# EIT Review Weiss, Mechanics of Materials Page 1

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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)
• 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
∑ ∑
∆•==
N
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
δ
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
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

Shear Stress and Strain
Resultant
Force (R)
Normal
Component (F)
R
F
V
Shear
Component (V)
90°
A
V
AVE

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

x

y

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
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

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

2

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

2

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