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Structural Analysis and Trigonometry Concepts for Engineers

Learn structural analysis principles using trigonometry and vectors for equilibrium calculations, reactions, and stability assessment in complex structures. Apply Pythagorean Theorem, sine, and cosine methods.

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Structural Analysis and Trigonometry Concepts for Engineers

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  1. Structural Analysis I • Structural Analysis • Trigonometry Concepts • Vectors • Equilibrium • Reactions • Static Determinancy and Stability • Free Body Diagrams • Calculating Bridge Member Forces

  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?

  5. c a b Pythagorean Theorem • In a right triangle, the length of the sides are related by the equation: a2 + b2 = c2

  6. θ2 c a θ1 Opposite a Hypotenuse c b Sine (sin) of an Angle • In a right triangle, the angles are related to the lengths of the sides by the equations: sinθ1 = = Opposite b Hypotenuse c sinθ2 = =

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

  8. θ2 c a θ1 b This Truss Bridge is Built from Right Triangles

  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

  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.

  11. y F = 5N Θ = 40o x Vector Example • Suppose you hit a billiard ball with a force of 5 newtons at a 40o angle • This is represented by a force vector

  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.

  13. y F = 5N x y F = 5N Fy θ x Fx Vector Component Example • The billiard ball hit of 5N/40o can be represented by two vector components, Fx and Fy

  14. Opposite Hypotenuse Fy 5N F = 5N Fy Θ=40o Fx Fy Component Example To calculate Fy, sinθ = sin40o = 5N * 0.64 = Fy 3.20N = Fy

  15. Adjacent Hypotenuse Fx 5N F = 5N Fy Θ=40o Fx Fx Component Example To calculate Fx, cosθ = cos40o = 5N * 0.77 = Fx 3.85N = Fx

  16. y y x F = 5N Fx = 3.85N Θ=40o Fy=3.20N x What does this Mean? 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. y F = 5N Θ=40o x Vector Component Summary

  18. How do I use these? 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? He pulls with 150 pound force

  19. Boat Pull Solution y • Represent the boat as a point at the (0,0) location • Represent the pulling forces with vectors Fm = 150 lb Ff = 100 lb Θm = 50o Θf = 70o x

  20. y Ff = 100 lb Θf = 70o -x x Boat Pull Solution (cont) Separate force Ff into x and y components 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

  21. y Fm = 150 lb Θm = 50o x 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

  22. y 100 lb 70o x y 150 lb 50o x Boat Pull Solution (cont)

  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 -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’ :

  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 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?

  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

  28. Determinancy and Stability • Statically determinate truss: 2j = m + 3 • Statically indeterminate truss: 2j < m + 3 • Unstable truss: 2j > m + 3

  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

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