1. Structural Analysis I Structural Analysis Trigonometry Concepts
Vectors
Equilibrium
Reactions
Static Determinancy and Stability
Free Body Diagrams
Calculating Bridge Member Forces
The first sections in white text will be covered in this first PowerPoint section. If the students are well versed in this material, proceed to Structural Analysis II covering the last sections. Some students have trouble relating math concepts outside of the math classroom, so this review may be necessary.
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 Define structural analysis
Calculate using the Pythagoreon Theorem, sin, and cos
Calculate the components of a force vector
Add two force vectors together
Understand the concept of equilibrium
Calculate reactions
Determine if a truss is stable

3. Structural Analysis
Structural analysis is a mathematical examination of a complex structure
Analysis breaks a complex system down to individual component parts
Uses geometry, trigonometry, algebra, and basic physics

4. How Much Weight Can This Truss Bridge Support?
A structural analysis is used to determine the load carrying capability of this truss bridge.
This is pretty close to the bridge the students just constructed. Challenge the students to identify if there are any differences in this bridge and the one they just completed. (The two center diagonals are in opposite directions than the manila folder bridges)

5. Pythagorean Theorem a b c
In a right triangle, the length of the sides are related by the equation:
a2 + b2 = c2
This simple theorem will come in handy when doing the truss bridge analysis. The truss bridge is made completely of right triangles.

6. Sine (sin) of an Angle
In a right triangle, the angles are related to the lengths of the sides by the equations:
sinθ1 = = a b c θ1 θ2
Opposite a
Hypotenuse c
Might do some examples to find out if all students are up on this concept. This is a fundamental concept to structural analysis and all students need to be comfortable.
Opposite b
Hypotenuse c
sinθ2 = =

7. Cosine (cos) of an Angle
In a right triangle, the angles are related to the lengths of the sides by the equations:
cosθ1 = = a b c θ1 θ2
Adjacent b
Hypotenuse c
Same message as sine on the preceding page.
Adjacent a
Hypotenuse c
cosθ2 = =

8. This Truss Bridge is Built from Right Trianglesc θ1 θ2
Stress the simple concepts stated before will be used to complete the structural analysis of the truss bridge. It is made of right triangles, an object they have seen over and over again in math class.
Ask students how many geometric shapes they can recognize. There are right triangles, isosceles triangles, rectangles, parallelograms, trapezoids, etc. Also ask them to identify angle bisectors, parallel and perpendicular lines, and right angles. If you get bold, ask how many triangles they can find.

9. Trigonometry Tips for Structural Analysis
A truss bridge is constructed from members arranged in right triangles
Sin and cos relate both lengths AND magnitude of internal forces
Sin and cos are ratios
Some tips that might help the students.
Point 2 is important: sin and cos work on both lengths and magnitude of forces.

10. Vectors Mathematical quantity that has both magnitude and direction
Represented by an arrow at an angle θ
Establish Cartesian Coordinate axis system with horizontal x-axis and vertical y-axis.
Definition of vectors.

11. Vector Example
Θ = 40o
F = 5N y x
Suppose you hit a billiard ball with a force of 5 newtons at a 40o angle
This is represented by a force vector
This assumes the ball was initially at the origin, or the (0,0) point. The arrow indicates the direction of the applied 5 newton force. The 40o angle gives the direction of the applied force.
This concept of a force vector is the same that will be used to calculate the internal forces on each bridge member.

12. Vector Components
Every vector can be broken into two parts, one vector with magnitude in the x-direction and one with magnitude in the y-direction.
Determine these two components for structural analysis.
Breaking a vector into its x and y components is the next exercise. During the truss bridge structural analysis, all forces not along the x or y axis must be divided into x and y components to calculate the internal member forces. Here is where we will use the concept of sin and cos.

13. Vector Component Example
F = 5N x y
The billiard ball hit of 5N/40o can be represented by two vector components, Fx and Fy
F = 5N Fx Fy θ x y
This tries to show how the single original vector (white dotted line), is made of a component in the x-direction and a component in the y-direction (gray solid line).

14. Fy Component Example Opposite To calculate Fy, sinθ = Hypotenuse
sin40o =
5N * 0.64 = Fy
3.20N = Fy Fy 5N
F = 5N Fx Fy
Θ=40o
Using sin with the vector data yields the magnitude of the x-direction component and the y-direction component.

15. Fx Component Example Adjacent To calculate Fx, cosθ = Hypotenuse
cos40o =
5N * 0.77 = Fx
3.85N = Fx Fx 5N
F = 5N Fx Fy
Θ=40o
Using cos with the vector data yields the magnitude of the x-direction component and the y-direction component.

16. What does this Mean? Fx = 3.85N Fy=3.20N x y F = 5N Θ=40o x y
If one person hit the ball in the x-direction with a 3.85N force and another person simultaneously hit the ball in the y-direction with a 3.20N force, the ball would travel exactly as the first vector indicates, a 5N force at a 40o angle.
These x and y components make it possible to add, subtract, and compare forces during the structural analysis.
Your 5N/40o hit is represented by this vector
The exact same force and direction could be achieved if two simultaneous forces are applied directly along the x and y axis

17. Vector Component Summary
Force Name
5N at 40°
Free Body
Diagram
x-component
5N * cos 40°
y-component
5N * sin 40°
F = 5N
Θ=40o x y
This tabular format will be used throughout Structural Analysis I and Structural Analysis II. This format makes it easy for the students to see the components of each vector and makes the addition concept more apparent.

18. How do I use these? Calculate net forces on an object
She pulls with 100 pound force
Calculate net forces on an object
Example: Two people each pull a rope connected to a boat. What is the net force on the boat?
This example is to show summing of vectors. As its drawn, the white rope is longer than the black rope, but the white rope’s magnitude of force is lower than the black rope’s. Remind students that we are analyzing forces, not the length of the rope.
He pulls with 150 pound force

19. Boat Pull Solution Represent the boat as a point at the (0,0) locationy
Represent the boat as a point at the (0,0) location
Represent the pulling forces with vectors
Fm = 150 lb
Ff = 100 lb
Note the vector length represents the magnitude of the force, not the length of the rope.
The angles are drawn to the x-axis, but they could as easily been drawn to the y-axis. The solution will remain the same.
Θm = 50o
Θf = 70o x

20. Boat Pull Solution (cont)
Separate force Ff into x and y components
Θf = 70o
Ff = 100 lb -x x y
First analyse the force Ff
x-component = -100 lb * cos70°
x-component = lb
y-component = 100 lb * sin70°
y-component = 93.9 lb
Students will have understand the vector components to calculate the resultant force.
Note the x-direction is negative, as defined by our placement of the coordinate axis. The y-direction is positive since the vector is pointing in the +y direction.

21. Boat Pull Solution (cont)
Separate force Fm into x and y components
Next analyse the force Fm
x-component = 150 lb * cos50°
x-component = 96.4 lb
y-component = 150 lb * sin50°
y-component = lb
Θm = 50o
Fm = 150 lb x y
Students will have understand the vector components to calculate the resultant force.
Note the x-direction and y-direction are positive since the vector is pointing in the +x and the +y direction.

22. Boat Pull Solution (cont)
Force Name Ff Fm
Resultant (Sum)
Vector
Diagram
(See next slide)
x- component
-100lb*cos70
= lb
150lb*cos50
= 96.4 lb
62.2 lb
y-component
100lb*sin70
= 93.9 lb
150lb*sin50
= lb
208.8 lb
50o
150 lb x y
70o
100 lb x y
This table summarizes the vector components found on the prior slides. This makes it easier for the students to see the components and the calculation for the Resultant Vector components. The actual diagram for the Resultant vector is on the next slide.

23. Boat Pull Solution (end)y
White represents forces applied directly to the boat
Gray represents the sum of the x and y components of Ff and Fm
Yellow represents the resultant vector
FTotalY Fm Ff
Since the magnitudes of FTotalY and FTotalX are known, students should be able to use the Pythagoreon Theorem to calculate the magnitude of the yellow resultant vector and the angle theta. -x x
FTotalX

24. Equilibrium Total forces acting on an object is ‘0’
Important concept for bridges – they shouldn’t move!
Σ Fx = 0 means ‘The sum of the forces in the x direction is 0’
Σ Fy = 0 means ‘The sum of the forces in the y direction is 0’ :
Equilibrium is a simple and powerful concept for structures. If they shouldn’t be moving, the algebraic sum of all forces on the structure should be 0. This is mathematically represented by the sigma (Σ) symbol. This will be another new symbol for many students but saves a lot of time.

25. Reactions
Forces developed at structure supports to maintain equilibrium.
Ex: If a 3kg jug of water rests on the ground, there is a 3kg reaction (Ra) keeping the bottle from going to the center of the earth.
3kg
Reactions are external forces applied to the structures supports. For example, if a bridge is supported on each end by an abutment, the total reactions at the abutments are equal and opposite to the gravitational forces of the bridge. This equation is validated by the fact that the bridge isn’t sinking into the earth or flying away.
Ra = 3kg

26. Reactions
A bridge across a river has a 200 lb man in the center. What are the reactions at each end, assuming the bridge has no weight?
Since the man is in the center, the 200 lb force downwards is evenly distributed between the two supports, so each support has a 100 lb reaction upwards.

27. Determinancy and Stability
Statically determinant trusses can be analyzed by the Method of Joints
Statically indeterminant bridges require more complex analysis techniques
Unstable truss does not have enough members to form a rigid structure
Our truss bridge project is a statically determinate truss so the students can be successful at calculating the internal bridge member forces.
A statically indeterminate bridge is not bad but it requires advanced college level mathematics to solve.

28. Determinancy and Stability
Statically determinate truss: 2j = m + 3
Statically indeterminate truss: 2j < m + 3
Unstable truss: 2j > m + 3
If you take four craft sticks and pin the joints (4 members, 4 joints so 2j>m+3) so it makes a square, the students will see the instability of this structure. By adding another member across either set of diagonal joints (5 members, 4 joints, so 2j=m+3), the structure becomes stable and statically determinate. We could calculate the forces on each member by using the Method of Joints. By adding yet another member across the remaining diagonal (8 members, 5 joints so 2j<m+3), the structure remains stable but becomes statically indeterminate. This means we cannot use the Method of Joints to calculate the internal force on each member.

29. 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 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.