Materials

3/22/2017

FE REVIEW COURSE – SPRING 2017

Derek Gaw- TDOT M&T Concrete Section derek.gaw@tn.gov

Tyler Lacy- TDOT M&T Asphalt Section tyler.lacy@tn.gov

2

Materials Knowledge

• 4 – 6 problems • Mix Design

▫ Concrete & asphalt • Test methods & specifications

▫ Steel, concrete, aggregates, asphalt, wood • Physical & mechanical properties of concrete, ferrous & nonferrous

metals, masonry, wood, engineered materials, & asphalt • Sections in FE Reference Handbook

▫ Mechanics of Materials ▫ Electrical & Computer Engineering ▫ Materials Science/Structure of Matter ▫ Mechanical Engineering ▫ Thermodynamics

3

Material Properties & Testing

Materials Selection

Materials Science

Electrical Properties

Thermal Properties

Mechanical Properties

Classification of Materials

Engineering Stress & Strain

Stress-Strain Curves

Points Along the Stress-Strain

Curve

Allowable Stress Design

Ultimate Stress Design

Ductile & Brittle Behavior

Crack Propagation in

Brittle Materials

Fatigue

Toughness Charpy Test

Ductile-Brittle Transition

Creep Test

Hardness Testing

4

FE Reference Handbook 9.4 P. 84

5

FE Reference Handbook 9.4 P. 61

6

FE Reference Handbook 9.4 P. 84

Mechanical Properties

7

Classification of Materials Strong material • High ultimate strength

Weak material • Low ultimate strength

Tough material • Yield greatly before breaking

Brittle material • Will not yield greatly before breaking • Strain at fracture is less than ≈ 0.5%

Hard material • High modulus of elasticity

Soft material • Low modulus of elasticity

8

Engineering Stress & Strain

• Engineering stress, σ ▫ Load per unit original area ▫ MPa (typically)

• Engineering strain, ε ▫ Elongation of the test

specimen expressed as a percentage or decimal fraction of the original length

P. 80

P. 62

9

Stress-Strain Curves

• Hooke’s law

P. 62

10

Stress-Strain Curve Example

11

Stress-Strain Curve Example – Answer

12

Crack Propagation in Brittle Materials

• Modes of crack propagation 1. Opening or tensile

– Forces act perpendicular to the crack, which pulls the crack open

2. In-plane shear or sliding – Forces act parallel to the

crack, which causes the crack to slide along itself

3. Out-of-plane shear or pushing (pulling) – Forces act perpendicular to

the crack, tearing the crack apart

13

Crack Propagation in Brittle Materials

• Fracture toughness ▫ Stress intensity fracture

P. 62

P. 63

14

Crack Propagation in Brittle Materials Example

15

Crack Propagation in Brittle Materials Example – Answer

16

Fatigue

• Endurance limit modifying factors

P. 243

17

Fatigue Example

18

Fatigue Example – Answer

19

Concrete Moisture Correction Example

• The following preliminary concrete mix has been designed assuming that the aggregates are in oven-dry condition:

▫ Water= 36.6 gal

▫ Cement= 693 𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

▫ Coarse aggregate (SSD)= 1,674 𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

▫ Fine aggregate (SSD)= 1,100 𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

• Find: The amount of water (gal) that will actually be used in the final mix?

Property Coarse Aggregate

Fine Aggregate

Absorption 0.5% 0.7%

Moisture Content

2.0% 6.0%

20

Concrete Moisture Correction Example

• STEP 1: Determine the Aggregate Moisture Corrections

𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪 𝑪𝑪𝑪𝑪 =𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑪 − 𝑪𝑪𝑨𝑨𝑴𝑴𝑴𝑴𝑴𝑴𝑨𝑨𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑪𝑪𝑪𝑪𝑪

𝟏𝟏𝟏𝟏𝟏𝟏∗ 𝑾𝑾(𝑺𝑺𝑺𝑺𝑺𝑺)𝑪𝑪𝑪𝑪

=2𝑪 − 0.5𝑪

100∗ 1,674

𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

= 𝟐𝟐𝟐𝟐.𝟏𝟏𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪 𝑭𝑭𝑪𝑪 =𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑭𝑭𝑪𝑪 − 𝑪𝑪𝑨𝑨𝑴𝑴𝑴𝑴𝑴𝑴𝑨𝑨𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑪𝑭𝑭𝑪𝑪

𝟏𝟏𝟏𝟏𝟏𝟏∗𝑾𝑾(𝑺𝑺𝑺𝑺𝑺𝑺)𝑭𝑭𝑪𝑪

=6𝑪− 0.7𝑪

100∗ 1,100

𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

= 𝟐𝟐𝟓𝟓.𝟑𝟑𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

21

Concrete Moisture Correction Example

• STEP 2: Determine the Water Moisture Correction 𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴(𝒈𝒈𝑾𝑾𝒈𝒈) =

𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑪𝑪𝑪𝑪 + 𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑭𝑭𝑪𝑪

𝟓𝟓.𝟑𝟑𝟑𝟑 𝒈𝒈𝑨𝑨𝑴𝑴𝒈𝒈𝑾𝑾𝒈𝒈

𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 =25.1 𝑙𝑙𝑙𝑙

𝑦𝑦𝑦𝑦3 + 58.3 𝑙𝑙𝑙𝑙𝑦𝑦𝑦𝑦3

8.34 𝑙𝑙𝑙𝑙𝑙𝑙𝑔𝑔𝑔𝑔𝑙𝑙= 𝟏𝟏𝟏𝟏.𝟏𝟏 𝒈𝒈𝑾𝑾𝒈𝒈

• STEP 3: Determine the Actual Mix Water 𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑾𝑾𝒈𝒈 𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 𝑩𝑩𝑾𝑾𝑴𝑴𝑪𝑪𝑩𝑩𝑴𝑴𝑩𝑩 = 𝑺𝑺𝑴𝑴𝑴𝑴𝑴𝑴𝒈𝒈𝑪𝑪 𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 𝒈𝒈𝑾𝑾𝒈𝒈 −𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑪𝑪𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴(𝒈𝒈𝑾𝑾𝒈𝒈) 𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑾𝑾𝒈𝒈 𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 𝑩𝑩𝑾𝑾𝑴𝑴𝑪𝑪𝑩𝑩𝑴𝑴𝑩𝑩 = 36.6 𝑔𝑔𝑔𝑔𝑙𝑙 − 10.0 𝑔𝑔𝑔𝑔𝑙𝑙 = 𝟐𝟐𝟐𝟐.𝟐𝟐 𝐠𝐠𝐠𝐠𝐠𝐠

22

Concrete Mix Design Example

• Given: ▫ Type I Cement= 423 𝒈𝒈𝑨𝑨

𝒚𝒚𝑩𝑩𝟑𝟑

▫ Class F Fly Ash= 141 𝒈𝒈𝑨𝑨𝒚𝒚𝑩𝑩𝟑𝟑

▫ w/c ratio= 0.45 ▫ 6% Air Entrainment ▫ 42% Fine Aggregate of Total Aggregate Volume ▫ Specific Gravities:

GCement= 3.15 GFly Ash= 2.65 GCA= 2.68 GFA= 2.59

• Determine the Unit Weight of this Concrete Mix Design?

23

Concrete Mix Design Example

• STEP 1: Find the Volume of the Paste 𝑽𝑽𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴 =

𝑾𝑾𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴

𝑮𝑮𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴 ∗ 𝑼𝑼=

423 𝑙𝑙𝑙𝑙𝑙𝑙

3.15 ∗ 62.4 𝑙𝑙𝑙𝑙𝑓𝑓𝑓𝑓3

= 𝟐𝟐.𝟏𝟏𝟐𝟐 𝒇𝒇𝑴𝑴𝟑𝟑

𝑽𝑽𝑭𝑭𝒈𝒈𝒚𝒚 𝑪𝑪𝑴𝑴𝑩𝑩 =𝑾𝑾𝑭𝑭𝒈𝒈𝒚𝒚 𝑪𝑪𝑴𝑴𝑩𝑩

𝑮𝑮𝑭𝑭𝒈𝒈𝒚𝒚 𝑪𝑪𝑴𝑴𝑩𝑩 ∗ 𝑼𝑼=

141 𝑙𝑙𝑙𝑙𝑙𝑙

2.65 ∗ 62.4 𝑙𝑙𝑙𝑙𝑓𝑓𝑓𝑓3

= 𝟏𝟏.𝟓𝟓𝟐𝟐𝒇𝒇𝑴𝑴𝟑𝟑

𝑾𝑾𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 =𝒘𝒘𝑪𝑪∗𝑾𝑾𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑴𝑴𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑾𝑾𝒈𝒈𝑴𝑴 = 0.45 ∗ 423 𝑙𝑙𝑙𝑙𝑙𝑙 + 141 𝑙𝑙𝑙𝑙𝑙𝑙 = 𝟐𝟐𝟐𝟐𝟑𝟑 𝒈𝒈𝑨𝑨𝑴𝑴

𝑽𝑽𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 =𝑾𝑾𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴

𝑼𝑼=

254 𝑙𝑙𝑙𝑙𝑙𝑙

62.4 𝑙𝑙𝑙𝑙𝑓𝑓𝑓𝑓3

= 𝟑𝟑.𝟏𝟏𝟎𝟎𝒇𝒇𝑴𝑴𝟑𝟑

𝑽𝑽𝑪𝑪𝑴𝑴𝑴𝑴 =𝑪𝑪𝑴𝑴𝑴𝑴 𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑴𝑴𝑪𝑪𝑴𝑴𝑪

𝟏𝟏𝟏𝟏𝟏𝟏∗ 𝑽𝑽𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 =

6100

∗ 27.0 𝑓𝑓𝑓𝑓3 = 𝟏𝟏.𝟐𝟐𝟐𝟐 𝒇𝒇𝑴𝑴𝟑𝟑

𝑽𝑽𝑷𝑷𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 = 𝑽𝑽𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴+𝑽𝑽𝑭𝑭𝒈𝒈𝒚𝒚 𝑪𝑪𝑴𝑴𝑩𝑩+𝑽𝑽𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴+𝑽𝑽𝑪𝑪𝑴𝑴𝑴𝑴= 2.15 𝑓𝑓𝑓𝑓3+ 0.85𝑓𝑓𝑓𝑓3+ 4.07𝑓𝑓𝑓𝑓3+ 1.62𝑓𝑓𝑓𝑓3= 𝟓𝟓.𝟐𝟐𝟔𝟔𝒇𝒇𝑴𝑴𝟑𝟑

24

Concrete Mix Design Example

• STEP 2: Find Aggregate Volumes 𝑽𝑽𝑪𝑪𝒈𝒈𝒈𝒈 = 𝑽𝑽𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 − 𝑽𝑽𝑷𝑷𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 = 27.0 𝑓𝑓𝑓𝑓3 − 8.69 𝑓𝑓𝑓𝑓3 = 𝟏𝟏𝟓𝟓.𝟑𝟑𝟏𝟏 𝒇𝒇𝑴𝑴𝟑𝟑 𝑽𝑽𝑭𝑭𝑪𝑪 = 𝑪𝑭𝑭𝑪𝑪 ∗ 𝑽𝑽𝑪𝑪𝒈𝒈𝒈𝒈 = 0.42 ∗ 18.31 𝑓𝑓𝑓𝑓3 = 𝟎𝟎.𝟐𝟐𝟔𝟔 𝒇𝒇𝑴𝑴𝟑𝟑 𝑽𝑽𝑪𝑪𝑪𝑪 = 𝑽𝑽𝑪𝑪𝒈𝒈𝒈𝒈 − 𝑽𝑽𝑭𝑭𝑪𝑪 = 18.31 𝑓𝑓𝑓𝑓3 − 7.69 𝑓𝑓𝑓𝑓3 = 𝟏𝟏𝟏𝟏.𝟐𝟐𝟐𝟐 𝒇𝒇𝑴𝑴𝟑𝟑

25

Concrete Mix Design Example

• STEP 3: Find Aggregate Weights

𝑾𝑾𝑪𝑪𝑪𝑪 = 𝑽𝑽𝑪𝑪𝑪𝑪 ∗ 𝑮𝑮𝑪𝑪𝑪𝑪 ∗ 𝑼𝑼 = 10.62 𝑓𝑓𝑓𝑓3 ∗ 2.68 ∗ 62.4𝑙𝑙𝑙𝑙𝑙𝑙𝑓𝑓𝑓𝑓3

= 𝟏𝟏𝟎𝟎𝟎𝟎𝟐𝟐 𝒈𝒈𝑨𝑨𝑴𝑴

𝑾𝑾𝑭𝑭𝑪𝑪 = 𝑽𝑽𝑭𝑭𝑪𝑪 ∗ 𝑮𝑮𝑭𝑭𝑪𝑪 ∗ 𝑼𝑼 = 7.69 𝑓𝑓𝑓𝑓3 ∗ 2.59 ∗ 62.4𝑙𝑙𝑙𝑙𝑙𝑙𝑓𝑓𝑓𝑓3 = 𝟏𝟏𝟐𝟐𝟑𝟑𝟑𝟑 𝒈𝒈𝑨𝑨𝑴𝑴

26

Concrete Mix Design Example

• Step 4: Find the Unit Weight of the Concrete

𝑾𝑾𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 = 𝑾𝑾𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴𝑪𝑪𝑴𝑴 + 𝑾𝑾𝑭𝑭𝒈𝒈𝒚𝒚 𝑪𝑪𝑴𝑴𝑩𝑩 + 𝑾𝑾𝑾𝑾𝑾𝑾𝑴𝑴𝑴𝑴𝑴𝑴 + 𝑾𝑾𝑪𝑪𝑪𝑪 + 𝑾𝑾𝑭𝑭𝑪𝑪 𝑾𝑾𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 = 423 𝑙𝑙𝑙𝑙𝑙𝑙 + 141 𝑙𝑙𝑙𝑙𝑙𝑙 + 254 𝑙𝑙𝑙𝑙𝑙𝑙 + 1776 𝑙𝑙𝑙𝑙𝑙𝑙 + 1243 𝑙𝑙𝑙𝑙𝑙𝑙 = 𝟑𝟑𝟓𝟓𝟑𝟑𝟎𝟎 𝒈𝒈𝑨𝑨𝑴𝑴

𝑼𝑼𝑾𝑾 =𝑾𝑾𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑽𝑽𝑪𝑪𝑴𝑴𝑪𝑪𝑪𝑪𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴

=3837 𝑙𝑙𝑙𝑙𝑙𝑙27.0 𝑓𝑓𝑓𝑓3 = 𝟏𝟏𝟑𝟑𝟐𝟐.𝟏𝟏

𝒈𝒈𝑨𝑨𝑴𝑴𝒇𝒇𝑴𝑴𝟑𝟑

27

Advice: What You Need To Do To Pass

1. Identify the sections you are confident with at the moment.

2. Identify the sections you do not think you will ever be confident in.

3. Study the remaining sections that you can make yourself confident in.

These sections should still be studied, but can be done quickly as a refresher.

These sections are probably not worth your time. Study the absolute basics from these topics.

These sections are the difference between passing and failing. The bulk of your studying should be these topics.

You have approximately 2 minutes and 54 seconds per question.

28

Review

• Which process makes an entire steel part more brittle? ▫ A. Annealing ▫ B. Cold Working ▫ C. Hot Working ▫ D. Carburizing

• Answer: B

▫ If a steel piece is cold-worked the yield point changes and the area under the stress-strain diagram decreases.

▫ Annealing actually results in a more ductile piece of steel. ▫ Carburizing only effects the surface and leaves the core unaltered. ▫ Hot-worked steel would not see a change is properties when done

properly.

29

Review

• If a steel part was required to have a hard surface with a ductile core, which process could be used to achieve this requirement?

▫ A. Annealing ▫ B. Tempering ▫ C. Hot Working ▫ D. Gas Carburization

• Answer: D

▫ Gas Carburization increases the carbon content of the surface and leaves the core unaltered. This allows the core to remain ductile while hardening the surface of the steel.

30

Review

• Of the types of atomic bonding, which is the weakest? ▫ A. Metallic ▫ B. Ionic ▫ C. Covalent ▫ D. Secondary

• Answer: D

▫ Van der Waals. Attraction attributed to electric dipoles. In other words, the charges are local forces.

31

Review

• No permanent deformation will take place as long as the steel is kept below what?

▫ A. Facture Point ▫ B. Endurance Limit ▫ C. Elastic Limit ▫ D. Ultimate Strength

• Answer: C

▫ Stress levels in the elastic range will not result in plastic deformation of the steel; thus, allowing it to return to its original configuration.

32

Review

• Specimens A and B are tested in tension until failure. Specimen B has gone through a higher reduction in area than Specimen A. Which of the following is true?

▫ A. Specimen A is tougher. ▫ B. Specimen B is more ductile. ▫ C. Specimen B is more rigid. ▫ D. Specimen A is resistant to fatigue.

• Answer: B

▫ High ductility will allow for more necking to occur. This will allow Specimen B to decrease the area more than Specimen A will prior to fracture.

33

Review

• A liquid phase is cooled slowly and solidifies into two solid phases. Which reaction is taking place?

▫ A. Eutectic Reaction ▫ B. Hypoeutectic Reaction ▫ C. Peritectic Reaction ▫ D. Monotectic Reaction

• Answer: A

34

Review

• A material has a half-life of 12 days. If 80 milligrams of this material is measured after 24 days, how much of this material remains?

▫ A. 60 milligrams ▫ B. 40 milligrams ▫ C. 20 milligrams ▫ D. 10 milligrams

• Answer: C

▫ Two half-life cycles pass. 80/2 = 40 milligrams. 40/2 = 20 milligrams.

35

Review

• What is found by determining the difference in mass between an asphalt specimen in Saturated Surface Dry (SSD) condition and the specimen submerged in 25°C water?

▫ A. Mass of a Dry Specimen ▫ B. Volume ▫ C. Air Voids ▫ D. Maximum Specific Gravity

• Answer: B

▫ This difference accounts for the amount of water displaced by the asphalt specimen. 1 gram of water equals 1 cubic centimeter of water.

36

• Structural A36 steel alloy yields at 36 ksi. Given a factor of safety of 2.0, what is the allowable stress for this specimen?

▫ A. 18 ksi ▫ B. 54 ksi ▫ C. 72 ksi ▫ D. 90 ksi

• Answer: A

▫ 𝐴𝐴𝑙𝑙𝑙𝑙𝐴𝐴𝐴𝐴𝑔𝑔𝑙𝑙𝑙𝑙𝐴𝐴 𝑆𝑆𝑓𝑓𝑆𝑆𝐴𝐴𝑙𝑙𝑙𝑙 = 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑙𝑙 𝑆𝑆𝑀𝑀𝑀𝑀𝑀𝑀𝑆𝑆𝑆𝑆𝑀𝑀𝑆𝐹𝐹𝑀𝑀𝐹𝐹𝑀𝑀𝐹𝐹𝑀𝑀 𝐹𝐹𝑜𝑜 𝑆𝑆𝑀𝑀𝑜𝑜𝑀𝑀𝑀𝑀𝑦𝑦

= 36 𝑘𝑘𝑘𝑘𝑀𝑀2.0

= 18 ksi

Review

37

Review

• A 500 mm long metal rod with a circular cross section (1.963 x 10-3 m2) is tested in tension. What is most nearly the modulus of elasticity given that a tensile force of 40 kN is applied and 0.17 mm of elongation occurs?

▫ A. 40 x 109 Pa ▫ B. 60 x 109 Pa ▫ C. 80 x 109 Pa ▫ D. 95 x 109 Pa

• Answer: B

▫ 𝐸𝐸 = 𝐹𝐹𝐹𝐹0𝐴𝐴0Δ𝐹𝐹

= (40 𝑘𝑘𝑘𝑘 × 1000 𝑁𝑁𝑘𝑘𝑁𝑁 × 500 𝑚𝑚𝑚𝑚)

(0.001963 𝑚𝑚2 ×0.17 𝑚𝑚𝑚𝑚)

= 200000000.000334

𝑘𝑘𝑚𝑚2 = 60 x 109 Pa

38

• A copper wire has a diameter of 1.8 mm. What is most nearly the resistance of a 200 m length of wire at 0°C? (Given: Electrical resistivity (ρ) of 1.55 x 10-8 Ωm at 0°C)

▫ A. 0.9 Ω ▫ B. 1.0 Ω ▫ C. 1.1 Ω ▫ D. 1.2 Ω

• Answer: D

▫ 𝑅𝑅 = ρ𝐹𝐹𝐴𝐴

= 𝟏𝟏.𝟐𝟐𝟐𝟐 ×𝟏𝟏𝟏𝟏−𝟓𝟓 Ω𝑪𝑪 × 𝟐𝟐𝟏𝟏𝟏𝟏 𝑪𝑪 ×(𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑪𝑪𝑪𝑪𝑪𝑪

𝟐𝟐)

(π𝟑𝟑 × 𝟏𝟏.𝟓𝟓 𝑪𝑪𝑪𝑪 𝟐𝟐)=1.22 Ω

Review

39

• What is most nearly the equilibrium percentage of β in an alloy of 30% Ag, 70% Cu at 800°C?

▫ A. 33 % ▫ B. 59 % ▫ C. 67 % ▫ D. 75 %

• Answer: C

▫ β = 𝑥𝑥−𝑥𝑥𝐹𝐹𝑥𝑥β−𝑥𝑥𝐹𝐹

× 100𝑪

▫ β = 70−3488−34

× 100𝑪

▫ β = 66.7 %

Review

THANK YOU