MLE1101 - Tutorial 4 - Suggested Solutions

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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) Engineering Strain (in./in.) Engineering Stress (kpsi) Engineering Strain (in./in.) 0 0 60 0.0035 15 0.0005 66 0.004 30 0.001 70 0.006 40 0.0015 72 0.008 50 0.0020 Solution: (a)

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

(c) From figure in part (a), tensile elastic modulus 30 1000 psi 3.0 10 psi7 0.001

E  



2 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:









Assume no volume change, A l0 0 A li i,

(a) ₂ 0 25000lb 125,000 psi 0.200in f F A     i 0  0   2   2 0 i 0.200in 1 1 0.058 0.189in l l A l A (b) 25000lb₂ 133,000 psi 0.189in f T i F A       i  0  2  2 0 0.200in ln ln ln 0.057 0.189in T i A l l A

3. (a) Why do pure FCC metals like Ag and Cu have low values of critical resolved shear stress c?

(b) 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.

3 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 54.7 3

a a

   

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

(b) Referring to figures (i) and (ii) below,  = 90°, cos 54.7 3

a a

   

4 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) Calculate the total percent cold work that the wire undergoes.

(b) Estimate the wire’s tensile and yield strengths and elongation from Figure (a) in the appendix.

Solution:

(a) % cold reduction change in cross-sectional area 100% original cross-sectional area

 

















total % cold work 100% 38.75%

3.13mm     _        

(b) From fig. (a) in appendix, for cold work ≈ 39%,

Tensile Strength  76 kpsi; Yield Strength  64 kpsi and Elongation  7 %

6. 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.

5 Recognise that the sum of weight fractions of liquid and solid phases is equal to 1

1 l S X X 

Considering the weight balance of B in the alloy as a whole and the sum of B in the two phases, we arrive at:





0 0

0

1

wt fraction of solid phase, l l S S S l S S l S S l w X w X w w X w X w w w X w w          Similarly,   

wt fraction of liquid phase, S o l

S l

w w

X

w w

7. 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     

6 (c) at 780oC + T,

Phases present: Liquid beta

Composition of phase: 71.9 wt% Ag in liquid phase 91.2 wt% Ag in beta phase Amount of each phase:

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

Composition of phase: 18 wt% Os in liquid phase 40 wt% Os in alpha 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

7 (b) at 2665oC - T,

Phases present: Liquid alpha

Composition of phase: 23.0 wt% Os in liquid phase 43.0 wt% Os in alpha 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 beta

Composition of phase: 23.0 wt% Os in liquid phase 61.5 wt% Os in beta 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. 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 wt% of phase 100% 66.7% 4.3 4             4.3 4.2 wt% of phase 100% 33.3% 4.3 4