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.

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

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

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)

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.

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 = =

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 = =

This Truss Bridge is Built from Right Triangles c 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.

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.

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.

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.

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.

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).

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.

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.

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

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.

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

Boat Pull Solution Represent the boat as a point at the (0,0) location y 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

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 = -34.2 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.

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 = 114.9 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.

Boat Pull Solution (cont) Force Name Ff Fm Resultant (Sum) Vector Diagram (See next slide) x- component -100lb*cos70 = -34.2 lb 150lb*cos50 = 96.4 lb 62.2 lb y-component 100lb*sin70 = 93.9 lb 150lb*sin50 = 114.9 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.

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

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.

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

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.

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.

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.

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.