# Note #2 - Unix.eng.ua.edu - The University of Alabama

Urban and Civil

Nov 15, 2013 (4 years and 7 months ago)

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1

ME

350

Static Machine Components

Lecture Note #2

Free
-

Dr. Y.B. Guo

Mechanical Engineering

The University of Alabama

2

Topics

Free
-
Body

Diagram

Equilibrium

Equation

Force

Analysis

Beam

3

Free
-
Body Diagram

Free
-
Body

A conceptually isolated portion of a structure

Free
-
Body Diagram

The isolated portion with all forces and moments
acting on it

4

Equilibrium Equation

Force Equilibrium

Moment Equilibrium

0
0
0
z
y
x
F
F
F

0
0
0
z
y
x
M
M
M
x

y

z

5

Force Analysis

Sign

Convention

Force

up

(
+
)
;

down

(
-
)

Moment
:

CW

(
+
)
;

CCW

(
-
)

Common

Constraints

P

R
1

R
2

M

x

y

6

Force Analysis

Case

1

Fa
M
M
a
F
M
a
a
o
0
0

F
F
F
F
F
a
a
x
0
0
F
a

M
a

o

x

y

z

7

Force Analysis

Case

2

Fb
M
M
b
F
M
b
b
x

0
0

F
F
F
F
F
c
c
z
0
0
Fa
M
M
a
F
M
a
a
y

0
0
x

z

F
c
M
a
M
b
y

8

Force Analysis

Case

3

)
(
30
30
)
(
60
0
42
12
32
42
12
42
N
Sin
F
V
F
Cos
F
H
N
F
M
o

x

y

o

9

Shear

Force

&

Bending

Moment

Sign

convention

Count

from

left

to

right

Case

1
:

10

Case

2
:

y

x

Z

x
-
y

plane

x
-
y

plane

x
-
y

plane

z
-
y

plane

11

Shear

Force

and

Bending

Moment

Shear

force

~

bending

moment

relationship

q(x)
:

distribution

B
A
B
A
B
A
B
A
M
M
A
B
x
x
V
V
x
x
A
B
M
M
Vdx
dM
V
V
qdx
dV
x
q
dx
M
d
dx
dV
dx
dM
V
)
(
2
2
x
A

x
B

12

Case

Study

(exercise)

Find

support

reaction

forces

M
1

and

R
1

Sketch

shear

force

and

moment

diagrams

A

B

C

D

R
1

M
1

240 lb
-
in

x

y

3 in

7 in

10 in

20 lb/in

13

Case

Study

Find

reaction

forces

)
(
160
0
240
)
2
4
3
(
)
20
4
(
)
(
160
0
240
)
3
(
)
20
(
0
)
(
80
0
4
20
0
1
1
4
0
1
1
1
1
in
lb
M
M
or
in
lb
M
x
dx
M
M
lb
R
R
F
A
y

A

B

C

D

R
1

M
1

240 lb
-
in

x

y

4 in

20 lb/in

x

dx

3 in

14

Case

Study

Find

shear

force

diagram

A

B

C

D

R
1

M
1

240 lb
-
in

x

y

4 in

20 lb/in

x

3 in

)
(
0
:
)
3
(
)
(
0
4
20
80
,
4
,
int
@
20
80
:
)
2
(
80
:
)
1
(
segment
CD
on
no
V
V
segment
CD
in
lb
V
x
C
po
x
V
segment
BC
lb
V
segment
AB
C
C

15

Case

Study

Find

moment

diagram

in
lb
M
M
D
po
in
lb
M
dx
M
M
segment
CD
in
lb
M
x
C
po
x
x
dx
x
M
M
segment
BC
in
lb
M
x
B
po
x
dx
M
M
segment
AB
Vdx
M
M
Since
c
D
x
C
C
C
x
B
B
x
A
x
x
A
B
B
A

240
,
int
@
)
(
240
0
:
)
3
(
)
(
240
4
10
4
80
80
,
4
,
int
@
10
80
80
)
20
80
(
:
)
2
(
)
(
80
3
80
160
,
3
,
int
@
80
160
80
:
)
1
(
0
2
2
0
0