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Structural Analysis II

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Nutcracker truss formed by tied ends ... Nutcracker truss. with 8N load. Step 2: Draw and label all loads on the structure ... – PowerPoint PPT presentation

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Title: Structural Analysis II


1
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

3
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

4
Structural Analysis Problem
  • Calculate the internal member forces on this
    nutcracker truss if the finger is pushing down
    with a force of eight newtons.

5
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
6
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
7
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.

8
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

9
Structural Analysis Solution Method of Joints
  • Step 1 Isolate one joint
  • Step 2 Draw the free body diagram

10
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

11
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

12
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

13
Summary of Force Components, Node a
y
x
Fab
14
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

15
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

16
Structural Analysis Solution Method of Joints
Tabulated Force Solutions
17
Structural Analysis Solution Method of Joints
Step 5 Repeat for other joints Step 1 Isolate
one joint Step 2 Draw the free body diagram
18
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

19
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

20
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)

21
Summary of Force Components, Node c
c
Fac
20o
a
22
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

23
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

24
Structural Analysis Solution Method of Joints
Tabulated Force Solutions
25
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
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