 49AAPCh. 6.13 - A 0.505-in.-diameter aluminum alloy test bar is...Ch. 6.13 - A 20-cm-long rod with a diameter of 0.250 cm is...Ch. 6.13 - Prob. 52AAPCh. 6.13 - Prob. 53AAPCh. 6.13 - Prob. 54AAPCh. 6.13 - Prob. 55AAPCh. 6.13 - Prob. 56AAPCh. 6.13 - A specimen of commercially pure titanium has a...Ch. 6.13 - Prob. 58AAPCh. 6.13 - Prob. 59AAPCh. 6.13 - Prob. 60AAPCh. 6.13 - Prob. 61AAPCh. 6.13 - Prob. 62AAPCh. 6.13 - Prob. 63AAPCh. 6.13 - Prob. 64AAPCh. 6.13 - Prob. 65SEPCh. 6.13 - Prob. 66SEPCh. 6.13 - A 20-mm-diameter, 350-mm-long rod made of an...Ch. 6.13 - Prob. 68SEPCh. 6.13 - Prob. 69SEPCh. 6.13 - Consider casting a cube and a sphere on the same...Ch. 6.13 - When manufacturing complex shapes using cold...Ch. 6.13 - Prob. 74SEPCh. 6.13 - Draw a generic engineering stress-strain diagram...Ch. 6.13 - (a) Draw a generic engineering stress-strain...Ch. 6.13 - Prob. 77SEPCh. 6.13 - Prob. 78SEPCh. 6.13 - Prob. 79SEPCh. 6.13 - The material for a rod of cross-sectional area...Ch. 6.13 - What do E, G, v, Ur, and toughness tell you about...Ch. 6.13 - A cylindrical component is loaded in tension until...Ch. 6.13 - Referring to Figures 6.20 and 6.21 (read the...Ch. 6.13 - (a) Show, using the definition of the Poissons...Ch. 6.13 - A one-inch cube of tempered stainless steel (alloy...Ch. 6.13 - Prob. 87SEPCh. 6.13 - Prob. 88SEPCh. 6.13 - Prob. 89SEPCh. 6.13 - Prob. 90SEPCh. 6.13 - Prob. 91SEPCh. 6.13 - Prob. 92SEPCh. 6.13 - Prob. 93SEPCh. 6.13 - Prob. 94SEPCh. 6.13 - Starting with a 2-in.-diameter rod of brass, we...Ch. 6.13 - Prob. 96SEPCh. 6.13 - Prob. 97SEPCh. 6.13 - Prob. 98SEPCh. 6.13 - The cupro-nickel substitutional solid solution...Ch. 6.13 - Prob. 100SEP
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- 5. Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [102] direction. If slip occurs on a (111) plane and in a [101] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 3.42 MPa.arrow_forwardIf a series of dislocations produced by a single FrankRead source encounter an impassible barrier, a back stress is created. What is the reason for this back stress, and why is it very effective for dislocations produced by a single F-R source?arrow_forwardHelp me pleasearrow_forward
- 2 . Write the slip systems of the metals with crystal structures of BCC, FCC and HCP. Calculate the total number of slip systems for each case. Which among the BCC, FCC and HCP is the most ductile and why?arrow_forwardConsider an iron single crystal with CCC crystal structure oriented in such a way that a tensile stress is applied to the along one direction [010]. a) Calculate the shear stress resolved along a plane (110) and a direction when a tensile stress of 52 MPa is applied. The normal to the slip plane forms 45° with traction direction. the direction of slip forms 55º with the traction direction. b) If the slip occurs in one plane (110) and in one direction and the critical resolved shear stress is 30 MPa, calculate the magnitude of tensile stress that must be applied to initiate the flow.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_forward
- A single crystal of iron (BCC) is pulled along [123]. Which is the first slip system to operate? Use {110}<111> slip systems. Hint: Use Schmid’s law to solve the problem. Also, you need to use Weiss zone law to identify the slip systems Please follow the approach as shown below. In BCC, there are six non-duplicating planes of {110} type: (011), (101), (110), (1-10), (01-1), and (10-1). You need to use then the appropriate single slip directions under <111> family of directions to get the valid slip system combinations (twelve of them). Note that there are more than one possible single slip directions under <111> family of directions. So, you definitely need the zone law to ascertain which slip direction(s) satisfy in relation to each possible slip plane (as given above) and then only you can find out the 12 independent slip systems of {110}<111> type. Then you can apply Schmid's law to solve the problem.arrow_forward1. 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.arrow_forwardA copper-zine alloy has the properties shown in the table below. Grain Diameter (mm) Strength (MPa) 0.015 170 MPa 0.025 158 MPa 0.035 151 MPa 0.050 145 MPa Determine (a) the constants in the Hall-Petch equa- tion; and (b) the grain size required to obtain a strength of 200 MPa.arrow_forward
- exaplin with illustrations, slip via dislocation motion (edge abd screw dislocation) and twinningarrow_forwardMoving to another question will save this response. uestion 37 Materials with dislocations have lower shear strengths than materials with no dislocations. Is this true or false? Moving to another question will save this response. Larrow_forwardWhat process will result in Ferrite plus Pearlite microstructure? List the reasons why this structure will have low strength.arrow_forward
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