1. Structural Analysis II
Trigonometry Concepts
Vectors
Equilibrium
Reactions
Static Determinancy and Stability
Free Body Diagrams
Calculating Bridge Member Forces
This PowerPoint presentation will cover the last sections in white text. The section on calculating bridge member forces is a challenge for many students.
In some cases, the material in Structural Analysis I and Structural Analysis II may require more time than is allotted in the six week module. You may have to adjust the timeframes according to your class.

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
Keep this concept simple. The truss bridge has a friendly, repeatable free body diagram. We will do a simple nutcracker truss as a first example.

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.
The following slides will give a step by step example of how to create the free body diagram and then use the method of joints to calculate the internal forces on each truss member. This technique is the exact same as used to calculate the internal forces on any truss bridge.

5. Structural Analysis Solution Draw the Free Body Diagram
Step 1: Draw simple sketch with dimensions, angles, and x-y coordinate system
70o
40o a b
12 cm c x y
This is Step 1 of how to draw a free body diagram. A similar example can be found in Learning Activity #1.
The node names are arbitrary and important.
Nutcracker truss formed by tied ends
Corresponding sketch

6. Structural Analysis Solution Draw the Free Body Diagram
Step 2: Draw and label all loads on the structure
70o
40o 8N a b c
12 cm x y
The green arrow represents the external force on the free body diagram. In this case it is a finger applying a downward (negative y-direction) on the peak of the nutcracker truss.
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
70o
40o 8N Ra Rb a b c
12 cm x y
The truss is in equilibrium so there must reactions at the two supports. They are named Ra and Rb.
The students will be able to guess the magnitude to be 4N for each reaction and their direction to be pointed up towards node c.
Structural analysis verifies their intuition. By using vertical symmetry, Ra = Rb. And because the structure is in equilibrium, ΣFy = 0, so Ra + Rb - 8N = 0. Solving yields Ra = Rb = 4N in a positive y direction.

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
After the entire structure’s free body diagram is complete, systematically apply this Method of Joints to solve the internal forces on each member. This takes some effort and some students may become discouraged, but engineering is rigorous and methodical. To offer some hope, this analysis can be computerized. In fact, the simulator they used is based on these equations.

9. Structural Analysis Solution Method of Joints
Step 1: Isolate one joint
Step 2: Draw the free body diagram
70o
40o 8N
Ra = 4N
Rb = 4N a b c x y
12 cm
70o
Ra = 4N a x y b c
Fac
Fab
Start by isolating joint a.
Note the lengths are not important since the angles are known. When working the Method of Joints, we will be working with internal forces. The concepts of sin and cos are still valid when dealing with forces. Students sometimes have trouble grasping this concept.
Internal forces Fac and Fab are vectors drawn away from the isolated node. Fab is made of vector components with a positive x-direction and no magnitude in the y-direction. Fac has vector components in a positive x-direction and a positive y-direction.
A vector pointed away from the analyzed joint assumes the internal force to be in tension. If the math works out negative, then the assumption is incorrect and the force is actually in compression. Be consistent throughout the analysis and carry all signs.

10. Structural Analysis Solution Method of Joints
Step 3: Separate every force and reaction into x and y components a x y
Ra = 4N
First analyse Ra
x-component = 0N
y-component = 4N
Students will have understand the vector components to calculate the member forces. The next three slides break the three vectors into components one at a time. The first two components (Ra and Fab) are fairly obvious as they only have one orientation in either the x or y direction.

11. Structural Analysis Solution Method of Joints
Step 3: Separate every force and reaction into x and y components a x y b
Fab
Next analyse Fab
x-component = Fab
y-component = 0N
Students will have understand the vector components to calculate the member forces. This slide breaks down Fab into its x and y components which allows vector math in the next slide.

12. Structural Analysis Solution Method of Joints
Step 3: Separate every force and reaction into x and y components
70o a x y c
Fac
Lastly, analyse Fac
x-component = Fac*cos70˚ N
y-component = Fac*sin70˚ N
Students will have understand the vector components to calculate the member forces. This slide breaks down Fac into its x and y components which allows vector math.

13. Summary of Force Components, Node ‘a’
Force Name Ra
Fab
Fac
Free Body Diagram
x- component 0N
Fac * cos70˚ N
y-component 4N
Fac * sin70˚ N
70o a x y c
Fac a x y
Ra = 4N y x
Fab
This table summarizes the vector components found on the prior slides. This makes it easier for the students to see the components for the equilibrium equations. Those are done on the next slides.

14. 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
Data from the table in the prior slide is used to solve the y-axis equilibrium equation.
Internal forces are drawn away from the isolated node. This direction assumes the force to be in tension. If the math works out negative, then the assumption is incorrect and the force is actually in compression. Be consistent throughout the analysis.
We started from the y-direction because it has only one unknown, Fac.
Fac resulted in a negative number, -4.26N. This means the magnitude of the internal force is 4.26N and the negative sign means the force is in compression. To keep these concepts separate, emphasize the fact that variable Fac = -4.26N for future math calculations and “Internal Fac” is the actual internal force on the member and is equal to 4.26N in compression. Because Fac is used twice, students tend to become confused and they tend to drop the negative sign.
Students need to apply a final sanity check to their answer. You might go back to the system free body diagram and ask if the magnitude and direction solution makes sense. Examination of the forces and reactions make it reasonable that this member is in compression. And since the 8N load is shared by two members, it is also reasonable that the force is close to ½ of that force.

15. 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
Internal Fab has magnitude 1.45N in tension
Data from the table in the prior slide is used to solve the y-axis equilibrium equation.
Most confusion occurs when substituting for variable Fac. This is a mathematical equation and the variable value needs to be substituted as it was solved.
The solution comes out to Internal Fab = 1.45N in tension. A sanity check verifies the stress makes sense to be in tension since the two bottom ends will seek to separate from each other. The magnitude is not intuitive.

16. Structural Analysis Solution Method of Joints
Tabulated Force Solutions
Member
Force Magnitude AB
4.26N, compression BC
1.45N, tension AC
(not yet calculated)
Summarizes the calculations from the node ‘a’ analysis. Member AC has yet to be calculated.

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
70o
40o 8N Ra Rb a b c
12 cm x y b
40o 8N c a
Fac = -4.26N
Fbc x y
We’ll do one more joint, isolating joint c and drawing the free body diagram. Before beginning, students will probably state the force will be the same. It may be comforting to the doubters when the math works out and proves them correct.
Note internal forces Fac and Fbc are drawn as vectors away from joint c. We will carry our solution with variable Fac = -4.26N.

18. Structural Analysis Solution Method of Joints
Step 3: Separate every force and reaction into x and y components 8N c x y
First analyse Rc
y-component is -8N
x-component is 0N
Students will have understand the vector components to calculate the member forces. The next three slides break the three vectors into components one at a time. This first component is fairly obvious as it only has one orientation in the y direction.

19. Structural Analysis Solution Method of Joints
Step 3: Separate every force and reaction into x and y components
20o c a
Fac x y
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
Students will have understand the vector components to calculate the member forces. This slide breaks down Fac into its x and y components which allows vector math. Since we already know the value of Fac, we can calculate the two components.

20. Structural Analysis Solution Method of Joints
Step 3: Separate every force and reaction into x and y components
20o c
Fbc b x y
Lastly analyse Fbc
y-component = –(Fbc * cos20˚)
x-component = (Fbc * sin20˚)
Students will have understand the vector components to calculate the member forces. This slide breaks down Fbc into its x and y components which allows vector math.

21. 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
20o c
Fbc b 8N c
Fac c
20o a
This table summarizes the vector components found on the prior slides. This makes it easier for the students to see the components for the equilibrium equations. Those are done on the next slides.

22. 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 N - Fbc * cos20˚ = 0
Fbc = -4.26N
Internal Fbc has magnitude 4.26N in compression
Data from the table is used to solve the y-axis equilibrium equation.
Internal forces are drawn away from the isolated node. This direction assumes the force to be in tension. If the math works out negative, then the assumption is incorrect and the force is actually in compression. Be consistent throughout the analysis.
Fbc resulted in a negative number, -4.26N. This is one data point that verifies the symmetry property as we had found that Fac = -4.26N when isolating node ‘a’

23. 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 = N
This verifies the ΣFy = 0 equilibrium equation and also the symmetry property
Data from the prior table is used to solve the x-axis equilibrium equation.
This serves as another verification that the force solution is correct for member Fac and Fbc.

24. Structural Analysis Solution Method of Joints
Tabulated Force Solutions
Member
Force Magnitude AB
4.26N, compression BC
1.45N, tension AC
Summarizes the calculations from the node ‘a’ and node ‘c’ analyses.

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
This Civil Structures module is heavily based on this book which is found in the documents directory of the module.
This module does not address Learning Activities #4 and #5, but you are encouraged to complete them if time allows.