REMINDERS
•  HW#11 posted. Try these method of
sections problems by Thursday’s
class
•  Read section 6.6 Frames and
Machines
•  Frames this Thursday
•  Machines next Tuesday (along w/
quiz #6)
REMAINING SCHEDULE
•  Finish chapter six by Nov 10
•  Review session in class Nov 12
•  Test #3 Monday Nov 16, 6 – 8 pm Ball Hall 210
•  Chapters 5 & 6
•  Nov 17: 7.1 Internal Forces
•  Nov 19: 8.1 & 8.2 Dry Friction
•  Nov 24: 9.1 Centers of Gravity, Mass, Centroids
•  Dec 1: 9.2 Composite Bodies
•  Dec 3: 10 Moments of Inertia
•  Dec 8: Final Exam Review
•  Dec 15: Final exam scheduled for 3 – 6 pm
AGENDA
•  Quiz #5
•  Finish 6.4 Method of Sections
•  Section 6.6 Frames--maybe
QUIZ #5
Please take everything off your desk except a pencil and
calculator.
BACK TO METHOD OF
SECTIONS, LESSON 16
FRAMES
Today’s Objectives:
Students will be able to:
a) Draw the free body diagram of a frame
and its members.
b) Determine the forces acting at the joints
and supports of a frame
READING QUIZ
1. Frames and machines are different as compared to trusses since they have
___________.
A)  Only two-force members
B)  Only multiforce members
C)  At least one multiforce member
D)  At least one two-force member
2. Forces common to any two contacting members act with _______ on the
other member.
A) Equal magnitudes but opposite sense
B) Equal magnitudes and the same sense
C) Different magnitudes and the opposite sense
D) Different magnitudes and the same sense
APPLICATIONS
Frames are commonly
used to support various
external loads.
How is a frame different than
a truss?
To be able to design a frame,
you need to determine the
forces at the joints and
supports.
APPLICATIONS (continued)
“Machines,” like those above, are used in a variety of
applications. How are they different from trusses and
frames?
You must determine the loads at the joints and supports.
These forces and moments are required when designing
the machine’s members.
FRAMES AND MACHINES: DEFINITIONS
Frame
Machine
Frames and machines are composed of pin-connected
members, that have at least one multi-force member.
(Recall that trusses have nothing but two-force members).
Frames are generally stationary and support external
loads.
Machines contain moving parts and are designed to alter
the effect of forces.
STEPS FOR ANALYZING A FRAME OR MACHINE
1. Draw a FBD of the frame or machine and its
members, as necessary.
Hints:
a) Identify any two-force members,
b) Note that forces on contacting surfaces
(usually between a pin and a member) are equal
and opposite, and,
FAB
c) For a joint with more than two members or
an external force, it is advisable to draw a FBD
of the pin.
STEPS FOR ANALYZING A FRAME OR MACHINE
2.
Develop a strategy to apply the equations of
equilibrium to solve for the unknowns. Look
for ways to form single equations and single
unknowns.
3.
Use the E of E to solve for the unknown
forces
Problems are going to be challenging since
there are usually several unknowns. A lot of
practice is needed to develop good strategies
and ease of solving these problems.
FAB
EXAMPLES
To the overhead projector…
EXAMPLE
Given: The frame supports an
external load and moment
as shown.
Find: The horizontal and vertical
components of the pin
reactions at C and the
magnitude of reaction at B.
Plan:
a) Draw FBDs of the frame member BC. Why pick
this part of the frame?
b) Apply the equations of equilibrium and solve for
the unknowns at C and B.
EXAMPLE (continued)
800 N m
400
N
CX
CY
1m
1m
2m
B
45°
FBD of member BC
FAB
Please note that member AB is a two-force member.
Equations of Equilibrium: Start with ∑ MC since it yields one unknown.
+ ∑ MC = FAB cos 45° (1) – FAB sin 45° (3) + 800 N m + 400 (2) = 0
FAB = 1131 N
EXAMPLE (continued)
800 N m
400
N
CX
CY
1m
1m
2m
B
45°
FBD of member BC
FAB
Now use the x and y-direction Equations of Equilibrium:
→  + ∑ FX = – CX + 1131 cos 45° = 0
CX = 800 N
↑ + ∑ FY = – CY + 1131 sin 45° – 400 = 0
CY = 400 N
CONCEPT QUIZ
1. The figures show a frame and its FBDs. If an additional couple moment is
applied at C, how will you change the FBD of member BC at B?
A)
B)
C)
D)
No change, still just one force (FAB) at B.
Will have two forces, BX and BY, at B.
Will have two forces and a moment at B.
Will add one moment at B.
GROUP PROBLEM SOLVING
Given: A frame supports
a 50 lb load as shown.
Find: The reactions
exerted by the pins on
the frame members at
B and C.
Plan:
a) Draw a FBD of member BC and another one for AC.
b) Apply the equations of equilibrium to each FBD to
solve for the four unknowns. Think about a strategy to
easily solve for the unknowns.
GROUP PROBLEM SOLVING (continued)
CY
FBDs of members BC and AC
CX
50 lb
3.5 ft
6 ft
8 ft
AX
Note how forces at
“C” are shown in
opposite directions
in the 2 diagrams
AY
Applying E-of-E to member AC:
+∑ MA = – CY (8) + CX (6) + 50 (3.5) = 0
CX = [CY(8) - 50 (3.5)]/6
GROUP PROBLEM SOLVING (continued)
FBDs of members BC and AC
CY
CX
50 lb
3.5 ft
6 ft
8 ft
AX
Applying E-of-E to member BC:
+ ∑ MB = – 50 (2) – 50 (3.5) + CY (8) = 0 ;
AY
CY = 34.38 = 34.4 lb; You can use previous eq. to find CX:
CX= [CY(8) - 50 (3.5)]/6 ;
CX = 16.67 = 16.7 lb
→ + ∑ FX = 16.67 + 50 – BX = 0 ; BX = 66.7 lb
↑+ ∑ FY = BY – 50 + 34.38 = 0 ; BY = 15.6 lb