Section 6.6
ENGR-1100 Introduction to Engineering Analysis
Lecture 23
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 1.1-1.6
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FRAMES
In-Class Activities:
Reading Quiz
Applications
Analysis of a Frame
Concept Quiz
Group Problem Solving
Attention Quiz
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.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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APPLICATIONS
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.
Frames are commonly used to support various external loads.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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FRAMES AND MACHINES: DEFINITIONS
Frames are generally stationary and support external loads.
Machines contain moving parts and are designed to alter the effect of forces.
Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members).
Frame
Machine
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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STEPS FOR ANALYZING A FRAME
1. Draw a FBD of the frame 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,
c) For a joint with more than two members or an external force, it is advisable to draw a FBD of the pin.
FAB
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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STEPS FOR ANALYZING A FRAME
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.
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
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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EXAMPLE
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.
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:
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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EXAMPLE (continued)
Note that member AB is a two-force member.
FBD of member BC
(Note AB is a 2-force member!)
CX
CY
B
45°
FAB
400 N
1 m
2 m
1 m
800 N m
Equations of Equilibrium: Start with MC since it yields one unknown.
+ MC = FAB sin45° (1) – FAB cos45° (3) + 800 N m + 400 (2) = 0
FAB = 1131 N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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EXAMPLE (continued)
FBD of member BC
(Note AB is a 2-force member!)
CX
CY
B
45°
FAB
400 N
1 m
2 m
1 m
800 N m
+ FY = – CY + 1131 cos 45° – 400 = 0
CY = 400 N
+ FX = – CX + 1131 sin 45° = 0
CX = 800 N
Now use the x and y-direction Equations of Equilibrium:
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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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
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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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) No change, still just one force (FAB) at B.
B) Will have two forces, BX and BY, at B.
C) Will have two forces and a moment at B.
D) Will add one moment at B.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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2. The figures show a frame and its FBDs. If an additional force is applied at D, then how will you change the FBD of member BC at B?
A) No change, still just one force (FAB) at B.
B) Will have two forces, BX and BY, at B.
C) Will have two forces and a moment at B.
D) Will add one moment at B.
CONCEPT QUIZ (continued)
D
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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ATTENTION QUIZ
2. For the above problem, imagine that you have drawn a FBD of member BC. What will be the easiest way to write an equation involving unknowns at B?
A) MC = 0 B) MB = 0
C) MA = 0 D) FY = 0
1. When determining the reactions at joints A, B and C, what is the total number of unknowns in solving this problem?
A) 6 B) 5
C) 4 D) 3
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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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:
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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GROUP PROBLEM SOLVING
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.
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:
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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GROUP PROBLEM SOLVING (continued)
FBDs of members BC and AC
Applying E-of-E to member AC:
+ MA = – CY (8) + CX (6) + 50 (3.5) = 0 (1)
+ FX = -CX + AX = 0
+ FY = 50 + AY – CY = 0
CY
CX
AX
AY
3.5 ft
8 ft
6 ft
50 lb
Starting with this piece is not super useful! Why not?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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GROUP PROBLEM SOLVING (continued)
FBDs of members BC and AC
Applying E-of-E to member BC:
+ MB = – 50 (2) – 50 (3.5) + CY (8) = 0 ; CY = 34.38 = 34.4 lb
From Eq (1), CX can be determined; 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
If you start with this piece, then the equations for AC will be simpler!
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6
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