MLE1101 - Tutorial 4 - Suggested Solutions

MLE1101 Tutorial 4 - Suggested Solutions 1.

The following engineering stress-strain data were obtained at the beginning of a tensile test for a 0.2%C plain-carbon steel. (a) Plot the engineering stress-strain curve for these data. (b) Determine the 0.2% offset yield stress for this steel. (c) Determine the tensile elastic modulus of this steel. (Note that these data only give the beginning part of the stress-strain curve.) (Leave your answers in terms of pounds and inches.) Engineering Stress (kpsi) 0 15 30 40 50

Engineering Strain (in./in.) 0 0.0005 0.001 0.0015 0.0020

Engineering Stress Engineering (kpsi) Strain (in./in.) 60 0.0035 66 0.004 70 0.006 72 0.008

Solution: (a)

(b)

From figure in part (a), 0.2 percent yield stress = 66 kpsi.

(c)

From figure in part (a), tensile elastic modulus E 

 30  1000 psi   3.0  107 psi  0.001

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2.

A 0.505-in.-diameter aluminum alloy test bar is subjected to a load of 25,000 lbf. If the diameter of the bar is 0.490 in. at this load, determine (a) the engineering stress and strain and (b) the true stress and strain. (Leave your answers in terms of pounds and inches.) Solution: A0  Ai 

 d0 2 4

 di 2

  0.505in

2



4

  0.490in

 0.200in2

2



 0.189in2

4 4 Assume no volume change, A0 l0  Ai li ,

(a)

(b)



F 25000lb f   125,000 psi A0 0.200in2



li  l0 A0 0.200in2  1   1  0.058 l0 Ai 0.189in2

T 

F 25000lb f   133,000 psi Ai 0.189in2

 T  ln

3.

(a) (b)

A li 0.200in2  ln 0  ln  0.057 l0 Ai 0.189in2

Why do pure FCC metals like Ag and Cu have low values of critical resolved shear stress c? What is believed to be responsible for the high values of c for HCP titanium?

Solution: (a) For FCC metals like Ag and Cu, slip takes place on the close-packed {111} octahedral planes and in the  110  close packed directions. Lower shear stress is required for slip to occur in densely packed planes. (b)

The high c values associated with HCP titanium is attributed to the mixed covalent and metallic bonding with the atomic lattice structure.

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4.

A stress of 75 MPa is applied in the [001] direction on an FCC single crystal. Calculate (a) the resolved shear stress acting on the (111)[101] slip system and (b) the resolved shear stress acting on the (111)[110] slip system. Solution: (a)

Referring to figures (i) to (iii) below,  = 45°, cos  

a    54.7 3a

resolved shear stress,  r   cos  cos   75cos45 cos54.7  30.6MPa

(b)

Referring to figures (i) and (ii) below,  = 90°, cos  

a    54.7 3a

resolved shear stress,  r   cos  cos   75cos90 cos54.7  0MPa

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5.

A 70%Cu–30%Zn brass wire is cold-drawn 20 percent to a diameter of 2.80 mm. The wire is then further cold-drawn to a diameter of 2.45 mm. Definition of cold reduction is given as: change in cross-sectional area % cold reduction   100% original cross-sectional area (a) (b)

Calculate the total percent cold work that the wire undergoes. Estimate the wire’s tensile and yield strengths and elongation from Figure (a) in the appendix.

Solution: (a)

% cold reduction 

20% 

 4

change in cross-sectional area  100% original cross-sectional area

d12  4  2.80mm

2

 4

d12

 100%

0.20d12  d12  7.84mm2  d1  3.13mm    3.13mm2    2.45mm2  4 total % cold work   4   100%  38.75%  3.13mm  2   4

(b)

6.

From fig. (a) in appendix, for cold work ≈ 39%, Tensile Strength  76 kpsi; Yield Strength  64 kpsi and Elongation  7 %

Derive the lever rule for the amount in weight percent of each phase in two-phase regions of a binary phase diagram. Use a phase diagram in which two elements are completely soluble in each other. Solution:

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Recognise that the sum of weight fractions of liquid and solid phases is equal to 1 Xl  X S  1 Considering the weight balance of B in the alloy as a whole and the sum of B in the two phases, we arrive at: w0  X l w l  X S w S w0   1  X S  w l  X S w S

wt fraction of solid phase, X S 

w0  wl w S  wl

Similarly, wt fraction of liquid phase, Xl 

7.

w S  wo w S  wl

Consider the binary eutectic copper-silver phase diagram in the appendix (Figure b). Make phase analyses of an 88wt%Ag–12wt%Cu alloy at the temperatures (a) 1000oC, (b) 800oC, (c) 780oC + T, (d) 780oC – T. T is assumed to be less than 1oC. In the phase analyses, include: (i) The phases present; (ii) The chemical compositions of the phases; (iii) The amounts of each phase; (iv) Sketch the microstructure by using 2-cm-diameter circular fields. Solution: (a) at 1000oC, Phases present: Liquid Composition of phase: 88 wt% Ag Amount of each phase: wt% of liquid phase = 100%

(b) at 800oC, Phases present:

Liquid

beta

Composition of phase:

78 wt% Ag in liquid phase

93 wt% Ag in beta phase

Amount of each phase: 93  88 wt% of liquid phase   100%  33.3% 93  78 88  78 wt% of beta phase   100%  66.7% 93  78

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(c) at 780oC + T, Phases present: Composition of phase: Amount of each phase:

Liquid 71.9 wt% Ag in liquid phase

beta 91.2 wt% Ag in beta phase

91.2  88  100%  16.6% 91.2  71.9 88  71.9 wt% of beta phase   100%  83.4% 91.2  71.9

wt% of liquid phase 

(d) at 780oC - T, Phases present:

alpha

beta

Composition of phase:

7.9 wt% Ag in alpha phase

91.2 wt% Ag in beta phase

Amount of each phase: 91.2  88 wt% of alpha phase   100%  3.8% 91.2  7.9 88  7.9 wt% of beta phase   100%  96.2% 91.2  7.9

8.

Consider the binary peritectic iridium-osmium phase diagram in the appendix (Figure c). Make phase analyses of a 70wt%Ir–30wt%Os alloy at the temperatures (a) 2600oC, (b) 2665oC – T, (c) 2665oC + T. T is assumed to be less than 1oC. In the phase analyses, include: (i) The phases present; (ii) The chemical compositions of the phases; (iii) The amounts of each phase; (iv) Sketch the microstructure by using 2-cm-diameter circular fields. Solution: (a) at 2600oC, Phases present: Liquid Composition of phase: 18 wt% Os in liquid phase Amount of each phase: 40  30 wt% of liquid phase   100%  45.4% 40  18 30  18 wt% of alpha phase   100%  54.6% 40  18

alpha 40 wt% Os in alpha phase

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(b) at 2665oC - T, Phases present: Liquid Composition of phase: 23.0 wt% Os in liquid phase Amount of each phase: 43  30 wt% of liquid phase   100%  65.0% 43  23 30  23 wt% of alpha phase   100%  35.0% 43  23

(c) at 2665oC + T, Phases present: Liquid Composition of phase: 23.0 wt% Os in liquid phase Amount of each phase: 61.5  30 wt% of liquid phase  100%  81.8% 61.5  23 30  23 wt% of beta phase   100%  18.2% 61.5  23

9.

alpha 43.0 wt% Os in alpha phase

beta 61.5 wt% Os in beta phase

Consider an Fe – 4.2wt%Ni alloy (figure d in appendix) that is slowly cooled from 1550oC to 1450oC. Determine the amount of  phase and  phase, respectively, at a temperature of 1517oC – T. Assume T to less than 1oC. Solution:

4.2  4  100%  66.7% 4.3  4 4.3  4.2 wt% of  phase   100%  33.3% 4.3  4

wt% of  phase 

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