cive.3100 e m assignment #7: interpret tensile test...

CIVE.3100 ENGINEERING MATERIALS ASSIGNMENT #7: INTERPRET TENSILE TEST RESULTS,

PHASE DIAGRAMS, AND RULE OF MIXTURES

CIVE.3100 2016 Assignment 07 Solution of 12

PROBLEM #1: Given: You have been provided with the results of a tensile test in the Specimen_RawData_1.csv file on the course website. Required: Using the data in this file, determine the following:

The Modulus of Elasticity (E) The Yield Stress at 0.2% offset (Y) The Ultimate Tensile Stress (ult) Percent Elongation at Fracture Percent Reduction in Area at Fracture Using this data, determine the material via http://matweb.com.

Solution: Data results using Excel see Figures 1, 2, and 3 for determination of ult, y, and E.

Figure 1. Plot of Total Zeroed Stress vs. Strain Data.

050

100150200250300350400450500550600650

0.00 0.20 0.40 0.60 0.80

Str

ess

(MP

a)

Strain (mm/mm)

ult = 608 MPa




Figure 2. Plot of Zeroed Stress vs. Strain Data to determine y.

Figure 3. Plot of Zeroed Stress vs. Strain Data to determine E.

0

50

100

150

200

250

300

0.000 0.005 0.010

Str

ess

(MP

a)

Strain (mm/mm)

y = 236 MPa

y = 219849xR² = 0.9969

0

50

100

150

200

250

0.0000 0.0002 0.0004 0.0006 0.0008 0.0010

Str

ess

(MP

a)

Strain (mm/mm)




Percent Elongation at Fracture = Strain at Fracture x 100 = 73.5% Percent Area Reduction at Fracture = (40.26mm2 – 3.50mm2)/40.26mm2 x 100 = 91.3% ANSWERS:

E = 220000 MPa y = 236 MPa ult = 608 MPa Percent Elongation at Fracture = 73.5% Percent Reduction in Area at Fracture = 91.3% Material: 304 Stainless Steel. You probably came up with something

different. Which is OK. This part of the problem was to show you that a lot of metals have similar properties.

PROBLEM #2: Problem 8.3 in the textbook. Given:

Figure 8.2 in Textbook.




Required: Determine how triple points are there in the pure iron pressure-temperature equilibrium phase diagram of Figure. 8.2 and what phases are in equilibrium at each of the triple points. Solution: Three triple points can be identified having the following phases in equilibrium:

1. vapor, liquid and δ Fe 2. vapor, δ Fe, and γ Fe 3. vapor, γ Fe, α Fe

PROBLEM #3: Problem 8.5 in the textbook. Given:

Figure 8.1 in Textbook.

Required: Determine the following:

(a) How many degrees of freedom are there at the triple point?




(b) How many degrees of freedom are there along the freezing line? Solution: (a) At the triple point, there are zero degrees of freedom. (b) Along the freezing line of pure water, there is one degree of freedom. PROBLEM #4: Problem 8.21 in the textbook. Given: Consider an alloy containing 70 wt % Ni and 30 wt % Cu shown in Figure 8.5.

Figure 8.5 in Textbook (note tie line and subsequent wl and ws values are slightly off

due to limitations in Word). Required: (a) At 1350C make a phase analysis assuming equilibrium conditions. In the phase

analysis include the following: (i) What phases are present? (ii) What is the chemical composition of each phase? (iii) What amount of each phase is present?

(b) Make a similar phase analysis at 1500°C. Solution: (a)

(i) The phases present are the liquid and solid (L + α).




(ii) The chemical composition of liquid is wl = 62 wt % Ni while that of the solid is ws = 74 wt % Ni.

(iii) The weight percent of solid and liquid are: 74 70

Wt % of liquid phase 100%74 62

33.3%

70 62Wt % of solid phase 100%

74 62

66.67%

(b) At 1500ºC, the alloy is 100% liquid. PROBLEM #5: Problem 8.22 in the textbook. Given: Consider the binary eutectic copper-silver phase diagram in Figure 8.22. You have an 88 wt % Ag–12 wt % Cu alloy.

Figure 8.22 in Textbook (note tie lines not shown).




Required: Make phase analyses at the following temperatures: (a) 1000°C, (b) 800°C, (c) 780°C + ΔT , and (d) 780°C - ΔT . In the phase analyses, include: (i) The phases present (ii) The chemical compositions of the phases (iii) The amounts of each phase Solution: (a) At 1000ºC:

Phases present: liquid Compositions of phases: 100%

(b) At 800ºC,

Phases present: liquid beta Compositions of phases: 78% Ag in liquid phase 93% Ag in β phase Amounts of phases:

93 88

Wt % liquid phase 100%93 78

88 78Wt % beta phase 100%

93 78

33.3%

66.6%

(c) At 780 C + To ,

Phases present: liquid beta Compositions of phases: 71.9% Ag in liquid phase 91.2% Ag in β phase Amounts of phases:

91.2 88

Wt % liquid phase 100%91.2 71.9

88 71.9Wt % beta phase 100%

91.2 71.9

16.6%

83.4%




(d) At 780 C T, o

Phases present: alpha beta Compositions of phases: 7.9% Ag in α phase 91.2% Ag in β phase Amounts of phases:

91.2 88

Wt % alpha phase 100%91.2 7.9

88 7.9Wt % beta phase 100%

91.2 7.9

3.84%

96.16%

PROBLEM #5: Problem 8.23 in the textbook. Given: A 500 g of a 40 wt % Ag–60 wt % Cu alloy that is slowly cooled from 1000°C to just below 780°C and the phase diagram in Figure 8.22.

Figure 8.22 in Textbook (note tie lines not shown).




Required: Determine the following: (a) How many grams of liquid and proeutectic alpha are present at 850°C? (b) How many grams of liquid and proeutectic alpha are present at 780°C + T ? (c) How many grams of alpha are present in the eutectic structure at 780°C - T ? (d) How many grams of beta are present in the eutectic structure at 780°C - T ? (a) At 850ºC,

40 7.9

Wt % liquid 100% 72.8%52 7.9

52 40Wt % proeutectic 100% 27.2%

52 7.9

Weight of liquid phase 500 g 0.728Weight of proeutectic 500 g 0.272

364 g136 g

(b) In the eutectic structure at 780 C + To ,

40 7.9Wt % liquid 100% 50.2%

71.9 7.9

71.9 40Wt % proeutectic 100% 49.8%

71.9 7.9

Weight of liquid phase 500 g 0.502Weight of proeutectic 500 g 0.498

251 g249 g

(c) In the eutectic structure at 780 C T, o the number of grams of α present is,

91.2 40Wt % total 100% 61.5%

91.2 7.9

Weight of total 500 g 0.615

307.5 g

(d) In the eutectic structure at 780 C T, o the number of grams of β present is,




40 7.9Wt % total 100% 38.5%

91.2 7.9

Weight of 500 g 0.385

192.5 g

PROBLEM #7: Given: A 14 inch square concrete column reinforced by four (4) #6 steel reinforcing bars. The reinforcing steel is placed at 2 inches from each side (see diagram provided in class). The column is 10 feet in total length. The concrete has a compressive strength of 4,000 pounds per square inch (psi). Required: Determine the composite elastic modulus for this column and determine the change of length of the column for loads ranging from 0 to 100 kips. Solution:

57 (ACI 318-14) where E is in ksi and concrete compressive strength is in

psi. Therefore, Econc = 3605 ksi Es = 29,000 ksi (see textbook and AISC Manual of Steel Design). Asteel = 4 x A#6 bar = 4 x (0.44in2) = 1.76 in2. Note A#6 bar is from ASTM A615/A615M-16. Table 1 from ASTM A615/A615M-16.




Acomp = 14in x 14in = 196 in2 Therefore: Ecomp = (29000ksi)(1.76in2/196in2)+3605ksi(194.24in2/196in2) Ecomp = 3833 ksi

Deflection =

Plotting from P = 0 kips to 100 kips and using E=3833 ksi:

0

10

20

30

40

50

60

70

80

90

100

0.000 0.005 0.010 0.015

Lo

ad (

kip

s)

Deflection (in)




PROBLEM #8: Given: You are conducting a wave equation analysis of piles (WEAP) for a single acting Delmag D30-32 diesel hammer. The contractor tells you that he would like to use three (3) layers of 1 inch thick Micarta (EM = 225 ksi) and two (2) layers of ½ inch thick aluminum (EAL = 10,000 ksi) as the hammer cushion. These layers have a diameter of 22.5 inches as they fit into the helmet. Required: Determine the elastic modulus of the composite hammer cushion. Does it affect the composite modulus if he changes the order in which he places the aluminum and Micarta layers together? Briefly explain why or why not. Solution:

Acushion = 398 in2 Vtotal = Acushion x Thickness = (398in2)(4inches) = 1592 in3 VAl = (Acushion x TAl)/Vtotal = (398in2)(1in)/1592in3 = 0.25 VM = Acushion x TM= (398in2)(3in)/1592in3 = 0.75 Therefore Ecomp = 298 ksi. No. It does not affect the composite modulus. Is the order of the composite materials a factor in the relevant equation? No, it is not.

cive.3100 e m assignment #7: interpret tensile test...

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