# Dynamics

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16 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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1

Load and Stress
Analysis

Section III

2

Introduction about stresses

Shearing force and bending moment
diagrams

Bending, Transverse, & Torsional stresses

Compound stresses and Mohr

s circle

Stress concentration

Stresses in pressurized cylinders, rotating
rings, curved beams, & contact

Talking Points

3

Assume downward force as negative and upward
force as positive; and counterclockwise moment as
positive and clockwise as negative.

Loads may act on multiple planes.

Introduction about stresses

Body Diagram
-
i. Static Equilibrium and Free

0

F
0

M
0

T
4

The load is applied along
the axis of the bar
(
perpendicular to the
cross
-
sectional area
) and it
is uniformly distributed
across the cross
-
sectional
area of the bar.

The normal stress can be
tensile (+) or compressive
(
-
) depending on the
direction of the applied
load P.

The stress unit in N/m
2

or
Pa or multiple of this unit,
i.e. MPa, GPa.

Introduction about stresses

Cont.

ii. Direct Normal Stress & Strain

A
P

E

Assuming elasticity

o
L
L
A
P

E
E

A
L
P
L
o

o
L
L

Hooke

s Law

5

Sometimes, a body is subjected
to a number of forces acting on
its outer edges as well as at
some other sections, along the
length of the body. In such case,
the forces are split up, and their
effects are considered on
individual sections. The resulting
deformation of the body is equal
to the algebraic sum of the
deformation of the individual
sections. Such a principle of
finding out the resultant
deformation is called the
principle of superposition
.

Introduction about stresses

Cont.

n
E
1
o
A
L
P
L

Principle of Superposition:

L
1

L
2

L
3

d
1

d
2

d
3

L
3

L
2

L
1

d
3

d
2

d
1

P
1

P
2

P
3

P
4

6

Introduction about stresses

Cont.

Example on Principle of Superposition:

A brass bar, having cross sectional area of
10
cm
2

is subjected to axial forces as
shown in the figure. Find the total elongation of the bar (

L). Take
E

=
80
GPa.

L

=
-
150
m
m

7

For engineering materials,
n

=
0.25
to
0.33
.

For a rounded bar, the lateral strain is equal to the reduction
in the bar diameter divided by the original diameter.

Introduction about stresses

Cont.

s Ratio

iii. Poisson

Strain

Axial
Strain

Lateral

Ratio

s
Poisson'

n
x
z
x
y

n

or

s Law:

From Hooke

E
x
x

E
x
z
y
n

For
1
D stress system ( )

1
D stress
system

0

z
y

For
2
D stress system ( )

0
,
0

z
y

y
x
x
E
n

1

x
y
y
E
n

1
and

8

Introduction about stresses

Cont.

s Ratio:

Example on Poisson

A
500
mm long,
16
mm diameter rod made of a homogenous, isotropic material is
observed to increase in length by
300
m
m, and to decrease in diameter by
2.4
m
m
when subjected to an axial
12
kN load. Determine the modulus of elasticity and
Poisson

s ratio of the material.

E
=
99.5
GPa

n

〮㈵

9

Introduction about stresses

Cont.

iii. Direct Shear Stress & Strain

Assuming elasticity

The load, here,
is applied in a
direction
parallel to the
cross
-
sectional
area of the bar.

A
Q

G

Strain
Shear
G

is known as
modulus of rigidity

Single & Double Shear

The rivet is subjected
to single shear

The rivet is subjected
to double shear

A
Q
2

n

1
2
G
E
Relation between
n
Ⱐ慮搠

E

Q

Q

10

Shearing Force (S.F.) and Bending
Moment (B.M.) Diagrams

Simply supported
beam

Cantilever beam

Sign Convention

Relationship between shear force
and bending moment

dx
dM
Q

Qdx
M
Or

diagram

force
shear

under the

area

The

M
11

Shearing Force (S.F.) and Bending
Moment (B.M.) Diagrams
-

Examples

i. Concentrated Load Only:

12

Shearing Force (S.F.) and Bending
Moment (B.M.) Diagrams
-

Examples

ii. Distributed Load Only:

13

Shearing Force (S.F.) and Bending
Moment (B.M.) Diagrams
-

Examples

iii. Combination of Concentrated and Distributed Load:

14

Shearing Force (S.F.) and Bending
Moment (B.M.) Diagrams
-

Examples

iv. If Couple or Moment is Applied:

15

Bending, Transverse, &
Torsional stresses

I
y
M

i. Bending Stress

where,
M

is the applied bending moment (B.M.) at a transverse

section,
I

is the second moment of area of the beam cross
-
section

about the neutral axis (N.A.), i.e. ,

is the stress

at distance
y

from the N.A. of the beam cross
-
section.

dA
y
I

2
16

ii. Transverse Stress

Bending, Transverse, & Torsional
stresses

Cont.

Ib
y
A
Q

where
Q

is the applied vertical shear force at that section;
A

is the area of cross
-
section

above

y
, i.e. the area between
y

and the outside of the section, which may be above or below

the neutral axis (N.A.);
y
is the distance of the centroid of

area
A

from the N.A.;
I
is the second moment of area of the

complete cross
-
section; and
b
is the breadth of the section at

position y.

or

dA
y
Ib
Q

d

b

R

17

iii. Torsional Stress

Bending, Transverse, & Torsional
stresses

Cont.

J
T

where
T

is the applied external torque;

is the radial direction

from the shaft center
;
J

is the polar second moment of area of

shaft cross
-
section;
r

is the shaft radius; and

is the shear

stress at radius

.

J
r
T

max

4

2
1
r
J

4
4
2
1
i
o
r
r
J

Solid
section

Hollow shaft

when torsion is present

Note:

Ductile materials tends to break in a plane perpendicular
to its longitudinal axis; while brittle material breaks along
surfaces perpendicular to direction where tension is
maximum; i.e. along surfaces forming
45
o

angle with
longitudinal axis.

Ductile material

Brittle material

18

Compound stresses and Mohr

s
circle

Machine Design

involves among other
considerations, the proper sizing of a machine
member to safely withstand the maximum
stress which is induced within the member
when it is subjected separately or to any
combination of bending, torsion, axial, or
transverse load.

19

Compound stresses and
Mohr

s circle

Cont.

Maximum & Minimum Normal Stresses

2
2
(min)
2
2
(max)
2
2
2
2
xy
y
x
y
x
n
xy
y
x
y
x
n

Stress State

3
D General
Stress State

2
D Stress
State

D Case:
2
For

Where:

x

is a stress at a critical point in tension or compression normal to the
cross section under consideration, and
may be either bending or axial
load, or a combination of the two.

y

is a stress at the same critical point and in direction normal to the

x

stress.

xy
is the shear stress at the same critical point acting in the plane normal
to the Y axis (which is the XZ plane) and in a plane normal to the X axis
(which is the YZ plane). This shear stress
may be due to a torsional
moment, transverse load, or a combination of the two.

n
(max)

and

n
(min)

are called principal stresses and occurs on planes that
are at
90
°

to each other, called principle planes also planes of zero shear.

Note:

x
,

y
,

z

all

+ve,

xy
,

yx
,

zy
,

yz
,

xz
,

zx

all

+ve.
Due to static balance,

xy

=

yx
,

zy

=

yz
, and

xz

=

zx
.

Counterclockwise (CCW)

Clockwise (CW)

20

Compound stresses and
Mohr

s circle

Cont.

y
x
xy

2
2
tan

max

at the critical point being investigated is equal to half of the greatest difference of
any of the three principal stresses. For the case of two
-
dimensional loading on a particle
causing a two
-
dimensional stresses; The planes of maximum shear are inclined at
45
°

with the principal planes.

2
1
2
min
max
2
2
max
n
n
xy
y
x

)
max

Maximum Shear Stresses (

The planes of maximum shear are inclined at
45
°

with the principal planes.

The angle between the principal plane and the X
-
Y plane is defined by:

21

Compound stresses and Mohr

s
circle

Cont.

s Circle

Mohr

It is a graphical method to find the maximum and minimum normal stresses
and maximum shear stress of any member.

From the diagram:

x

= OA,

xy

= AB,

y

=OC, and

yx

= CD. The line BED

is the diameter of Mohr's circle with center at E on the

axis. Point B represents the stress coordinates

x
,

xy

on

the X faces and point D the stress coordinates

y
,

yx

on

the Y faces. Thus EB corresponds to the X
-
axis and ED to

the Y
-
axis. The maximum principal normal stress

max

occurs at F, and the minimum principal normal stress

min

at G. The two extreme
-
value shear stresses one

clockwise and one counterclockwise, occurs at H and I,

respectively. We can construct this diagram with

compass and scale and find the required information

with the aid of scales. A semi
-
graphical approach is

easier and quicker and offer fewer opportunities for

error.

2
-
D

22

Compound stresses and
Mohr

s circle

Cont.

Principal Element

True views on the various faces of the principal element

Max

Min

max

is equal to half of the greatest
difference of any of the three principal
stresses. In the case of the below figure:

3
-
D

2
1
3
1
13
max

where,

3
2
23
2
1
12
2
1

,
2
1

23

Example:

A machine member
50
mm diameter by
250
mm long is supported at one end as a
cantilever. In this example note that

y

=
0
at the critical point.

Compound stresses and Mohr

s
circle

Examples.

: Axial load only:
1
Case

: Bending only:
2
Case

In this case
all points in the member are subjected to

the same stress
.

MPa

83
.
3
2

MPa,

65
.
7
0
MPa

65
.
7
10
96
.
1
10
15
A
P
m
10
96
.
1
10
50
4
A
(max)
(max)
(max)
3
3
2
3
2
3

n
n
xy
x

(Shear)

MPa

60
.
30
2

on)
(Compressi

MPa

1
.
61

,
0
MPa

1
.
61
:
B
point
At
(Shear)

MPa

60
.
30
2

0

(Tension),

MPa

1
.
61
MPa

1
.
61
64
10
50
10
25
10
250
10
3
:
A
point
At
(max)
(max)
(min)
(max)
(max)
(max)
(min)
(max)
4
3
3
3
3

n
n
n
x
n
n
n
x
I
y
M
I
y
M

24

Compound stresses and Mohr

s
circle

Examples.

: Torsion only:
3
Case

:

: Bending & Axial Load
4
Case

In this case
the critical point occur along the outer

surface of the member
.

(Shear)

MPa

7
.
26
2
5
.
53

on)
(Compressi

MPa

5
.
53

,
0

on)
(Compressi

MPa

5
.
53
1
.
61
65
.
7
:
B
point
At
(Shear)

MPa

4
.
34
2
8
.
68

0

(Tension),

MPa

8
.
68

(Tension)

MPa

8
.
68
1
.
61
65
.
7
:
A
point
At
(max)
(min)
(max)
(max)
(min)
(max)

n
n
x
n
n
x
I
y
M
A
P
I
y
M
A
P

(Shear)

MPa

7
.
40

on)
(Compressi

MPa

7
.
40

(Tension)

MPa

7
.
40
MPa

7
.
40
32
10
50
10
25
10
1
0
(max)
(min)
(max)
4
3
3
3

n
n
xy
x
J
r
T
25

Compound stresses and Mohr

s
circle

Examples.

: Bending & Torsion:
5
Case

:

: Torsion & Axial Load
6
Case

(Shear)

MPa

9
.
50
2

on)
(Compressi

MPa

4
.
81

(Tension),

MPa

3
.
20

MPa

7
.
40

MPa

1
.
61
:
B
point
At
(Shear)

MPa

9
.
50
2

on)
(Compressi

MPa

3
.
20
7
.
40
2
1
.
61
2
1
.
61

(Tension),

MPa

4
.
81
7
.
40
2
1
.
61
2
1
.
61

MPa

7
.
40

MPa

1
.
61
:
A
point
At
(min)
(max)
(max)
(min)
(max)
(min)
(max)
(max)
2
2
(min)
2
2
(max)

n
n
n
n
xy
x
n
n
n
n
xy
x
J
r
T
I
y
M

(Shear)

MPa

9
.
40
2

on)
(Compressi

MPa

1
.
37
7
.
40
2
65
.
7
2
65
.
7

(Tension),

MPa

7
.
44
7
.
40
2
65
.
7
2
65
.
7
MPa

7
.
40
MPa

65
.
7
A
P
(min)
(max)
(max)
2
2
(min)
2
2
(max)

n
n
n
n
xy
x
J
r
T

26

Compound stresses and Mohr

s
circle

Examples.

: Bending, Axial Load, and Torsion:
7
Case

(Shear)

MPa

7
.
48
2

on)
(Compressi

MPa

5
.
75

(Tension),

MPa

9
.
21

MPa

7
.
40

MPa

3
.
53
1
.
61
65
.
7
:
B
point
At
(Shear)

MPa

3
.
53
2

on)
(Compressi

MPa

19
7
.
40
2
8
.
68
2
8
.
68

(Tension),

MPa

7
.
87
7
.
40
2
8
.
68
2
8
.
68

MPa

7
.
40

MPa

8
.
68
1
.
61
65
.
7
:
A
point
At
(min)
(max)
(max)
(min)
(max)
(min)
(max)
(max)
2
2
(min)
2
2
(max)

n
n
n
n
xy
x
n
n
n
n
xy
x
I
y
M
A
P
J
r
T
I
y
M
A
P

27

s circle:

Example on Mohr

The stress element shown in figure has

x

=
80
MPa
and

xy
, =
50
MPa (CW). Find the principal stresses
and directions.

Compound stresses and Mohr

s
circle

Examples.

Locate

x

=
80
MPa along the

axis. Then from

x
,

locate

xy

=
50
MPa in the (CW) direction of the

axis to

establish point A. Corresponding to

y
=
0
, locate

yx

=
50

MPa in the (CCW) direction along the

axis to obtain point

D. The line AD forms the diameter of the required circle

which can now be drawn. The intersection of the circle

with the

axis defines

max
and

min

as shown.

3
.
51
40
50
tan
2

:
is

CW to

axis
-
X

the
from

2

angle

The
MPa

24
64
40

MPa,

104
64
40
MPa

64
40
50
1
-
max
(min)
(max)
2
2
(max)

28

Stress Concentration

Occurs when there is sudden changes in cross
-
sections of members
under consideration. Such as holes, grooves, notches of various
kinds.

The regions of these sudden changes are called areas of
stress
concentration
.

Stress
-
concentration factor (
K
t
or
K
ts
)

The analysis of geometric shapes to determine stress
-
concentration
factors is a difficult problem, and not many solutions can be found.

o
ts
o
t
K
K

max
max

Theoretically

29

Stresses in pressurized
cylinders, rotating
rings, curved beams,
& contact