Statics - Internal Forces

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2/1/2012
1
STATICS: CE201
Chapter 7
Internal Forces
Notes are prepared based on: Engineering Mechanics, Statics by R. C.
Hibbeler
, 12E Pearson
Dr M.
Touahmia
& Dr M.
Boukendakdji
Civil Engineering Department, University of Hail
(Fall 2010)
7
.
Internal Forces
_____________________________________________________________________________________________________________________________
___
________________
Chapter
Objective
:
Use
the
method
of
sections
to
determine
internal
forces
in
2
-
D
load
cases
.
Contents:
7.1
Internal Forces Development in Structural Members
7.2
Shear and Moment Equations and Diagram
7.3
Cables
Chapter 7: Internal Forces
2/1/2012
2
7.1
Internal Forces Development in Structural
Members

The
design
of
any
structural
member
requires
to
know
both
the
external
loads
acting
on
the
member
and
the
internal
forces
acting
within
the
member
in
order
to
be
sure
the
material
can
resist
these
loading
.
Chapter 7: Internal Forces

The
concrete
supporting
a
bridge
has
fractured
:
What
might
have
caused
it
to
do
this?

Is
it
because
of
the
internal
forces
?

If
so,
what
are
they
and
how
can
we
design
these
structures
to
make
them
safer?
7.1
Internal Forces Development in Structural
Members

If
a
coplanar
force
system
acts
on
a
member,
then
in
general
a
resultant
internal
normal
force
N
(acting
perpendicular
to
the
section),
shear
force
V
(acting
along
the
surface),
and
bending
moment
M
will
act
at
any
cross
section
along
the
member
.
Chapter 7: Internal Forces
2/1/2012
3
Steps for Determining Internal Loadings:

Internal
Loadings
can
be
determined
by
using
the
method
of
section
.
The
following
example
explains
the
steps
that
we
should
follow
to
determine
the
internal
forces
acting
on
the
cross
section
at
point
C
.
1.
Before
the
member
is
sectioned,
it
is
first
necessary
to
determine
its
support
reactions
by
drawing
a
FBD
of
the
entire
structure
and
solving
for
the
unknown
reactions
.
Chapter 7: Internal Forces
Steps for Determining Internal Loadings:
3.
Pass
an
imaginary
section
a
-
a
perpendicular
to
the
axis
of
the
beam
through
point
C
and
separate
the
beam
into
2
segments
.
The
internal
loadings
acting
at
C
will
be
exposed
and
become
external
on
the
FBD
of
each
segment
:
Then,
decide
which
resulting
section
or
piece
will
be
easier
to
analyze
.
Chapter 7: Internal Forces
2/1/2012
4
Steps for Determining Internal Loadings:
3.
Draw
a
FBD
of
the
piece
of
the
structure
you’ve
decided
to
analyze
.
Remember
to
show
the
N,
V,
and
M
loads
at
the
“cut”
surface
.
4.
Apply
the
Equations
of
Equilibrium
to
the
FBD
(drawn
in
step
3
)
and
solve
for
the
unknown
internal
loads
.
Chapter 7: Internal Forces
0


x
F
0


y
F
0


C
M
Sign Convention:
Chapter 7: Internal Forces
2/1/2012
5
Example 1

Determine
the
normal
force
,
shear
force
and
bending
moment
acting
just
to
the
left,
point
B
,
and
just
to
the
right,
point
C
,
of
the
6
kN
force
on
the
beam
.
Chapter 7: Internal Forces
Solution 1
1.
Support
Reactions
:
Can
be
determined
from
the
free
-
body
diagram
of
the
beam
.
Chapter 7: Internal Forces
0


D
M






0
m

9
m

6
kN

6

kN.m

9



y
A
kN

5

y
A
2/1/2012
6
Solution 1
2.
Free
-
Body
Diagrams
:
The
free
-
body
diagrams
of
the
left
segments
AB
and
AC
of
the
beam
are
:
3.
Equations
of
Equilibrium
:
Segment
AB
:
Chapter 7: Internal Forces
0


x
F
0

B
N
0


y
F
0
kN

5


B
V
kN

5

B
V
0


B
M




0
m

3
kN

5



B
M
kN.m

15

B
M
Solution 1
Segment
AC
:
Chapter 7: Internal Forces
0


x
F
0


y
F
0


B
M
0

C
N
0
kN

6
kN

5



C
V
kN

1


C
V




0
m

3
kN

5



C
M
kN.m

15

C
M
2/1/2012
7
Example 2

Determine
the
normal
force
,
shear
force
and
bending
moment
at
point
C
of
the
beam
.
Chapter 7: Internal Forces
Solution 2

Note
:
It
is
not
necessary
to
find
the
support
reactions
at
A
since
segment
BC
of
the
beam
can
be
used
to
determine
the
internal
loadings
at
C
.

Free
-
Body
Diagram
:
The
distributed
load
acting
on
segment
BC
can
be
replaced
by
its
resultant
force
:
Chapter 7: Internal Forces


N/m

600
m

3
m

5
.
1
N/m

1200








C
w
2/1/2012
8
Solution 2

Equations
of
Equilibrium
:

Note
:
The
negative
sign
indicates
that
M
C
acts
in
the
opposite
sense
to
that
shown
on
the
free
body
diagram
.
Chapter 7: Internal Forces
0


x
F
0

C
N
0


y
F
0
N

450


C
V
N

450

C
V
0


C
M




0
m

5
.
0
N

450



C
M
N.m

225


C
M
7.2 Shear and Moment Diagrams

Beams
are
designed
to
support
loads
perpendicular
to
their
axes
.

The
design
of
a
beam
requires
a
detailed
knowledge
of
the
variation
of
the
internal
shear
force
V
and
bending
moment
M
acting
at
each
point
along
the
axis
of
the
beam
.
Chapter 7: Internal Forces
2/1/2012
9
7.2 Shear and Moment Diagrams

To
construct
the
shear
and
moment
diagrams
,
it
is
necessary
to
section
the
member
at
an
arbitrary
point
,
located
at
distance
x
from
the
left
end
.

The
variations
of
V
and
M
as
functions
of
the
position
x
along
the
beam’s
axis
can
be
obtained
using
the
method
of
sections
.

The
graphical
variations
of
V
and
M
as
functions
of
x
are
termed
the
shear
diagram
and
bending
moment
diagram
respectively
.
Chapter 7: Internal Forces
7.2 Shear and Moment Diagrams

shear
diagram
and
bending
moment
diagram
of
the
beam
:
Chapter 7: Internal Forces
2/1/2012
10
Example 3

Draw
the
shear
and
moment
diagrams
for
the
shaft
shown
in
the
figure
below
.
The
support
at
A
is
a
thrust
bearing
and
the
support
at
C
is
a
journal
bearing
.
Chapter 7: Internal Forces
Solution 3

Support
Reactions
:

Shear
and
Moment
Functions
:
The
shaft
is
sectioned
at
an
arbitrary
distance
x
from
point
A
,
extending
within
the
segment
AB
.
The
FBD
of
the
left
segment
is
:

Applying
the
equations
of
equilibrium
:
Chapter 7: Internal Forces
0


C
M




0
m

2

kN

5
m

4



y
A
kN

5
.
2

y
A
0


y
F
0
kN

5
.
2
kN

5




y
C
kN

5
.
2

y
C
0


y
F
(1)

kN

5
.
2

V
0


M


0
kN

5
.
2



M
x
(2)

kN.m

5
.
2
x
M

2/1/2012
11
Solution 3

A
free
body
diagram
for
the
left
segment
of
the
shaft
extending
a
distance
x
within
the
region
BC
is
shown
:

Applying
the
equilibrium
equations
yields
:
Chapter 7: Internal Forces
0


y
F
0


M
0
kN

5
kN

5
.
2



V




0
kN

5
.
2
m
2
kN

5




x
x
M


(4)

kN.m

5
.
2
10
x
M


(3)

kN

5
.
2


V
Solution 3

Shear
and
Moment
Diagrams
:
When
equations
(
1
)
to
(
4
)
are
plotted
within
the
regions
in
which
they
are
valid
,
the
shear
and
moment
diagrams
are
obtained
.

The
shear
diagram
indicates
that
the
internal
shear
force
is
always
2
.
5
kN
(positive)
within
segment
AB
.
Just
to
the
right
of
point
B
,
the
shear
force
changes
sign
and
remains
at
a
constant
value
of
-
2
.
5
kN
for
segment
BC
.

The
moment
diagram
starts
at
zero,
increases
linearly
to
point
B
at
x
=
2
m,
where
M
max
=
2
.
5
kN
(
2
m)
=
5
kN
.
m,
and
therefore
decreases
back
to
zero
.
Chapter 7: Internal Forces
2/1/2012
12
Example 4

Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in
the
figure
below
.
Chapter 7: Internal Forces
Solution 4

Support
Reactions
:
The
support
reactions
are
shown
on
the
beam’s
free
-
body
diagram
.

Shear
and
Moment
Functions
:
Chapter 7: Internal Forces
0


y
F
0
3
1
9
2



V
x
(1)

kN

3
9
2










x
V
0


M
0
9
3
3
1
2









x
x
x
M
(2)

kN.m

9
9
3










x
x
M
2/1/2012
13
Solution 3

Shear
and
Moment
Diagrams
:
The
shear
and
moment
diagrams
are
obtained
by
plotting
Eqs
.
(
1
)
(
2
)
.

The
point
of
zero
shear
can
be
found
using
Eq
.
(
1
)
:

Maximum
moment
can
be
found
using
the
value
of
x
=
5
.
2
m
in
Eq
.
(
2
)
:
Chapter 7: Internal Forces
m

20
.
5

,
0
3
9
2




x
x
V




kN.m

2
.
31
9
2
.
5
2
.
5
9
3
max











M
QUIZ

Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in
the
figure
below
.
Chapter 7: Internal Forces
2/1/2012
14
QUIZ

Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in
the
figure
below
.
Chapter 7: Internal Forces
7.3 Cables

Flexible
cables
and
chains
combine
strength
with
lightness
and
often
are
used
in
structures
for
support
and
to
transmit
loads
from
one
member
to
another
.

Cables
are
used
to
support
suspension
bridges
and
trolley
wheels
.
Chapter 7: Internal Forces
2/1/2012
15
7.3 Cables

2
assumptions
are
considered
in
the
force
analysis
of
these
systems
:
1.
The
weight
of
the
cable
is
negligible
.
2.
The
cable
is
perfectly
flexible
and
inextensible
.
Cable
Subjected
to
Concentrated
loads
:
Chapter 7: Internal Forces

When
a
cable
of
negligible
weight
supports
several
concentrated
loads,
the
cable
takes
the
form
of
several
straight
line
segments,
each
of
which
is
subjected
to
a
constant
tensile
.
7.3 Cables

The
equilibrium
analysis
is
performed
by
writing
down
a
sufficient
number
of
equilibrium
equations
and
equations
describing
the
geometry
of
the
cable
to
solve
for
all
the
unknowns
leading
to
a
description
of
the
tension
in
each
segment
of
the
cable
.

Example
:
Consider
the
cable
shown
in
the
figure,
where
h
,
L
1
,
L
2
,
L
3
,
P
1
and
P
2
are
known
:
Chapter 7: Internal Forces

The
problem
here
is
to
determine
the
nine
unknowns
:
Tension
in
each
of
the
three
segments,
the
four
components
of
reaction
at
A
and
B
and
the
two
sags
y
C
and
y
D
at
point
C
and
D
.
2/1/2012
16
Example 5

Determine
the
tension
in
each
segment
of
the
cable
shown
in
the
figure
below
:
Chapter 7: Internal Forces
Solution 5

Unknowns
:
There
are
4
unknown
external
reactions
(
A
x
,
A
y
,
E
x
,
E
y
),
4
unknown
cable
tensions
(one
in
each
cable
segment)
and
2
unknown
sags
(
y
B
and
y
D
)
.

Consider
the
free
-
body
diagram
for
the
entire
cable
:
Chapter 7: Internal Forces
0


x
F
0



x
x
E
A
x
x
E
A

0


E
M








0
m

2
kN

3
m

10
kN

15
m

15
kN

4
m

18





y
A
kN

12

y
A
2/1/2012
17
Solution 5

Consider
the
leftmost
segment,
which
cuts
cable
BC
:

Thus
:
Chapter 7: Internal Forces
0


C
M






0
m

5
kN

4
m

8
kN

12
m

12



x
A
kN

33
.
6


x
x
E
A
0


x
F
0
kN

33
.
6
cos


BC
BC
T

0


y
F
0
sin
kN

4
kN

12



BC
BC
T


6
.
51

BC

kN

2
.
10

BC
T
Solution 5

Proceeding
now
to
analyze
the
equilibrium
of
point
A
,
C
,
and
E
in
sequence,
we
have
:

Point
A
:

Point
C
:
Chapter 7: Internal Forces
0


x
F
0
kN

33
.
6
cos


AB
AB
T

0


y
F
0
kN

12
sin



AB
AB
T


2
.
62

AB

kN

6
.
13

AB
T
0


x
F
0


y
F
0
kN

6
.
51
cos
2
.
12
cos



CD
CD
T

0
kN

15
kN

6
.
51
sin
2
.
10
sin




CD
CD
T


9
.
47

CD

kN

44
.
9

CD
T
2/1/2012
18
Solution 5

Point
E
:

Sags
y
B
and
y
D
:
Chapter 7: Internal Forces
0


x
F
0


y
F
0
cos
kN

33
.
6


ED
ED
T

0
sin
kN

10


ED
ED
T


7
.
57

ED

kN

8
.
11

ED
T
m

69
.
5
2
.
62
tan
3



B
y
m

16
.
3
7
.
57
tan
2



D
y