Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 6.13, Problem 77SEP
a)
To determine
Relationship between the true strain and engineering strain has to be derived.
b)
To determine
The relationship between the trues stress and engineering strain has to be derived.
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Bronze alloy, the following true stresses produce the corresponding plastic true strains, before to necking: On the basis of this information, compute the true stress necessary to produce a true strain of 0.25 and then find the engineering stress and strain at this point. True stresses (MPa) True strain 354 0121 70 018
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Question 4
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You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic
strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version
expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as
✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is
✓ MPa for 0.1 plastic strain,
proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k=
MPa. Hence, we can predict that we need a prior plastic strain of approximately
True Stress (MPa)
True Stress (MPa)
True Stress (MPa)
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Ep = 0.1
0.01
Ep=0.2
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Ep=0.3
0.01
0.02
True Strain
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True Strain
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True Stress (MPa)
0.05 0.000 0.001
True Stress (MPa)
Ep=0.1
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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
- 1) Draw (using a normal graph paper) a conventional stress-strain diagram for ANY metallic material (e.g. steel, aluminium, copper, brass, iron, tungsten). The diagram should be as accurate as possible using a suitable scale (e.g. 1cm: 10 N). 2) Calculate the Modulus of Elasticity, Modulus of Toughness and Modulus of Resilience for the material from the stress-strain diagram. Show your calculations in detail on a separate A4 piece of paper.arrow_forwardAt higher temperature, strength and strain hardening are increased, whereas, ductility is decreased which permits greater plastic .deformation True O False O .The strength constant (C) is increased with increasing of temperature True O False O The metal is becoming weaker as strain increases, this is because of .strain hardening (work hardening) property نقطة واحدة True O False O The engineering stress and strain are defined relative to the .instantaneous area and length of test specimen True O False aaly ihi In sheetmetal working processes, the surface area-to-volume ratio of .w.p. is low True O False ialy ihii Determine the value of the strain-hardening exponent for a metal that will cause the average flow stress to be 70% of the final flow stress after .deformation 0.444 0.421 0.422 0.428 aals i For pure copper (annealed), the strength coefficient = 330 MPa and strain-hardening exponent = 0.52 in the flow curve equation. Determine the average flow stress that the metal experiences…arrow_forwardDefine normal strain?arrow_forward
- What is the transformation of strain at a point?arrow_forwardAfter explaining the difference between the true stress/strain and the nominal stress/strain, show the true stress-strain diagram for the nominal stress-strain diagram (Figure 1) to correspond points 1, 2, 3, and 4.arrow_forwardDraw two schematic graphs using pencil showing a typical stress-strain curve for aluminum. The first graph should show engineering stress vs engineering strain, and the second graph should show true stress vs true strain. Label the showing: (i) elastic modulus (ii) proportional limit (iii) yield stress (iv)yield strain (v) fracture stress (vi) fracture strain on each graph. You may showboth graphs on one plot. Explain the difference between engineering stress and true stress.arrow_forward
- Write the definitions for engineering stress, true stress,engineering strain, and true strain for loading along a singleaxis.arrow_forwardThe engineering stress-engineering strain tensile curve for a stainless steel type 304 alloy is shown. Calculate the constants K and n for the true stress and true strain curve before necking (or = K (&)"). Engineering Stress (MPa) 1000 900- 800- 700- 600- 500- 400- 300- 200- 100- of 0.0 02 0.4 Enginering Strain (mm/mm) 0.6 0.8 1.0arrow_forwardQuestion 4 You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.1 plastic strain, ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= ✓ MPa. Hence, we can predict that we need a prior plastic strain of approximately ✓to obtain a hgth of 120 MPa True Stress (MPa) 250 200 100 85 95 50 105 115 125 135 145 300 350 0.12 0.14 150 0.16 0.18 155 165 175 200 250 0.1 True Stress (MPa) Ep = 0.1 Zoomed version of left plotarrow_forward
- Strain of Silly Putty: Initial length (L0): 2.85 cm Final length (L): 8.50 cm Strain: % (show calculation below): (Please type answer no write by hend)arrow_forwardAfter the yield region in the stress-strain curve the strength up to ultimate value because: -Molecules orientation in the direction of applied stress -Molecules fails to withstand the applied stress -Molecules orthogonal to the axis of applied force The following figure shows typical engineering tensile stress vs. strain curves, where: * Stress Strain - Points A, C, E and F correspond to the tensile strength and elongation at yield - Points A, B, D and F correspond to the tensile strength and elongation at break Points A, C, E and F correspond to the tensile strength and elongation at breakarrow_forwardDraw the tensile curves of the following materials, taking into account the appropriate elongation at break, modulus of elasticity and tensile stress, on the side stress-strain graph. a)Low carbon steel (Ecelik=200Gpa),(Syield=300MPa),(Stensile=400MPa) b)High carbon steel (Esteel=200GPa),(Stensile=700MPa) c)Aluminum (Ealuminum=70GPa), (Syield=200MPa),(Stensile=300MPa) d)Cast iron (Ecast iron=120GPa),(Stensile=200MPa)arrow_forward
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