Frames and Machines
Trusses
1) Consists of pins
2) Straight 2-force members
3) Forces directed along the member
Frames
-Structures that have at least one multi-force member
*3 or more forces
-Forces usually will not be directed along the member
* Thus, should be represented by 2 unknown components (for 2-D).
Frames
1) Support loads
2) Usually stationary
3) Fully constrained
Machines
1) Transmit and modify forces
2) May or may not be stationary
3) Always contain moving parts
Analysis of a frame (rigid)
Let's take a look at the crane
D
G
F.B.D.
D
B
C E F
W
A
A x
T
B
C
A
E F
W
A y

M

F x
A


0
0


T
A x

F y

0

A y
Lecture17.doc 1

Internal Forces
C y
-BE
C x
T
C x
-BE
C y
BE
W
BE
B
A x
A

M
C

M
E

F x



0
0
0



BE
C
C x y
A y
Analysis of a frame (non-rigid)
The crane we looked at could keep its shape without the help of its supports A, and G.
P
C
Q
A B
Many frames will collapse if detached from their supports, for example:
Should be considered as 2 rigid bodies
C
P Q
A x
B x
4 unknowns
3 equations
A y
B y
C y
P
C x
-C x
-C y
Q
A x
B x
A y
4 unknowns
3 equations
6 unknowns
6 equations
B y
2 unknowns
3 equations
Lecture17.doc 2

Analysis of a machine
Let's look at a pair of cutting pliers
P
-Q
Q
-P
This forms a nonrigid structure, thus we must use one of the component parts to determine the unknown forces.
P
A y
A x a b
Q

Pa



M
A
Qb


0
0
Pa

Qb

F x
A x


0
0

F y
A y

0

P

Q
A y

P

Q

0
Lecture17.doc 3

Examples
1). Given: The following frame
D
200 Nm
800 mm
800 mm
C
1600 mm
800 mm
A B
800 mm
Find: Forces at B and C
200 Nm
C x
C y
C y
C x
B x
B y
  
M
B
200

C x

0
( 0 .
8 )
C x

250 N

0

F x
B
B x x

0

C x

250

0
N
A x
A y

C y
 
M
A

( 1 .
386 )

C y
( 1 .
386 )

0
C x
( 0 .
8 )
250 ( 0 .
8 )

0
C y

144 .
3 N

F y
B y
B y

0

C y

0

144 .
3 N
Lecture17.doc 4

2). Given: The following frame
A
310 N
30 o r = 1.4m
B
1.92 m
C
D
0.56 m
Find: Force in BD and reaction at C
Note: BD is a 2 force member, thus the force it exerts on ABC is directed along BD.
A
310 N
0.7 m
B
1.4 m F
BD
1.92 m
C x
C y
1.4 m

F x
C x

0

310

C x
 
205
C x

205 N
0 .
56
2
(

375 )
N


0
0.56 m
 
M
C

0
310 (
F
BD
2 .

1 )


375
0 .
56
2
N
F
BD
( 1 .
4 )
F
BD

375 N C

1 .
92
2
F
BD
( 1 .
4 )

0

F y
C y
C y

0


1 .
92
2
(


360
375
N
)

0
C y

360 N

Lecture17.doc 5

3). Given: A force P of magnitude 2.4 kN applied to the piston of the engine system shown.
P
C
250 mm
B
M
100 mm
A
75 mm
Find: Determine the moment M required to hold the system in equilibrium
FBD Piston
P
F
BCy
C
F
BC
N

F y

0
F
BCy

P

0
F
BCy

P
BC

75
2 
250
2 
261
F
BCy

F
BC
F
BCx

F
BC
250
261
75
261


F
F
BC

F
BCy
BCy
261
250
261

75
250
 
261

P
75
250

0 .
3 P
B
F
BCx
F
BD
is a 2 force member
F.B.D. Crank AB
F
BCx M
A x
F
BCy
A y
  
M
A

0
F
BCx
( 0 .
1 )

F
BCy
( 0 .
075 )

M
M

0 .
3 P ( 0 .
1 )

P
M

0 .
105 P
For P = 2.4 kN
M

0 .
105 ( 2 .
4 )
M

252 Nm cw
( 0 .
075 )

0
Lecture17.doc 6