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6'5 Frames

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... structures, such as the frame of a car & the human structure of bones, tendons ... Draw the free-body diagram of the entire frame ... – PowerPoint PPT presentation

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Title: 6'5 Frames

1
6.5 Frames Machines

Many structures, such as the frame of a car the
human structure of bones, tendons muscles are
not composed entirely of 2-force members thus
cannot be modeled as trusses

Such structures are called
Frames if they are designed to remain stationary
support loads
Machines if they are designed to move apply
loads

2
6.5 Frames Machines

When trusses are analyzed by cutting members to
obtain free-body diagrams of joints or sections,
the internal forces acting at the cuts are
simple axial forces
This is generally not true for frames or machines
a different method of analysis is necessary
Instead of cutting members, you isolate the
entire member, or in some cases combinations of
members, from the structure

3
6.5 Frames Machines

To begin analyzing a frame or machine, draw a
free-body diagram of the entire structure (i.e.
treat the structure as a single object)
determine the reactions at its supports
In some cases the entire structure will be
statically indeterminate but it is helpful to
determine as many of the reactions as possible
We then draw the free-body diagrams of individual
members or selected combinations of members
apply the equilibrium equations to determine the
forces couples acting on them

4
6.5 Frames Machines

E.g. consider the stationary structure
Member BE is a 2-force member but the other 3
members ABC, CD DEG are not
This structure is a frame
Our objective is to determine the forces on its
members

5
6.5 Frames Machines

Analyzing the Entire Structure
Draw the free-body diagram of the entire frame
It is statically indeterminate there are 4
unknown reactions, Ax, Ay, Gx Gy, whereas we
can only write 3 independent equilibrium
equations
However, notice that the lines of action of 3 of
the unknown reactions intersect at A

6
6.5 Frames Machines

Summing the moments about A yields
S Mpoint A (2 m)Gx (1 m)(8 kN) ? (3 m)(6
kN) 0
and we obtain the reaction Gx 5 kN
Then from the equilibrium equation
S Fx Ax Gx 8 kN 0
we obtain the reaction Ax ?13 kN
Although we cannot determine Ay or Gy from the
free-body diagram of the entire structure, we can
do so by analyzing the individual members

7
6.5 Frames Machines

Analyzing the Members
The next step is to draw the free-body diagrams
of the members
To do so, we treat the attachment of a member to
another member just as if it were a support
Looked at in this way, we can think of each
member as a supported object of the kind analyzed
in Chapter 5
Furthermore, the forces couples the members
exert on 1 another are equal in magnitude
opposite in direction

8
6.5 Frames Machines

A simple demonstration is instructive
If you clasp your hands and exert a force on your
left hand with your right hand, your left hand
exerts an equal opposite force on your right
hand
Similarly, if you exert a couple on your left
hand, your left hand exerts an equal opposite
couple on your right hand

9
6.5 Frames Machines

We disassemble the frame draw the free-body
diagrams of its members

10
6.5 Frames Machines

Observe that the forces exerted on one another by
members are equal opposite
E.g. at point C on the free-body diagram of
member ABC, the force exerted by member CD is
denoted by the components Cx Cy
The forces exerted by member ABC on member CD at
point C must be equal opposite as shown

11
6.5 Frames Machines

2-Force Members
Member BE is a 2-force member we have taken
this into account in drawing its free-body
diagram
The force T is the axial force in member BE an
equal opposite force is subjected on the member
ABC at B on member GED at E
Recognizing the 2-force members in frames
machines drawing free-body diagrams as we have
done will reduce the number of unknowns will
greatly simplify the analysis

12
6.5 Frames Machines

In our example, if we did not treat member BE as
a 2-force member, its free-body diagram would
have 4 unknown force (Fig. a)
By treating it as a 2-force member (Fig. b), we
reduce the number of unknown forces by 3

13
6.5 Frames Machines

Loads Applied at Joints
When a load is applied at a joint, where does the
load appear on the free-body diagrams of the
individual members?
You can place the load on any 1 of the members
attached at the joint
In the same example

14
6.5 Frames Machines

The 6-kN load acts at the joint where members ABC
CD are connected
In drawing the free-body diagrams of the
individual members, we assumed that the 6-kN load
acted on the member ABC
The force components Cx Cy on the free-body
diagram of member ABC are the forces exerted by
the member CD

15
6.5 Frames Machines

To explain why we can draw the free-body diagrams
in this way
Assume that the 6-kN force acts on the pin
connecting members ABC CD draw separate
free-body diagrams of the pin the 2 members

The force components Cx Cy are the forces
exerted by the pin on member ABC Cx Cy are
the forces exerted by the pin on member CD

16
6.5 Frames Machines

If we superimpose the free-body diagrams of the
pin member ABC, we obtain the 2 free-body
diagrams, which is the way we drew them
Alternatively, by superimposing the free-body
diagrams of the pin member CD, we obtain the 2
free-body diagrams

17
6.5 Frames Machines

Thus if a load acts at a joint, it can be placed
on any 1 of the members attached at the joint
when drawing the free-body diagrams of the
individual members
Make sure not to place it on more than 1 member
To detect errors in the free-body diagrams of the
members, it is helpful to reassemble them
The forces at the connections between the members
cancel (they are internal forces once the members
are reassembled) the free-body diagram of the
entire structure is recovered

18
6.5 Frames Machines
19
6.5 Frames Machines

The final step is to apply the equilibrium
equations to the free-body diagrams of the
members

In 2 dimensions, we can obtain 3 individual
independent equilibrium equations from the
free-body diagram of each member of a structure
that we do not treat as a 2-force member

20
6.5 Frames Machines

By assuming that the forces on a 2-force member
are equal opposite axial forces, we have
already used the 3 equilibrium equations for that
member
In this example, there are 3 members in addition
to the 2-force member, so we can write (3)(3) 9
independent equilibrium equations there are 9
unknown forces Ax, Ay, Cx, Cy, Dx, Dy, Gx,
Gy T

21
6.5 Frames Machines

The equilibrium equations we obtained from the
free-body diagram of the entire structure are not
independent of the equilibrium equations obtained
from the free-body diagrams of the members but by
using them to determine Ax Gx, we get a had
start on solving the equations for the members

22
6.5 Frames Machines

Consider the free-body diagram of member ABC
Because we know Ax, we can determine Cx from the
equation
S Fx Ax Cx 0
Obtaining Cx ?Ax 13 kN.
Consider the free-body diagram of GED
We can determine Dx from the equation
S Fx Gx Dx 0
Obtaining Dx ?Gx ?5 kN.

23
6.5 Frames Machines

Consider the free-body diagram of CD
Because we know Cx, we can determine Cy
by summing moments about D
S Mpoint D (2 m)Cx ? (1 m)Cy
? (1 m)(8 kN) 0
We obtain Cy 18 kN.
Then, from the equation
S Fy ?Cy ? Dy 0
We find that Dy ?Cy ?18 kN.

24
6.5 Frames Machines

Now we can return to the free-body diagrams of
members ABC GED to determine Ay Gy
Summing moments about point B of member ABC
yields
S Mpoint B ?(1 m)Ay (2 m)Cy ? (2 m)(6 kN)
0
and we obtain Ay 2Cy ? 12 kN 24 kN.
Then, summing moments about point E of member
GED, we have
S Mpoint E (1 m)Dy ? (1 m)Gy 0
from which we obtain Gy Dy ? 18 kN.

25
6.5 Frames Machines

Finally, from the free-body diagram of member
GED
We use the equilibrium equation
S Fy Dy Gy T 0
which gives us the result T ?Dy ? Gy 36 kN.
The forces on the members are

26
6.5 Frames Machines

As this example demonstrates, determination of
the forces on the members can often be simplified
by carefully choosing the order in which the
equations are solved
Determination of the forces couples on the
members of the frames machines involves 2 steps

27
6.5 Frames Machines

1.Determine the reactions at the supports draw
the free-body diagram of the entire structure
determine the reactions at its supports. Although
this step is not essential, it can greatly
simplify your analysis of the members. If the
free-body diagram is statically indeterminant,
determine as many of the reactions as possible.

28
6.5 Frames Machines

2.Analyze the members draw free-body diagrams
of the members apply the equilibrium equations
to determine the forces acting on them. You can
simplify this step by identifying 2-force
members. If a load acts at a joint of the
structure, you can place the load on the
free-body diagram of any 1 of the members
attached at that point.

29
Example 6.6 Analyzing a Frame

The frame in Fig. 6.36 is subjected to a 200
N-m couple. Determine the forces couples on its
members.

30
Example 6.6 Analyzing a Frame

Strategy
1st, draw a free-body diagram of the entire
frame, treating it as a single object attempt
to determine the reactions at the supports. Then
draw the free-body diagrams of the individual
members use the equilibrium equations to
determine the forces couples acting on them.

31
Example 6.6 Analyzing a Frame

Solution
Determine the Reactions at the Supports
Draw the free-body diagram of the entire frame
The term MA is the couple exerted by the fixed
support.

32
Example 6.6 Analyzing a Frame

Solution
From the equilibrium equations
S Fx Ax 0
S Fy Ay C 0
S Mpoint A MA ? 200 N-m (1 m)C 0
we obtain the reaction Ax 0. We cant determine
Ay, MA or C from this free-body diagram.

33
Example 6.6 Analyzing a Frame

Solution
Analyze the Members
Disassemble the frame to obtain the free-body
diagrams of the members

34
Example 6.6 Analyzing a Frame

Solution
The equilibrium equations for member BC are
S Fx ?Bx 0
S Fy ?By C 0
S Mpoint A ? 200 N-m (0.4 m)C 0
Solving these equations, we obtain Bx 0, By
500 N
C 500N.

35
Example 6.6 Analyzing a Frame

Solution
The equilibrium equations for member AB are
S Fx Ax Bx 0
S Fy Ay By 0
S Mpoint B MA (0.6 m)By 0
Because we already know Ax, Bx By, we can solve
these equations for Ay MA.
The results are Ay ?500 N MA ?300 N-m.

36
Example 6.6 Analyzing a Frame

Solution
This completes the solution

37
Example 6.6 Analyzing a Frame

Critical Thinking
We were able to solve the equilibrium equations
for member BC without having to consider the
free-body diagram of member AB
We were then able to solve the equilibrium
equations for member AB
By choosing the members with the fewest unknowns
to analyze 1st, you will often be able to solve
the sequentially
But in some cases you will have to solve the
equilibrium equations for the members
simultaneously

38
Example 6.6 Analyzing a Frame

Critical Thinking
Even though we were unable to determine the 4
reactions Ax, Ay, MA C with the 3 equilibrium
equations obtained from the free-body diagram of
the entire frame, we were able to determine them
from the free-body diagrams of the individual
members
By drawing the free-body diagrams of the members,
we gained 3 equations because we obtained 3
equilibrium equations from each member but only 2
new unknowns Bx By

39
Example 6.7 Determining Forces on Members of a
Frame

The frame in Fig. 6.37 supports a suspended
weight W 200 N. Determine the forces on members
ABCD CEG.

40
Example 6.7 Determining Forces on Members of a
Frame

Strategy
Draw a free-body diagram of the entire frame
attempt to determine the reactions at the
supports. Then draw free-body diagrams of the
individual members use the equilibrium
equations to determine the forces couples
acting on them. In doing so, we can take
advantage of the fact that the bar BE is a
2-force member.

41
Example 6.7 Determining Forces on Members of a
Frame

Solution
Determine the Reactions at the Supports
Draw the free-body diagram of the entire frame

42
Example 6.7 Determining Forces on Members of a
Frame

Solution
From the equilibrium equations
S Fx Ax ? D 0
S Fy Ay ? 200 N 0
S Mpoint A (0.36 m)D ? (0.38 m)(200 N) 0
we obtain the reactions Ax 211 N, Ay 200 N
D 211 N.

43
Example 6.7 Determining Forces on Members of a
Frame

Solution
Analyze the Members
We obtain the free-body diagram of the members

44
Example 6.7 Determining Forces on Members of a
Frame

Solution
Notice that BE is a 2-force member. The angle
? arctan (6/8) 36.9
The free-body diagram of the pulley has only 2
unknown forces. From the equilibrium equations
S Fx Gx ? 200 N 0
S Fy Gy ? 200 N 0
we obtain Gx 200 N Gy 200 N.

45
Example 6.7 Determining Forces on Members of a
Frame

Solution
There are now only 3 unknown forces on the
free-body diagram of member CEG.
From the equilibrium equations
S Fx ?Cx ? R cos ? ? 200 N 0
S Fy ?Cy ? R sin ? ? 200 N 0
S Mpoint C ?(0.16 m) R sin ? ? (0.32 m)(200 N)
0
we obtain Cx 333.3 N, Cy 200 N R ?666.7 N.

46
Example 6.7 Determining Forces on Members of a
Frame

Solution
The complete solution

47
Example 6.7 Determining Forces on Members of a
Frame

Critical Thinking
In problems of this kind, the reactions on the
individual members of the frame can be determined
from the free-body diagrams of the members
Why did we draw the free-body diagram of the
entire frame solve the associated equilibrium
equations?
It gave us a head start on solving the
equilibrium equations for the members

48
Example 6.7 Determining Forces on Members of a
Frame

Critical Thinking
In this example when we drew the free-body
diagrams of the members we already knew the
reactions at A D, which simplified the
remaining analysis
Analyzing the entire frame can also provide a
check on your work
Notice that we did not use the equilibrium
equations for member ABCD

49
Example 6.7 Determining Forces on Members of a
Frame