Structural Analysis II

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Structural Analysis II

• Structural Analysis
• Trigonometry Concepts
• Vectors
• Equilibrium
• Reactions
• Static Determinancy and Stability
• Free Body Diagrams
• Calculating Bridge Member Forces

2
Learning Objectives

• Generate a free body diagram
• Calculate internal member forces using the Method
  of Joints

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Free Body Diagram

• Key to structural analysis
• Draw a simple sketch of the isolated structure,
  dimensions, angles and x-y coordinate system
• Draw and label all loads on the structure
• Draw and label reactions at each support

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Structural Analysis Problem

• Calculate the internal member forces on this
  nutcracker truss if the finger is pushing down
  with a force of eight newtons.

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Structural Analysis SolutionDraw the Free Body
Diagram
Step 1 Draw simple sketch with dimensions,
angles, and x-y coordinate system

• Nutcracker truss formed by tied ends

Corresponding sketch
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Structural Analysis SolutionDraw the Free Body
Diagram
Step 2 Draw and label all loads on the structure

• Nutcracker truss
• with 8N load

Added to free body diagram
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Structural Analysis Solution Draw the Free Body
Diagram
Step 3 Draw and label all reactions at each
support

• The truss is in equilibrium so there must
  reactions at the two supports. They are named Ra
  and Rb.

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Structural Analysis Solution Method of Joints

• Use the Method of Joints to calculate the
  internal member forces of the truss
• Isolate one joint from the truss
• Draw a free body diagram of this joint
• Separate every force and reaction into x and y
  components
• Solve the equilibrium equations
• Repeat for all joints

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Structural Analysis Solution Method of Joints

• Step 1 Isolate one joint
• Step 2 Draw the free body diagram

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Structural Analysis Solution Method of Joints
Step 3 Separate every force and reaction into x
and y components

• First analyse Ra
• x-component 0N
• y-component 4N

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Structural Analysis Solution Method of Joints
Step 3 Separate every force and reaction into x
and y components

• Next analyse Fab
• x-component Fab
• y-component 0N

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Structural Analysis Solution Method of Joints
Step 3 Separate every force and reaction into x
and y components

• Lastly, analyse Fac
• x-component Faccos70 N
• y-component Facsin70 N

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Summary of Force Components, Node a
Force Name Ra Fab Fac
Free Body Diagram
x- component 0N Fab Fac cos70 N
y-component 4N 0N Fac sin70 N
y
x
Fab
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Structural Analysis Solution Method of Joints
Step 4 Solve y-axis equilibrium equations

• The bridge is not moving, so SFy 0
• From the table,
• SFy 4N Fac cos70 0
• Fac ( -4N / cos70 ) -4.26N
• Internal Fac has magnitude 4.26N in compression

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Structural Analysis Solution Method of Joints
Step 4 Solve x-axis equilibrium equations

• The bridge is not moving, so SFx 0
• From the table,
• SFx Fab Fac sin70 0
• Fab - ( -4.26N / sin70 ) 1.45N
• Internal Fab has magnitude 1.45N in tension

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Structural Analysis Solution Method of Joints
Tabulated Force Solutions
Member Force Magnitude
AB 4.26N, compression
BC 1.45N, tension
AC (not yet calculated)
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Structural Analysis Solution Method of Joints
Step 5 Repeat for other joints Step 1 Isolate
one joint Step 2 Draw the free body diagram
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Structural Analysis Solution Method of Joints
Step 3 Separate every force and reaction into x
and y components

• First analyse Rc
• y-component is -8N
• x-component is 0N

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Structural Analysis Solution Method of Joints
Step 3 Separate every force and reaction into x
and y components

• Next analyse Fac
• x-component is (Fac sin20)
• - (-4.26N 0.34)
• 1.46N
• y-component is (Fac cos20)
• - (-4.26N 0.94)
• 4.00N

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Structural Analysis Solution Method of Joints
Step 3 Separate every force and reaction into x
and y components

• Lastly analyse Fbc
• y-component (Fbc cos20)
• x-component (Fbc sin20)

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Summary of Force Components, Node c
Force Name Rc Fac Fbc
Free Body Diagram
x- component 0.00 N 1.46 N Fbc sin20 N
y-component -8.00 N 4.00 N -Fbc cos20 N
c
Fac
20o
a
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Structural Analysis Solution Method of Joints
Step 4 Solve y-axis equilibrium equations

• The bridge is not moving, so SFy 0
• From the table,
• SFy -8.00N 4.00N - Fbc cos20 0
• Fbc -4.26N
• Internal Fbc has magnitude 4.26N in compression

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Structural Analysis Solution Method of Joints
Step 4 Solve x-axis equilibrium equations

• The bridge is not moving, so SFx 0
• From the table,
• SFx 1.46N Fbc sin20 0
• Fbc -4.26N
• This verifies the SFy 0 equilibrium equation
  and also the symmetry property

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Structural Analysis Solution Method of Joints
Tabulated Force Solutions
Member Force Magnitude
AB 4.26N, compression
BC 1.45N, tension
AC 4.26N, compression
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Acknowledgements

• This presentation is based on Learning Activity
  3, Analyze and Evaluate a Truss from the book by
  Colonel Stephen J. Ressler, P.E., Ph.D.,
  Designing and Building File-Folder Bridges