EMA 3702 Mechanics & Materials Science (Mechanics of ... · PDF fileEMA 3702 Mechanics &...

EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 1 Introduction

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 1 Introduction

Course Information

Instructor: Dr. Zhe Cheng

Phone: 305-348-1973

Email: [email protected]

Office: EC3441

Prerequisites:

EGN3311 Statics


More About Dr. Zhe Cheng

Education & Experiences:

PhD in Materials Sci. & Eng., Georgia Tech 2008

Research scientist at DuPont 2008-2013

Research group: https://ac.fiu.edu


Textbook and Other Materials

Mechanics of Materials, 7th ed, Beer, Johnston,

Dewolf, and Mazurek, McGraw Hill Education,

2015, ISBN 9780073398235

(Other earlier editions are OK!)

Via MHEducation.com:

http://www.mheducation.com/highered/product/

mechanics-materials-beer-johnston-jr/0073398233.html

Via Amazon:

https://www.amazon.com/Mechanics-Materials-7th-Ferdinand-eer/dp/0073398233

Course website:

https://ac.fiu.edu/teaching/ema3702/

WhatsApp chat group link (Supports Google Chrome, Firefox, Microsoft

Edge, Safari, etc.):

https://chat.whatsapp.com/7rUHtjl2bke9gZ5c2eufQc

http://www.mheducation.com/highered/product/mechanics-materials-beer-johnston-jr/0073398233.html






























Grading Policy

Homework (10 points)

Attendance (10 points, in the form of random class exercises)

Three exams (25 points each for exam 1 & 2, 30 points for final exam)

Overall grade: A:>=85; A-:81-84.9; B+:77-80.9; B:71-76.9; B-:67-70.9;

C+:63-66.9; C:55-62.9; D: 50-54.9; F:<50

NOTE: At the beginning of the semester, if you perceive your other

obligations (e.g., outside employment/job, family responsibilities) will

prevent you from attending a significant number of classes (e.g.,

missing 30% or more of the classes), you may request to have the

overall grade calculated based only on homework and exams.

Past grade statistics

2016 sum: Average 60.4/C; Median 64.7/C+; Highest 94.1/A; <C including drop: 7 out of 34

2017 sum: Average 67.6/C+; Median 70.5/B-; Highest 92.6/A;<C including drop: 9 out of 30

Example:

mid-term ~25/100; final ~45/100; almost perfect (44/45) in hw/participation;

overall – almost pass, but not quite: should he/she get C or D? Dr. Cheng choice: D


Grade vs. Work Hours

Grade = -0.249t + 76.444

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Time on Job-Employment (hour/week)

2017 summer data

Median # courses taken: 3


Grade vs. Number of Courses Taken

y = -4.8202x + 85.608

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2017 summer data



Grade vs. Class Attendance

y = 2.2913x + 28.314

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Class Participation/Lectures Attended

2017 summer data



Grade vs. Homework Submission

y = 2.8983x + 20.1370

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Homework Submission

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Introduction

Statics Mechanics of Materials

Rigid body Practical engineering materials

w/ Elasticity & Plasticity

No failure Failure (Fracture, Fatigue, Creep)

Important foundation for advanced courses such as

• Mechanical Design I & II (EML 3350 & 4501)

• Finite Element Analysis (EGM4350)

• Synthesis of Engineering Mechanics (EGM5615)

“Mechanics of Materials” is a branch of

Mechanics that develops relationships between :

The external

loads

Internal forces intensity (stress) &

resulting deformation/shape

change (strain) and even failure

https://en.wikipedia.org/wiki/Strength_of_materials

“Strength of materials, also called mechanics of

materials, is a subject which deals with the behavior of

solid objects subject to stresses and strains”

Statics: Equilibrium

Mechanics of Materials:

1. Stress & strain in practical engineering component

under load

2. Statically indeterminate situations

3. Deformation (deflection) of engineering components

under load

4. Yielding or failure of engineering components

Balance of

• Forces

• Moments of force or torque

0 xF 0 yF

0 oM

Stress & Normal Stress

Convention:

• Positive for tension

• Negative for compression

To describe intensity of load,

introduce the concept of Stress

For axial loading, define

Normal Stress

In statics: 10 N vs. 106 N

In reality:

• Can a design (materials and/or

geometry) sustain the load?

A

P

Beer et al. (2015)

Force and corresponding

stress perpendicular to

the area of interest

Units of (normal) stress

SI Units for

stress

English Units

for stress:

psi (pounds per square inch) = lb/in2 = 6895 Pa

ksi = 103 psi

Pa = l N/m2

kPa = 103 N/m2

MPa = 106 N/m2

GPa = 109 N/m2

Load P in unit of N (SI) or lb of force (English)

Cross-section area in unit of m2 (SI) or in2 (English) A

P

Average Stress & Local Stress

(Normal) stress may vary within

the same cross-section due to

variations in local conditions (e.g.

material properties, clamping)

Local stress at Q:

Average normal stress A

P

A

F

A

lim

0

Stress distribution generally

NOT uniform

For a small area:

A

dAdFP

External load P is related to

local stress σ by:

For simple axial loading, if loads

pass through centroid or centric

loading, normal stress can be

assumed to be uniform if not too

close to the ends

dAdF

For eccentric (i.e., off-center) loading, the distribution of

the internal stress will not be uniform (see Chapter 4)


Class Exercise on Stress (1)

For a cylindrical sample

with tensile force of 10000 N

loaded along its long axis, if

the initial cross-section area is

10 cm2, please calculate the

average normal stress σ.

18

A

F

MPaPam

N

m

N

cm

N

A

F101010

1010

10

10

10000 7

2

7

24

4

2


Class Exercise on Stress (2)

For a straight metal bar with

a square cross-section, if

knowing the tensile stress is

along its long axis and it is

10 MPa and the square cross-

section has initial edge length

of 1 cm, please calculate the

tensile force applied.

19

1 cm 1 cm

AF

NmPacmcmPabAF 10001010)11(1010 24762

Shearing Stress

A

F

Strictly, as defined here is also an average stress

Force and corresponding

stress parallel to the

area of interest

Situation of Double Shear

If cross-section

area of bolt is A

A

F

A

F

A

Pave

2

2/


Shearing Stress Class Exercise (1)

For the pin as illustrated, if

cross-section area is 50 cm2,

and load of 50 kN is applied,

please calculate the (average)

shearing stress

MPaPam

N

m

N

cm

N

A

F101010

10

10

50

1050 7

2

7

24

3

2

3

A

F


Shearing Stress Class Exercise (2)

For the pin as illustrated, if

cross-section area is 20 cm2,

and load of 40 kN is applied,

please calculate the (average)

shearing stress

MPaPam

N

m

N

cm

N

A

F101010

10

10

20

2/)1040( 7

2

7

24

3

2

3

Double shear situation


Bearing Stress in Connections

Complex condition for

bearing surfaces

An approximation:

Bearing Stress

Bearing surface

td

P

A

Pb t Thickness

d Diameter

General Method of Problem Solution 1. Draw free body diagrams (FBD)

2. Apply equations based on equilibrium of force &

moment and additional geometry considerations (if

applicable)

3. Solve for reactions (force/load), and resulting stress,

strain, and deformation (e.g., deflections)

4. Check answers including units

By convention,

• Use 4 figures to record numbers beginnings with “1”

• Use 3 figures in all other cases.

Examples:

A force of 40 lb. should be read as 40.0 lb

A force of 15 lb. should be read as 15.00 lb

Accuracy criteria

• Accuracy of the given data

• Accuracy of the computations performed

For engineering, accuracy better than 0.2% is rare

Numerical Accuracy

(1.12)

Stress on an Oblique Plane under Axial

Loading P

cosPF

sinPV

For a plane at angle

from the normal plane

Normal force for that plane

Shearing force for that plane

(Average) normal stress &

shearing stress for that plane

A

F

A

V

Aθ Area of that plane

θ

θ P

F

V

Relationship between the area of the two planes:

Normal cross-section A0 and Oblique plane Aθ

cos0 AA

2

00

coscos/

cos

A

P

A

P

A

F

cossincos

sin

00 A

P

A

P

A

V

Average normal stress & shearing stress for that plane

(at angle from the normal cross-section) will be

θ

θ P

F

V


Maximum Normal & Shearing Stress

Max shearing

stress at = 45o

Max normal

stress at = 0o 0

maxA

P

0

max2A

P

2

0

cosA

P

cossin0A

P


Class Example

Two wooden pieces of uniform square cross-section with

edge length of 1 cm are joined by gluing along oblique

interface, as shown. Knowing tensile load of 2 kN. Please

calculate the normal and shearing stress along the glued

interface as illustrated. Knowing Sin30o=Cos60o=0.5,

Cos30o=Sin60o=0.866.

P P

NNPCosFN 10002

1200060

NNPSinV 17322

3200060

30o

In this case, θ = 90o - 30o = 60o !!!!

2

00 2260

cmACos

AA


Class Example (continued)

Two wooden pieces of uniform square cross-section with

edge length of 1 cm are joined by gluing along oblique

interface, as shown. Knowing tensile load of 2 kN. Please

calculate the normal and shearing stress along the glued

interface as illustrated.

Pam

N

cm

N

A

FN 6

242105

10500

2

1000

Pam

N

cm

N

A

V 6

2421066.8

10866

2

1732

P P

30o

Stress under General Loading Conditions:

Components of Stress

The plane is ┴ to x-axis

The vector is // to y - direction

Notation:

A

F x

Ax

lim

0

A

V x

y

Axy

lim

0

A

V x

z

Axz

lim

0

x

yV

The surface is ┹ to

x - direction or axis

The direction of the

component, i.e., shear

stress is // to y - direction

Notation for

Shearing Stress

Three normal stress

components

Six shearing stress

components

xy

For a FBD diagram, all the forces in a system must

fulfill the equations of equilibrium:

0 xF

0 xM

0 yF 0 zF

0 yM 0 zM

Equation of equilibrium

around z axis

0)()( aAaA yxxy

Therefore,

Similarly,

0 zM

yxxy

zxxz zyyz

Implication #2: only six independent components are

required to uniquely define the stress state at a given

point:

Implication #1: if there is

shearing in one plane, there

must be shear on another

plane perpendicular to the

first one

yxxy

σx, σy, σz, τxy, τyz, and τxz

For small cube with = 0o

max normal stress and

zero shearing stress

For small cube w/ = 45o

Max shearing stress and

same magnitude of

normal stress

Same Loading – Different Stress Interpretations?

Addressed in Chapter 7

Concept of Factor of Safety:

Design Considerations

ultimate load

allowable load

Factor of Safety = F.S.

stress Allowable

stress Ultimate

load Allowable

load Ultimate


Class Exercise

For a long cylinder component with

designed cross-section area of 1 cm2 to

bear maximum axial load of

2104 N, if factor of safety (FS) is at

least 3, which material listed in the

table could be used?

Material Tensile strength

(MPa)

Steel 1090 mild 841

Aluminum 2014-T6 483

Cu, 99.9% 220

Cast iron, 4.5% C, ASTM-A-48

200

MPaPam

N

cm

Nall 200102

10

102

1

102 8

24

4

2

4

3all

TSFS

MPaMPaallTS 60020033

Only 1090 mild steel could be used

Class Exercise For the loading condition on the right, if AB and BC cross-section area is both 4cm2, determine the normal stress in beam AB and BC Since AB is a two-force beam, reaction at A must be horizontal, i.e., FAy = 0 Consider moment around C

mmkNmmFAx 80030600 kNFAx 40 kNFF AxAB 40

MPaPam

N

cm

NAB 100101

10

1010

4

1040 8

24

3

2

3

Class Exercise Similarly, BC is also a two-force beam. For point B, the three force balance is below

kNkNkNFBC 50)40()30( 22

MPaPam

N

cm

NBC 1251025.1

10

105.12

4

1050 8

24

3

2

3

30kN

40kN

FBC


Homework 1.0

Read textbook section 1.1 to 1.5 (you may skip the sections named

“sample problem”) and give an honor statement confirm reading.


Homework 1.1

Two solid cylinder rods DE and EF are bounded together at E and bonded to base at F. Knowing diameter d1 = 30 mm and d2=50 mm. Given the load condition as illustrated, calculate the average normal stress at the midsection of

(a) Rod DE

(b) Rod EF

F1=30kN

F2=62.5kN F3=62.5kN

d1

D

E

F

d2


Homework 1.2

Link FH consists of a single bar 1 inch wide and 1 inch thick. Each pin has a 0.375 inch diameter. Determine the value of the maximum average normal stress in bar FH if

(a) θ = 0 (Pushing to left)

(b) θ = 90o (Pushing down) Hint: Have to consider

that stress at the cross-section where the connection pin is located will be higher than that at center of the bar (see textbook examples)

12in

6in

F1= 4 kips

θ

30o E

F

G

H

Pins with

diameter of

0.375 inch


Homework 1.3

Determine the diameter of the largest circular hole that can be

punched to a sheet of polystyrene 4 mm thick. Punch force applied

will be 22.5 kN and the average shearing stress of 55 MPa is needed

to cause the polystyrene material used to fail (i.e., for the punch to

cut through and make the hole)


Homework 1.4

Two rectangular wood parts are bonded by glue as illustrated. The

wood parts are 2 inch long and 1 inch wide. Shearing force are applied

to the two bonded pieces. When shearing force P reaches 1600 lb, the

two pieces started to separate or fail along the bonding interface,

please calculate the average shearing stress at the time of failure

P

P

2 in 1 in


Homework 1.5

Two wooden pieces of uniform square cross-section with edge length of

1 cm are joined by gluing along oblique interface, as shown. Knowing

tensile load of 10 KN. Please calculate the normal and shearing stress

along the glued interface as illustrated.

60o P P


Homework 1.6

Members of DE, EF, and DF of the truss shown are made of the same

material. It is known that the material has ultimate tensile strength

U = 200 MPa. If safety factor is 3.0 is to be achieved for both DE and

DF bars, please determine the required minimum cross-section area

of bar DE and bar DF

0.75 m

0.4 m

1.4 m

10 kN

D

E

F

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