Structural Analysis II
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Structural Analysis
Trigonometry Concepts
Vectors
Equilibrium
Reactions
Static Determinancy and Stability
Free Body Diagrams
Calculating Bridge Member Forces
Learning Objectives
• Generate a free body diagram
• Calculate internal member forces using the
Method of Joints
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
Structural Analysis Problem
• Calculate the internal
member forces on this
nutcracker truss if the
finger is pushing down
with a force of eight
newtons.
Structural Analysis Solution
Draw the Free Body Diagram
Step 1: Draw simple sketch with dimensions, angles, and
x-y coordinate system
c
y
x
40o
70o
a
Nutcracker truss formed
by tied ends
70o
b
Corresponding
sketch
Structural Analysis Solution
Draw the Free Body Diagram
Step 2: Draw and label all
loads on the structure
8N
y
c
x
40o
70o
a
Nutcracker truss
with 8N load
70o
b
Added to free
body diagram
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.
8N
y
c
x
40o
70o
a
Ra
70o
b
Rb
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 truss
2. Draw a free body diagram of this joint
3. Separate every force and reaction into x
and y components
4. Solve the equilibrium equations
5. Repeat for all joints
Structural Analysis Solution
Method of Joints
y
Step 1: Isolate one joint
Step 2: Draw the free body diagram
8N
x
c
y
40o
70o
a
Ra = 4N
70o
b
Rb = 4N
Fac
a
c
70o
Fab b
Ra = 4N
x
Structural Analysis Solution
Method of Joints
Step 3: Separate every force and reaction into x and y
components
y
First analyse Ra
• x-component = 0N
•
y-component = 4N
a
x
Ra = 4N
Structural Analysis Solution
Method of Joints
Step 3: Separate every force and reaction into x and y
components
y
Next analyse Fab
• x-component = Fab
•
y-component = 0N
a
Fab b
x
Structural Analysis Solution
Method of Joints
Step 3: Separate every force and reaction into x and y
components
y
Lastly, analyse Fac
• x-component = Fac*cos70˚ N
•
y-component = Fac*sin70˚ N
Fac
a
c
70o
x
Summary of Force
Components, Node ‘a’
Force Name
Free Body
Diagram
x- component
y-component
Ra
y
a
Fab
Fac
y c
y
x
Ra = 4N
0N
4N
x
Fab
Fab
0N
Fac
a
70o
x
Fac * cos70˚ N
Fac * sin70˚ N
Structural Analysis Solution
Method of Joints
Step 4: Solve y-axis equilibrium equations
•
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
Structural Analysis Solution
Method of Joints
Step 4: Solve x-axis equilibrium equations
•
The bridge is not moving, so ΣFx = 0
•
From the table,
ΣFx = Fab + Fac * sin70˚ = 0
Fab = - ( -4.26N / sin70˚ ) = 1.45N
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Internal Fab has magnitude 1.45N in tension
Structural Analysis Solution
Method of Joints
Tabulated Force Solutions
Member
Force Magnitude
AB
4.26N, compression
BC
1.45N, tension
AC
(not yet calculated)
Structural Analysis Solution
Method of Joints
Step 5: Repeat for other joints
Step 1: Isolate one joint
Step 2: Draw the free body diagram
y
8N
x
y
8N
c
c
40o
Fac = -4.26N
70o
a
70o
40o
b
a
Ra
Fbc
x
Rb
b
Structural Analysis Solution
Method of Joints
Step 3: Separate every force and reaction into x and y
components
y
First analyse Rc
• y-component is -8N
• x-component is 0N
8N
c
x
Structural Analysis Solution
Method of Joints
Step 3: Separate every force and reaction into x and y
components
y
Next analyse Fac
• x-component is –(Fac * sin20˚)
= - (-4.26N * 0.34)
= 1.46N
c
Fac
20o
•
y-component is –(Fac * cos20˚)
= - (-4.26N * 0.94)
= 4.00N
a
x
Structural Analysis Solution
Method of Joints
Step 3: Separate every force and reaction into x and y
components
y
Lastly analyse Fbc
• y-component = –(Fbc * cos20˚)
c
x
Fbc
•
x-component = (Fbc * sin20˚)
20o
b
Summary of Force
Components, Node ‘c’
Force Name
Free Body
Diagram
Rc
8N
c
Fac
Fbc
c
c
Fac
20o
a
x- component
y-component
0.00 N
-8.00 N
1.46 N
4.00 N
Fbc
20o
b
Fbc * sin20˚ N
-Fbc * cos20˚ N
Structural Analysis Solution
Method of Joints
Step 4: Solve y-axis equilibrium equations
•
The bridge is not moving, so ΣFy = 0
•
From the table,
ΣFy = -8.00N + 4.00N - Fbc * cos20˚ = 0
Fbc = -4.26N
•
Internal Fbc has magnitude 4.26N in
compression
Structural Analysis Solution
Method of Joints
Step 4: Solve x-axis equilibrium equations
•
The bridge is not moving, so ΣFx = 0
•
From the table,
ΣFx = 1.46N + Fbc * sin20˚ = 0
Fbc = -4.26N
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This verifies the ΣFy = 0 equilibrium
equation and also the symmetry property
Structural Analysis Solution
Method of Joints
Tabulated Force Solutions
Member
Force Magnitude
AB
4.26N, compression
BC
1.45N, tension
AC
4.26N, compression
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