EIT Review Weiss, Mechanics of Materials Page 1

EIT Review Session

Mechanics of Materials Review Session

For the Fundamentals of Engineering Exam

Using examples from Fundamentals of Engineering by Potter

and Beer and Johnson Mechanics of Materials

or Hibbler Mechanics of Materials

Conducted By:

W. Jason Weiss

Room G 215A

wjweiss@ecn.purdue.edu

Topics Covered

• Basic Definitions

• Stress and Strain (Mohrs Circle, Hooke’s Law)

• Uni-Axial Deformations (Load, Thermal)

• Torsion

• Beam Behavior (Shear, Bending)

• Combined Stresses

• Composites

• Columns

• Material Behavior

• Cylindrical Pressure Vessels

Review of Basics

Types of Problems

• Internal Equilibrium

(stresses)

• Geometry of

Deformation (strains)

• Mechanical and

Thermal

Assumptions

• Homogeniety

• Isotropy

• Elastic Behavior

• Isotropic - Elastic Properties Constant in All Directions

• Homogenous - Same physical properties

• Elastic - resumes initial shape after load is removed

EIT Review Weiss, Mechanics of Materials Page 2

Stress and Strain

L

L

or

L

commonly

x

u

X

∆

=

∆

∆

=

δ

ε

P

P

∆F

∆A

P

A

B

u

u+∆u

∆x

u - Rigid Body Movement

∆u - Change in Length of Element ∆x

δ or ∆L

L

A

F

Lim

A

∆

∆

=

→∆ 0

σ

∫ ∫

==

A

dAdFP σ

∑ ∑

∆•==

N

AdFP σ

Force Deformation Relationship

and Hookes Law

F

δ

F

δ

k

σ

ε

E

( )

εσ

εσ

εσ

δ

⋅=

⋅

=

⋅=

=

E

A

Lk

LkA

kF

Uniaxial Deformation

P

L

O

L

A

A

E

PL

L

E

A

P

E

O

O

=

⋅=

⋅=

δ

δ

ε

σ

Change eTemperatur

Length

/105.6 Steel

)/( COTE

6

−∆

−

−

∆⋅⋅=

−

T

L

Fininx

Cll

TL

o

o

TEMP

α

α

δ

δ

TEMP

Temperature Deformations

Loading Deformations

δ

Load

EIT Review Weiss, Mechanics of Materials Page 3

Example Problem #1

A composite bar is made of aluminum and steel. An axial

load is applied at the positions shown. Determine the

average stress in each section.

Steel A = 1.2 in

2

8000 lb

Aluminum, A = 1.6 in

2

Steel

A = 1.2 in

2

3000 lb

8000 lb

Aluminum, A = 1.6 in

2

11,000 lb

Steel

A = 1.2 in

2

3000 lb

8000 lb

Example Problem #2

Find the Allowable Load in kN of a 1 meter long rod that has

a 10 cm diameter if its elongation can not exceed 1 mm. E =

207 GPa

P

L

O

L

A

( )

( )

( )

MNP

m

mxm

P

L

AE

P

AE

PL

m

N

O

O

62.1

1

001.01020705.0*

2

922

=

=

=

=

π

δ

δ

δ

Load

Example Problem #3

A Steel Bridge Spans 500 m. What is the Change of Length

When the Temperature is changed from 30°C to -20 °C.

(Note the COTE of steel is 12E-6 /°C)

( )

( )

( )

( )

m

CCmx

TL

mC

C

3.0

30205001012

006

0

0

−=

−−=

∆=

⋅

−

δ

δ

α

δ

δ

TEMP

Original Length

Apply Temperature Change

EIT Review Weiss, Mechanics of Materials Page 4

Example Problem #4

An aluminum bar is placed between two rigid walls and

heated by 50°. What is the stress in the aluminum bar at the

end of the test if the COTE of the aluminum is 23E-6 /°C and

E is 69 GPa)

TL∆=

α

δ

A

E

PL

=δ

Shear Stress and Strain

Resultant

Force (R)

Normal

Component (F)

R

F

V

Shear

Component (V)

90°

A

V

AVE

=τ

xy

γ

−90

( )

ν

γτ

+

==

12

,

E

GG

Example Problem #5

A wood block is rigidly attached to the horizontal surface as

shown below. The block is subjected to a 1000 kN horizontal

force as shown. Determine the shear stress in a typical

horizontal plane of the block and the horizontal displacement

of the top edge of the block AB (a=0.2m, b=0.6m, c=0.8m). G

= 4.1 GPa

mm

GPa

kPa

m

G

bb

kPa

mm

kN

A

V

xy

xy

AVE

91.0

1.4

6250

6.0

250,6

2.08.0

1000

====

=

•

==

τ

γδ

τ

a

c

b

1000kN

EIT Review Weiss, Mechanics of Materials Page 5

Poisson’s Ratio

x

y

ε

ε

ν −==

strain axial

strain lateral

ε

x

ε

y

Example Problem #6

A 500 mm long, 16 mm diameter rod made of a homogenous

isotropic materials is observed to increase in length by 300

µm, and to decrease in diameter by 2.4 µm when subjected

to an axial 12 kN load. Determine the modulus of elasticity

and Poisson’s ratio of the material

From Beer and Johnston

25.0

5.99

10300201

5001012

500

300.0

16

0024.0

6

3

=

−

−=−=−=

=

⋅

⋅

==

−

L

D

x

y

x

y

GPa

x

x

A

PL

E

δ

δ

ε

ε

ν

δ

Hooke’s Law In Three Dimensions

[ ]

[ ]

[ ]

G

G

G

EE

EE

EE

XZ

XZ

YZ

YZ

XY

XY

XY

Z

Z

ZX

Y

Y

ZY

X

X

τ

γ

τ

γ

τ

γ

σσ

νσ

ε

σσ

νσ

ε

σσ

ν

σ

ε

=

=

=

+−=

+−=

+−=

EIT Review Weiss, Mechanics of Materials Page 6

Mohr’s Circle

σ

1

τ

σ

3

2α

(σ

x

,σ

y

,τ

xy

)

( )

2

22

31

2

3,1

σσ

τ

τ

σσσσ

σ

−

=

+

−

±

+

=

MAX

xy

yxyx

σ

α

Cylindrical Pressure Vessels

Internal Pressure

Ir

IO

IO

Il

P

RR

RR

P −=

−

+

= σσ ,

22

22

L

P

D

Cylindrical Pressure Vessels

External Pressure

Or

IO

IO

Ol

P

RR

RR

P −=

−

+

−= σσ ,

22

22

L

P

EIT Review Weiss, Mechanics of Materials Page 7

Example Problem #7

A 1.5 m diameter pipe is made of wooden planks that are

held together with steel hoops. If the cross section of the

hoops is 300 mm2 each and the pressure is 350 kPa how far

apart can the hoops spaced to prevent failure?

mL

xxLE

P

pDL

149.0

101301030025.13350

2

66

=

⋅⋅=⋅⋅

=

−

P

P

pDL

Torsion

T

J

TR

J

Tr

O

MAX

=

=

τ

τ

G

J

TL

=φ

φ

L

Section Dynamics

in theGiven =J

Example Problem #8

What is the maximum shearing stress that exists in a 6cm

diameter shaft that is subjected to a 200 N-m torque

Pax

J

Tr

6

4

1072.4

32

06.0

03.0200

=

⋅

⋅

==

π

τ

EIT Review Weiss, Mechanics of Materials Page 8

Internal Shear Forces, Axial Forces,

and Bending Moments

0.5L

4F

3F

F

0.25L

L

( ) ( )

FLM

MLFLFM

FV

VFFF

PF

Cut

Y

AXX

5.0

25.045.030

430

0

=

++−==

−=

−−==

==

∑

∑

∑

3F

0.5L

4F

0.25L

M

P

AX

V

Beam Theory - Sign Convention

M

V

M

V

Negative Bending

Positive Bending

M

V

Positive

Shear

Negative

Shear

Beam Theory - Example #9

10 ft

6 ft

2 ft

2 ft

200 lb

60 lb/ft

EIT Review Weiss, Mechanics of Materials Page 9

Basic Shear and Moment Diagrams

P

0

P/2

-P/2

Shear

Moment

-PL/4

w

0

wL/2

-wL/2

Shear

Moment

-wL

2

/8

L

L

Beam Bending Stress Distribution

( )

I

My

x =σ

b

I

Mc

MAX

=σ

h

12

3

bh

I =

Page 27

( )

ρ

σ

y

x =

y

Beam Shear Stress Distribution

area theofportion theof Area-A

area theofportion theof centroid the todistance−

=

=

y

AyQ

Ib

VQ

t

y

y

A

EIT Review Weiss, Mechanics of Materials Page 10

Example Problem #10

2000 lb

8 in

1 in

NA

1 in

5 in

I=60.67in

4

Compute the Maximum Tensile Stress

the Maximum Compressive Stress

the Maximum Shearing Stress

Example Continued

Beam Deflections

( ) ( )

dxxVxM

∫

=

( ) ( )

∫

= dxxwxV

( )

( )

( )

( )

dxxM

EI

xy

dxxM

EI

x

∫∫

∫

=

=

1

1

θ

Use boundary conditions to obtain constants

EIT Review Weiss, Mechanics of Materials Page 11

Beam Tables

Guidebook

tables of

formulas

Superposition

P

w

L

L

P

w

L

=

Superposition

P

w

L

δ

Max

Find the Maximum Deflection

EIT Review Weiss, Mechanics of Materials Page 12

Maximum Principle (Normal) Stress

• Whenever the maximum or minimum

principle stress exceeds the uni-axial

strength of the material

- assume the material yields σ

1

>σ

2

>σ

3

- compression positive

S

T

is the tensile strength and

S

C

is the compressive strength

Failure Limits

C

T

S

S

−≤

≥

3

1

σ

σ

σ

1

τ

σ

3

τ

Max

Safe

Region

Maximum Shear

• Whenever the maximum or minimum

shear stress exceeds the maximum shear

strength of a material

- assume the material yields σ

1

>σ

2

>σ

3

- compression positive

S

Y

is the

yield strength

Failure Limits

2

Y

MAX

S

≥τ

S

y

τ

τ

Max

=S

y

/2

S

y

Envelope

Envelope

Safe Design

Distortion Energy Theory

• Energy Based Failure Criteria - when the

energy of a volume equals the energy of a

a uniaxial specimen at yield

( ) ( ) ( )

[ ]

Y

S=

−+−+−

2

2

31

2

32

2

21

σσσσσσ

EIT Review Weiss, Mechanics of Materials Page 13

Elastic Strain Energy

P

P

E

AL

U

u

PWU

W

U

2

VolumePer Unit Energy Strain

DoneWork

StoredEnergy

Energ

y

Potential

2

2

1

σ

δ

==

==

=

=

δ

Composite Sections

V

A

V

P

P

P

Columns

( )

2

2

2

2

k

l

CR

e

CR

e

E

A

P

l

EI

P

π

π

=

=

Pinned-pinned l

e

= l

pinned-fixed l

e

=0.7 l

fixed-fixed l

e

=0.5 l

Radius of gyration

EIT Review Weiss, Mechanics of Materials Page 14

Column Example

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