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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
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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
1) Draw a simple sketch of the isolated structure, dimensions, angles and x-y coordinate system
2) Draw and label all loads on the structure
3) 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
Nutcracker truss formed by tied ends
Step 1: Draw simple sketch with dimensions, angles, and x-y coordinate system
Corresponding sketch
70o
40o
a b70o
12 cm
c
x
y
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Structural Analysis SolutionDraw the Free Body Diagram
Nutcracker truss
with 8N load
Step 2: Draw and label all loads on the structure
Added to free body diagram
70o 70o
40o
8N
a b
c
12 cmx
y
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Structural Analysis Solution Draw the Free Body Diagram
• The truss is in equilibrium so there must reactions at the two supports. They are named Ra and Rb.
Step 3: Draw and label all reactions at each support
70o 70o
40o
8N
Ra Rb
a b
c
12 cmx
y
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Structural Analysis Solution Method of Joints
• Use the Method of Joints to calculate the internal member forces of the truss
1. Isolate one joint from the truss2. Draw a free body diagram of this joint3. Separate every force and reaction into x
and y components4. Solve the equilibrium equations5. Repeat for all joints
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Structural Analysis Solution Method of Joints
Step 1: Isolate one jointStep 2: Draw the free body diagram
70o 70o
40o
8N
Ra = 4N Rb = 4N
a b
cx
y
12 cm
70o
Ra = 4Na x
y
b
c
Fac
Fab
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Structural Analysis Solution Method of Joints
First analyse Ra
• x-component = 0N
• y-component = 4N
Step 3: Separate every force and reaction into x and y components
a x
y
Ra = 4N
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Structural Analysis Solution Method of Joints
Next analyse Fab
• x-component = Fab
• y-component = 0N
Step 3: Separate every force and reaction into x and y components
a x
y
bFab
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Structural Analysis Solution Method of Joints
Lastly, analyse Fac• x-component = Fac*cos70˚ N
• y-component = Fac*sin70˚ N
Step 3: Separate every force and reaction into x and y components
70o
a x
yc
Fac
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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 0NFac * sin70˚ N
a x
y
Ra = 4N
x
y
Fab
70o
a x
y cFac
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Structural Analysis Solution Method of Joints
• The bridge is not moving, so ΣFy = 0
• From the table,ΣFy = 4N + Fac * cos70˚ = 0
• Fac = ( -4N / cos70˚ ) = -4.26N
• Internal Fac has magnitude 4.26N in compression
Step 4: Solve y-axis equilibrium equations
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Structural Analysis Solution Method of Joints
• The bridge is not moving, so ΣFx = 0
• From the table,ΣFx = Fab + Fac * sin70˚ = 0
Fab = - ( -4.26N / sin70˚ ) = 1.45N
• Internal Fab has magnitude 1.45N in tension
Step 4: Solve x-axis equilibrium equations
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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 jointsStep 1: Isolate one jointStep 2: Draw the free body diagram
70o 70o
40o
8N
Ra Rb
a b
c
12 cm
x
y
b
40o
8N
c
a
Fac = -4.26N Fbc
x
y
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Structural Analysis Solution Method of Joints
First analyse Rc
• y-component is -8N• x-component is 0N
Step 3: Separate every force and reaction into x and y components
8N
c x
y
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Structural Analysis Solution Method of Joints
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
Step 3: Separate every force and reaction into x and y components
20o
c
a
Fac
x
y
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Structural Analysis Solution Method of Joints
Lastly analyse Fbc
• y-component = –(Fbc * cos20˚)
• x-component = (Fbc * sin20˚)
Step 3: Separate every force and reaction into x and y components
20o
c
Fbc
b
x
y
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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
8Nc
20o
c
a
Fac
20o
cFbc
b
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Structural Analysis Solution Method of Joints
• The bridge is not moving, so ΣFy = 0
• From the table,ΣFy = -8.00N + 4.00N - Fbc * cos20˚ =
0Fbc = -4.26N
• Internal Fbc has magnitude 4.26N in compression
Step 4: Solve y-axis equilibrium equations
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Structural Analysis Solution Method of Joints
• The bridge is not moving, so ΣFx = 0
• From the table,ΣFx = 1.46N + Fbc * sin20˚ = 0Fbc = -4.26N
• This verifies the ΣFy = 0 equilibrium equation and also the symmetry property
Step 4: Solve x-axis equilibrium equations
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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